Process Dynamics and Control - Dale E (solução)

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1-1 
 
��������	�
 
 
 
 
 1.1 
 
 
a) True 
b) True 
c) True 
d) False 
e) True 
 
 
 
 1.2 
 
 
 
 
 Controlled variable- T (house interior temperature) 
 
 Manipulated variable- Q (heat from the furnace) 
 
 
 
Solution Manual for Process Dynamics and Control, 2nd edition, 
Copyright © 2004 by Dale E. Seborg, Thomas F. Edgar and Duncan A. Mellichamp. 
TC
ON/OFF SWITCH
TQL Q
Solution Manual: Process Dynamics and Control (Second Edition)
Dale E. Seborg, Thomas F. Edgar, Duncan A. Mellichamp
2004, John Wiley and Sons Inc.
1-2 
 
 Disturbance variable- QL (heat lost to surroundings); other possible 
 sources of disturbances are the loss of gas pressure and the outside door 
 opening. 
 
Specific disturbances include change in outside temperature, change in 
outside wind velocity (external heat transfer coefficient), the opening of 
doors or windows into the house, the number of people inside (each one 
generating and transmitting energy into the surrounding air), and what 
other electric lights and appliances of any nature are being used. 
 
 
 
 1.3 
 
 
The ordinary kitchen oven (either electric or gas), the water heater, and the 
furnace (Ex. 1.2) all work similarly, generally using a feedback control 
mechanism and an electronic on-off controller. For example, the oven uses 
a thermal element similar to a thermocouple to sense temperature; the 
sensor's output is compared to the desired cooking temperature (input via 
dial or electronic set-point/display unit); and the gas or electric current is 
then turned on or off depending on whether the temperature is below or 
above the desired value. Disturbances include the introduction or removal 
of food from the oven, etc. A non-electronic household appliance that 
utilizes built-in feedback control is the water tank in a toilet. Here, a float 
(ball) on a lever arm closes or opens a valve as the water level rises and 
falls above the desired maximum level. The float height represents the 
sensor; the lever arm acting on the valve stem provides actuation; and the 
on-off controller and its set point are built into the mechanical assembly. 
 
 
 
 1.4 
 
 
No, a microwave oven typically uses only a timer to operate the oven for a 
set (desired) period of time and a power level setting that turns the power 
on at its maximum level for a fixed fraction of the so-called duty cycle, 
generally several seconds. 
Thus setting the Power Level at 6 (60% of full power) and the Cook Time 
to 1:30 would result in the oven running for a total of one and one-half 
minutes with the power proportioned at 60% (i.e., turned on 100% for 6 
seconds and off for 4 seconds, if the fixed duty cycle is 10 seconds long). 
This type of control is sometimes referred to as programmed control, as it 
utilizes only time as the reference variable . 
 
1-3 
 
The big disadvantage of such an approach is that the operator (here the 
cook) has to estimate what settings will achieve the desired food 
temperature or will cook the food to the desired state. This can be 
dangerous, as many people can attest who have left a bag of popcorn in 
the oven too long and set the bag on fire, or embarrassing, as anyone 
knows who has served a frozen meal that did not quite thaw out, let alone 
cook. What good cooks do is provide a measure of feedback control to the 
microwave cooking process, by noting the smell of the cooking food or 
opening the door and checking occasionally to make sure it is heating 
correctly. However, anyone who has used a microwave oven to cook fish 
filets, for example, and blown them all over the oven, learns to be very 
conservative in the absence of a true feedback control mechanism. [Note 
that more expensive microwaves do come equipped with a temperature 
probe that can be inserted into the food and a controller that will turn off 
the oven when the temperature first reaches the desired (set point) value. 
But even these units will not truly control the temperature.] 
 
 
 
 1.5 
 
 
a) In steering a car, the driver's eyes are the sensor; the drivers hands and the 
steering system of the car serve as the actuator; and the driver's brain 
constitutes the controller (formulates the control action i.e., turning the 
steering wheel to the right when the observed position of the car within its 
desired path is too far to the left and vice versa). Turns in the road, 
obstructions in the road that must be steered around, etc. represent 
disturbances. 
 
b) In braking and accelerating, a driver has to estimate mentally (on a 
practically continuous basis) the distance separating his/her car from the 
one just ahead and then apply brakes, coast, or accelerate to keep that 
distance close to the desired one. This process represents true feedback 
control where the measured variable (distance of separation) is used to 
formulate an appropriate control response and then to actuate the 
brakes/accelerator according to the driver's best judgment. Feedforward 
control comes into the picture when the driver uses information other than 
the controlled variable (separation distance) that represents any measure of 
disturbance to the ongoing process; included would be observations that 
brake lights on preceding vehicle(s) are illuminating, that cars are arriving 
at a narrowing of the road, etc. Most good drivers also pay close attention 
to the rate of change of separation distance, which should remain close to 
zero. Later we will see that use of this variable, the time derivative of the 
controlled variable, is just another element in feedback control because a 
function of the controlled variable is involved. 
1-4 
 
 1.6 
 
 
a) Feedback Control : Measured variable: y 
 Manipulated variable: D,R, or B(schematic shows D) 
 
 
 
 
 
 
b) Feedforward Control: Measured variable: F 
 Manipulated variable: D (shown), R or B 
 
 
 
 
 
1-5 
 
 1.7 
 
 
Both flow control loops are feedback control systems. In both cases, the 
controlled variable (flow) is measured and the controller responds to that 
measurement. 
 
 
 1.8 
 
 
 a) 
 
 
 
LTTT
TC LC
FILTER
CITY SUPPLY
GAS
p(T)
Q(t)
TpGR
A
V
E
L
TG X
Ta
QL
leak
F
L
PUMP
Tw , Fw
HEATER
AIR
ON/OFF
VALVE
 
 
 Outputs: Tp, L(level) 
Inputs: Q(t), Fw 
Disturbances: Tw, Ta 
 
b) Either Tw or Ta or both can be measured in order to add feedforward 
control. 
 
c) Steady-state energy balance 
 
)()()()( wpwGpGap TTCF
x
TT
kTTUAtQ −ρ+
∆
−
+−= 
 
1-6 
 
Notice that, at steady state, Fw = F (from material balance.) 
 
Here, A is the area of water surface exposed to the atmosphere 
 ρ is the density of supply water 
 C is the specific heat of supply water. 
 
The magnitudes of the terms UA(Tp-Ta) and FwρC(Tp-Tw) relative to the 
magnitude of Q(t) will determine whether Ta or Tw (or both/neither) is the 
important disturbance variable. 
 
d) Determine which disturbance variable is important as suggested in part c) 
and investigate the economic feasibility of using its measurement for 
feedforward control 
 
 
2-1 
��������	�
��
��������	�
 
 
 
 
 2.1 
 
 
a) Overall mass balance: 
 
 321
)(
www
dt
Vd
−+=
ρ
 (1) 
 
Energy balance: 
 
 
( )
3
1 1 2 2
3 3
� � � ( ) ( )ref
ref ref
ref
d V T T
C w C T T w C T T
dt
w C T T
 − 
= − + −
− −
 (2) 
 Because ρ = constant and VV = = constant, Eq. 1 becomes: 
 
 213 www += (3) 
 
b) From Eq. 2, substituting Eq. 3( ) ( )
3 3
1 1 2 2
1 2 3
( )� � � � � �ref ref ref
ref
d T T dTCV CV w C T T w C T T
dt dt
w w C T T
−
= = − + −
− + −
 (4) 
 
Constants C and Tref can be cancelled: 
 
 3212211
3 )( TwwTwTw
dt
dTV +−+=ρ (5) 
 
The simplified model now consists only of Eq. 5. 
 
Degrees of freedom for the simplified model: 
 
Parameters : ρ, V 
 
Solution Manual for Process Dynamics and Control, 2nd edition, 
Copyright © 2004 by Dale E. Seborg, Thomas F. Edgar and Duncan A. Mellichamp. 
2-2 
Variables : w1, w2, T1, T2, T3 
NE = 1 
NV = 5 
 
Thus, NF = 5 – 1 = 4 
 
Because w1, w2, T1 and T2 are determined by upstream units, we assume 
they are known functions of time: 
 
 w1 = w1(t) 
w2 = w2 (t) 
T1 = T1(t) 
T2 = T2(t) 
 
Thus, NF is reduced to 0. 
 
 
 
 2.2 
 
 
 Energy balance: 
 
� � �
( ) ( ) ( )refp p i ref p ref s a
d V T T
C wC T T wC T T UA T T Q
dt
 − 
= − − − − − + 
 
Simplifying 
 
� � �p p i p s adTVC wC T wC T UA T T Qdt = − − − + 
� � � � �p p i s adTVC wC T T UA T T Qdt = − − − + 
 
 
b) T increases if Ti increases and vice versa. 
 
T decreases if w increases and vice versa if (Ti – T) < 0. In other words, if 
Q > UAs(T-Ta), the contents are heated, and T >Ti. 
 
 
 2.3 
 
 
a) Mass Balances: 
 
2-3 
 321
1
1 wwwdt
dhA −−=ρ (1) 
 
 2
2
2 wdt
dhA =ρ (2) 
 
Flow relations: 
 
 Let P1 be the pressure at the bottom of tank 1. 
 
 Let P2 be the pressure at the bottom of tank 2. 
 
 Let Pa be the ambient pressure. 
 
Then )( 21
22
21
2 hhRg
g
R
PP
w
c
−
ρ
=
−
= (3) 
 
 1
33
1
3 hRg
g
R
PP
w
c
a ρ
=
−
= (4) 
 
b) Seven parameters: ρ, A1, A2, g, gc, R2, R3 
 
Five variables : h1, h2, w1, w2, w3 
 
Four equations 
 
Thus NF = 5 – 4 = 1 
 
 1 input = w1 (specified function of time) 
 4 outputs = h1, h2, w2, w3 
 
 
 
 2.4 
 
 
Assume constant liquid density, ρ . The mass balance for the tank is 
 
)()( qq
dt
mAhd
i
g
−ρ=
+ρ
 
 
Because ρ, A, and mg are constant, this equation becomes 
 
2-4 
qq
dt
dhA i −= (1) 
 
The square-root relationship for flow through the control valve is 
 
2/1




−
ρ
+= a
c
gv Pg
ghPCq (2) 
 
From the ideal gas law, 
 
)(
)/(
hHA
RTMm
P gg
−
= (3) 
 
where T is the absolute temperature of the gas. 
 
Equation 1 gives the unsteady-state model upon substitution of q from Eq. 2 and 
of Pg from Eq. 3: 
 
1/ 2( / )
( )
g
i v a
c
m M RTdh ghA q C P
dt A H h g
ρ 
= − + − 
− 
 (4) 
 
Because the model contains Pa, operation of the system is not independent of Pa. 
For an open system Pg = Pa and Eq. 2 shows that the system is independent of Pa. 
 
 
 
 2.5 
 
 
 a) For linear valve flow characteristics, 
 
 
a
d
a R
PP
w 1
−
= , 
b
b R
PP
w 21
−
= , 
c
f
c R
PP
w
−
=
2
 (1) 
Mass balances for the surge tanks 
 
 ba wwdt
dm
−=
1
, cb wwdt
dm
−=
2
 (2) 
 
where m1 and m2 are the masses of gas in surge tanks 1 and 2, 
respectively. 
 
If the ideal gas law holds, then 
 
2-5 
 1
1
11 RTM
mVP = , 2222 RTM
mVP = (3) 
 
where M is the molecular weight of the gas 
 T1 and T2 are the temperatures in the surge tanks. 
 
Substituting for m1 and m2 from Eq. 3 into Eq. 2, and noticing that V1, T1, 
V2, and T2 are constant, 
 
 ba wwdt
dP
RT
MV
−=
1
1
1
 and cb wwdt
dP
RT
MV
−=
2
2
2
 (4) 
 
The dynamic model consists of Eqs. 1 and 4. 
 
 b) For adiabatic operation, Eq. 3 is replaced by 
 
 C
m
V
P
m
V
P =



=


 γγ
2
2
2
1
1
1 , a constant (5) 
or 
γγ




=
/1
11
1 C
VP
m and 
γγ




=
/1
22
2 C
VP
m (6) 
 
Substituting Eq. 6 into Eq. 2 gives, 
 
 ba wwdt
dPP
C
V
−=



γ
γγ−
γγ
1/)1(
1
/1
11
 
 cb wwdt
dPP
C
V
−=



γ
γγ−
γγ
2/)1(
2
/1
21
 
 
as the new dynamic model. If the ideal gas law were not valid, one would 
use an appropriate equation of state instead of Eq. 3. 
 
 
 2.6 
 
 
a) Assumptions: 
 
1. Each compartment is perfectly mixed. 
2. ρ and C are constant. 
3. No heat losses to ambient. 
 
 Compartment 1: 
2-6 
 Overall balance (No accumulation of mass): 
 
 0 = ρq − ρq1 thus q1 = q (1) 
 
 Energy balance (No change in volume): 
 
 
1
1 1 1 2� � � � � �idTV C qC T T UA T Tdt = − − − (2) 
 
 Compartment 2: 
 
 Overall balance: 
 
 0 = ρq1 − ρq2 thus q2 = q1= q (3) 
 
 Energy balance: 
 
 
2
2 1 2 1 2 2� � � � � � � �c c cdTV C qC T T UA T T U A T Tdt = − + − − − (4) 
 
b) Eight parameters: ρ, V1, V2, C, U, A, Uc, Ac 
Five variables: Ti, T1, T2, q, Tc 
Two equations: (2) and (4) 
 
Thus NF = 5 – 2 = 3 
 
 2 outputs = T1, T2 
 3 inputs = Ti, Tc, q (specify as functions of t) 
 
 
c) Three new variables: ci, c1, c2 (concentration of species A). 
Two new equations: Component material balances on each compartment. 
c1 and c2 are new outputs. ci must be a known function of time. 
 
 
 
 2.7 
 
Let the volume of the top tank be γV, and assume that γ is constant. 
 
Then, an overall mass balance for either of the two tanks indicates that the 
flow rate of the stream from the top tank to the bottom tank is equal to 
q +qR. Because the two tanks are perfectly stirred, cT2 = cT. 
 
2-7 
Component balance for chemical tracer over top tank: 
 
 
1
1� � �T Ti R T R T
dcV qc q c q q c
dt
= + − + (1) 
 
Component balance on bottom tank: 
 
 
2
1(1 �� � �T R T R T TdcV q q c q c qcdt− = + − − 
or 
 
 1(1 �� � �� �T R T TdcV q q c cdt− = + − (2) 
 
Eqs. 1 and 2 constitute the model relating the outflow concentration, cT, to 
inflow concentration, cTi. Describing the full-scale reactor in the form of 
two separate tanks has introduced two new parameters into the analysis, qR 
and γ. Hence, these parameters will have to be obtained from physical 
experiments. 
 
 
 2.8 
 
 
Additional assumptions: 
 
 (i) Density of the liquid, ρ, and density of the coolant, ρJ, are constant. 
 (ii) Specific heat of the liquid, C, and of the coolant, CJ, are constant. 
 
Because V is constant, the mass balance for the tank is: 
 
 0=−=ρ qq
dt
dV
F ; thus q = qF 
 
Energy balance for tank: 
 
 )()( 8.0 JJFF TTAKqTTCqdt
dTVC −−−ρ=ρ (1) 
 
Energy balance for the jacket: 
 
 )()( 8.0 JJJiJJJJJJJ TTAKqTTCqdt
dTCV −+−ρ=ρ (2) 
 
2-8 
where A is the heat transfer area (in ft2) between the process liquid and the 
coolant. 
 
Eqs.1 and 2 comprise the dynamic model for the system. 
 
 
 
 2.9 
 
 
 Additional assumptions: 
 
i. The density ρ and the specific heat C of the process liquid are 
constant. 
ii. The temperature of steam Ts is uniform over the entire heat 
transfer area 
iii. Ts is a function of Ps , Ts = f(Ps) 
 
Mass balance for the tank: 
 
 qq
dt
dV
F −= (1) 
Energy balance for the tank: 
 
 
( )� � � � � � �
( )
ref
F F ref ref
s
d V T T
C q C T T q C T T
dt
UA T T
 
− 
= − − −
+ −
 (2) 
 
where: Tref is a constant reference temperature 
 A is the heat transfer areaEq. 2 is simplified by substituting for (dV/dt) from Eq. 1, and replacing Ts 
by f(Ps), to give 
 
 [ ]TPfUATTCq
dt
dTVC sFF −+−ρ=ρ )()( (3) 
 
Then, Eqs. 1 and 3 constitute the dynamic model for the system. 
 
 
 
 
 
 
 
2-9 
 2.10 
 
 
Assume that the feed contains only A and B, and no C. Component 
balances for A, B, C over the reactor give. 
 
 
1 /
1
E RTA
i Ai A A
dcV q c qc Vk e c
dt
−
= − − (1) 
 
 
1 2/ /
1 2( )E RT E RTB i Bi B A B
dcV q c qc V k e c k e c
dt
− −
= − + − (2) 
 
 
2 /
2
E RTC
C B
dcV qc Vk e c
dt
−
= − + (3) 
 
An overall mass balance over the jacket indicates that qc = qci because the 
volume of coolant in jacket and the density of coolant are constant. 
 
Energy balance for the reactor: 
 
( ) ( ) ( )A A A B B B C C C i Ai A A i Bi B B id Vc M S Vc M S Vc M S T q c M S q c M S T Tdt
 + + 
= + −
 
 
1 2/ /
1 1 2 2( ) ( ) ( )E RT E RTc A BUA T T H Vk e c H Vk e c− −− − + −∆ + −∆ (4) 
 
where MA, MB, MC are molecular weights of A, B, and C, respectively 
 SA, SB, SC are specific heats of A, B, and C. 
 U is the overall heat transfer coefficient 
 A is the surface area of heat transfer 
 
Energy balance for the jacket: 
 
 � � � � � �cj j j j j ci ci c cdTS V S q T T UA T Tdt = − + − (5) 
 
where: 
 
 ρj, Sj are density and specific heat of the coolant. 
 Vj is the volume of coolant in the jacket. 
 
Eqs. 1 - 5 represent the dynamic model for the system. 
 
 
 
2-10 
 2.11 
 
 
 Model (i) : 
 
 Overall mass balance (w=constant=w): 
 
 1 2
( )d V dhA w w w
dt dt
ρ ρ= = + − (1) 
 
A component balance: 
 
 xww
dt
Vxd
−=
ρ
1
)(
 
 
 or 1
( )d hxA w wx
dt
ρ = − (2) 
 
 Note that for Stream 2, x = 0 (pure B). 
 
 
 Model (ii) : 
 
 Mass balance: 
 
 1 2
(� �d V dhA w w w
dt dt
= ρ = + − (3) 
 
Component balance on component A: 
 
 wxw
dt
Vxd
−=
ρ
1
)(
 
 
 or 1
( )d hxA w wx
dt
ρ = − (4) 
 
 
 
 
 
 
 
 
 
 
2-11 
 2.12 
 
 
 a) 
 
 
Note that the only conservation equation required to find h is an overall 
mass balance: 
 
1 2
( )dm d Ah dhA w w w
dt dt dt
ρ
= = ρ = + − (1) 
 
 
 Valve equation: w = hCh
g
gC v
c
v =
ρ
′
 (2) 
 where 
c
vv g
gCC ρ′= (3) 
 
 Substituting the valve equation into the mass balance, 
 
 )(1 21 hCwwAdt
dh
v−+ρ
= (4) 
 
 Steady-state model : 
 
 0 = hCww v−+ 21 (5) 
 
b) 1 2 1/2
2.0 1.2 3.2 kg/s2.13
1.52.25 mv
w wC
h
+ +
= = = = 
 
 
 
 c) Feedforward control 
 
2-12 
 
 
Rearrange Eq. 5 to get the feedforward (FF) controller relation, 
 
 12 whCw Rv −= where 2.25 mRh = 
 
112 2.3)5.1)(13.2( www −=−= (6) 
 
 Note that Eq. 6, for a value of w1 = 2.0, gives 
 
w2 = 3.2 –1.2 = 2.0 kg/s which is the desired value. 
 
If the actual FF controller follows the relation, 12 1.12.3 ww −= (flow 
transmitter 10% higher), 2w will change as soon as the FF controller is 
turned on, 
 
 w2 = 3.2 –1.1 (2.0) = 3.2 – 2.2 = 1.0 kg/s 
 
 (instead of the correct value, 1.2 kg/s) 
 
Then 0.10.213.2 +== hhCv 
 
 or 408.1
13.2
3
==h and h = 1.983 m (instead of 2.25 m) 
 
 
 Error in desired level = 
2.25 1.983 100% 11.9%
2.25
−
× = 
 
The sensitivity does not look too bad in the sense that a 10% error in flow 
measurement gives ~12% error in desired level. Before making this 
2-13 
conclusion, however, one should check how well the operating FF 
controller works for a change in w1 (e.g., ∆w1 = 0.4 kg/s). 
 
 
 
 2.13 
 
 
a) Model of tank (normal operation): 
 
1 2 3
dhA w w w
dt
ρ = + − (Below the leak point) 
2
2(2) 3.14 m
4
A pi pi= = = 
 
(800)(3.14) 20200100120 =−+=
dt
dh
 
 
20 0.007962 m/min(800)(3.14)
dh
dt
= =
 
 
Time to reach leak point (h = 1 m) = 125.6 min. 
 
 
b) Model of tank with leak and 321 ,, www constant: 
 
4� �� �� �������� 	dhA q hdt δ= − = − − = 20 − 20 1−h , h ≥ 1 
 
To check for overflow, one can simply find the level hm at which dh/dt = 
0. That is the maximum value of level when no overflow occurs. 
 
0 = 20 − 20 1−mh or hm = 2 m 
 
Thus, overflow does not occur for a leak occurring because hm < 2.25 m. 
 
 
 2.14 
 
 Model of process 
 
 Overall material balance: 
 
2-14 
 321 wwwdt
dhAT −+=ρ = hCww v−+ 21 (1) 
 
 Component: 
 
 332211
3 )( xwxwxw
dt
hxdAT −+=ρ 
 
 3322113
3 xwxwxw
dt
dh
xA
dt
dxhA TT −+=ρ+ρ 
 
 Substituting for dh/dt (Eq. 1) 
 
 =−++ρ )( 32133 wwwxdt
dxhAT 332211 xwxwxw −+ 
 
 )()( 3223113 xxwxxwdt
dxhAT −+−=ρ (2) 
 
 or [ ])()(1 3223113 xxwxxwhAdt
dx
T
−+−
ρ
= (3) 
 
a) At initial steady state , 
 
Kg/min220100120213 =+=+= www 
Cv = 3.166
75.1
220
= 
 
b) If x1 is suddenly changed from 0.5 to 0.6 without changing flowrates, then 
level remains constant and Eq.3 can be solved analytically or numerically 
to find the time to achieve 99% of the x3 response. From the material 
balance, the final value of x3 = 0.555. Then, 
 
 [ ]3 3 31 120(0.6 ) 100(0.5 )(800)(1.75)
dx
x x
dt
= − + −
pi
 
 
 [ ]31 (72 50) 220 )(800)(1.75) x= + −pi 
 
 30.027738 0.050020x= − 
 
 Integrating, 
2-15 
 
3
3
3
3 00.027738 0.050020
f
o
x t
x
dx dt
x
=
−
∫ ∫ 
 
 where x3o=0.5 and x3f =0.555 – (0.555)(0.01) = 0.549 
 
 Solving, 
 
 t = 47.42 min 
 
c) If w1 is changed to 100 kg/min without changing any other input variables, 
then x3 will not change and Eq. 1 can be solved to find the time to achieve 
99% of the h response. From the material balance, the final value of the 
tank level is h =1.446 m. 
 
800pi 100 100 v
dh C h
dt
= + − 
 
 
1 200 166 3
800
dh
. h
dt
 = − pi 
 
 0 079577 0 066169. . h= − 
 
 where ho=1.75 and hf =1.446 + (1.446)(0.01) = 1.460 
 
 By using the MATLAB command ode45 , 
 
 t = 122.79 min 
 
Numerical solution of the ode is shown in Fig. S2.14 
0 50 100 150 200 250 300
1.4
1.5
1.6
1.7
1.8
time (min)
 h(m)
 
 Figure S2.14. Numerical solution of the ode for part c) 
2-16 
d) In this case, both h and x3 will be changing functions of time. Therefore, 
both Eqs. 1 and 3 will have to be solved simultaneously. Since 
concentration does not appear in Eq. 1, we would anticipate no effect on 
the h response. 
 
 
 
2.15 
 
 
 a) The dynamic model for the chemostat is given by: 
 
 Cells: FXVr
dt
dXV g −= or XV
F
r
dt
dX
g 


−= (1) 
 
 Product: FPVr
dt
dPV p −= or PV
F
r
dt
dP
p 


−= (2) 
 
 Substrate: )( SSF
dt
dSV f −=
/ /
1 1
g P
X S P S
Vr Vr
Y Y
− − 
 
 or 
 )( SS
V
F
dt
dS
f −


= P
SP
g
SX
r
Y
r
Y //
11
−− (3) 
 
b) At steady state, 
 
 0=
dt
dX
 ∴ DXrg = 
 then, 
 DXX =µ ∴ D=µ (4) 
 
 A simple feedback strategy can be implemented where the growth rate 
 is controlled by manipulating the mass flow rate, F. 
 
 c) Washout occurs if dX/dt = 0 is negative for an extended period of time; 
 that is, 
 
 0<− DXrgor D<µ 
 
 Thus, if D<µ the cells will be washed out. 
 
d) At steady state, the dynamic model given by Eqs. 1, 2 and 3 becomes: 
 
2-17 
 DXrg −=0
 
(5) 
 DPrp −=0 (6) 
 )(0 SSD f −= P
SP
g
SX
r
Y
r
Y //
11
−− (7) 
 
 From Eq. 5, 
 grDX = (8) 
 From Eq. 7 
 P
SP
SX
fSXg rY
Y
DSSYr
/
/
/ )( +−= (9) 
 Substituting Eq. 9 into Eq. 8, 
 
 p
SP
SX
fSX rY
Y
DSSYDX
/
/
/ )( +−= (10) 
 From Eq. 6 and the definition of YP/S in (2-92), 
 
 )(/ SSDYDPr fSPp −== 
 
 From Eq. 4 
 
D
DKS S
−µ
=
max
 
 
 Substituting these two equations into Eq. 10, 
 
 D
D
DKSYDX SfSX 



−µ
−=
max
/2 
 
2-18 
0 0.05 0.1 0.15 0.2 0.25
0
0.2
0.4
0.6
0.8
1
D (1/h)
DX
 
(g/
L.
h)
 MAXIMUM 
PRODUCTION 
WASHOUT 
 
 Figure S2.15. Steady-state cell production rate DX as a function of dilution rate D. 
 
 From Figure S2.15, washout occurs at D = 0.18 h-1 while the maximum 
 production occurs at D = 0.14 h-1. Notice that maximum and washout points 
 are dangerously close to each other, so special care must be taken when 
 increasing cell productivity by increasing the dilution rate. 
 
 
2.16 
 
 
a) We can assume that ρ and h are approximately constant. The dynamic 
model is given by: 
 
 sd kAcdt
dM
r =−= (1) 
 Notice that: 
 VM ρ= ∴ 
dt
dV
dt
dM ρ= (2) 
 
 hrV 2pi= ∴ 
dt
drA
dt
dr
rh
dt
dV
=pi= )2( (3) 
2-19 
 Substituting (3) into (2) and then into (1), 
 
 skAcdt
drA =ρ− ∴ skcdt
dr
=ρ− 
 Integrating, 
 
 
0�o
r ts
r
kcdr dt= −∫ ∫ ∴ tkcrtr so ρ−=)( (4) 
 
 Finally, 
 
 
2hrVM ρpi=ρ= 
 
 then 
 
2
)( 


ρ
−ρpi= tkcrhtM so
 
 
b) The time required for the pill radius r to be reduced by 90% is given by 
Eq. 4: 
 
 t
kc
rr soo ρ
−=1.0 ∴ 54)5.0)(016.0(
)2.1)(4.0)(9.0(9.0
==
ρ
=
s
o
kc
r
t min 
 
 Therefore, 54 min .t = 
 
 
2.17 
 
For V = constant and F = 0, the simplified dynamic model is: 
 
 X
SK
S
r
dt
dX
s
g +
µ== max 
 
 X
SK
SYr
dt
dP
s
XPp +
µ== max/ 
 
 P
XP
g
SX
r
Y
r
Ydt
dS
//
11
−−= 
 
 Substituting numerical values: 
 
 
S
SX
dt
dX
+
=
1
2.0 
 
2-20 
 
S
SX
dt
dP
+
=
1
)2.0)(2.0( 
 
 


−−
+
=
1.0
2.0
5.0
1
1
2.0
S
SX
dt
dS
 
 
By using MATLAB, this system of differential equations can be solved. 
The time to achieve a 90% conversion of S is t = 22.15 h. 
 
 Figure S2.17. Fed-batch bioreactor dynamic behavior. 
3-1 
��������	�
 
 
 
 
 3.1 
 
a) � [ ] ∫∫ ∞ +−∞ −−− ω=ω=ω
0
)(
0
sinsinsin dtetdtetete tbsstbtbt 
 
[ ] ∞+− 


ω++
ωω−ω+−
=
0
22
)(
)(
cossin)(
bs
ttbs
e tbs 
 22)( ω++
ω
=
bs
 
b) � [ ] ∫∫ ∞ +−∞ −−− ω=ω=ω
0
)(
0
coscoscos dtetdtetete tbsstbtbt 
 
[ ] ∞+− 


ω++
ωω+ω+−
=
0
22
)(
)(
sincos)(
bs
ttbs
e tbs 
 22)( ω++
+
=
bs
bs
 
 
 
 
 3.2 
 
 
a) The Laplace transform provided is 
 
4643
4)( 234 ++++= sssssY 
 
We also know that only sin ωt is an input, where ω = 2 . Then 
 
( ) 2
2
2
2)( 22222 +=+
=
ω+
ω
=
sss
sX 
 
Since Y(s) = D-1(s) X(s) where D(s) is the characteristic polynomial (when 
all initial conditions are zero), 
 
Solution Manual for Process Dynamics and Control, 2nd edition, 
Copyright © 2004 by Dale E. Seborg, Thomas F. Edgar and Duncan A. Mellichamp. 
3-2 
)2(
2
)23(
22)( 22 +++= ssssY 
 
and the original ode was 
 
ty
dt
dy
dt
yd 2sin22232
2
=++ with 0)0()0( ==′ yy 
 
b) This is a unique result. 
 
c) The solution arguments can be found from 
 
)2()2)(1(
222)( 2 ++++= ssssY 
 
which in partial fraction form is 
 
221
)( 2 2121 +
+
+
+
α
+
+
α
=
s
asa
ss
sY 
 
Thus the solution will contain four functions of time 
 
 e-t , e-2t , sin 2 t , cos 2 t 
 
 
 
 3.3 
 
 
a) Pulse width is obtained when x(t) = 0 
 
Since x(t) = h – at 
 
 tω : h − atω = 0 or tω = h/a 
 
 b) 
 x(t) = hS(t) – atS(t) + a(t -tω) S(t-tω) 
x(t) x(t)
h
slope = a
slope = -a
slope = -a
3-3 
c) 222
1)(
s
e
s
h
s
ae
s
a
s
h
sX
stst
−
+=+−=
ωω −−
 
 
d) Area under pulse = h tω/2 
 
 
 
 3.4 
 
 
a) f(t) = 5 S(t) – 4 S(t-2) – S(t- 6) 
 =)(sF = ( )6s- 2s- e -4e - 51
s
 
 
 b) 
 
 x(t) = x1(t) + x2(t) + x3(t) + x4(t) 
 
 = at – a(t − tr)S(t − tr) − a(t −2tr)S(t − 2tr) + a(t − 3tr)S(t − 3tr) 
 
 
 following Eq. 3-101. Thus 
 
 X(s) = [ ]ststst rrr eee
s
a 32
2 1
−−− +−− 
 
 by utilizing the Real Translation Theorem Eq. 3-104. 
 
 
x(t)
 -a -a
x2 x3
tr 2tr 3tr
x1
x4
 a
 a
3-4 
 3.5 
 
 
 T(t) = 20 S(t) + 
30
55
t S(t) – 
30
55
(t-30) S(t-30) 
( )ss e
ss
e
sss
sT 302
30
22 1
1
30
55201
30
551
30
5520)( −− −+=−+= 
 
 
 
 3.6 
 
a) 
432)4)(3)(2(
)1()( 321
+
α
+
+
α
+
+
α
=
+++
+
=
ssssss
ss
sX 
 
 1)4)(3(
)1(
2
1 =++
+
=α
−=s
ss
ss
 
 
 6)4)(2(
)1(
3
2 −=++
+
=α
−=s
ss
ss
 
 
 6)3)(2(
)1(
4
3 =++
+
=α
−=s
ss
ss
 
 
 
 
4
6
3
6
2
1)(
+
+
+
−
+
=
sss
sX 2 3 4and ( ) 6 6t t tx t e e e− − −= − + 
 
b) )2)(2)(3)(2(
1
)4)(3)(2(
1)( 2 jsjsss
s
sss
s
sX
−+++
+
=
+++
+
= 
 
 js
j
js
j
ss
sX
2232
)( 333321
+
β−α
+
+
β+α
+
+
α
+
+
α
= 
 
 
8
1
)4)(3(
1
2
21 −=++
+
=α
−=s
ss
s
 
 
13
2
)4)(2(
1
3
22 =++
+
=α
−=s
ss
s
 
 
208
113
840
21
)2)(3)(2(
1
2
33
j
j
j
jsss
sj
js
+−
=
−−
−
=
−++
+
=β+α
−=
 
3-5 
 tteetx tt 2sin
208
1122cos
208
32
13
2
8
1)( 32 


+

 −
++−= −− 
 
 ttee tt 2sin
104
112cos
104
3
13
2
8
1 32 +−+−= −− 
 
 
c) 2212 )1(1)1(
4)(
+
α
+
+
α
=
+
+
=
sss
s
sX (1) 
 
 3)4( 12 =+=α −=ss 
 
In Eq. 1, substitute any s≠-1 to determine α1. Arbitrarily using s=0, Eq. 1 
gives 
 
 1or
1
3
11
4
12
1
2 =α+
α
= 
 
 2)1(
3
1
1)(
+
+
+
=
ss
sX and 3t tx( t ) e te− −= + 
d) ( ) 2222
1
4
3
2
1
1
1
1)(
ω++
=
+


+
=
++
=
bs
s
ss
sX 
1 3
where and
2 2
b = ω = 
 
tetetx
t
bt
2
3
sin
3
2
sin1)( 2−− =ω
ω
= 
 
e) X(s) = se
sss
s 5.0
)3)(2(
1
−
++
+
 
 
To invert, we first ignore the time delay term. Using the Heaviside 
expansion with the partial fraction expansion, 
 
32)3)(2(
1)(ˆ
+
+
+
+=
++
+
=
s
C
s
B
s
A
sss
s
sX 
 
Multiply by s and let s → 0 
 
3-6 
A = 
6
1
)3)(2(
1
= 
Multiply by (s+2) and let s→ −2 
 
B = 
2
1
)1)(2(
1
)32)(2(
12
=
−
−
=
+−−
+−
 
 
Multiply by (s+3)and let s→-3 
 
C = 
3
2
)1)(3(
2
)23)(3(
13
−=
−−
−
=
+−−
+−
 
 
Then 
 
3
32
2
2161)(ˆ
+
−
+
+
+=
sss
sX 
 
tt eetx 32
3
2
2
1
6
1)(ˆ −− −+= 
 
Imposing shift theorem 
 
)5.0(3)5.0(2
3
2
2
1
6
1)5.0(ˆ)( −−−− −+=−= tt eetxtx 
 
 for t ≥ 0.5 
 
 
 
 3.7 
 
 a) 22122
6
)1(
)1(6)(
sssss
s
sY
α
+
α
==
+
+
= 
 
 066 1
0
2
2
2 =α==α
=ss
s 
 2
6
)(
s
sY = 
 
b) 
9)9(
)2(12)( 2 3212 +
α+α
+
α
=
+
+
=
s
s
sss
s
sY 
 
3-7 
Multiplying both sides by s(s2+9) 
 
 
))(()9()2(12 3221 ssss α+α++α=+ or 
 13
2
21 9)(2412 α+α+α+α=+ sss 
 
 Equating coefficients of like powers of s, 
 
 s
2: α1 + α2 = 0 
 s
1: α3 = 12 
 s
0: 9α1 = 24 
 
 Solving simultaneously, 
 
 12,
3
8
,
3
8
321 =α
−
=α=α 
 
9
12
3
8
1
3
8)( 2 +



+−
+=
s
s
s
sY 
 
c) 
654)6)(5)(4(
)3)(2()( 321
+
α
+
+
α
+
+
α
=
+++
++
=
ssssss
ss
sY 
 
 1)6)(5(
)3)(2(
4
1 =++
++
=α
−=s
ss
ss
 
 
 6)6)(4(
)3)(2(
5
2 −=++
++
=α
−=s
ss
ss
 
 
 6)5)(4(
)3)(2(
6
3 =++
++
=α
−=s
ss
ss
 
 
 
6
6
5
6
4
1)(
+
+
+
−
+
=
sss
sY 
 
d) [ ] )2()22(
1
)2(1)1(
1)( 2222 +++=+++
=
sssss
sY 
 
 = 
2)22(22
5
22
43
2
21
+
α
+
++
α+α
+
++
α+α
sss
s
ss
s
 
 
3-8 
 Multiplying both sides by )2()22( 22 +++ sss gives 
 
1 = α1s4 + 4α1s3 + 6α1s2 +4α1s + α2s3 +4α2s2 +6α2s +4α2 + α3s2 +2α3s + 
α4s + 2α4 + α5s4 + 4α5s3 + 8α5s2 + 8α5s + 4α5 
 
 Equating coefficients of like power of s, 
 
 s
4
 : α1 + α5 = 0 
 
 s
3
 : 4α1 + α2 + 4α5 = 0 
 
 s
2 : 6α1 + 4α2 + α3 + 8α5 = 0 
 
 s
1
 : 4α1 + 6α2 + 2α3 + α4 + 8α5 = 0 
 
 s
0
 : 4α2 + 2α4 + 4α5 = 1 
 
 Solving simultaneously: 
 
 α1 = -1/4 α2 = 0 α3=-1/2 α4=0 α5 = ¼ 
 
 
2
4/1
)22(
2/1
22
4/1)( 222 ++++
−
+
++
−
=
sss
s
ss
s
sY 
 
 
 
 3.8 
 
 
a) From Eq. 3-100 
 
� )(1)(
0
** sF
s
dttf
t
=

∫ 
we know that � 


τ∫ τ−t de
0
=
s
1
�� [ ]τ−e )1(
1
+
=
ss
 
 
∴ Laplace transforming yields 
 
s
2X(s) + 3X(s) + 2X(s) = )1(
2
+ss
 
or (s
2
 + 3s + 1) X(s) = )1(
2
+ss
 
3-9 
 X(s) = )2()1(
2
2 ++ sss
 
 
and x(t) = 1 − 2te-t − e-2t 
 
b) Applying the final Value Theorem 
 
2)2()1(
2lim)(lim)(lim 200 =++== →→∞→ ssssXtx sst 
 
 [ Note that Final Value Theorem is applicable here] 
 
 
 
 3.9 
 
a) )4)(5)(4(
)2(6
)4)(209(
)2(6)( 2 +++
+
=
+++
+
=
sss
s
sss
s
sX 
 
 0)4)(5(
)2(6lim)0( 2 =


++
+
=
∞→ ss
ss
x
s
 
0)4)(5(
)2(6lim)( 20 =


++
+
=∞
→ ss
ss
x
s
 
 
x(t) is converging (or bounded) because [sX(s)] does not have a limit at 
s = −4, and s = −5 only, i.e., it has a limit for all real values of s ≥ 0. 
 
x(t) is smooth because the denominator of [sX(s)] is a product of real 
factors only. See Fig. S3.9a. 
 
b) )2)(23()23(
310
)2)(106(
310)(
2
2
2
+−−+−
−
=
++−
−
=
sjsjs
s
sss
s
sX 
 
 10)2)(106(
310lim)0( 2
3
=


++−
−
=
∞→ sss
ss
x
s
 
 
Application of final value theorem is not valid because [sX(s)] does not 
have a limit for some real s ≥ 0, i.e., at s = 3±2j. For the same reason, x(t) 
is diverging (unbounded). 
 
x(t) is oscillatory because the denominator of [sX(s)] includes complex 
factors. See Fig. S3.9b. 
3-10 
c) )3()3(
516
)9(
516)( 2 jsjs
s
s
s
sX
−+
+
=
+
+
= 
16)9(
516lim)0( 2
2
=


+
+
=
∞→ s
ss
x
s
 
 
Application of final value theorem is not valid because [sX(s)] does not 
have a limit for real s = 0. This implies that x(t) is not diverging, since 
divergence occurs only if [sX(s)] does not have a limit for some real value 
of s>0. 
 
x(t) is oscillatory because the denominator of [sX(s)] is a product of 
complex factors. Since x(t) is oscillatory, it is not converging either. See 
Fig. S3.9c 
 
0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 5
-0.05
0
0.05
0.1
0.15
0.2
0.25
0.3
0.35
0.4
Time
x
(t)
 
 Figure S3.9a. Simulation of X(s) for case a) 
 
 
0 0.5 1 1.5 2 2.5
-12000
-10000
-8000
-6000
-4000
-2000
0
2000
4000
6000
8000
Time 
x(t
) 
 
 Figure S3.9b. Simulation of X(s) for case b) 
3-11 
 
0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 5
-20
-15
-10
-5
0
5
10
15
20
Time
x
(t)
 
 Figure S3.9c. Simulation of X(s) for case c) 
 
 
The Simulink block diagram is shown below. An impulse input should be 
used to obtain the function’s behavior. In this case note that the impulse 
input is simulated by a rectangular pulse input of very short duration. (At 
time t = 0 and t =0.001 with changes of magnitude 1000 and –1000 
respectively). The MATLAB command impulse might also be used. 
 
Figure S3.9d. Simulink block diagram for cases a), b) and c). 
3-12 
 3.10 
 
 
 a) 
i) Y(s) = 
4)4(
2
)4(
2
222 +
++=
+
=
+ s
C
s
B
s
A
sssss
 
 
∴ y(t) will contain terms of form: constant, t, e-4t 
 
ii) Y(s) = 
31)3)(1(
2
)34(
2
2 +
+
+
+=
++
=
++ s
C
s
B
s
A
ssssss
 
 
∴ y(t) will contain terms of form: constant, e-t, e-3t 
 
iii) 
2)2()2(
2
)44(
2)( 222 ++++=+=++= s
C
s
B
s
A
sssss
sY 
 
∴ y(t) will contain terms of form: constant, e-2t , te-2t 
 
iv) )84(
2)( 2 ++= ssssY 
 
2222 2)2()48()44(84 ++=−+++=++ sssss 
]2)2[(
2)( 22 ++= sssY 
 
∴ y(t) will contain terms of form: constant, e-2t sin2t, e-2tcos2t 
 
b) 2222222 22)2(
)1(2
)4(
)1(2)(
+
+
+
+=
+
+
=
+
+
=
s
C
s
Bs
s
A
ss
s
ss
s
sY 
A =
2
1
)4(
)1(2lim 20 =+
+
→ s
s
s
 
 
2(s+1) = A(s2+4) + Bs(s) + Cs 
2s+2 = As2 + 4A + Bs2 + Cs 
 
Equating coefficients on like powers of s 
 s
2: 0 = A + B → B = −A = − 
2
1
 
 s
1: 2 = C → C = 2 
 s
0: 2 = 4A → A = 
2
1
 
3-13 
∴ Y(s) 2222 2
2
2
)21(21
+
+
+
−
+=
ss
s
s
 
 
y(t) = tt 2sin
2
22cos
2
1
2
1
+− 
 
y(t) = tt 2sin)2cos1(
2
1
+− 
 
 
 3.11 
 
 
Since convergent and oscillatory behavior does not depend on initial 
conditions, assume 0)0()0()0(2
2
=== x
dt
dx
dt
dx
 
a) Laplace transform of the equation gives 
 
3 2 3( ) 2 ( ) 2 ( ) ( )s X s s X s sX s X s
s
+ + + = 
)
2
3
2
1)(
2
3
2
1)(1(
3
)122(
3)( 23
jsjsssssss
sX
−++++
=
+++
= 
Denominator of [sX(s)] contains complex factors so that x(t) is oscillatory, 
and denominator vanishes at real values of s= −1 and -½ which are all <0 
so that x(t) is convergent. See Fig. S3.11a. 
b) 
1
2)()(2
−
=−
s
sXsXs 
)1()1(
2
)1)(1(
2)( 22 +−=−−= sssssXThe denominator contains no complex factors; x(t) is not oscillatory. 
The denominator vanishes at s=1 ≥0; x(t) is divergent. See Fig. S3.11b. 
 
c) 
1
1)()( 23 +=+ ssXsXs 
)
2
3
2
1)(
2
3
2
1)(1)()((
1
)1)(1(
1)( 32
jsjssjsjsss
sX
−−+−+−+
=
++
=
 
 
The denominator contains complex factors; x(t) is oscillatory. 
The denominator vanishes at real s = 0, ½; x(t) is not convergent. See Fig. 
S3.11c. 
3-14 
d) 
s
ssXsXs 4)()(2 =+ 
)1(
4
)(
4)( 22 +=+= ssssssX 
 
The denominator of [sX(s)] contains no complex factors; x(t) is not 
oscillatory. 
The denominator of [sX(s)] vanishes at s = 0; x(t) is not convergent. See 
Fig. S3.11d. 
 
0 1 2 3 4 5 6 7 8 9 10
-0.5
0
0.5
1
1.5
2
2.5
3
3.5
time
x
(t)
 
 Figure S3.11a. Simulation of X(s) for case a) 
 
0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 5
0
100
200
300
400
500
600
700
time
x
(t)
 
 Figure S3.11b. Simulation of X(s) for case b) 
3-15 
0 1 2 3 4 5 6 7 8 9 10
-40
-20
0
20
40
60
80
time
x
(t)
 
 Figure S3.11c. Simulation of X(s) for case c) 
 
0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 5
0
2
4
6
8
10
12
14
16
18
time 
x
(t) 
 
 Figure S3.11d. Simulation of X(s) for case d) 
 
 
 3.12 
 
 
Since the time function in the solution is not a function of initial 
conditions, we Laplace Transform with 
 
 0)0()0( ==
dt
dx
x 
 
 τ1τ2s
2X(s) + (τ1+τ2)sX(s) + X(s) = KU(s) 
3-16 
 )(
1)()( 21221
sU
ss
K
sX
+τ+τ+ττ
= 
 
 Factoring denominator 
 
 )()1)(1()( 21
sU
ss
K
sX
+τ+τ
= 
 
a) If u(t) = a S(t) then U(s)=
s
a
 
 
)1)(1()( 21 +τ+τ
=
sss
Ka
sX a 21 τ≠τ 
 
xa(t) = fa( S(t), e-t/τ1, e– t/τ2) 
 
b) If u(t) = be-t/τ then U(s) = 
1+τ
τ
s
b
 
 
)1)(1)(1()( 21 +τ+τ+τ
τ
=
sss
Kb
sX b 21 τ≠τ≠τ 
 
xb(t) = fb(e-t/τ , e-t/τ1, e– t/τ2) 
 
 
c) If u(t) =ce-t/τ where τ = τ1 , then U(s) = 11 +τ
τ
s
c
 
)1()1()( 221 +τ+τ
τ
=
ss
Kc
sX c 
 
xc(t) = fc(e– t/τ1, t e– t/τ1, e– t/τ2) 
 
 d) If u(t) = d sin ωt then U(s) = 22 ω+
ω
s
d
 
 
 )1)(1)(()( 2122 +τ+τω+
=
sss
Kd
sX d 
 
xd(t) = fd(e– t/τ1, e– t/τ2, sin ωt, cos ωt) 
 
 
 
3-17 
 3.13 
 
a) 
3 2
3 2
(0) (0)4 with (0) 0tdx d x dxx e x
dt dt dt
+ = = = = 
 
 Laplace transform of the equation, 
 
 
1
14
−
=+
s
X(s)X(s)s 3 
 
 )37.179.0)(37.179.0)(59.1)(1(
1
)4)(1(
1)( 3 jsjssssssX −−+−+−=+−= 
 
 js
j
js
j
ss 37.179.037.179.059.11
333321
−−
β−α
+
+−
β+α
+
+
α
+
−
α
= 
 
 
5
1
)4(
1
1
31 =+
=α
=s
s
 
 
6.19
1
)37.179.0)(37.179.0)(1(
1
59.1
2 −=
−−+−−
=α
−=s
jsjss 
 jjsssj js
59.074.0)37.179.0)(59.1)(1(
1
37.179.0
33 −−=
−−+−
=β+α
−=
 
 
X(s) js
j
js
j
ss 37.179.0
059.0074.0
37.179.0
059.0074.0
59.1
6.19
1
1
5
1
−−
+−
+
+−
−−
+
+
−
+
−
= 
 
)37.1sin059.037.1cos074.0(2
6.19
1
5
1)( 79.059.1 tteeetx ttt +−−= − 
b) 12 sin 3 with (0) 0dx x t x
dt
− = = 
 
 
9
312(s) 2 +=− sX(s)sX 
 
 )12)(3)(3(
3
)12)(9(
3)( 2
−−+
=
−+
=
sjsjssssX 
 
 
1233
31111
−
α
+
−
β−α
+
+
β+α
=
sjs
j
js
j
 
3-18 
 jjsjsj js 102
4
102
1
7218
3
)12)(3(
3
3
11 −−=+−
=
−−
=β+α
−=
 
 
 3 2
12
3 1
9 51
s
( s )
=
α = =
+
 
 
 
12
51
1
3
102
4
102
1
3
102
4
102
1
)(
−
+
−
+−
+
+
−−
=
sjs
j
js
j
sX 
 
 
tetttx 12
51
1)3sin43(cos
51
1)( ++−= 
 
c) 
2
2
(0)6 25 with (0) 0td x dx dxx e x
dt dt dt
−+ + = = = 
 
2
2
1 16 25 or
1 1 6 25
s X( s ) sX( s ) X( s ) X( s ) X ( s )
s ( s )( s s )+ + + = =+ + + +
 
js
j
js
j
sjsjsssX 43431)43)(43)(1(
1)( 22221
−+
β−α
+
++
β+α
+
+
α
=
−++++
= 
20
1
)256(
1
1
21 =++
=α
−=s
ss
 
 
jjssj js 80
1
40
1
)43)(1(
1
43
22 −−=
−++
=β+α
−−=
 
 
 js
j
js
j
s
sX
43
80
1
40
1
43
80
1
40
1
1
20
1
)(
−+
−−
+
++
−−
+
+
= 
 
 )4sin
40
14cos
20
1(
20
1)( 3 tteetx tt +−= −− 
 
d) Laplace transforming (assuming initial conditions = 0, since they do not 
affect results) 
 
sY1(s) + Y2(s) = X1(s) (1) 
 
sY2(s) – 2Y1(s) + 3 Y2(s) = X2(s) (2) 
 
3-19 
 From (2), 
 
 (s+3) Y2(s) = X2(s) + 2Y1(s) 
 
 )(
3
2)(
3
1)( 122 sY
s
sX
s
sY
+
+
+
= 
 
 Substitute in Eq.1 
 
sY1(s) + )(3
2)(
3
1
12 sY
s
sX
s +
+
+
= X1(s) 
 
 We neglect X2(s) since it is equal to zero. 
 
 [ ] )()3()(2)3( 11 sXssYss +=++ 
 
 )()3()()23( 112 sXssYss +=++ 
 
 )()2)(1(
3)(
23
3)( 1121 sXss
s
sX
ss
s
sY
++
+
=
++
+
= 
 
 Now if x1(t) = e-t then X1(s) = 1
1
+s
 
∴ 
2)1()1()2()1(
3)( 221 +++++=++
+
=
s
C
s
B
s
A
ss
s
sY 
 
so that y1(t) will contain e-t/τ , te-t/τ, e–2t functions of time. 
 
 For Y2(s) 
 
 
2)1()1()2()1(
2)( 222 +++++=++= s
C
s
B
s
A
ss
sY 
 
so that y2(t) will contain the same functions of time as y1(t) (although 
different coefficients). 
 
 
 
 
 
 
 
 
 
3-20 
 3.14 
 
 )2()2(4)()(3)(2
2
−−
−
=++ tx
dt
xd
ty
dt
tdy
dt
tyd
 
 
 Taking the Laplace transform and assuming zero initial conditions, 
 
 s
2Y(s) + 3sY(s) + Y(s) = 4 e-2ssX(s) −e-2sX(s) 
 
 Rearranging, 
 
 
13
)41()()(
)(
2
2
++
−−
==
−
ss
es
sG
sX
sY s
 (1) 
 
a) The standard form of the denominator is : τ2s2 + 2ζτs + 1 
 
From (1) , τ = 1 , ζ = 1.5 
 
Thus the system will exhibit overdamped and non-oscillatory response. 
 
b) Steady-state gain 
 
1)(lim
0
−==
→
sGK
s
 (from (1)) 
 
c) For a step change in x 
X(s) = 
s
5.1
 and Y(s) =
sss
es s 5.1
)13(
)41(
2
2
++
−−
−
 
 
Therefore )(ˆ ty = −1.5 + 1.5e-1.5t cosh(1.11t) + 7.38e-1.5t sinh(1.11t) 
Using MATLAB-Simulink, y(t)= )2(ˆ −ty is shown in Fig. S3.14 
 
0 5 10 15 20 25 30
-1.5
-1
-0.5
0
0.5
1
1.5
 
Figure S3.14. Output variable for a step change in x of magnitude 1.5 
3-21 
 3.15 
 
 
 )/1()()( hthSthStf −−= 
 [ ])/1()(4 htStShx
dt
dx
−−=+ , x(0)=0 
 
 Take Laplace transform, 
 
 



−=+
−
s
e
s
hsXssX
hs /1)(4)( 
 


+
α
+
α
−=
+
−=
−−
4
)1()4(
1)1()( 21//
ss
eh
ss
ehsX hshs 
 
4
1
4
1
0
1 =+
=α
=ss
 , 
4
11
4
2 −==α
−=ss
 
 
 


+
−−=
−
4
11)1(
4
)( /
ss
e
h
sX hs 
 
 


+
+
+
−−=
−−
44
11
4
//
s
e
ss
e
s
h hshs
 
 
 0 t <0 
 =)(tx )1(
4
4te
h
−
− 0 < t <1/h 
 
[ ]tht eeh 4)/1(4
4
−−−
− t > 1/h 
0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 2
0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1
time
x
(t)
h=1
h=10
h=100
 
 Figure S3.15. Solution for values h= 1, 10 and 100 
3-22 
3.16 
 
 
 a) Laplace transforming 
 
 
[ ] [ ]
1
)(9)0()(6)0()0()( 22 +=+−+′−− s
s
sYyssYysysYs 
 
 (s2 + 6s + 9)Y(s) − s(1) − 2 –(6)(1)=
12 +s
s
 
 
 (s
2 
+ 6s + 9)Y(s) = 
12 +s
s
+ s + 8 
 
 (s
2 
+ 6s + 9)Y(s) =
1
88
2
23
+
++++
s
ssss
 
 
 Y(s) = )1()3(
828
22
23
++
+++
ss
sss
 
 
 To find y(t) we have to expand Y(s) into its partial fractions 
 
 
113)3()( 222 +++++++= s
D
s
Cs
s
B
s
A
sY 
 
 y(t) = Ate-3t + Be-3t + C cost + D sint 
 
 
b) Y(s)= )84(
1
2 ++
+
sss
s
 
 
Since 8
4
42
< we know we will have complex factors. 
 
∴ complete square in denominator 
 
s
2 
+ 4s + 8 = s
2 
+ 4s + 4 + 8−4 
 
 = s
2 
+ 4s + 4 + 4 = (s+2)
2 
+ (2)
2
 { b = 2 , ω=2} 
 
∴ Partial fraction expansion gives 
 
3-23 
)84(
1
8484
)2()( 222 ++
+
=
++
+
++
+
+=
sss
s
ss
C
ss
sB
s
A
sY 
 
Multiply by s and let s→0 
 
 A=1/8 
 
 Multiply by s(s2+4s+8) 
 
 A(s2+4s+8) + B(s+2)s + Cs = s + 1 
 
 As2 + 4As + 8A + Bs2 + 2Bs + Cs = s + 1 
 
 
s
2: A + B = 0 → B = −A = − 
8
1
 
s
1: 4A + 2B + C = 1 → C = 1 + 2 


8
1
− 4 


8
1
=
4
3
 
s
0: 8A = 1 → A = 
8
1
 (This checks with above result) 
 
 
( )
2222 2)2(
4/3
2)2(
)2(8/18/1)(
++
+
++
+−
+=
ss
s
s
sY 
 
 y(t) = 


8
1
 − 


8
1
e-2t cos 2t + 


8
3
e-2t sin 2t 
 
 
 
 3.17 
 
 
 V iqCqCdt
dC
=+ 
 
 Since V and q are constant, we can Laplace Transform 
 
 sVC(s) + qC(s) = q Ci(s) 
 
 Note that c(t = 0) = 0 
 
 Also, ci(t) = 0 , t ≤0 
 ci(t) = ic , t > 0 
3-24 
 
 Laplace transforming the input function, a constant, 
 
 
s
c
sC ii =)( 
 so that 
 sVC(s) + qC(s) = q
s
ci
 or C(s) = 
sqsV
cq i
)( + 
 
 Dividing numerator and denominator by q 
 
 C(s) = 
ss
q
V
ci




+1
 
 
 Use Transform pair #3 in Table 3.1 to invert (τ =V/q) 
 
 c(t) = ic 



−
− t
q
V
e1 
 
Using MATLAB, the concentration response is shown in Fig. S3.17. 
(Consider V = 2 m3, Ci=50 Kg/m3 and q = 0.4 m3/min) 
 
 
 
0 5 10 15 20 25 30
0
5
10
15
20
25
30
35
40
45
50
Time
c(t
)
 
 Figure S3.17. Concentration response of the reactor effluent stream. 
 
 
3-25 
 3.18 
 
a) If Y(s) = )( 22 ω+
ω
ss
KA
 
and input U(s) = )( 22 ω+
ω
s
A
 = � {A sin ωt} 
 
then the differential equation had to be 
 
 )(tKu
dt
dy
= with y(0) = 0 
 
b) Y(s) = )( 22 ω+
ω
ss
KA
= 22
3
22
21
ω+
ωα
+
ω+
α
+
α
ss
s
s
 
 
ω
=
ω+
ω
=α
→
KA
s
KA
s 0
221 
 
Find α2 and α3 by equating coefficients 
 
 KAω= α1(s2+ω2) + α2s2+α3ωs 
 
 KAω = α1s2 + α1ω2 + α2s2 + α3ωs 
s
2
 : 0 = α1 + α2 → α2 = −α1 = 
ω
− KA
 
 
 s: 0 = α3 ω → α3 = 0 
 
 
∴ Y(s) = )( 22 ω+
ω
ss
KA
= 22
)/(/
ω+
ω
−
ω
s
sKA
s
KA
 
y(t) = )cos1( tKA ω−
ω
 
 
 
 
 
 
 
 
 
 
 
3-26 
 c) 
A 
-A 
 0
 Time
 y(t)
u(t) 
2KA/ωωωω
 
 
i) We see that y(t) follows behind u(t) by 1/4 cycle = 2pi/4= pi/2 rad. 
which is constant for all ω 
 
ii) The amplitudes of the two sinusoidal quantities are: 
 
y : KA/ω 
u: A 
 
 Thus their ratio is K/ω, which is a function of frequency. 
 
 
 
4-1 
��������	�
 
 
 
 
 4.1 
 
 
a) iii 
b) iii 
c) v 
d) v 
 
 
 4.2 
 
 
a) 5 
 
b) 10 
c) )110(
10)(
+
=
ss
sY 
From the Final Value Theorem, y(t) = 10 when t→∞ 
 
 d) y(t) = 10(1−e−t/10) , then y(10) = 6.32 = 63.2% of the final value. 
 
e) 
s
e
s
sY
s )1(
)110(
5)(
−
−
+
= 
 From the Final Value Theorem, y(t) = 0 when t→∞ 
 
f) 1)110(
5)(
+
=
s
sY 
 From the Final Value Theorem, y(t)= 0 when t→∞ 
 
g) )9(
6
)110(
5)( 2 ++= sssY then 
 
y(t) = 0.33e-0.1t − 0.33cos(3t) + 0.011sin(3t) 
 
The sinusoidal input produces a sinusoidal output and y(t) does not have a 
limit when t→∞. 
 
Solution Manual for Process Dynamics and Control, 2nd edition, 
Copyright © 2004 by Dale E. Seborg, Thomas F. Edgar and Duncan A. Mellichamp. 
4-2 
 By using Simulink-MATLAB, above solutions can be verified: 
 
 
0 5 10 15 20 25 30 35 40 45 50
0
1
2
3
4
5
6
7
8
9
10
time
y(t
)
 
0 5 10 15 20 25 30 35 40 45 50
0
0.05
0.1
0.15
0.2
0.25
0.3
0.35
0.4
0.45
0.5
time
y(t
)
 
 Fig S4.2a. Output for part c) and d) Fig S4.2b. Output for part e) 
 
 
0 5 10 15 20 25 30 35 40 45 50
0
0.5
1
1.5
2
2.5
3
3.5
4
4.5
5
time
y(t
)
 
0 2 4 6 8 10 12 14 16 18 20
-0.3
-0.2
-0.1
0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
time
y(t
)
 
 
 Fig S4.2c. Output for part f) Fig S4.2d. Output for part g) 
 
 
 
 
 4.3 
 
 
a) The dynamic model of the system is given by 
 
)(1 ww
dt
dV
i −ρ
= (2-45) 
CV
QTT
V
w
dt
dT
i
i
ρ
+−
ρ
= )( (2-46) 
 
Let the right-hand side of Eq. 2-46 be f(wi,V,T), 
 
( ) T
T
fV
V
f
w
w
fTVwf
dt
dT
ss
i
si
i ′


∂
∂
+′


∂
∂
+′



∂
∂
== ,, (1) 
4-3 
)(1 TT
Vw
f
i
si
−
ρ
=



∂
∂
 
 
01)( 22 =


−=
ρ
−−
ρ
−=


∂
∂
s
i
i
s dt
dT
VCV
QTT
V
w
V
f
 
 
ρ
−=


∂
∂
V
w
T
f i
s
 
 
=
dt
dT
ii wTTV
′−
ρ
)(1 T
V
wi
′
ρ
− , 
dt
Td
dt
dT ′
= 
 
 Taking Laplace transform and rearranging 
 
 
1
/)(
)(
)(
+


 ρ
−
=
′
′
s
w
V
wTT
sW
sT
i
ii
i
 (2) 
 
Laplace transform of Eq. 2-45 gives 
s
sW
sV i
ρ
′
=′
)()( (3) 
 
If 
sV
f 


∂
∂
were not zero, then using (3) 
 
 
1
1)(
)(
)(
+


 ρ


 


∂
∂
+
−
=
′
′
s
w
V
sV
f
w
V
w
TT
sW
sT
i
sii
i
i
 (4) 
 
Appelpolscher guessed the incorrect form (4) instead of the correct form 
(2) because he forgot that 
sV
f 


∂
∂
would vanish. 
 
b) From Eq. 3, 
 
 
ssW
sV
i ρ
=
′
′ 1
)(
)(
 
 
 
 
4-4 
 4.4 
 
 
1
KY( s ) G( s )X( s )
s( s )= = τ + 
 
 
 G(s) Interpretation U(s) Interpretation 
 of G(s) of u(t) 
 
 )1( +τss
K
 2nd order process * 1 δ(0) [ Delta function] 
 
 
1+τs
K
 1st order process 
s
1
 S(0) [Unit step function] 
 
 
 
s
K
 Integrator 
1+τs
K
 
τ−
τ
/1te [Exponential input] 
 
 
 K Simple gain )1(
1
+τss
 
τ−
−
/1 te 
 (i.e no dynamics) [Step + exponential input] 
 
 
* 2nd order or combination of integrator and 1st order process 
 
 
 
 
 4.5 
 
 a) 2
dt
d 1y
 
= -2y1 – 3y2 + 2u1 (1) 
 
dt
d 2y = 4y1 – 6y2 + 2u1 + 4u2 (2) 
 
 Taking Laplace transform of the above equations and rearranging, 
 
 (2s+2)Y1(s) + 3Y2(s) = 2U1(s) (3) 
 
 -4 Y1(s) + (s+6)Y2(s)=2U1(s) + 4U2(s) (4) 
 
 Solving Eqs. 3 and 4 simultaneously for Y1(s) and Y2(s), 
4-5 
 Y1(s) = )4)(3(2
)(U12)(U)3(2
24142
)(U12)(U)62( 21
2
21
++
−+
=
++
−+
ss
sss
ss
sss
 
 
 Y2(s) = )4)(3(2
)(U)1(8)(U)3(4
24142
)(U)88()(U)124( 21
2
21
++
+++
=
++
+−+
ss
ssss
ss
ssss
 
 
 Therefore, 
 
 
4
1
)(U
)(Y
1
1
+
=
ss
s
 , )4)(3(
6
)(U
)(Y
2
1
++
−
=
sss
s
 
 
 
4
2
)(U
)(Y
1
2
+
=
ss
s
 , )4)(3(
)1(4
)(U
)(Y
2
2
++
+
=
ss
s
s
s
 
 
 
 
 4.6 
 
 
 The physical model of the CSTR is (Section 2.4.6) 
 
 AAAi
A Vkcccq
dt
dcV −−= )( (2-66) 
 
 )()()( TTUAVkcHTTwC
dt
dTCV cAi −+∆−+−=ρ (2-68) 
 
 where: k = ko e-E/RT (2-63) 
 
 These equations can be written as, 
 
 ),(1 Tcfdt
dc
A
A
= (1) 
 
 ),,(2 cA TTcfdt
dT
= (2) 
 
Because both equations are nonlinear, linearization is required. After 
linearization and introduction of deviation variables, we could get an 
expression for )(scA′ / )(sT ′ . 
 
4-6 
But it is not possible to get an expression for )(sT ′ / )(sTc′ from (2) due to 
the presence of cA in (2). Thus the proposed approach is not feasible 
because the CSTR is an interacting system. 
 
Better approach: 
 
After linearization etc., solve for )(sT ′ from (1) and substitute into the 
linearized version of (2). Then rearrange to obtain the desired, AC ( s )′ / 
)(sTc′ (See Section 4.3) 
 
 
 4.7 
 
 
a) The assumption that H is constant is redundant. For equimolal overflow, 
 
 LLL == 10 , VVV == 21 
 
01120 =−−+= VLVLdt
dH
 , i.e., H is constant. 
 
 The simplified stage concentration model becomes 
 
 )()( 12101 yyVxxLdt
dxH −+−= (1) 
 y1 = a0 + a1x1 + a2x12 +a3x13 (2) 
 
 
 b) Let the right-hand side of Eq. 1 be f(L, x0, x1, V, y1, y2) 
 
 
 == ),,,,,( 21101 yyVxxLfdt
dxH 1
1
0
0
x
x
f
x
x
fL
L
f
sss
′



∂
∂
+′



∂
∂
+′


∂
∂
 
 
 2
2
1
1
y
y
fy
y
fV
V
f
sss
′



∂
∂
+′



∂
∂
+′


∂
∂
+ 
 
 Substituting for the partial derivatives and noting that 
dt
xd
dt
dx 11 ′
= 
 =
′
dt
xdH 1 12121010 )()( yVyVVyyxLxLLxx ′−′+′−+′−′+′− (3) 
 
4-7 
 Similarly, 
 
 1
2
131211
1
11 )32()( xxaxaax
x
g
xgy
s
′++=′



∂
∂
==′ (4) 
 
 c) For constant liquid and vapor flow rates, 0=′=′ VL 
 
 Taking Laplace transform of Eqs. 3 and 4, 
 
 )()()()()( 12101 sYVsYVsXLsXLsXHs ′−′+′−′=′ (5) 
 
 )()32()( 12131211 sXxaxaasY ′++=′ (6) 
 
 
 From Eqs. 5 and 6, the desired transfer functions are 
 
 
1)(
)(
0
1
+τ
τ
=
′
′
s
H
L
sX
sX
 , 
1)(
)(
2
1
+τ
τ
=
′
′
s
H
V
sY
sX
 
 
1
)32(
)(
)(
2
13121
0
1
+τ
τ++
=
′
′
s
H
L
xaxaa
sX
sY
 
 
1
)32(
)(
)(
2
13121
2
1
+τ
τ++
=
′
′
s
H
V
xaxaa
sY
sY
 
 
 where 
 
)32( 213121 xaxaaVL
H
+++
=τ 
 
 
 
 4.8 
 
 
 From material balance, 
 
 
5.1)( Rhw
dt
Ahd
i −=
ρ
 
 
 
5.11 h
A
R
w
Adt
dh
i ρ
−
ρ
= 
4-8 
 We need to use a Taylor series expansion to linearize 
 
 )(5.1)(11
5.0
5.1 hh
A
hR
ww
A
h
A
R
w
Adt
dh
iii −ρ
−−
ρ
+


ρ
−
ρ
= 
 
 Since the bracketed term is identically zero at steady state, 
 
 h
A
hR
w
Adt
hd
i ′ρ
−
′
ρ
=
′
5.05.11
 
 
 Rearranging 
 
 iwhR
h
dt
hd
hR
A
′=′+
′ρ
5.05.0 5.1
1
5.1
 
 
 
 Hence 
1)(
)(
+τ
=
′
′
s
K
sW
sH
i
 
 
 where 
====
w
h
hR
h
hR
K
5.15.15.1
1
5.15.0
[ ]
[ ]flowrate
height
 
=
ρ
=
ρ
=
ρ
=τ
w
V
hR
hA
hR
A
5.15.15.1 5.15.0
[ ]
[ ] [ ]timetimemass
mass
=
/
 
 
 
 
 4.9 
 
 
a) The model for the system is given by 
 
)()( TTAhTTwC
dt
dT
mC wppi −+−= (2-51) 
 
 )()( TTAhTTAh
dt
dTCm wppwssswww −−−= (2-52) 
 
 
Assume that m, mw, C, Cw, hp, hs, Ap, As, and w are constant. Rewriting the 
above equations in terms of deviation variables, and noting that 
 
4-9 
dt
Td
dt
dT ′
= 
dt
Td
dt
dT ww ′
= 
 
 )()( TTAhTTwC
dt
Td
mC wppi ′−′+′−′=
′
 
 )()0( TTAhTAh
dt
TdCm wppwsswww ′−′−′−=
′
 
 
 Taking Laplace transforms and rearranging, 
 
 )()()()( sTAhsTwCsTAhwCmCs wppipp ′+′=′++ (1) 
 
 )()()( sTAhsTAhAhsCm ppwppssww ′=′++ (2) 
 
 Substituting in Eq. 1 for )(sTw′ from Eq. 2, 
 
 ppipp AhsTwCsTAhwCmCs +′=′++ )()()( )()( sTAhAhsCm
Ah
ppssww
pp
′
++
 
 
 Therefore, 
 
=
′
′
)(
)(
sT
sT
i
2)())((
)(
ppppsswwpp
ppssww
AhAhAhsCmAhwCmCs
AhAhsCmwC
−++++
++
 
 
 
 b) The gain is =


′
′
=0)(
)(
si
sT
sT
ppssppss
ppss
AhAhAhAhwC
AhAhwC
++
+
)(
)(
 
 
c) No, the gain would be expected to be 1 only if the tank were insulated so 
that hpAp= 0. For heated tank the gain is not 1 because heat input changes 
as T changes. 
 
 
 
 4.10 
 
 
 Additional assumptions 
 
1) perfect mixing in the tank 
2) constant density, ρ , and specific heat, C. 
3) Ti is constant. 
4-10 
 
Energy balance for the tank, 
 
)()()( ai TTAbvUQTTwCdt
dTVC −+−+−=ρ 
 
Let the right-hand side be f(T,v), 
 
==ρ ),( vTf
dt
dTVC v
v
fT
T
f
ss
′


∂
∂
+′


∂
∂
 (1) 
 
 wC
T
f
s
−=


∂
∂ AvbU )( + 
 
 =


∂
∂
sv
f )( aTTbA −− 
 
 Substituting for the partial derivatives in Eq. 1 and noting that 
dt
dT
=
dt
Td ′
 
 =
′ρ
dt
TdVC [ ] vTTbATAvbUwC a ′−−′++− )()( 
 
 Taking the Laplace transform and rearranging 
 
 [ ] )()( sTAvbUwCVCs ′+++ρ = vTTbA a ′−− )( (s) 
 
 
1)(
)(
)(
)(
)(
+


++
ρ



++
−−
=
′
′
s
AvbUwC
VC
AvbUwC
TTbA
sv
sT
a
 
 
 
 4.11 
 
 
a) Mass balances on surge tanks 
 
21
1 ww
dt
dm
−= (1) 
 
32
2 ww
dt
dm
−= (2) 
4-11 
 Ideal gas law 
 RT
M
mVP 111 = (3) 
 RT
M
mVP 222 = (4) 
 
 Flows (Ohm's law is 
Resistance
ForceDriving
==
R
EI ) 
 )(1 1
1
1 PPR
w c −= (5) 
 )(1 21
2
2 PPR
w −= (6) 
 )(1 2
3
3 hPPR
w −= (7) 
 
 Degrees of freedom: 
 
 number of parameters : 8 (V1, V2, M, R, T, R1, R2, R3) 
 
 number of variables : 9 (m1, m2, w1, w2, w3, P1, P2, Pc, Ph) 
 
 number of equations : 7 
 
 ∴ number of degrees of freedom thatmust be eliminated = 9 − 7 = 2 
 
Since Pc and Ph are known functions of time (i.e., inputs), NF = 0. 
 
 
b) Development of model 
 
Substitute (3) into (1) : 21
11 ww
dt
dP
RT
MV
−= (8) 
 
Substitute (4) into (2) : 32
22 ww
dt
dP
RT
MV
−= (9) 
 
Substitute (5) and (6) into (8) : 
 
 )(1)(1 21
2
1
1
11 PP
R
PP
Rdt
dP
RT
MV
c −−−= 
 2
2
1
211
11 1)11()(1 P
R
P
RR
tP
Rdt
dP
RT
MV
c ++−= (10) 
4-12 
Substitute (6) and (7) into (9): 
 
 )(1)(1 2
3
21
2
22
hPPR
PP
Rdt
dP
RT
MV
−−−= 
 
 )(1)11(1
3
2
32
1
2
22 tP
R
P
RR
P
Rdt
dP
RT
MV
h++−= (11) 
 
 Note that ),( 2111 PPfdt
dP
= from Eq. 10 
 
 ),( 2122 PPfdt
dP
= from Eq. 11 
 
This is exactly the same situation depicted in Figure 6.13, therefore the 
two tanks interact. This system has the following characteristics: 
 
i) Interacting (Eqs. 10 and 11 interact with each other ) 
ii) 2nd-order denominator (2 differential equations) 
iii) Zero-order numerator (See example 4.4 in text) 
iv) No integrating elements 
v) Gain of )(
)(3
sP
sW
c
′
′
 is not equal to unity. (Cannot be because 
the units on the two variables are different). 
 
 
 
 
 4.12 
 
 
 
a) 2/1hCq
dt
dhA vi −= 
 
Let f = 2/1hCq vi − 
 
Then f ≈ )(
2
1 2/12/1 hhhCqqhCq viivi −−−+−
−
 
 
so h
h
C
q
dt
hdA vi ′−′=
′
2/12
 because 02/1 ≡− hCq vi 
 
4-13 
)()(
2 2/1
sQsH
h
C
sA iv ′=′


+ 
 
2/12
1
)(
)('
h
C
sAsQ
sH
vi +
=
′
 Note: Not a standard form 
 
12
/2
)(
)('
2/1
2/1
+
=
′
s
C
hA
Ch
sQ
sH
v
v
i
 
 
where 
vC
hK
2/12
= and 
vC
hA 2/12
=τ 
 
b) Because 2/1hCq v= 
h
K
h
h
ChhCq vv ′=′=′=′ −
1
22
1
2/1
2/1
 
 
 ∴ 
KsH
sQ 1
)(
)(
=
′
′
 , 
1
1
)(
)(
)(
)(
+τ
=
′
′
′
′
s
K
KsQ
sH
sH
sQ
i
 
 
and 
1
1
)(
)(
+τ
=
′
′
ssQ
sQ
i
 
 
 c) For a linear outlflow relation 
 
 hCq
dt
dhA vi
*
−= Note that vv CC ≠
*
 
 
hCq
dt
hdA vi ′−′=
′ *
 
 
iv qhCdt
hdA ′=′+′ * or i
vv
q
C
h
dt
hd
C
A
′=′+
′
**
1
 
 
Multiplying numerator and denominator by h on each side yields 
 
i
vv
q
hC
hh
dt
hd
hC
hA
′=′+
′
**
 
 
4-14 
or i
ii
q
q
hh
dt
hd
q
V
′=′+
′
 
 
iq
V
=τ∗ 
iq
hK =∗ q.e.d 
 
 To put τ and K in comparable terms for the square root outflow form of 
the transfer function, multiply numerator and denominator of each by 
2/1h . 
 
*
2/12/1
2/12/1
2222 K
q
h
hC
h
h
h
C
hK
ivv
==== 
 
*
2/12/1
2/12/1
2222 τ====τ
ivv q
V
hC
hA
h
h
C
hA
 
 
Thus level in the square root outflow TF is twice as sensitive to changes in 
qi and reacts only ½ as fast (two times more slowly) since τ = 2 ∗τ . 
 
 
 
 
 4.13 
 
 
a) The nonlinear dynamic model for the tank is: 
 
 ( )1( ) i vdh q C hdt D h h= −pi − (1) 
 
 (corrected nonlinear ODE; model in first printing of book is incorrect) 
 
 To linearize Eq. 1 about the operating point ( , )i ih h q q= = , let 
 
 ( )
i vq C hf
D h h
−
=
pi −
 
 
 Then, 
 ( , )if h q ≈ i
s i s
f fh q
h q
 ∂ ∂ 
′ ′+   ∂ ∂    
where 
 
4-15 
 
1
( )i s
f
q D h h
 ∂
= ∂ pi −  
 ( ) ( )2
1 1 2
2 ( ) ( )
v
i v
s
Cf D hq C h
h D h hh D h h
 ∂ −pi + pi   = − + −   ∂ pi −  pi − 
 
 
Notice that the second term of last partial derivative is zero from the 
steady-state relation, and the term ( )D h hpi − is finite for all 0 h D< < . 
Consequently, the linearized model of the process, after substitution of 
deviation variables is, 
 
 
1 1 1
2 ( ) ( )
v
i
Cdh h q
dt D h h D h hh
 ′  
′ ′= − +   pi − pi −  
 
 
Since i vq C h= 
 
 
1 1 1
2 ( ) ( )
i
i
qdh h q
dt h D h h D h h
′    
′ ′= − +   pi − pi −    
 
or i
dh
ah bq
dt
′
′ ′= + 
 
 where 
 
1 1
2 ( )
i i
o
q q
a
h D h h V
 
= − = − pi −  , 
1
= ( )b D h h
  pi −  
 
 = volume at the initial steady stateoV 
 
 
b) Taking Laplace transform and rearranging 
 
( ) ( ) ( )is h s ah s bq s′ ′ ′= + 
 
Therefore 
 
( ) ( ) ( / )
 or ( ) ( ) ( ) ( 1/ ) 1i i
h s b h s b a
q s s a q s a s
′ ′ −
= =
′ ′
− − +
 
 
 Notice that the time constant is equal to the residence time at the initial 
 steady state. 
 
4-16 
 4.14 
 
 
 Assumptions 
 
1) Perfectly mixed reactor 
2) Constant fluid properties and heat of reaction. 
 
a) Component balance for A, 
 
AAA
A cTVkccq
dt
dcV
i
)()( −−= (1) 
Energy balance for the tank, 
 
( ) ( ) ( )i A
dTVC qC T T H Vk T c
dt
ρ = ρ − + − ∆ (2) 
 
Since a transfer function with respect to cAi is desired, assume the other 
inputs, namely q and Ti, to be constant. 
 
Linearize (1) and (2) and not that 
dt
cd
dt
dc AA ′
= , 
dt
Td
dt
dT ′
= , 
 
T
T
TkcVcTVkqcq
dt
cdV AAAiA ′−′+−′=
′
2
20000)())(( (3) 
 
T
T
TkcHVqC
dt
TdVC A ′

 ∆+ρ−=′ρ 220000)( + ( ) ( ) AH Vk T c′−∆ (4) 
 
 
 Taking the Laplace transforms and rearranging 
 
 
[ ] )(20000)()()()( 2 sTTTkcVsCqsCTVkqVs AAiA ′−′=′++ (5) 
 
 
2
20000( ) ( ) ( ) ( ) ( ) ( )A AVCs qC H Vc k T T s H Vk T C sT
 
′ ′ρ + ρ − −∆ = −∆   (6) 
 
 
Substituting for )(sCA′ from Eq. 5 into Eq. 6 and rearranging, 
 
 
4-17 
2 2
2 2
( ) ( )
20000 20000( ) ( ) ( ) ( ) ( ) ( )Ai A A
T s HVk T q
C s Vs q Vk T VCs qC H Vc k T H c V k T
T T
′
−∆
=
′   + + ρ + ρ − −∆ + −∆    
 
 (7) 
Ac is obtained from Eq. 1 at steady state, 
 
 0011546.0)( =+= TVkq
cq
c AiA mol/cu.ft. 
 
Substituting the numerical values of T , ρ, C, ( − ∆H), q, V, Ac into Eq. 7 
and simplifying, 
 
)150)(10722.0(
38.11
)(
)(
++
=
′
′
sssC
sT
Ai
 
b) The gain K of the above transfer function is 
0)(
)(
=



′
′
sAi sC
sT
, 
 
6 7
2 2
0.15766
3.153 10 13.84 4.364.10
1000 1000
A A
qK
c cq q
T T
=   
− × + +    
 (8) 
 
obtained by putting s=0 in Eq. 7 and substituting numerical values for ρ, 
C, ( − ∆H), V. Evaluating sensitivities, 
 
4
26
2
1050.6315301384.0
10
2
15766.0
−×−=


−+−=
T
cq
q
K
q
K
qd
dK A
 
 



 ××
−


 ××


+−= 3
7
3
62 10364.42210153.384.13
1000153.3 T
c
T
cqK
Td
dK AA
 
 
 
51057.2 −×−= 
 
 
Ai
A
AAi cd
cd
cd
dK
cd
dK
×= 
 
 



+


 ×
+


 ×


+−
−
=
13840
10364.410153.384.13
100015766.0 2
7
2
62
q
q
TT
q
q
K
 
 
31087.8 −×= 
4-18 
 4.15 
 
 
 From Example 4.4, system equations are: 
 
 1
1
1
1
1 h
R
q
dt
hdA i ′−′=
′
 , 1
1
1
1 h
R
q ′=′ 
 2
2
1
1
2
2
11 h
R
h
Rdt
hdA ′−′=
′
 , 2
2
2
1 h
R
q ′=′ 
 Using state space representation,BuAxx +=� 
 DuCxy += 
 where 


′
′
=
2
1
h
h
x , iqu = and 2qy ′= 
 then, 
 
 iq
A
h
h
ARAR
AR
dt
hd
dt
hd
′










+








′
′








−
−
=










′
′
0
1
11
01
1
2
1
2211
11
2
1
 
 








′
′



=′
2
1
2
2
10
h
h
R
q 
 
 Therefore, 
 
 0,10,
0
1
,11
01
2
1
2211
11
=


=










=








−
−
= E
R
C
A
B
ARAR
ARA 
 
 
 
 
 
 
 
 
 
4-19 
 4.16 
 
 
 Applying numerical values, equations for the three-stage absorber are: 
 
 
 21
1 539.0173.1881.0 xxy
dt
dx
f +−= 
 
 321
2 539.0173.1634.0 xxx
dt
dx
+−= 
 
 fxxxdt
dx 539.0173.1634.0 323 +−= 
 
 ii xy 72.0= 
 
 Transforming into a state-space representation form: 
 
 
 fy
x
x
x
dt
dx
dt
dx
dt
dx








+
















−
−
−
=












0
0
881.0
173.1634.00
539.0173.1634.0
0539.0173.1
3
2
1
3
2
1
 
 
 
 fy
x
x
x
y
y
y








+
















=








0
0
0
72.000
072.00
0072.0
3
2
1
3
2
1
 
 
 
 Therefore, because xf can be neglected in obtaining the desired transfer 
 functions, 
 
 
 
 








=








−
−
−
=
0
0
881.0
173.1634.00
539.0173.1634.0
0539.0173.1
BA 
 
4-20 
 








=








=
0
0
0
72.000
072.00
0072.0
DC 
 
 
 Applying the MATLAB function ss2tf , the transfer functions are: 
 
 
 
8123.0443.35190.3
6560.04881.16343.0
)(
)(
23
2
1
+++
++
=
′
′
sss
ss
sY
sY
f
 
 
 
8123.0443.35190.3
4717.04022.0
)(
)(
23
2
+++
+
=
′
′
sss
s
sY
sY
f
 
 
 
8123.0443.35190.3
2550.0
)(
)(
23
2
+++
=
′
′
ssssY
sY
f
 
 
 
4.17 
 
 
 Dynamic model: 
 
 DXXS
dt
dX
−µ= )( 
 
 )(/)( / SSDYXSdt
dS
fSX −−µ−= 
 
 Linearization of non-linear terms: (reference point = steady state point) 
 
 1. X
SK
S
XSXSf
s
m
+
µ
=µ= )(),(1 
 )()(),(),(
,
1
,
1
11 XXX
fSS
S
f
XSfXSf
XSXS
−
∂
∂
+−
∂
∂
+≈ 
 
 Putting into deviation form, 
 
 ''
)(
)(
),( 2
,
1
,
1
1 XSK
SSX
SK
SSK
X
X
fS
S
f
XSf
s
m
s
msm
XSXS




+
µ
+



+
µ−+µ
=′
∂
∂
+′
∂
∂
≈′′ 
 
 
4-21 
 Substituting the numerical values for SK sm ,,µ and X then: 
 
 '1.0'113.0),(1 XSXSf +≈′′ 
 
 
 2. )(),,(2 SSDSSDf ff −= 
 f
SSDfSSDSSD
f SS
fS
S
fD
D
fSSDf
fff
′
∂
∂
+
∂
∂
+
∂
∂
≈′′
,,
2
,,
2
,,
2
2 ''),,'( 
 fff SDSDDSSSSDf ′+−−≈′′ '')(),,'(2 
 
 '1.0'1.0'9),,'(2 ff SSDSSDf +−≈′′ 
 
 
 3. DXXDf =),(3 
 '1.0'25.2'')','(3 XDDXXDXDf +=+≈ 
 
 
 Returning to the dynamic equation: putting them into deviation form by 
 including the linearized terms: 
 
 '1.0'113.0
' XS
dt
dX
+= − '1.0'25.2 XD − 
 
 
' 0.113 0.1
' ' 9 ' 0.1 ' 0.1
0.5 0.5
f
dS S X D S S
dt
−
′= − − + − 
 
 
Rearranging: 
 
 '25.2'113.0
' DS
dt
dX
−= 
 
'
0.126 ' 0.2 ' 9 ' 0.1 f
dS S X D S
dt
′= − − − − 
 
 
 Laplace Transforming: 
 
 )('113.0)(' sSssX = − )('25.2 sD 
 
 '( ) 0.126 '( ) 0.2 '( ) 9 '( ) 0.1 ( )fsS s S s X s D s S s′= − − − − 
 
 
4-22 
 Then, 
 
 )('113.0)(' sS
s
sX = − )('25.2 sD
s
 
0.2 9 0.1
'( ) '( ) '( ) ( )
0.126 0.126 0.126
fS s X s D s S s
s s s
−
′= − −
+ + +
 
 
 or 
 
 =


+
+ )126.0(
0226.01)('
ss
sX 
 
1.017 0.0113 2.25
'( ) ( ) ( )
0.126 0.126
fD s S s D s
s s s
′ ′= − − −
+ +
 
 Therefore, 
 
 
 
0226.0126.0
25.23005.1
)('
)('
2 ++
−−
=
ss
s
sD
sX
 
 
5-1 
��������	�
 
 
 
 5.1 
 
 
a) xDP(t) = hS(t) – 2hS(t-tw) + hS(t-2tw) 
xDP (s) = 
s
h (1 − 2e-tws + e-2tws) 
 
b) Response of a first-order process, 
s
h
s
K
sY 


+τ
=
1
)( (1 − 2e-tw s + e-2tw s) 
or Y(s) = (1 − 2e-tw s + e-2tw s) 


+τ
α
+
α
1
21
ss
 
Kh
s
Kh
s
=
+τ
=α
=0
1 1
 τ−==α
τ
−=
Kh
s
Kh
s
1
2 
 Y(s) = (1 − 2e-tw s + e-2tw s) 


+τ
τ
−
1s
Kh
s
Kh
 
 
 Kh(1−e-t/τ) , 0 < t < tw 
 
 y(t) = Kh(–1 – e-t/τ + 2e-(t-tw)/τ) , tw < t < 2tw 
 
 Kh(–e
-t/τ
 + 2e
-(t-tw)/τ − e-(t-2tw)/τ ) , 2tw < t 
 
 Response of an integrating element, 
 
 
s
h
s
K
sY =)( (1 − 2e-tw s + e-2tw s) 
 
 Kht , 0 < t < tw 
 
 y(t) = Kh(-t + 2 tw) , tw < t < 2tw 
 
 0 , 2tw < t 
 
c) This input gives a response, for an integrating element, which is zero after 
a finite time. 
 
Solution Manual for Process Dynamics and Control, 2nd edition, 
Copyright © 2004 by Dale E. Seborg, Thomas F. Edgar and Duncan A. Mellichamp. 
5-2 
 5.2 
 
 
a) For a step change in input of magnitude M 
 
y(t) = KM (1- e-t/τ) + y(0) 
 
We note that KM = y(∞) – y(0) = 280 – 80 = 200°C 
 
Then K = 
Kw
C
5.0
200�
= 400 °C/Kw 
 
At time t = 4, y(4) = 230 °C 
 
Thus τ−−=
−
− /41
80280
80230
e or τ = 2.89 min 
 
∴ 
189.2
400
)(
)(
+
=
′
′
ssP
sT
 [°C/Kw] 
 
a) For an input ramp change with slope a = 0.5 Kw/min 
 
Ka = 400 ×0.5 = 200 °C/min 
 
This maximum rate of change will occur as soon as the transient has died 
out, i.e., after 
 
 5×2.89 min ≈ 15 min have elapsed. 
 
 
0 1 2 3 4 5 6 7 8 9 10
0
500
1000
1500
time(min)
T
'
 
 Fig S5.2. Temperature response for a ramp input of magnitude 0.5 Kw/min. 
5-3 
 5.3 
 
 
 The contaminant concentration c increases according to this expression: 
 
 c(t) = 5 + 0.2t 
 
 Using deviation variables and Laplace transforming, 
 
 ( ) 0.2c t t′ = or 2
2.0)(
s
sC =′ 
 Hence 
2
2.0
110
1)(
ss
sCm ⋅+
=′ 
 
 and applying Eq. 5-21 
 
 
/10( ) 2( 1) 0.2t
m
c t e t−′ = − + 
 
 As soon as ( ) 2 ppmmc t′ ≥ the alarm sounds. Therefore, 
 
 ∆t = 18.4 s (starting from the beginning of the ramp input) 
 
The time at which the actual concentration exceeds the limit (t = 10 s) is 
subtracted from the previous result to obtain the requested ∆t . 
 
 ∆t = 18.4 − 10.0 = 8.4 s 
 
0 2 4 6 8 10 12 14 16 18 20
0
0.5
1
1.5
2
2.5
time
c
' m
 
 Fig S5.3. Concentration response for a ramp input of magnitude 0.2 Kw/min. 
5-4 
 5.4 
 
 
a) Using deviation variables,the rectangular pulse is 
 
0 t < 0 
Fc′ = 2 0 ≤ t < 2 
0 2 ≤ t ≤ ∞ 
 
 Laplace transforming this input yields 
 
 ( )sF e
s
sc 212)( −−=′ 
 
The input is then given by 
 
 )12(
8
)12(
8)(
2
+
−
+
=′
−
ss
e
ss
sc
s
 
 
and from Table 3.1 the time domain function is 
 
 )1(8)1(8)( 2/)2(2/ −−− −−−=′ tt eetc S )2( −t (1) 
 
0 2 4 6 8 10 12 14 16 18 20
0
1
2
3
4
5
6
time
C
'
 
 Fig S5.4. Exit concentration response for a rectangular input. 
 
 
b) By inspection of Eq. 1, the time at which this function will reach its 
maximum value is 2, so maximum value of the output is given by 
5-5 
)1(8)1(8)2( 2/01 −− −−−=′ eec S )0( (2) 
 
 and since the second term is zero, 057.5)2( =′c 
 
c) By inspection, the steady state value of )(tc′ will be zero, since this is a 
first-order system with no integrating poles and the input returns to zero. 
To obtain )(∞′c , simplify the function derived in a) for all time greater 
than 2, yielding 
 
)(8)( 2/2/)2( tt eetc −−− −=′ (3) 
 
 which will obviously converge to zero. 
 
Substituting 05.0)( =′ tc in the previous equation and solving for t gives 
 
t = 9.233 
 
 
 
 5.5 
 
 
a) Energy balance for the thermocouple, 
 
mC )( TThA
dt
dT
s −= (1) 
 
 where m is mass of thermocouple 
 C is heat capacity of thermocouple 
 h is heat transfer coefficient 
 A is surface area of thermocouple 
 t is time in sec 
 
 Substituting numerical values in (1) and noting that 
 
 TTs = and dt
Td
dt
dT ′
= , 
 
 TT
dt
Td
s
′
−
′=
′
15 
 
 Taking Laplace transform, 
115
1
)(
)(
+
=
′
′
ssT
sT
s
 
 
5-6 
b) Ts(t) = 23 + (80 − 23) S(t) 
 
23== TTs 
 
From t = 0 to t = 20, 
 
=′ )(tTs 57 S(t) , 
s
sTs
57)( =′ 
)115(
57)(
115
1)(
+
=′
+
=′
ss
sT
s
sT s 
 
Applying inverse Laplace Transform, 
 
 )1(57)( 15/tetT −−=′ 
 
Then 
 
)1(5723)()( 15/teTtTtT −−+=+′= 
 
Since T(t) increases monotonically with time, maximum T = T(20). 
 
Maximum T(t) = T(20) = 23 + 57 (1-e-20/15) = 64.97 °C 
 
0 5 10 15 20 25 30 35 40 45 50
0
5
10
15
20
25
30
35
40
45
50
time
T
'
41.97 º 
 
 Fig S5.5. Thermocouple output for part b) 
5-7 
 5.6 
 
 
a) The overall gain of G is 0=sG 
 
= 21
2
2
1
1
1010
KKKK =
+×τ
⋅
+×τ
 
 
b) If the equivalent time constant is equal to τ1 + τ2 = 5 + 3 = 8, then 
 
y(t = 8)/KM = 0.632 for a first-order process. 
 
y(t = 8)/KM = 
35
351
3/85/8
−
−
−
−− ee
 = 0.599 ≠ 0.632 
 
 Therefore, the equivalent time constant is not equal to τ1 + τ2 
 
c) The roots of the denominator of G are 
 
 -1/τ1 and -1/τ2 
 
which are negative real numbers. Therefore the process transfer function 
G cannot exhibit oscillations when the input is a step function. 
 
 
 
 5.7 
 
 
Assume that at steady state the temperature indicated by the sensor Tm is 
equal to the actual temperature at the measurement point T. Then, 
 
 
( ) 1
( ) 1 1.5 1
mT s K
T s s s
′
= =
′ τ + +
 
 
 350
m
T T C= = � 
 
 ( ) 15sinmT t t′ = ω 
 
 where ω=2pi×0.1 rad/min = 0.628 rad/min 
 
At large times when t/τ >>1, Eq. 5-26 shows that the amplitude of the 
sensor signal is 
5-8 
 
2 2 1
m
AA =
ω τ +
 
 
 where A is the amplitude of the actual temperature at the measurement 
 point. 
 
 Therefore 2 215 (0.628) (1.5) 1A = + = 20.6°C 
 
 Maximum T T A= + =350 + 20.6 = 370.6 
 
 Maximum Tcenter = 3 (max T) – 2 Twall 
 
 = (3×370.6)−(2×200) = 711.8°C 
 
Therefore, the catalyst will not sinter instantaneously, but will sinter if 
operated for several hours. 
 
 
 
 
 5.8 
 
 
a) Assume that q is constant. Material balance over the tank, 
 
qqq
dt
dhA −+= 21 
 
Writing in deviation variables and taking Laplace transform 
 
)()()( 21 sQsQsHAs ′+′=′ 
 
 
AssQ
sH 1
)(
)(
1
=
′
′
 
 
b) =′ )(1 tq 5 S(t) – 5S(t-12) 
 
se
ss
sQ 121
55)( −−=′ 
se
s
A
s
A
sQ
As
sH 12221
/5/5)(1)( −−=′=′ 
 
5-9 
 t
A
th 5)( =′ S(t) − )12(5 −t
A
S(t-12) 
 
 
 4 + tt
A
177.045 += 0 ≤ t 12≤ 
 h(t) = 
 4 + 122.6125 =


×
A
 12 < t 
 
 
 
0 5 10 15 20 25 30 35 40 45 50
0
0.5
1
1.5
2
2.5
time
h
'(
t)
 
 Fig S5.8a. Liquid level response for part b) 
 
 
c) ft122.6=h at the new steady state t ≥ 12 
 
d) =′ )(1 tq 5 S(t) – 10S(t-12) + 5S(t-24) ; tw = 12 
 
( )ss ee
s
sQ 24121 21
5)( −− +−=′ 
ss e
s
A
e
s
A
s
A
sH 242
12
22
/5/10/5)( −− +−=′ 
 
h(t) = 4 + 0.177tS(t) − 0.354(t-12)S(t-12) + 0.177(t-24)S(t-24) 
 
For t 24≥ 
 
24ft4)24(177.0)12(354.0177.04 ≥=−+−−+= tatttth 
5-10 
 
0 5 10 15 20 25 30 35 40 45 50
-0.5
0
0.5
1
1.5
2
2.5
time
h
'(
t)
 
 Fig S5.8b. Liquid level response for part d) 
 
 
 
 
 5.9 
 
 
a) Material balance over tank 1. 
 
)33.8( hqC
dt
dhA i −= 
 
where A = pi× (4)2/4 = 12.6 ft2 
 
 C = 0.1337 
USGPM
/minft 3
 
 
)()33.8()()( sHCsQCsHAs i ′×−′=′ 
 
128.11
12.0
)(
)(
+
=
′
′
ssQ
sH
i
 
 
For tank 2, 
 
)( qqC
dt
dhA i −= 
5-11 
)()( sQCsHAs i′=′ , 
ssQ
sH
i
011.0
)(
)(
=
′
′
 
 
b) ssQi /20)( =′ 
 
For tank 1, 
128.11
1.274.2
)128.11(
4.2)(
+
−=
+
=′
ssss
sH 
 
 h(t) = 6 + 2.4(1 – e-t/11.28) 
 
For tank 2, 2/22.0)( ssH =′ 
 
 h(t) = 6 + 0.22t 
 
0 5 10 15 20 25 30 35 40
0
1
2
3
4
5
6
7
8
9
time
h
'(
t)
Tank 1
Tank2
 
 Fig S5.9. Transient response in tanks 1 and 2 for a step input. 
 
 
c) For tank 1, h(∞) = 6 +2.4 – 0 = 8.4 ft 
 
For tank 2, h(∞) = 6 + (0.22×∞) = ∞ ft 
 
 d) For tank 1, 8 = 6 + 2.4(1 – e
-t/11.28
) 
 h = 8 ft at t = 20.1 min 
 For tank 2, 8 = 6 + 0.22t 
 h = 8 ft at t = 9.4 min 
 
 Tank 2 overflows first, at 9.4 min. 
5-12 
 5.10 
 
 
 a) The dynamic behavior of the liquid level is given by 
 
 )(
2
tpChB
dt
hdA
dt
hd
′=′+
′
+
′
 
 
 where 
 A = 
ρ
µ
2
6
R
 B = 
L
g
2
3
 and C =
Lρ4
3
 
 
 Taking the Laplace Transform and assuming initial values = 0 
 
 )()()()(2 sPCsHBsHAssHs ′=′+′+′ 
 
 or )(
11
/)(
2
sP
s
B
A
s
B
BC
sH ′
++
=′ 
 
 We want the previous equation to have the form 
 
 )(
12
)( 22 sPss
K
sH ′
+ζτ+τ=′ 
 
 Hence K = C/B = 
gρ2
1
 
 
B
12
=τ then 
2/1
3
2/1 



==τ
g
LB 
 
B
A
=ζτ2 then 
2/1
2 3
23 



ρ
µ
=ζ
g
L
R
 
 
b) The manometer response oscillates as long as 0 < ζ < 1 or 
0 < 
2/1
2 3
23




ρ
µ
g
L
R
 < 1 
b) If ρ is larger , then ζ is smaller and the response would be more 
oscillatory. 
 
 If µ is larger, then ζ is larger and the response would be less oscillatory. 
 
 
5-13 
 5.11 
 
 
 Y(s) = )1()1(
2
2
1
2 +τ
+=
+τ ss
K
s
K
ss
KM
 
 
 K1τs + K1 + K2s = KM 
 
 K1 = KM 
 K2 = −K1τ = − KMτ 
 
 Hence 
Y(s) = )1(2 +τ
τ
−
ss
KM
s
KM
 
 or 
 y(t) = KMt − KMτ (1-e-t/τ)After a long enough time, we can simplify to 
 
 y(t) ≈ KMt - KMτ (linear) 
 
 slope = KM 
 intercept = −KMτ 
 
 That way we can get K and τ 
 
 
 Figure S5.11. Time domain response and parameter evaluation 
 
 
 
 
 
 
 
 
−ΚΜτ
Slope = KM
y(t)
5-14 
 5.12 
 
 
a) xyyKy =++ 4��� 
 
Assuming y(0) = 0)0( =y� 
 
 
125.025.0
25.0
4
1
)(
)(
22 ++
=
++
=
KssKsssX
sY
 
 
b) Characteristic equation is 
 
s
2
 + Ks + 4 = 0 
 
 The roots are s = 
2
162 −±− KK
 
 
 -10 ≤ K < -4 Roots : positive real, distinct 
 Response : A + B 1/ τte + C 2/ τte 
 
 K = -4 Roots : positive real, repeated 
 Response : A + Bet/τ + C et/τ 
 
 -4 < K < 0 Roots: complex with positive real part. 
 Response: A + eζt/τ (B cos 21 ζ−
τ
t
 + C sin 21 ζ−
τ
t ) 
 
 K = 0 Roots: imaginary, zero real part. 
 Response: A + B cos t/τ + C sin t/τ 
 
 
 0 < K < 4 Roots: complex with negative real part. 
 Response: A + e-ζt/τ (B cos 21 ζ−
τ
t
 + C sin 21 ζ−
τ
t ) 
 
 K = 4 Roots: negative real, repeated. 
 Response: A + Be-t/τ + C t e-t/τ 
 
 
 4 < K ≤ 10 Roots: negative real, distinct 
 Response: A + B 1/ τ−te + C 2/ τ−te 
 
Response will converge in region 0 < K ≤ 10, and will not converge in 
region –10 ≤ K ≤ 0 
5-15 
 5.13 
 
 
a) The solution of a critically-damped second-order process to a step change 
of magnitude M is given by Eq. 5-50 in text. 
 
y(t) = KM 

 


τ
+− τ− /11 tet 
 
 Rearranging 
 
 
τ−


τ
+−= /11 tet
KM
y
 
 
 
 
KM
y
e
t t
−=


τ
+ τ− 11 / 
 
When y/KM = 0.95, the response is 0.05 KM below the steady-state value. 
 
 
 05.095.011 / =−=


τ
+ τ−ts e
t
 
 
 00.3)05.0ln(1ln −==
τ
−


τ
+ ss
tt
 
 
 Let E = 31ln +
τ
−


τ
+ ss
tt
 
ts
KM
time
0.95KM
y
0
5-16 
and find value of 
τ
st that makes E 0≈ by trial-and-error. 
 
ts/τ E 
4 0.6094 
5 -0.2082 
4.5 0.2047 
4.75 -0.0008 
 
∴ a value of t = 4.75τ is ts, the settling time. 
 
b) Y(s) = 24322122 )1(1)1( +τ++τ++=+τ s
a
s
a
s
a
s
a
ss
Ka
 
 
We know that the a3 and a4 terms are exponentials that go to zero for large 
values of time, leaving a linear response. 
 
 a2 = Ka
s
Ka
s
=
+τ→ 20 )1(lim 
 
Define Q(s) = 2)1( +τs
Ka
 
 
 3)1(
2
+τ
τ−
=
s
Ka
ds
dQ
 
 
Then a1 = 


+τ
τ−
→ 30 )1(
2lim
!1
1
s
Ka
s
 
 
(from Eq. 3-62) 
 
 a1 = − 2 Kaτ 
 
∴ the long-time response (after transients have died out) is 
 
 )2(2)( τ−=τ−= tKaKaKatty
�
 
 )2( τ−= ta for K = 1 
 
and we see that the output lags the input by a time equal to 2τ. 
 
 
 
 
5-17 
 
 
 5.14 
 
 
 a) Gain = psi/mm20.0
psi15psi31
mm8mm2.11
=
−
−
 
 
 Overshoot = 47.0
mm8mm2.11
mm2.11mm7.12
=
−
−
 
 Overshoot = exp 47.0
1 2
=



ζ−
piζ−
 , ζ = 0.234 
 Period = sec3.2
1
2
2
=



ζ−
piτ
 
 τ = 2.3 sec sec356.0
2
234.01 2
=
−
×
pi
 
 
1167.0127.0
2.0
)(
)(
2 ++
=
′
′
sssP
sR
 (1) 
 
b) From Eq. 1, taking the inverse Laplace transform, 
 
P RRR ′=′+′+′ 0.2 0.167 0.127 ��� 
 
 158 P-P R-R RR RR =′=′=′=′ ������ 
 
 52016701270 P . R R . R . +=++ ��� 
 
 53957.1887311 . P R . R . R +=++ ��� 
yl
time
y
0
=a(t-2τ)x=at
2τ
actual response
5-18 
 5.15 
 
 
 
1)3)(7.0(2)3(
3
)(
)(
22 ++
=
′
′
sssT
sP
 [ºC/kW] 
 
 Note that the input change kw62026)( =−=′ tp 
 
Since K is 3 °C/kW, the output change in going to the new steady state 
will be 
 
 ( ) C18kW6)kW/3( �� ==′
∞→
CT
t
 
 
 a) Therefore the expression for T(t) is Eq. 5-51 
 














τ
−
−
+



−
−+=
−
ttetT
t 2
2
2
3
7.0 )7.0(1
sin
)7.0(1
7.0
3
)7.0(1
cos11870)( ��
 
 
0 5 10 15 20 25 30 35 40 45 50
0
5
10
15
20
25
time
T
'(
t)
 
 Fig S5.15. Process temperature response for a step input 
 
b) The overshoot can obtained from Eq. 5-52 or Fig. 5.11. From Figure 5.11 
we see that OS ≈ 0.05 for ζ=0.7. This means that maximum temperature is 
 
Tmax ≈ 70° + (18)(1.05) = 70 + 18.9 = 88.9° 
 
 From Fig S5.15 we obtain a more accurate value. 
5-19 
The time at which this maximum occurs can be calculated by taking 
derivative of Eq. 5-51 or by inspection of Fig. 5.8. From the figure we see 
that t / τ = 3.8 at the point where an (interpolated) ζ=0.7 line would be. 
 
 
∴ tmax ≈ 3.8 (3 min) = 11.4 minutes 
 
 
 
 5.16 
 
 
 For underdamped responses, 
 














τ
ζ−
ζ−
ζ
+



τ
ζ−
−=
τζ− tteKMty t
2
2
2
/ 1sin
1
1
cos1)( (5-51) 
 
a) At the response peaks, 
 
2 2
/
2
1 1
cos sin
1
tdy KM e t t
dt
−ζ τ
     
− ζ − ζζ ζ     = +    τ τ τ 
− ζ     
 
 
 
2 2 2
/ 1 1 1sin cos 0te t t−ζ τ
    
− ζ − ζ − ζζ     − − + =   τ τ τ τ     
 
 
 Since KM ≠ 0 and 0/ ≠τζ− te 
 
 



τ
ζ−




τ
ζ−
+ζ−τ
ζ
+



τ
ζ−


τ
ζ
−
τ
ζ
= tt
22
2
22 1
sin1
1
1
cos0 
 
 pi=



τ
ζ−
= nt sin
1
sin0
2
 , t
21 ζ−
piτ
= n 
 where n is the number of the peak. 
 
 Time to the first peak, 
21 ζ−
piτ
=pt 
b) Overshoot, OS = 
KM
KMty p −)(
 
5-20 
OS = 




piζ−
ζ
+pi


τ
ζ−
− )sin(
1
)cos(exp
2
t
 
 




ζ−
piζ−
=




ζ−τ
ζτpi−
=
22 1
exp
1
exp 
 
c) Decay ratio, DR = 
KMty
KMty
p
p
−
−
)(
)( 3
 
 where 
23 1
3)(
ζ−
piτ
=pty is the time to the third peak. 
 
 DR = 







ζ−
piτ
τ
ζ
−=


−
τ
ζ
−=
τζ−
τζ−
23/
/
1
2
exp)(exp
3
ppt
t
tt
eKM
eKM
p
p
 
 
2
2
2
exp (OS)
1
 
− piζ
= = 
− ζ  
 
 
d) Consider the trigonometric identity 
 
 sin (A+B) = sin A cos B + cos A sin B 
 
 Let B = 



τ
ζ−
t
21
 , sin A = 21 ζ− , cos A = ζ 
 
[ ]




 ζ+ζ−
ζ−
−=
τζ− BBeKMty t sincos1
1
11)( 2
2
/
 
 





+ζ−−=
τζ−
)sin(
1
1
2
/
BAeKM
t
 
 
 Hence for stt ≥ , the settling time, 
 
 05.0
1 2
/
≤ζ−
τζ− te
 , or ( )ζτζ−−≥ 2105.0lnt 
 Therefore, 



ζ−ζ
τ≥
21
20lnst 
 
 
 
5-21 
 5.17 
 
 
a) Assume underdamped second-order model (exhibits overshoot) 
 
 
gal/min
ft2.0
gal/min120140
f610
������
�������
=
−
−
==
tK 
 
 Fraction overshoot = 25.0
4
1
610
1011
==
−
−
 
 
 From Fig 5.11, this corresponds (approx) to ζ = 0.4 
 
 From Fig. 5.8 , ζ = 0.4 , we note that tp/τ ≈ 3.5 
 
 
 Since tp = 4 minutes(from problem statement), τ = 1.14 min 
 
 
 
191.031.1
2.0
14)2(0.4)(1.1(1.14)
0.2
 222 ++
=
++
=∴
ssss
(s)G p 
 
b) In Chapter 6 we see that a 2nd-order overdamped process model with a 
numerator term can exhibit overshoot. But if the process is underdamped, 
it is unique. 
 
 
 
 5.18 
 
 
a) Assuming constant volume and density, 
 
Overall material balances yield: q2 = q1 = q (1) 
 
Component material balances: 
 
 ( )111 ccqdt
dcV i −= (2) 
 ( )2122 ccqdt
dcV −= (3) 
 
b) Degrees of freedom analysis 
 
3 Parameters : V1, V2, q 
5-22 
3 Variables : ci, c1, c2 
 
2 Equations: (2) and (3) 
 
NF = NV − NE = 3 − 2 = 1 
 
 Hence one input must be a specified function of time. 
 
 2 Outputs = c1, c2 
 1 Input = ci 
 
 
c) If a recycle stream is used 
 
 
Overall material balances: 
 
 q1 = (1+r)q (4) 
 
 q2 = q1 = (1+r)q (5) 
 
 q3 = q2 – rq = (1+r)q − rq = q (6) 
 
Component material balances: 
 
V1 12
1 )1( qcrrqcqc
dt
dc
i +−+= (7) 
 
V2 21
2 )1()1( qcrqcr
dt
dc
+−+= (8) 
 
Degrees of freedom analysis is the same except now we have 
 
 4 parameters : V1, V2, q, r 
 
5-23 
d) If ∞→r , there will a large amount of mixing between the two tanks as a 
result of the very high internal circulation. 
 
Thus the process acts like 
 
 
Model : 
 (V1 +V2) )( 22 ccqdt
dc
i −= (9) 
 c1 = c2 (complete internal mixing) (10) 
 
 Degrees of freedom analysis is same as part b) 
 
 
 5.19 
 
 
a) For the original system, 
 
 
1
11
1 R
hCq
dt
dhA i −= 
 
2
2
1
12
2 R
h
R
h
dt
dhA −= 
 
where A1 = A2 = pi(3)2/4 = 7.07 ft2 
 
 C = 0.1337 
gpm
/minft 3
 
 R1 = R2 = /minft
ft187.0
1001337.0
5.2
3
1
=
×
=
iqC
h
 
 
ci
q
c2
q
Total Volume = V1 + V2
5-24 
Using deviation variables and taking Laplace transforms, 
 
 
132.1
025.0
11)(
)(
11
1
1
1
1
+
=
+
=
+
=
′
′
ssRA
CR
R
sA
C
sQ
sH
i
 
 
132.1
1
1
/
1
/1
)(
)(
22
12
2
2
1
1
2
+
=
+
=
+
=
′
′
ssRA
RR
R
sA
R
sH
sH
 
 2
2
)132.1(
025.0
)(
)(
+
=
′
′
ssQ
sH
i
 
 
For step change in qi of magnitude M, 
 
 Mh 025.0max1 =′ 
 Mh 025.0max2 =′ since the second-order transfer function 
 
2)132.1(
025.0
+s
 is critically damped (ζ=1), not underdamped 
 Hence Mmax = gpm100ft/gpm025.0
f5.2
=
t
 
 
For the modified system, 
 
 
R
hCq
dt
dhA i −= 
 
 
22 ft6.124/)4( =pi=A 
 V = V1 + V2 = 2 ft5ft07.7 2 ×× = 70.7ft3 
 hmax = V/A = 5.62 ft 
 
 R = 
/minft
ft21.0
1001337.0
62.55.0
3=×
×
=
iqC
h
 
 
 
164.2
0281.0
11)(
)(
+
=
+
=
+
=
′
′
sARs
CR
R
As
C
sQ
sH
i
 
 
 Mh 0281.0max =′ 
 Mmax = gpm100ft/gpm0281.0
f81.2
=
t
 
 
5-25 
Hence, both systems can handle the same maximum step disturbance in qi. 
 
b) For step change of magnitude M, 
s
M
sQi =′ )( 
 
For original system, 
 
 
s
M
s
sH
R
sQ 22
2
2 )132.1(
025.0
187.0
1)(1)(
+
=′=′ 
 
 


+
−
+
−= 2)132.1(
32.1
)132.1(
32.11134.0
sss
M 
 

 


+−=′ − 32.1/2 32.1
11134.0)( tetMtq 
 
For modified system, 
 
 


+
−=
+
=′=′
164.2
64.21134.0)164.2(
0281.0
21.0
1)(1)(
ss
M
s
M
s
sH
R
sQ 
 
 
[ ]64.2/1134.0)( teMtq −−=′ 
 
Original system provides better damping since )(2 tq′ < )(tq′ for t < 3.4. 
 
 
 
 5.20 
 
 
a) Caustic balance for the tank, 
 
wccwcw
dt
dCV −+=ρ 2211 
 
Since V is constant, w = w1 + w2 = 10 lb/min 
 
For constant flows, 
 
 )()()()( 2211 sCwsCwsCwsCVs ′−′+′=′ρ 
 
 
149
5.0
10)7)(70(
5
)(
)( 1
1 +
=
+
=
+ρ
=
′
′
sswVs
w
sC
sC
 
5-26 
 
1)(
)(
+τ
=
′
′
s
K
sC
sCm
 , K = (3-0)/3 = 1 , τ ≈ 6 sec = 0.1 min 
 (from the graph) 
 
 )149)(11.0(
5.0
)149(
5.0
)11.0(
1
)(
)(
1 ++
=
++
=
′
′
sssssC
sCm
 
 
b) 
s
sC 3)(1 =′ 
)149)(11.0(
5.1)(
++
=′
sss
sCm 



−
−
+=′ −− )491.0()1.049(
115.1)( 49/1.0/ ttm eetc 
 
c) )149(
5.13
)149(
5.0)(
+
=
+
=′
ssss
sCm 
 ( )49/15.1)( tm etc −−=′ 
 
 
d) The responses in b) and c) are nearly the same. Hence the dynamics of the 
conductivity cell are negligible. 
0 20 40 60 80 100 120 140 160 180 200
0
0.5
1
1.5
time
C
m
'(
t)
Part b)
Part c)
 
 Fig S5.20. Step responses for parts b) and c) 
 
 
5-27 
 5.21 
 
 
 Assumptions: 1) Perfectly mixed reactor 
2) Constant fluid properties and heat of reaction 
 
a) Component balance for A, 
 
AAiA
A cTVkccq
dt
dcV )()( −−= (1) 
 
Energy balance for the tank, 
 
ARi cTVkHTTqCdt
dTVC )()()( ∆−+−ρ=ρ (2) 
 
Since a transfer function with respect to cAi is desired, assume the other 
inputs, namely q and Ti, are constant. Linearize (1) and (2) and note that 
 
dt
cd
dt
dc AA ′
= , 
dt
Td
dt
dT ′
= , 
 
 T
T
TkcVcTVkqcq
dt
cdV AAiA
A
′−′+−′=
′
2
20000)())(( (3) 
 
 ARAR cTVkHTT
TkcVHqC
dt
TdVC ′∆−′

 ∆+ρ−=′ρ )(20000)( 2 (4) 
 
Taking Laplace transforms and rearranging 
 
[ ] )(20000)()()()( 2 sTTTkcVsCqsCTVkqVs AAiA ′−′=′++ (5) 
2
20000( ) ( ) ( ) ( ) ( ) ( )R A R AVCs qC H Vc k T T s H Vk T C sT
 
′ ′ρ + ρ − −∆ = −∆   (6) 
 
Substituting )(sCA′ from Eq. 5 into Eq. 6 and rearranging, 
 
2 2
2 2
( ) ( )( )
20000 20000( ) ( ) ( ) ( ) ( ) ( )
R
Ai
R A R A
H Vk T qT s
C s Vs q Vk T VCs qC H Vc k T H V c k T
T T
′ −∆
=
′   + + ρ + ρ − −∆ + −∆    
 
 (7) 
Ac is obtained from Eq. 1 at steady state, 
5-28 
 )(TVkq
cq
c AiA
+
= = 0.001155 lb mol/cu.ft. 
 
Substituting the numerical values of T , ρ, C, –∆HR, q, V, Ac into Eq. 7 
and simplifying, 
 
 )150)(10722.0(
38.11
)(
)(
++
=
′
′
sssC
sT
iA
 
 
For step response, ssCAi /1)( =′ 
 
 )150)(10722.0(
38.11)(
++
=′
sss
sT 
 
 


−
−
+=′ −− )500722.0()0722.050(
1138.11)( 50/0722.0/ tt eetT 
 
A first-order approximation of the transfer function is 
 
 
150
38.11
)(
)(
+
=
′
′
ssC
sT
iA
 
For step response, )150(
38.11)(
+
=′
ss
sT or [ ]50/138.11)( tetT −−=′ 
 
 The two step responses are very close to each other hence the 
approximation is valid. 
 
0 20 40 60 80 100 120 140 160 180 200
0
2
4
6
8
10
12
time
T
'(
t)
Using transfer function
Using first-order approximation
 
 Fig S5.21. Step responses for the 2nd order t.f and 1st order approx. 
 
5-29 
 5.22 
 
 
(τas+1)Y1(s) = K1U1(s) + Kb Y2(s) (1) 
(τbs+1)Y2(s) = K2U2(s) + Y1(s) (2) 
 
a) Since the only transfer functions requested involve U1(s), we can let U2(s) 
be zero. Then, substituting for Y1(s) from (2) 
 
Y1(s) = (τbs+1)Y2(s) (3) 
 
(τas+1)(τbs+1)Y2(s) =K1U1(s) + KbY2(s) (4) 
 
Rearranging (4) 
 
[(τas+1)(τbs+1) − Kb]Y2(s) =K1U1(s) 
 
∴ 
bKss
K
sU
sY
−+τ+τ
= )1)(1()(
)(
ba
1
1
2
 (5) 
Also, since 
 
 1)(
)(
b
2
1 +τ= s
sY
sY
 (6)From (5) and (6) 
 
bKss
sK
sY
sY
sU
sY
sU
sY
−+τ+τ
+τ
=×= )1)(1(
)1(
)(
)(
)(
)(
)(
)(
ba
b1
2
1
1
2
1
1
 (7) 
 
b) The gain is the change in y1(or y2) for a unit step change in u1. Using the 
FVT with U1(s) = 1/s. 
 
bb
s K
K
sKss
K
sty
−
=


−+τ+τ
=∞→
→ 1
1
)1)(1(lim)(
1
ba
1
02
 
 
This is the gain of TF Y2(s)/U1(s). 
 
Alternatively, 
 
K
bb
ss K
K
Kss
K
sU
sY
−
=


−+τ+τ
=


=
→→ 1)1)(1(lim)(
)(lim 1
ba
1
0
1
2
0
 
 
For Y1(s)/U1(s) 
5-30 
bb
b
s K
K
sKss
sK
sty
−
=


−+τ+τ
+τ
=∞→
→ 1
1
)1)(1(
)1(lim)( 1
ba
1
01
 
In other words, the gain of each transfer function is 
bK
K
−1
1
 
c) 
bKss
K
sU
sY
−+τ+τ
= )1)(1()(
)(
ba
1
1
2
 (5) 
 
Second-order process but the denominator is not in standard form, i.e., 
τ2s2+2ζτs+1 
 
Put it in that form 
 
bKss
K
sU
sY
−+τ+τ+ττ
=
1)()(
)(
ba
2
ba
1
1
2
 (8) 
 
Dividing through by 1- Kb 
 
1
1
)(
1
)1/(
)(
)(
ba2ba
1
1
2
+
−
τ+τ
+
−
ττ
−
=
s
K
s
K
KK
sU
sY
bb
b
 (9) 
 
Now we see that the gain K = K1/(1-Kb), as before 
 
 
bK−
ττ
=τ
1
ba2
 
bK−
ττ
=τ
1
ba
 (10) 
 
 
bK−
τ+τ
=ζτ
1
2 ba , then 
 
 
bK−
τ+τ
=ζ
12
1 ba
=
ττ
−
ba
1 bK




ττ
τ+τ
ba
ba
2
1
bK−1
1
 (11) 
 
Investigating Eq. 11 we see that the quantity in brackets is the same as ζ 
for an overdamped 2nd-order system (ζOD) [ from Eq. 5-43 in text]. 
 
 
bK−
ζ
=ζ
1
OD
 (12) 
 where 
ba
ba
OD 2
1
τ+τ
τ+τ
=ζ 
5-31 
 Since ζOD>1, 
 
 ζ>1, for all 0 < Kb < 1. 
 
In other words, since the quantity in brackets is the value of ζ for an 
overdamped system (i.e. for τa ≠ τb is >1) and bK−1 <1 for any positive 
Kb, we can say that this process will be more overdamped (larger ζ) if Kb 
is positive and <1. 
 
For negative Kb we can find the value of Kb that makes ζ = 1, i.e., yields a 
critically-damped 2nd-order system. 
 
 
1
OD
1
1
bK−
ζ
==ζ (13) 
 or 
1
2
OD
1
1
bK−
ζ
= 
 
 1 – Kb1 = ζOD2 
 
 Kb1 = 1 − ζOD2 (14) 
 
where 
 Kb1 < 0 is the value of Kb that yields a critically-damped process. 
 
Summarizing, the system is overdamped for 1 − ζOD2 < Kb < 1. 
 
Regarding the integrator form, note that 
 
bKss
K
sU
sY
−+τ+τ+ττ
=
1)()(
)(
ba
2
ba
1
1
2
 (8) 
 
For Kb = 1 
 
 [ ])()()(
)(
baba
1
ba
2
ba
1
1
2
τ+τ+ττ
=
τ+τ+ττ
=
ss
K
ss
K
sU
sY
 
 
 



+
τ+τ
ττ
τ+τ
=
1
)/(
ba
ba
ba1
ss
K
 
which has the form )1(
1
+τ′
′
=
ss
K
 ( s indicates presence of integrator) 
 
 
5-32 
d) Return to Eq. 8 
 
System A: 
 
164
1
164
2
5.01)12()1)(2()(
)(
22
1
2
1
1
2
++
=
++
=
−+++
=
ssss
K
ss
K
sU
sY
 
 
 τ2 = 4 → τ = 2 
 2ζτ = 6 → ζ = 1.5 
 
System B: 
For system 
132
1
)1)(12(
1
2 ++
=
++ ssss
 
 
 τ2
2
 = 2 → τ2 = 2 
 2ζ2τ2 = 3 → ζ2 = 
22
3
 = 
2
5.1
 ≈ 1.05 
 
Since system A has larger τ (2 vs. 2 ) and larger ζ (1.5 vs 1.05), it will 
respond slower. These results correspond to our earlier analysis. 
6-1 
��������	�
 
 
 
 
 6.1 
 
 
a) By using MATLAB, the poles and zeros are: 
 
Zeros: (-1 +1i) , (-1 -1i) 
Poles: -4.3446 
 (-1.0834 +0.5853i) 
 (-1.0834 –0.5853i) 
 (+0.7557 +0.5830i) 
 (+0.7557 −0.5830i) 
 
 These results are shown in Fig E6.1 
 
 Figure S6.1. Poles and zeros of G(s) plotted in the complex s plane. 
 
Solution Manual for Process Dynamics and Control, 2nd edition, 
Copyright © 2004 by Dale E. Seborg, Thomas F. Edgar and Duncan A. Mellichamp. 
6-2 
b) Process output will be unbounded because some poles lie in the right half 
plane. 
c) By using Simulink-MATLAB 
0 5 10 15 20 25 30
-16
-14
-12
-10
-8
-6
-4
-2
0
2
x 108
Time
O
ut
pu
t
 
 Figure E6.1b. Response of the output of this process to a unit step input. 
 
As shown in Fig. S6.1b, the right half plane pole pair makes the process 
unstable. 
 
 
 6.2 
 
a) Standard form = )1)(1(
)1(
21 +τ+τ
+τ
ss
sK a
 
b) 
Hence )12)(15.0(
)12(5.0)(
5
++
+
=
−
ss
es
sG
s
 
 
Applying zero-pole cancellation: 
 
 )15.0(
5.0)(
5
+
=
−
s
e
sG
s
 
 
c) Gain = 0.5 
Pole = −2 
Zeros = No zeros due to the zero-pole cancellation. 
6-3 
d) 1/1 Pade approximation: )2/51(
)2/51(5
s
s
e s
+
−
=
−
 
The transfer function is now 
 
)2/51(
)2/51(
15.0
5.0)(
s
s
s
sG
+
−
×
+
= 
 
Gain = 0.5 
Poles = −2, −2/5 
Zeros = + 2/5 
 
 
 6.3 
 
 
 )1(
)1(
)(
)(
1 +τ
+τ
=
s
sK
sX
sY a
 , 
s
M
sX =)( 
 
 From Eq. 6-13 
 
 y(t) = KM 


τ
τ−τ
+=






τ
τ
−−
τ−τ− 11 /
1
1/
1
111 tata eKMe 
a) KMKMy aa
11
11)0(
τ
τ
=


τ
τ−τ
+=+ 
 
b) Overshoot → y(t) > KM 
 
KMeKM ta >


τ
τ−τ
+ τ− 1/
1
11 
or τa − τ1 > 0 , that is, τa > τ1 
 
00)( 1/2
1
1 ><
τ
τ−τ
−=
τ− KMforeKMy ta� 
 
 c) Inverse response → y(t) < 0 
 
01 1/
1
1 <


τ
τ−τ
+ τ−ta eKM 
 
1/
1
1 τ+
−<
τ
τ−τ ta e or 01 1/
1
<−<
τ
τ τ+ta e at t = 0. 
 Therefore τa < 0. 
6-4 
 6.4 
 
 
 )1)(1(
)1(
)(
)(
21 +τ+τ
+τ
=
ss
sK
sX
sY a
 , τ1>τ2, X(s) = M/s 
 
 From Eq. 6-15 
 
 


τ−τ
τ−τ
−
τ−τ
τ−τ
+= τ−τ− 21 /
21
2/
21
11)( tata eeKMty 
 
a) Extremum → 0)( =ty� 
 
 0110 21 /
21
2
2
/
21
1
1
=






τ−τ
τ−τ
τ
+



τ−τ
τ−τ
τ
−
τ−τ− tata eeKM 
 
 1
/1
/1
21
11
1
2 ≥=
ττ−
ττ− 



τ
−
τ
−t
a
a e since τ1>τ2 
 
 
 b) Overshoot → )(ty > KM 
 
 KMeeKM tata >


τ−τ
τ−τ
−
τ−τ
τ−τ
+ τ−τ− 21 /
21
2/
21
11 
 012
11
2
1 >>
τ−τ
τ−τ 



τ
−
τ
−t
a
a e , therefore τa>τ1 
 
 
c) Inverse response → 0)( <ty� at t = 0+ 
 
 0110 21 /
21
2
2
/
21
1
1
<






τ−τ
τ−τ
τ
+



τ−τ
τ−τ
τ
−
τ−τ− tata eeKM at t = 0+ 
 011
21
2
221
1
1
<



τ−τ
τ−τ
τ
+



τ−τ
τ−τ
τ
−
aa
 
 
 0
11
21
12 <
−




−
ττ
ττ
τ a
 
6-5 
 Since τ1 > τ2, τa < 0. 
 
d) If an extremum in y exists, then from (a) 
 
=




τ
−
τ
−
21
11
t
e 



ττ−
ττ−
1
2
1
1
a
a
 
 



ττ−
ττ−
τ−τ
ττ
=
1
2
21
21
1
1ln
a
at 
 
 
 6.5 
 
 
 Substituting the numerical values into Eq. 6-15 
 
 Case (i) : y(t) = 1 (1 + 1.25e-t/10 − 2.25e-t/2) 
 
 Case (ii(a)) : y(t) = 1 (1 − 0.75e-t/10 − 0.25e-t/2) 
 
 Case (ii(b)) : y(t) = 1 (1 − 1.125e-t/10 +0.125e-t/2) 
 
 Case (iii) : y(t) = 1 (1 − 1.5e-t/10 + 0.5e-t/2) 
 
0 5 10 15 20 25 30 35 40 45 50
-0.2
0
0.2
0.4
0.6
0.8
1
1.2
1.4
1.6
Time
y(t
)
case(i)
case(ii)a
case(ii)b
case(iii)
 
 Figure S6.5. Step response of a second-order system with a single zero. 
6-6 
 Conclusions: 
 
 τa > τ1 gives overshoot. 
 
 0 < τa < τ1 gives response similar to ordinary first-order process 
 response. 
 
 τa < 0 gives inverse response. 
 
 
 
 
 6.6 
 
 )(
1
)(
1
)()( 2121 sU
s
K
s
K
sU
s
K
sU
s
K
sY 


+τ
+=
+τ
+= 
 
 )1(
)(
)1()(
)( 121211
+τ
++τ
=
+τ
++τ
=
ss
KsKK
ss
sKKsK
sU
sY
 
 
 Put in standard K/τ form for analysis: 
 
 )1(
1
)(
)()( 1
2
1
+τ



+



+τ
==
ss
s
K
K
K
sU
sY
sG 
 
a) Order of G(s) is 2 (maximum exponent on s in denominator is 2) 
 
b) Gain of G(s) is K1. Gain is negative if K1 < 0. 
 
c) Poles of G(s) are: s1 = 0 and s2 = –1/τ 
 
s1 is on imaginary axis; s2 is in LHP. 
 
d) Zero of G(s) is: 
 
21
1
1
2
1
KK
K
K
K
sa +τ
−
=




+τ
−
= 
 
If 0
21
1 <
+τ KK
K
, the zero is in RHP. 
 
6-7 
Two possibilities: 1. K1<0 and K1τ + K2 >0 
 
 2. K1 > 0 and K1τ + K2 < 0 
 
e) Gain is negative if K1 < 0 
 
Then zero is RHP if K1τ + K2 > 0 
 
This is the only possibility. 
 
f) Constant term and e-t/τ term. 
 
g) If input is M/s, the output will contain a t term, that is, it is not bounded. 
 
 
 
 6.7 
 
 
 a) )()24()( tStp −=′ , 
s
sP 2)( =′ 
ss
sP
s
sQ 2
120
3)(
120
3)(
+
−
=′
+
−
=′ 
 
)1(6)( 20/tetQ −−−=′ 
 
 
b) )()()( sPsQsR m′=′+′ 
 
)0()()()()( mmm ptptptqtr −=′=′+′ 
 
)1(612)()( 20/tm etptr −−+−=′ 
 
6
24
)01(61218
)0()(
)(
=
−
−+−
=
=−∞=
∞=′
=
tptp
trK 
 
Overshoot, 
 
 
15/ 20( 15) ( ) 27 12 6(1 ) 12OS 0.514( ) 12
r t r t e
r t
−
′ ′= − = ∞ − + − −
= = =
′ = ∞
 
 
 
6-8 
 514.0
1
exp
2
=



ζ−
piζ−
=OS , 2.0=ζ 
 
Period, T, for )(tr ′ is equal to the period for pm(t) since e-t/20 decreases 
monotonically. 
 
 Thus, T = 50 − 15 = 35 
 
 and 46.51
2
2
=ζ−
pi
=τ
T
 
 
c) 
112)(
)(
22 +τ′
′
+
+ζτ+τ=′
′
s
K
ss
K
sP
sPm
 
 ( )
)1)(12(
)()2(
22
22
+τ′+ζτ+τ
′++ζτ′+τ′+τ′
=
sss
KKsKKsK
 
 
d) Overall process gain = 
psi
%336)(
)(
0
=−=′+=
′
′
=
KK
sP
sP
s
m
 
 
 
 
 
 
 6.8 
 
 
a) Transfer Function for blending tank: 
 
1
)(
+
=
s
K
sG
bt
bt
bt τ
 where 1≠= ∑ i
in
bt q
q
K 
 min2
min/m1
m2
3
3
==τbt 
 
Transfer Function for transfer line 
 
( )51)( +τ= s
K
sG
tl
tl
tl where 1=tlK 
 min02.0
min/m15
m1.0
3
3
=
×
=τ tl 
 
6-9 
∴ 5)102.0)(12()(
)(
++
=
′
′
ss
K
sC
sC bt
in
out
 
 
a 6th-order transfer function. 
 
b) Since τbt >> τtl [ 2 >> 0.02] we can approximate 5)102.0(
1
+s
 by e-θ s 
where ∑
=
==θ
5
1
1.0)02.0(
i
 
 
∴ 
12)(
)( 1.0
+
≈
′
′
−
s
eK
sC
sC sbt
in
out
 
 
c) Since τbt ≈ 100 τtl , we can imagine that this approximate TF will yield 
results very close to those from the original TF (part (a)). We also note 
that this approximate TF is exactly the same as would have been obtained 
using a plug flow assumption for the transfer line. Thus we conclude that 
investing a lot of effort into obtaining an accurate dynamic model for the 
transfer line is not worthwhile in this case. 
 
[ Note that, if τbt ≈ τtl , this conclusion would not be valid] 
 
d) By using Simulink-MATLAB, 
0 5 10 15 20 25 30
-0.2
0
0.2
0.4
0.6
0.8
1
1.2
Time
O
ut
pu
t/K
bt
Exact model
Approximate model
 
 Fig S6.8. Unit step responses for exact and approximate model. 
 
6-10 
 6.9 
 
 
a), b) Represent processes that are (approximately) critically damped. A step 
response or frequency response in each case can be fit graphically or 
numerically. 
 
c) θ = 2, τ = 10 
 
d) Exhibits strong overshoot. Can’t approximate it well. 
 
e) θ = 0.5, τ = 10 
 
f) θ = 1, τ = 10 
 
g) Underdamped (oscillatory). Can’t approximate it well. 
 
h) θ = 2, τ = 0 
 
 
By using Simulink-MATLAB, models for parts c), e), f) and h) are 
compared: (Suppose K = 1) 
 
 Part c) 
 
0 5 10 15 20 25 30 35 40 45 50
0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1
time
Ou
tp
ut
Exact model
Approximate model
 
 Figure S6.9a. Unit step responses for exact and approximate model in part c) 
 
6-11 
Part e) 
 
0 5 10 15 20 25 30 35 40 45 50
0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1
Time
O
ut
pu
t
Exact model
Approximate model
 
Figure S6.9b. Unit step responses for exact and approximate model in part e) 
 
 
Part f) 
 
0 5 10 15 20 25 30 35 40 45 50
0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1
Time
O
ut
pu
t
Exact model
Approximate model
 
 Figure S6.9c. Unit step responses for exact and approximate model in part f) 
 
6-12 
 Part h) 
 
0 5 10 15 20 25 30 35 40 45 50
-1
-0.5
0
0.5
1
1.5
Time
O
u
tp
u
t
Exact model
Approximate model
 
 
 Figure S6.9d. Unit step responses for exact and approximate model in part h) 
 
 
 6.10 
 
 
a) The transfer function for each tank is 
 
1
1
)(
)(
1 +


=′
′
−
s
q
VsC
sC
i
i
 , i = 1, 2, …, 5 
 
where i represents the ith tank. 
 
 co is the inlet concentration to tank 1. 
 
 V is the volume of each tank. 
 
 q is the volumetric flow rate. 
 
∏
=
−



+
=


′
′
=
′
′
5
1
5
10
5
16
1
)(
)(
)(
)(
i i
i
ssC
sC
sC
sC
, 
 
Then, by partial fraction expansion, 
6-13 
2 3 4
/ 6
5
1 1 1( ) 0.60 0.15 1 1
6 2! 6 3! 6 4! 6
t t t t tc t e−
        
= − − + + + +                
 
 b) Using Simulink, 
0 5 10 15 20 25 30 35 40 45 50
0.44
0.46
0.48
0.5
0.52
0.54
0.56
0.58
0.6
time
Co
n
ce
n
tra
tio
n
c5
c4
c3
c2
c1
 
 Figure S6.10. Concentration step responses of the stirred tank. 
 
 The value of the expression for c5(t) verifies the simulation results above: 
 
2 3 4
5
5
5 5 5(30) 0.60 0.15 1 1 5 0.5161
2! 3! 4!
c e−
  
= − − + + + + =    
 
 
 
 6.11 
 
 
a) 
11
1)(
1
22
1 +τ
++=
+τ
+τ−
=
s
C
s
B
s
A
s
E
s
s
sY a 
 
We only need to calculate the coefficients A and B because 01/ →τ−tCe 
 for t >> τ1. However, there is a repeated pole at zero. 
 
6-14 
E
s
sE
B a
s
=


+τ
+τ−
=
→ 1
)1(lim
1
0
 
 
 Now look at 
 
 
2
11 )1()1()1( CssBsAssE a ++τ++τ=+τ− 
 
 
2
1
2
1 CsBsBAssAEsE a ++τ++τ=+τ− 
 
 Equate coefficients on s: 
 
 1τ+=τ− BAE a 
 
 )( 1τ+τ−= aEA 
 
 Then the long-time solution is 
 
 )()( 1τ+τ−≈ aEEtty 
 
 Plotting 
 
 
 
 
 
b) For a LHP zero, the apparent lag would be τ1 − τa 
 
c) For no zero, the apparent lagwould be τ1 
 
time
y =Et-E(τa+τ1)y(t)=Et
actual response
(τa+τ1)
-E(τa+τ1)
6-15 
 6.12 
 
 
a) Using Skogestad’s method 
 
( ) )15.4)(110(
5
1)5.04()110(
5)(
7.0)2.05.0(
++
=
+++
=
−+−
ss
e
ss
e
sG
ss
approx 
 
 
 
b) By using Simulink-MATLAB 
0 5 10 15 20 25 30 35 40 45 50
-1
0
1
2
3
4
5
Time
O
ut
pu
t
Exact model
Approximate model
 
 Figure S6.12a. Unit step responses for exact and approximate model. 
 
 
c) Using MATLAB and saving output data on vectors, the maximum error is 
 
Maximum error = 0.0521 at = 5.07 s 
 
This maximum error is graphically shown in Fig. S6.12b 
6-16 
0 1 2 3 4 5 6 7 8
-0.1
0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
Time
O
ut
pu
t
Exact model
Approximate model
 
Figure S6.12b. Maximum error between responses for exact and approximate 
 model. 
 
 
 
 6.13 
 
 
From the solution to Problem 2-5 (a) , the dynamic model for isothermal 
operation is 
 
 
ba
d
R
PP
R
PP
dt
dP
RT
MV 2111
1
1 −
−
−
= (1) 
 
c
f
b R
PP
R
PP
dt
dP
RT
MV −
−
−
=
2212
2
2
 (2) 
 
Taking Laplace transforms, and noting that 0)( =′ sPf since Pf is constant, 
 
 
1
)()()(
1
2
1 +τ
′+′
=′
s
sPKsPK
sP adb (3) 
 
1
)()(
2
1
2 +τ
′
=′
s
sPK
sP c (4) 
 where 
 
6-17 
 )/( baaa RRRK += 
 )/( babb RRRK += 
 
 )/( cbcc RRRK += 
 
 )(1
1
1
ba
ba
RR
RR
RT
MV
+
=τ 
 
 )(2
2
2
cb
cb
RR
RR
RT
MV
+
=τ 
 
 Substituting for )(1 sP′ from Eq. 3 into 4, 
 
 
1
11
1
)1)(1()(
)(
2122121
2
+



−
τ+τ
+



−
ττ




−
=
−+τ+τ
=
′
′
s
KK
s
KK
KK
KK
KKss
KK
sP
sP
caca
ca
cb
ca
cb
d
 (5) 
 
 Substituting for )(2 sP′ from Eq. 5 into 4, 
 
 
2
1
21 2 1 2
( 1)
1( )
( )
1
1 1
b
a c
d
a c a c
K
s
K KP s
P s
s s
K K K K
 
τ + 
−′  
=
′    τ τ τ + τ
+ +   
− −   
 (6) 
 
To determine whether the system is over- or underdamped, consider the 
denominator of transfer functions in Eqs. 5 and 6. 
 
 
 
caca KKKK −
τ+τ
=ζτ



−
ττ
=τ
1
2,
1
21212
 
 
 Therefore, 
 
 
)1(
1
2
1)1(
)1(
)(
2
1
1
2
2
1
21
21
ca
ca
ca KK
KK
KK
−




τ
τ
+
τ
τ
=
ττ
−
−
τ+τ
=ζ 
 
 Since x + 1/x ≥ 2 for all positive x, 
6-18 
 
)1(
1
ca KK−
≥ζ 
 
 Since KaKc ≥ 0, 
 
 1≥ζ 
 
 Hence the system is overdamped. 
 
 
 
 
 6.14 
 
 
 a) For 
s
M
sX =)( 
 
 
11)1)(1()( +τ+−+=+τ−= s
C
s
B
s
A
sss
KM
sY 
 
 KM
ss
KMA
s
=
+τ−
==
→ )1)(1(lim0 
 
 
1
lim ( 1) 1s
KM KMB
s s→
= = =
τ + τ +
 
 
 
2
1/
lim
1 1(1 ) 11s
KM KM KMC
s s→− τ
  − τ
= = = 
− τ +   
− +  τ τ  
 
 Then, 
 
 


+τ
τ
−
+τ
−=
τ− /
1 11
1)( t
t
e
eKMty 
 
 
 For M =2 , K = 3, and τ = 3, the Simulink response is shown: 
6-19 
 
0 2 4 6 8 10 12 14 16 18 20
-12
-10
-8
-6
-4
-2
0
x 108
Time
Ou
tp
u
t
 
 Figure S6.14a. Unit step response for part a). 
 
 b) If )1)(1()(
2
2 +τ−
=
−
ss
Ke
sG
s
 then, 
)2(
11
1)( /)2(
2
2 −


+τ
τ
−
+τ
−=
τ−−
−
tSeeKMty t
t
 
 
Note presence of positive exponential term. 
 
 c) Approximating G2(s) using a Padé function 
 
 )1)(1()1)(1)(1(
)1()(2 +τ+=−+τ+
−
=
ss
K
sss
sK
sG 
 
 Note that the two remaining poles are in the LHP. 
 
 d) For 
s
M
sX =)( 
 )1)(1()( +τ+= sss
KM
sY 
 
 Using Table 3.1 
 
 τ1 = 1 , τ2 = τ 
 
6-20 
 


τ−
−τ
+= τ−− )(
1
11)( /3 tt eeKMty 
 
 Note that no positive exponential term is present. 
 
e) Instability may be hidden by a pole-zero cancellation. 
 
f) By using Simulink-MATLAB, unit step responses for parts b) and c) are 
 shown below: (M = 2 , K = 3 , τ = 3) 
0 2 4 6 8 10 12 14 16 18 20
-10
-9
-8
-7
-6
-5
-4
-3
-2
-1
0
x 107
Time
O
ut
pu
t
 
 Figure S6.14b. Unit step response for part b). 
0 2 4 6 8 10 12 14 16 18 20
0
1
2
3
4
5
6
Time
O
ut
pu
t
 
 Figure S6.14c. Unit step response for part c) . 
6-21 
 6.15 
 
 
 From Eq. 6-71 and 6-72, 
 
 
21
12
11
22
22
11
2121
121122
2
1
2
1
2 AR
AR
AR
AR
AR
AR
AARR
ARARAR
+



+=
++
=ζ 
 
 Since 21 ≥+
x
x for all positive x and since R1, R2, A1, A2 are positive 
 
 
( ) 1
2
12
2
1
21
12 ≥+≥ζ
AR
AR
 
 
 
 
 
 6.16 
 
 
 a) If w1 = 0 and ρ = constant 
 
 20
2
2 wwdt
dhA −=ρ 
 
 2
2
2
1 h
R
w = 
 
 [ Note: could also define R2 by 2
2
222
2
2
1 h
R
qwh
R
q ρ=ρ=→= ] 
 
Substituting, 
 
2
2
0
2
2
1 h
R
w
dt
dhA −=ρ 
 
or 202
2
22 hwRdt
dhRA −=ρ 
 
Taking deviation variables and Laplace transforming 
 
 )()()( 022222 sWRsHsHsRA ′=′+′ρ 
6-22 
1)(
)(
22
2
0
2
+ρ
=
′
′
sRA
R
sW
sH
 
 
Since )(1)( 2
2
2 sHR
sW ′=′ 
 
 
1
1
1
1
)(
)(
2222
2
20
2
+ρ
=
+ρ
=
′
′
sRAsRA
R
RsW
sW
 
 
Let τ2 = ρA2R2 
 
 
1
1
)(
)(
20
2
+τ
=
′
′
ssW
sW
 
 
 
b) ρ = constant 
1
1
1 wdt
dhA −=ρ 21022 wwwdt
dhA −+=ρ 
 
)(1 21
1
1 hhR
w −= 2
2
2
1 h
R
w = 
 
c) Since this clearly is an interacting system, there will be a single zero. Also, 
we know the gain must be equal to one. 
 
∴ 
12
1
)(
)(
22
0
1
+ζτ+τ
+τ
=
′
′
ss
s
sW
sW a
 
12
1
)(
)(
22
0
2
+ζτ+τ=′
′
sssW
sW
 
 
or )1)(1(
1
)(
)(
210
1
+τ′+τ′
+τ
=
′
′
ss
s
sW
sW a
 )1)(1(
1
)(
)(
210
2
+τ′+τ′
=
′
′
sssW
sW
 
 
where 1τ′ and 2τ′ are functions of the resistances and areas and can 
only be obtained by factoring. 
 
f) Case b will be slower since the interacting system is 2nd-order, "including" 
the 1st-order system of Case a as a component. 
 
 
 
 
 
 
 
6-23 
 6.17 
 
 
 The input is ttTi ω=′ sin12)( 
 where 1hr262.0
hours24
radians2
−
=
pi
=ω 
 
 The Laplace transform of the input is from Table 3.1, 
 
 22
12)(
ω+
ω
=′
s
sTi 
 
 Multiplying the transfer function by the input transform yields 
 
 ))(15)(110(
)3672()( 22 ω+++
ω+−
=′
sss
s
sTi 
 
To invert, either (1) make a partial fraction expansion manually, or (2) use 
the Matlab residue function. The first method requires solution of a system 
of algebraic equations to obtain the coefficients of the four partial 
fractions. The second method requires that the numerator and denominator 
be defined as coefficients of descending powers of s prior to calling the 
Matlab residue function: 
 
 Matlab Commands 
 
>>b = [ 36*0.262 −72*0.262] 
 
b = 
 
 9.4320 −18.8640 
 
>> a = conv([10 1], conv([5 1], [1 0 0.262^2])) 
 
b = 
 
 50.0000 15.0000 4.4322 1.0297 0.0686 
 
>> [r,p,k] = residue(b,a) 
 
r = 
6.0865 − 4.9668i 
6.0865 + 4.9668i 
38.1989 
 −50.3718 
6-24 
p = 
 −0.0000 − 0.2620i 
 −0.0000 + 0.2620i 
 −0.2000 
 −0.1000 
k = 
 
 [] 
 
Note: the residue function recomputes all the poles (listed under p). These 
are, in reverse order: p1 = 0.1( )101 =τ , p2 = 0.2( )52 =τ , and the two 
purely imaginary poles corresponding to the sine and cosine functions. 
The residues (listed under r) are exactly the coefficients of the 
corresponding poles, in other words, the coefficients that would have been 
obtained via a manual partial fraction expansion. In this case, we are not 
interested in the real poles since both of them yield exponential functions 
that go to 0 as t→ ∞. 
 
The complex poles are interpreted as the sine/cosine terms using Eqs. 3-69 
and 3-74. From (3-69) we have: 
 
 α1= 6.0865, β1 = 4.9668, b = 0, and ω=0.262. 
 
Eq. 3-74 provides the coefficients of the periodic terms: 
 
 ...sin2cos2)( 11 +ωβ+ωα= −− tetety btbt 
 
Substituting coefficients (because b= 0, the exponential terms = 1) 
 
 ...sin)9668.4(2cos)068.6(2)( +ω+ω= ttty 
 
or ...sin9336.9cos136.12)( +ω+ω= ttty 
 
The amplitude of the composite output sinusoidal signal, for large values, 
of t is given by 
 
 7.15)9336.9()136.12( 22 =+=A 
 
Thus the amplitude of the output is 15.7° for the specified 12° amplitude 
input. 
 
 
 
 
 
6-25 
 6.18 
 
 
a) Taking the Laplace transform of the dynamic model in (2-7) 
 
[ ] )()()()(( 1 sCqsCqsCqqVs TRTiTR ′+′=′++γ (1) 
 
 [ ] )()()()()1( 1 sCqqsCqqVs TRTR ′+=′++γ− (2) 
 
 Substituting for )(sCT′ from (2) into (1), 
 
 
 
[ ]
[ ][ ] )()()1()(
)()1(
)(
)(1
RRRR
R
Ti
T
qqqqqVsqqVs
qqVsq
sC
sC
+−++γ−++γ
++γ−
=
′
′
 
 
 
 
1)(
)1(
1)(
)1(
2
2
+


+


+
γ−γ
+


+
γ−
=
s
q
V
s
qqq
V
s
qq
V
R
R
 (3) 
 
 
 Substituting for )(1 sCT′ from (3) into (2), 
 
 =
′
′
)(
)(
sC
sC
Ti
T
1)(
)1(
1
2
2
+


+


+
γ−γ
=
s
q
V
s
qqq
V
R
 (4) 
 
b) Case (i), 0→γ 
 =
′
′
)(
)(
sC
sC
Ti
T
1
1
+

=
s
q
V
 =
′
′
)(
)(1
sC
sC
Ti
T
1
1
+


+


+
=
s
q
V
s
qq
V
R
 
 
 
 Case (ii), 1→γ 
 
 )(
)(
sC
sC
Ti
T
′
′
1
1
+

=
s
q
V )(
)(1
sC
sC
Ti
T
′
′
= 
6-26 
 Case (iii), 0→Rq 
 
 =
′
′
)(
)(
sC
sC
Ti
T
1)1(
1
2
2
2
+


+

 γ−γ=
s
q
V
s
q
V
 , =
′
′
)(
)(1
sC
sC
Ti
T
1
1
+

 γ=
s
q
V
 
 
 Case (iv), ∞→Rq 
 
 )(
)(
sC
sC
Ti
T
′
′
1
1
+

=
s
q
V
 = )(
)(1
sC
sC
Ti
T
′
′
 
 
c) Case (i), 0→γ 
 
This corresponds to the physical situation with no top tank. Thus the 
dynamics for CT are the same as for a single tank, and TiT CC ′≈′1 for small 
qR. 
 
 Case (ii), 1→γ 
 
Physical situation with no bottom tank. Thus the dynamics for 1TC are the 
same as for a single tank, and 1TT CC = at all times. 
 
Case (iii), 0→Rq 
 
Physical situation with two separate non-interacting tanks. Thus, top tank 
dynamics, 1TC , are first order, and bottom tank, TC , is second order. 
 
Case (iv), ∞→Rq 
 
Physical situation of a single perfectly mixed tank. Thus, 1TT CC = , and 
both exhibit dynamics that are the same as for a single tank. 
 
d) In Eq.(3), 
 
0)(
)1(
≥


+
γ−
Rqq
V
 
 
Hence the system cannot exhibit an inverse response. From the 
denominator of the transfer functions in Eq.(3) and (4), 
 
6-27 
2
1
2
1
2
)1(4
)(
)(
)1(
2
1 


γ−γ
+
=


+
γ−γ
=ζ
−
q
qq
qqq
V
q
V R
R
 
 
Since )5.01)(5.0()1( −≤γ−γ for 10 ≤γ≤ , 
 
1)(
2
1
≥

 +
=ζ
q
qq R
 
 
Hence, the system is overdamped and cannot exhibit overshoot. 
 
e) Since 1≥ζ , the denominator of transfer function in Eq.(3) and (4) can be 
written as )1)(1( 21 +τ+τ ss where, using Eq. 5-45 and 5-46, 
 
2
1
2
1
2
1
2
1
1)1(4
)(
)1(4
)(
)(
)1(



−
γ−γ
+
−


γ−γ
+



+
γ−γ
=τ
q
qq
q
qq
qqq
V
RR
R
 
 
2
1
2
1
2
1
2
2
1)1(4
)(
)1(4
)(
)(
)1(



−
γ−γ
+
+


γ−γ
+



+
γ−γ
=τ
q
qq
q
qq
qqq
V
RR
R
 
 
 
 It is given that 
 
 [ ] ststTi ww e
s
h
s
h
e
s
h
sC −− −=−=′ 1)( 
 
 Then using Eq. 5-48 and (4) 
 
 
1 2/ /
1 2
1 2
( ) ( ) 1
t t
T
e e
c t S t h
− τ − τ τ − τ
= − 
τ − τ 
 
 
 


τ−τ
τ−τ
−−−
τ−−τ−−
21
/)(
2
/)(
1
21
1)(
ww tttt
w
eehttS 
 
6-28 
 And using Eq. 6-15 and (3) 
 
 


τ−τ
τ−τ
+
τ−τ
τ−τ
+= τ−τ− 21 /
12
2/
21
1
1 1)()( tataT eehtStC 
 
 
 


τ−τ
τ−τ
+
τ−τ
τ−τ
+−− τ−−τ−− 21 /)(
12
2/)(
21
11)( ww ttattaw eehttS 
 where 
 
 
( ) 


+
γ−
=τ )(
1
R
a qq
V
 
 
The pulse response can be approximated reasonably well by the impulse 
response in the limit as 0→wt , keeping htw constant. 
 
 
 
 6.19 
 
 
 Let VR = volume of each tank 
 A1 = ρ1Cp1VR 
 A2 = ρ2Cp2VR 
 B1 = w1Cp1 
 B2 = w2Cp2 
 K = UA 
 
 Then energy balances over the six tanks give 
 
 ( ) ( )8386282 TTKTTBdt
dTA −+−= (1) 
 ( ) ( )6564262 TTKTTBdt
dTA −+−= (2) 
 ( ) ( )4742242 TTKTTBdt
dTA −+−= (3) 
 
 ( ) ( )7475171 TTKTTBdt
dTA −+−= (4) 
 
 ( ) ( )5653151 TTKTTBdt
dTA −+−= (5) 
6-29 
 ( ) ( )3831131 TTKTTBdt
dTA −+−= (6) 
 
 Define vectors 
 
 
[ ]TsTsTsTsTsTsTsT )(),(),(),(),(),()( 345678 ′′′′′′=′ 
 
 


′
′
= )(
)()(
1
2*
sT
sT
sT 
 
Using deviation variables, and taking the Laplace transform of Eqs.1 to 6, 
we obtain an equation set that can be represented in matrix notation as 
 
 )()()( * sTBsTAsTIs +′=′ (7) 
 
 where I is the 6×6 identity matrix 
 
 
 




















−−
−−
−−
−−
−−
−−
=
1
1
1
2
2
2
1
1
1
1
1
2
2
22
2
11
1
1
1
22
2
2
2
0000
0000
000
000
000
000
A
BK
A
K
A
BK
A
K
A
B
A
BK
A
K
A
B
A
K
A
BK
A
K
A
B
A
BK
A
K
A
B
A
BK
A 
 
 
 
 








=
1
1
2
2
00000
00000
A
B
A
B
B 
 
 From Eq. 7, 
 
 )()()( *1 sTBAIssT −−=′ 
6-30 
 Then 
 
 


=


′
′
000010
000001
)(
)(
7
8
sT
sT )()( *1 sTBAIs −− 
 
 
 
 
 6.20 
 
 
 The dynamic model for the process is given by Eqs. 2-45 and 2-46, 
 which can be written as 
 
 )(1 ww
Adt
dh
i −ρ
= (1) 
 
 
AhC
QTT
Ah
w
dt
dTi
i
ρ
+−
ρ
= )( (2) 
 
 where h is the liquid-level 
 A is the constant cross-sectional area 
 
 System outputs: h , T 
 System inputs : w, Q 
 
Hence assume that wi and Ti are constant. In Eq. 2, note that the nonlinear 
term 


dt
dTh can be linearized as 
 
h
dt
Td
dt
Tdh ′+′ 
 
 or 
dt
Tdh ′ since 0=
dt
Td
 
 
Then the linearized deviation variable form of (1) and (2) is 
 
w
Adt
hd
′
ρ
−=
′ 1
 
 
Q
ChA
T
hA
w
dt
Td i
′
ρ
+′
ρ
−
=
′ 1
 
6-31 
Taking Laplace transforms and rearranging, 
 
s
K
sW
sH 1
)(
)(
=
′
′
 , 0)(
)(
=
′
′
sQ
sH
 , 0)(
)(
=
′
′
sW
sT
 , 
1)(
)(
2
2
+τ
=
′
′
s
K
sQ
sT
 
where 
A
K
ρ
−=
1
1 ; and Cw
K
i
1
2 = , 
iw
hAρ
=τ2 
 
Unit-step change in Q: hth =)( , )1()( 2/2 τ−−+= teKTtT 
 
Unit step change in w: tKhth 1)( += , TtT =)( 
 
 
 6.21 
 
 
 Additional assumptions: 
 
 (i) The density, ρ, and the specific heat, C, of the process liquid are 
 constant. 
 
 (ii) The temperature of steam, Ts, is uniform over the entire heat transfer 
 area. 
 
 (iii) The feed temperature TF is constant (not needed in the solution). 
 
 Mass balance for the tank is 
 
 qq
dt
dV
F −= (1) 
 
 Energy balance for the tank is 
 
 )()()()]([ TTUATTCqTTCq
dt
TTVd
C srefrefFF
ref
−+−ρ−−ρ=
−
ρ (2) 
 
 where Tref is a constant reference temperature 
 A is the heat transfer area 
 
 Eq. 2 is simplified by substituting for 
dt
dV
 from Eq. 1. Also, replace 
 V by hAT (where TA is the tank area) and replace A by Tp h 
 (where pT is the perimeter of the tank). Then, 
 
6-32 
 qq
dt
dhA FT −= (3) 
 
 ( ) ( )T F F T s
dTCA h q C T T Up h T T
dt
ρ = ρ − + − (4) 
 
 Then, Eqs. 3 and 4 constitute the dynamic model for the system. 
 
 a) Making Taylor series expansion of nonlinear terms in (4) and introducing 
 deviation variables, Eqs. 3 and 4 become: 
 
 qq
dt
hdA FT ′−′=
′
 (5) 
 
 ( ) ( )T F F F T
dTCA h C T T q Cq Up h T
dt
′
′ ′ρ = ρ − − ρ + 
 ( )T s T sUp hT Up T T h′ ′+ + − (6) 
 
 Taking Laplace transforms, 
 
 
1 1( ) ( ) ( )F
T T
H s Q s Q s
A s A s
′ ′ ′= − (7) 
 
 
( )1 ( ) ( )T F F
F T F T
CA h C T T
s T s Q s
Cq Up h Cq Up h
    ρ ρ −
′ ′+ =    ρ + ρ +     
 
( )( ) ( )T sT s
F T F T
Up T TUp h T s H s
Cq Up h Cq Up h
   
−
′ ′+ +   ρ + ρ +   
 (8) 
 
 Substituting for ( )H s′ from (7) into (8) and rearranging gives 
 
 [ ] ( )1 ( ) ( )T FT T F
F T F T
CA h C T TA s s T s A s Q s
Cq Up h Cq Up h
    ρ ρ −
′ ′+ =    ρ + ρ +     
 
 [ ]( )( ) ( ) ( )T sT T s F
F T F T
Up T TUp hA s T s Q s Q s
Cq Up h Cq Up h
   
−
′ ′ ′+ + −   ρ + ρ +   
 (9) 
 
 Let T
F T
CA h
Cq Up h
ρ
τ =
ρ +
 
 
 Then from Eq. 7 
 
6-33 
 
( ) 1
( )F T
H s
Q s A s
′
=
′
 , 
( ) 1
( ) T
H s
Q s A s
′
= −
′
 , 
( ) 0
( )s
H s
T s
′
=
′
 
 
 And from Eq. 9 
 
 
( )
( )F
T s
Q s
′
=
′ ( ) ( )
( ) ( ) 1( )
1
T s F T
F T T s
T
Up T T C T T A
s
Cq Up h Up T T
A s s
   
− ρ −
+   ρ + −    
τ +
 
 
 
( )
( )
T s
Q s
′
=
′ ( ) ( )
( )
1
T s
F T
T
Up T T
Cq Up h
A s s
 
−
−  ρ + 
τ +
 
 
 
( )
( )s
T s
T s
′
=
′ 1
T
F T
Up h
Cq Up h
s
  ρ + 
τ +
 
 
 
 Note: 
 
 2
( )
( )
F T
T s
C T T A
Up T T
ρ −
τ =
−
 is the time constant in the numerator. 
 Because 0FT T− < (heating) and 0sT T− > , 2τ is negative. 
 
 We can show this property by using Eq. 2 at steady state: 
 
 ( ) ( )F F T sCq T T Up h T Tρ − = − − 
 
 or 
( )( ) T sF
F
Up h T TC T T
q
− −ρ − = 
 Substituting 
 
 2
T
F
hA
q
τ = − 
 Let TV hA= so that 2
F
V
q
τ = − = − (initial residence time of tank) 
 For ( )( )F
T s
Q s
′
′
 and ( )( )
T s
Q s
′
′
 the “gain” in each transfer function is 
 
6-34 
 ( )
( )T s
T F T
Up T TK
A Cq Up h
 
− =
ρ +  
 
 
 and must have the units temp/volume . 
 
 (The integrator s has units of t-1). 
 
 To simplify the transfer function gain we can substitute 
 
 
( )( ) F FT s
Cq T TUp T T
h
ρ −
− = − 
 
 from the steady-state relation. Then 
 
 ( )
( )FT F
T F T
Cq T TK
hA Cq Up h
−ρ −
=
ρ +
 
 
 or 
1
F
T
F
T TK
Up hV
Cq
−
=  
+ ρ 
 
 
 and we see that the gain is positive since 0FT T− > . 
 
 Further, it has dimensions of temp/volume. 
 
 (The ratio T
F
Up h
Cqρ
 is dimensionless). 
 
 b) Fh q− transfer function is an integrator with a positive gain. Liquid level 
 accumulates any changes in Fq , increasing for positive changes and vice- 
 versa. 
 
 h q− transfer function is an integrator with a negative gain. h accumulates 
 changes in q, in opposite direction, decreasing as q increases and vice 
 versa. 
 
 sh T− transfer function is zero. Liquid level is independent of sT , and of 
 the steam pressure sP . 
 
 T q− transfer function is second-order due to the interaction with liquid 
 level; it is the product of an integrator and a first-order process. 
6-35 
 
 FT q− transfer function is second-order due to the interaction with liquid 
 level and has numerator dynamics since Fq affects T directly as well if 
 FT T≠ . 
 
 sT T− transfer function is simple first-order because there is no interaction 
 with liquid level. 
 
 c) Fh q− : h increases continuously at a constant rate. 
 
 h q− : h decreases continuously at a constant rate. 
 
 sh T− : h stays constant. 
 
 FT q− : for FT T< , T decreases initially (inverse response) and then 
 increases. After long times, T increases like a ramp function. 
 
 T q− : T decreases, eventually at a constant rate. 
 
 sT T− : T increases with a first-order response and attains a new steady 
 state. 
 
 
 
6.22 
 
 
a) The two-tank process is described by the following equations in deviation 
 variables: 
 
'
' ' '1
1 1 2
1
1 1 ( = − − ρ  
dh
w h h
dt A R
 (1) 
 
'
' '2
1 2
2
1 1 ( = − ρ  
dh h h
dt A R
 (2) 
 
 Laplace transforming 
 
 
' ' ' '
1 1 1 2( ) ( ) ( ) ( )ρ = − +iA RsH s RW s H s H s (3) 
 
 
' ' '
2 2 1 2( ) ( ) ( )ρ = −A RsH s H s H s (4) 
 
6-36 
 From (4) 
 
 
' '
2 2 1( 1) ( ) ( )ρ + =A Rs H s H s (5) 
 
or 
 
'
2
'
1 2 2
( ) 1 1
( ) 1 1= =ρ + τ +
H s
H s A Rs s
 (6) 
 
 where 2 2τ = ρA R 
 
Returning to (3) 
 
' ' '
1 1 2( 1) ( ) ( ) ( )ρ + − = iA Rs H s H s RW s (7) 
 
Substituting (6) with 1 1τ = ρA R 
 
 
' '
1 1
2
1( 1) ( ) ( )
1
 
τ + − = 
τ +  i
s H s RW s
s
 (8) 
 
 or 
 
2 ' '
1 2 1 2 1 2( ) ( ) ( ) ( 1) ( ) τ τ + τ τ = τ +  is s H s R s W s (9) 
[ ]
'
1 2
'
1 1 2 1 2
( ) ( 1)
( ) ( )
τ +
=
τ τ + τ + τ
H s R s
W s s s
 (10) 
 
 Dividing numerator and denominator by 1 2( )τ + τto put into standard form 
 
 
'
1 1 2 2
'
1 1 2
1 2
( ) [ /( )]( 1)
( )
1
τ + τ τ +
=  τ τ
+ τ + τ 
H s R s
W s
s s
 (11) 
 
 Note that 
 
 
1 2 1 2 1 2
1 1
( )= = = =τ + τ ρ + ρ ρ + ρ
R RK
A R A R A A A
 (12) 
 
 since 1 2= +A A A 
 
 Also, let 
 
6-37 
 
2 2
1 2 1 2 1 2
1 2 1 2( )
τ τ ρ ρ
τ = = =
τ + τ ρ +s
R A A RA A
R A A A
 (13) 
 
 so that 
 
 
'
1 2
'
3
( ) ( 1)
( ) ( 1)
τ +
=
τ +i
H s K s
W s s s
 (14) 
 
 and 
 
 
' ' '
2 2 1 2
' ' '
1 2 3
( ) ( ) ( ) ( 1)1
( ) ( ) ( ) ( 1) ( 1)
τ +
= =
τ + τ +i i
H s H s H s K s
W s H s W s s s s
 
 
3( 1)
=
τ +
K
s s
 (15) 
 
 
 Transfer functions (6), (14) and (15) define the operation of the two-tank 
 process. 
 
 The single-tank process is described by the following equation in 
 deviation variables: 
 
 
'
'1
=
ρ i
dh
w
dt A
 (16) 
 
 Note that ω , which is constant, subtracts out. 
 
 
 Laplace transforming and rearranging: 
 
 
'
'
( ) 1/
( )
ρ
=
i
H s A
W s s
 (17) 
 Again 
 
1
=
ρ
K
A
 
 
'
'
( )
( ) =i
H s K
W s s
 (18) 
 
 which is the expected integral relationship with no zero. 
 
 
 
6-38 
b) For 1 2 / 2A A A= = 
 
 
2
3
/ 2
/ 4
τ = ρ 
τ = ρ 
AR
AR
 (19) 
 
 Thus 2 32τ = τ 
 
 We have two sets of transfer functions: 
 
 One-Tank Process Two-Tank Process 
 
 
'
'
( )
( ) =i
H s K
W s s
 
'
3
'
3
( ) (2 1)
( ) ( 1)
τ +
=
τ +
i
i
H s K s
W s s s
 
 
 
'
2
'
3
( )
( ) ( 1)= τ +i
H s K
W s s s
 
 
 Remarks: 
 
- The gain ( 1/ )= ρK A is the same for all TF’s. 
- Also, each TF contains an integrating element. 
- However, the two-tank TF’s contain a pole 3( 1)τ +s that will “filter 
out” changes in level caused by changing wi(t). 
- On the other hand, for this special case we see that the zero in the first 
tank transfer function ' '( ( ) / ( ))i iH s W s is larger than the pole 
 2 3 3τ > τ 
and we should make sure that amplification of changes in h1(t) caused 
by the zero do not more than cancel the beneficial filtering of the pole 
so as to cause the first compartment to overflow easily. 
Now look at more general situations of the two-tank case: 
 
'
1 2 2
'
1 2 3
( ) ( 1) ( 1)
( ) ( 1)1
ρ + τ +
= =ρ τ + 
+  i
H s K A Rs K s
RA AW s s s
s s
A
 (20) 
'
2
'
3
( )
( ) ( 1)= τ +i
H s K
W s s s
 (21) 
 
For either 1 20 or 0→ →A A , 
 
 1 23 0
ρ
τ = →
RA A
A
 
6-39 
Thus the beneficial effect of the pole is lost as the process tends to 
“look” more like the first-order process. 
 
c) The optimum filtering can be found by maximizing 3τ with respect 
 to A1 (or A2) 
 
 1 2 1 13
( )ρ ρ −
τ = =
RA A RA A A
A A
 
 Find max [ ]33 1 1
1
: ( ) ( 1)∂τ ρτ = − + −
∂
R A A A
A A
 
 
 Set to 0: 1 1 0− − =A A A 
 
 12 =A A 
 
 1 / 2=A A 
 Thus the maximum filtering action is obtained when 1 2 / 2.= =A A A 
 
 The ratio of 2 3/τ τ determines the “amplification effect” of the zero on 
 1( ).h t 
 
 2 2
1 23 1
τ ρ
= =ρτ
A R A
RA A A
A
 
As 1A goes to 0, 2
3
τ
→ ∞
τ
 
 
Therefore the influence of changes in 1( ) on ( )iw t h t will be very large, 
leading to the possibility of overflow in the first tank. 
 
Summing up: 
 
The process designer would like to have 1 2 / 2= =A A A in order to obtain 
 the maximum filtering of 1 2( ) and ( ).h t h t However, the process response 
 should be checked for typical changes in ( )iw t to make sure that 1h does 
 not overflow. If it does, the area 1A needs to be increased until that is not 
a problem. 
 
Note that 2 3τ = τ when 1 =A A , thus someone must make a careful study 
 (simulations) before designing the partitioned tank. Otherwise, leave well 
 enough alone and use the non-partitioned tank. 
6-40 
 6.23 
 
 
 The process transfer function is 
 
 )124()11.0()()(
)(
22 +++
==
sss
K
sG
sU
sY
 
 
 where K = K1K2 
 
We note that the quadratic term describes an underdamped 2nd-order 
system since 
 42 =τ → 2=τ 
 
 22 =ζτ → 5.0=ζ 
a) For the second-order process element with τ2 = 2 and this degree of 
underdamping )5.0( =ζ , the small time constant, critically damped 2nd-
order process element (τ1 = 0.1 ) will have little effect. 
 
In fact, since 0.1 << τ2 (= 2) we can approximate the critically damped 
element as 12τ−e so that 
 
 
124
)( 2
2.0
++
≈
−
ss
Ke
sG
s
 
 
b) From Fig. 5.11 for 5.0=ζ , 15.0≈OS or from Eq. 5-53 
 
 Overshoot = exp 163.0
1 2
=



ζ−
piζ−
 
 
 Hence ymax = 0.163 KM + KM = 0.163 (1) (3) + 3 = 3.5 
 
 
c) From Fig. 5.4, ymax occurs at t/τ = 3K or tmax = 6.8 for underdamped 2nd-
order process with 5.0=ζ . 
 
Adding in effect of time delay t ′= 6.8 + 0.2 = 7.0 
 
 
d) By using Simulink-MATLAB 
 
 
 
6-41 
τ1 = 0.1 
0 5 10 15 20 25 30
-0.5
0
0.5
1
1.5
2
2.5
3
3.5
Time
O
ut
pu
t
Exact model
Approximate model
 
Fig S6.23a. Step response for exact and approximate model ; τ1 = 0.1 
 
 
 
 
τ1 = 1 
0 5 10 15 20 25 30
-0.5
0
0.5
1
1.5
2
2.5
3
3.5
Time
O
u
tp
u
t
Exact model
Approximate model
 
 Fig S6.23b. Step response for exact and approximate model ; τ1 = 1 
 
 
 
6-42 
τ1 = 5 
0 5 10 15 20 25 30
-0.5
0
0.5
1
1.5
2
2.5
3
3.5
Time
O
u
tp
u
t
Exact model
Approximate model
 
 Fig S6.23c. Step response for exact and approximate model ; τ1 = 5 
 
As noted in plots above, the smaller τ1 is, the better the quality of the 
approximation. For large values of τ1 (on the order of the underdamped 
element's time scale), the approximate model fails. 
 
 
 
 6.24 
 
 
0 50 100 150 200 250 300 350 400
-1.4
-1.2
-1
-0.8
-0.6
-0.4
-0.2
0
Time
O
ut
pu
t
 
 Figure S6.24. Unit step response in blood pressure. 
6-43 
Transport
Delay1= 30 s
Transport
Delay = 75 s
-0.4
40s+1
Transfer Fcn1
-1
40s+1
Transfer Fcn
Step1
Step
Scope
 The Simulink-MATLAB block diagram is shown below 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 It appears to respond approx. as a first-order or overdamped second-order 
 process with time delay. 
7-1 
��������	�
 
 
 
 
 7.1 
 
 
In the absence of more accurate data, use a first-order transfer function as 
 
 
'( )
'( ) 1
s
i
T s Ke
Q s s
−θ
=
τ +
 
 
o( ) (0) (124.7 120) F0.118
540 500 gal/mini
T TK
q
∞ − −
= = =
∆ −
 
 θ = 3:09 am – 3:05 am = 4 min 
 
Assuming that the operator logs a 99% complete system response as “no 
change after 3:34 am”, 5 time constants elapse between 3:09 and 3:34 am. 
 
5τ = 3:34 min − 3:09 min = 25 min 
 τ = 25/5 min = 5 min 
 
Therefore, 
 
 
4'( ) 0.188
'( ) 5 1
s
i
T s e
Q s s
−
=
+
 
 
To obtain a better estimate of the transfer function, the operator should log 
more data between the first change in T and the new steady state. 
 
 
7.2 
 
 Process gain, 2
(5.0) (0) (6.52 5.50) min0.336
30.4 0.1 fti
h hK
q
− −
= = =
∆ ×
 
a) Outputat 63.2% of the total change 
 
 = 5.50 + 0.632(6.52-5.50) = 6.145 ft 
 
 Interpolating between h = 6.07 ft and h = 6.18 ft 
 
Solution Manual for Process Dynamics and Control, 2nd edition, 
Copyright © 2004 by Dale E. Seborg, Thomas F. Edgar and Duncan A. Mellichamp. 
7-2 
 
(0.8 0.6)0.6 (6.145 6.07) min 0.74 min(6.18 6.07)
−
τ = + − =
−
 
 b) 
 
 
0
(0.2) (0) 5.75 5.50 ft ft1.25
0.2 0 0.2 min mint
dh h h
dt
=
− −
≈ = =
−
 
 
 Using Eq. 7-15, 
 
 
0t
KM
dh
dt
=
τ = =   
min84.0
25.1
)1.04.30(347.0
=
××
 
c) The slope of the linear fit between ti and 


−∞
−
−≡ )0()(
)0()(1ln
hh
hth
z ii gives an 
 approximation of (-1/τ) according to Eq. 7-13. 
 
Using h(∞) = h(5.0) = 6 .52, the values of zi are 
 
ti zi ti zi 
0.0 0.00 1.4 -1.92 
0.2 -0.28 1.6 -2.14 
0.4 -0.55 1.8 -2.43 
0.6 -0.82 2.0 -2.68 
0.8 -1.10 3.0 -3.93 
1.0 -1.37 4.0 -4.62 
1.2 -1.63 5.0 - ∞ 
 
Then the slope of the best-fit line, using Eq. 7-6 is 
 
2
131
13 ( )
tz t z
tt t
S S S
slope
S S
− 
= − = τ −  (1) 
 
where the datum at ti = 5.0 has been ignored. 
 
 Using definitions, 
 
 0.18=tS 4.40=ttS 
 5.23−=zS 1.51−=tzS 
 
 Substituting in (1), 
 
 
1 1.213 − = − 
τ  0.82 minτ = 
7-3 
 d) 
 
0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 5
5.4
5.6
5.8
6
6.2
6.4
6.6
6.8
Experimental data
Model a)
Model b)
Model c)
 
 Figure S7.2. Comparison between models a), b) and c) for step response. 
 
 
 
 7.3 
 
 
 a) 
 
1 1
1
( )
( ) 1
T s K
Q s s
′
=
′ τ +
 
2 2
1 2
( )
( ) 1
T s K
T s s
′
=
′ τ +
 
 
2
2 1 2 1 2
1 2 1
( )
( ) ( 1)( 1) ( 1)
sT s K K K K e
Q s s s s
−τ
′
= ≈
′ τ + τ + τ +
 (1) 
 
where the approximation follows from Eq. 6-58 and the fact that τ1>τ2 as 
revealed by an inspection of the data. 
 
667.2
8285
0.100.18)0()50( 11
1 =
−
−
=
∆
−
=
q
TT
K 
 75.0
0.100.18
0.200.26
)0()50(
)0()50(
11
22
2 =
−
−
=
−
−
=
TT
TT
K 
 
Let z1, z2 be the natural log of the fraction incomplete response for T1,T2, 
respectively. Then, 
7-4 


 −
=


−
−
=
8
)(18ln)0()50(
)()50(ln)( 1
11
11
1
tT
TT
tTT
tz 


 −
=


−
−
=
6
)(26ln)0()50(
)()50(ln)( 2
22
22
2
tT
TT
tTT
tz 
A graph of z1 and z2 versus t is shown below. The slope of z1 versus t line 
is –0.333 ; hence (1/-τ1)=-0.333 and τ1=3.0 
 
From the best-fit line for z2 versus t, the projection intersects z2 = 0 at 
t≈1.15. Hence τ2 =1.15. 
 
 
13
667.2
)('
)('1
+
=
ssQ
sT
 (2) 
 
115.1
75.0
)('
)('
1
2
+
=
ssT
sT
 (3) 
 Figure S7.3a. z1 and z2 versus t 
 
b) By means of Simulink-MATLAB, the following simulations are obtained 
 
0 2 4 6 8 10 12 14 16 18 20 22
10
12
14
16
18
20
22
24
26
28
time
T 1
 
 
,
 
 
 
T 2
T1
T2
T1 (experimental)
T2 (experimental)
 
 Figure S7.3b. Comparison of experimental data and models for step change 
-8.0
-7.0
-6.0
-5.0
-4.0
-3.0
-2.0
-1.0
0.0
0 5 10 15 20
time,t
z
1,
z
2
7-5 
7.4 
 
 
ssss
sXsGsY 5.1)1)(13)(15(
2)()()( ×
+++
== 
 
Taking the inverse Laplace transform 
 
( ) -75/8*exp(-1/5*t)+27/4*exp(-1/3*t)-3/8*exp(-t)+3y t = (1) 
 
a) Fraction incomplete response 
 



−=
3
)(1ln)( tytz 
 
Figure S7.4a. Fraction incomplete response; linear regression 
 
 From the graph, slope = -0.179 and intercept ≈ 3.2 
 
 Hence, 
 
 -1/τ = -0.179 and τ = 5.6 
 
θ = 3.2 
 
3.22( )
5.6 1
seG s
s
−
=
+
 
 
 b) In order to use Smith’s method, find t20 and t60 
 
 y(t20)= 0.2 × 3 =0.6 
 y(t60)= 0.6 × 3 =1.8 
 
 Using either Eq. 1 or the plot of this equation, t20 = 4.2 , t60 = 9.0 
 Using Fig. 7.7 for t20/ t60 = 0.47 
 ζ= 0.65 , t60/τ= 1.75, and τ = 5.14 
z(t) = -0.1791 t + 0.5734
-9.0
-8.0
-7.0
-6.0
-5.0
-4.0
-3.0
-2.0
-1.0
0.0
0 10 20 30 40 50
time,t
z(t
)
7-6 
 
168.64.26
2)( 2 ++≈ sssG 
 
 The models are compared in the following graph:
 
0 5 10 15 20 25 30 35 40
0
0.5
1
1.5
2
2.5
time,t
y(t
) Third-order model
First order model
Second order model
 
 Figure S7.4b. Comparison of three models for step input 
 
 
7.5 
 
 
 The integrator plus time delay model is 
G(s) sK e
s
−θ
 
 In the time domain, 
 y(t) = 0 t < 0 
y(t)= K (t-θ) t ≥ 0 
 
Thus a straight line tangent to the point of inflection will approximate the 
step response. Two parameters must be found: K and θ (See Fig. S7.5 a) 
 
1.- The process gain K is found by calculating the slope of the straight 
line. 
 K = 074.0
5.13
1
= 
 
2.- The time delay is evaluated from the intersection of the straight line 
and the time axis (where y = 0). 
 θ = 1.5 
7-7 
Therefore the model is G(s) = se
s
5.1074.0 −
 
 
 
 Figure S7.5a. Integrator plus time delay model; parameter evaluation 
 
From Fig. E7.5, we can read these values (approximate): 
 
Time Data Model 
0 0 -0.111 
2 0.1 0.037 
4 0.2 0.185 
5 0.3 0.259 
7 0.4 0.407 
8 0.5 0.481 
9 0.6 0.555 
11 0.7 0.703 
14 0.8 0.925 
16.5 0.9 1.184 
30 1 2.109 
 
Table.- Output values from Fig. E7.5 and predicted values by model 
 
 A graphical comparison is shown in Fig. S7.5 b 
 
Figure S7.5b. Comparison between experimental data and integrator plus time 
delay model. 
Slope = KM
y(t)
θ
0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1
0 5 10 15 20 25 30
Time
O
u
tp
u
t
Experimental data
Integrator plus time delay model
7-8 
 7.6 
 
 
a) Drawing a tangent at the inflection point which is roughly at t ≈5, the 
intersection with y(t)=0 line is at t ≈1 and with the y(t)=1 line at t ≈14. 
 Hence θ =1 , τ = 14−1=13 
 
113
)(1 +≈
−
s
e
sG
s
 
 
b) Smith’s method 
 
 From the graph, t20 = 3.9 , t60 = 9.6 ; using Fig 7.7 for t20/ t60 = 0.41 
 
 ζ = 1.0 , t60/τ= 2.0 , hence τ = 4.8 and τ1 = τ2 = τ = 4.8 
 
 2)18.4(
1)(
+
≈
s
sG 
 
 Nonlinear regression 
 
From Figure E7.5, we can read these values (approximated): 
 
 
 
 
 
 
 
 
 
Table.- Output values from Figure E7.5 
 
In accounting for Eq. 5-48, the time constants were selected to minimize 
the sum of the squares of the errors between data and model predictions. 
Use Excel Solver for this Optimization problem: 
 
τ1 =6.76 and τ2 = 6.95 
 
))176.6)(195.6(
1)(
++
≈
ss
sG 
 
The models are compared in the following graph: 
 
Time Output
0.0 0.0
2.0 0.1
4.0 0.2
5.0 0.3
7.0 0.4
8.0 0.5
9.0 0.6
11.0 0.7
14.0 0.8
17.5 0.9
30.0 1.0
7-9 
0 5 10 15 20 25 30 35 40 45 50
0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1
time
O
ut
pu
t
Non linear regression model
First-order plus time delay model
Second order model (Smith's method)
 
 Figure S7.6. Comparison of three models for unit step input 
 
 
 
 7.7 
 
 
a) From the graph, time delay θ = 4.0 min 
 
 Using Smith’s method, 
 
 from the graph, 20 5.6t + θ ≈ , 60 9.1t + θ ≈ 
 
 6.120 =t , 1.560 =t , 314.01.5/6.1/ 6020 ==ttFrom Fig.7.7 , 1.63ζ = , 60 / 3.10t τ = , 1.645τ = 
 
 Using Eqs. 5-45, 5-46, 1 4.81τ = , 2 0.56τ = 
 
 
b) Overall transfer function 
 
 
4
1 2
10( ) ( 1)( 1)
seG s
s s
−
=
τ + τ +
 , 1 2τ > τ 
 
 Assuming plug-flow in the pipe with constant-velocity, 
 
7-10 
 ( ) pspipeG s e−θ= , 
3 1 0.1min
0.5 60p
θ = × = 
 
 Assuming that the thermocouple has unit gain and no time delay 
 
 
2
1( ) ( 1)TCG s s= τ + since 2 1τ << τ 
 
 Then 
 
 
3
1
10( ) ( 1)
s
HE
eG s
s
−
=
τ +
 , so that 
 
3
0.1
1 2
10 1( ) ( ) ( ) ( ) ( )
1 1
s
s
HE pipe TC
eG s G s G s G s e
s s
−
−
   
= =    
τ + τ +    
 
 
 
 7.8 
 
 
a) To find the form of the process response, we can see that 
 
2( ) ( )( 1) ( 1) ( 1)
K K M K MY s U s
s s s s s s s
= = =
τ + τ + τ +
 
 
Hence the response of this system is similar to a first-order system with a 
ramp input: the ramp input yields a ramp output that will ultimately cause 
some process component to saturate. 
 
b) By applying partial fraction expansion technique, the domain response for 
this system is 
 
Y(s) = 2 1
A B C
s s s
+ +
τ +
 hence y(t) = -KMτ + KMt − KMτe-t/τ 
In order to evaluate the parameters K and τ, important properties of the 
above expression are noted: 
 
1.- For large values of time (t>>τ) , y(t) ≈ ( )y t′ = KM (t-τ) 
2.- For t = 0, (0)y′ = −KMτ 
 
These equations imply that after an initial transient period, the ramp input 
yields a ramp output with slope equal to KM. That way, the gain K is 
7-11 
obtained. Moreover, the time constant τ is obtained from the intercept in 
Fig. S7.8 
 
 
 Figure S7.8. Time domain response and parameter evaluation 
 
 
 
 
7.9 
 
 
 For underdamped responses, 
 














τ
ζ−
ζ−
ζ
+



τ
ζ−
−=
τζ− tteKMty t
2
2
2
/ 1sin
1
1
cos1)( (5-51) 
 
a) At the response peaks, 
 
2 2
/
2
1 1
cos sin
1
tdy KM e t t
dt
−ζ τ
     
− ζ − ζζ ζ     = +    τ τ τ 
− ζ     
 
 
 
2 2 2
/ 1 1 1sin cos 0te t t−ζ τ
    
− ζ − ζ − ζζ     − − + =   τ τ τ τ     
 
 
 Since KM ≠ 0 and 0/ ≠τζ− te 
 
 



τ
ζ−




τ
ζ−
+ζ−τ
ζ
+



τ
ζ−


τ
ζ
−
τ
ζ
= tt
22
2
22 1
sin1
1
1
cos0 
 
−ΚΜτ
Slope = KM
y(t)
7-12 
 pi=



τ
ζ−
= nt sin
1
sin0
2
 , t
21 ζ−
piτ
= n 
 where n is the number of peak. 
 
 Time to the first peak, 
21 ζ−
piτ
=pt 
 
b) Graphical approach: 
 
Process gain, 
 
 
( ) (0) 9890 9650 lb80 hr95 92 psig
D Dw wK
Ps
∞ − −
= = =
∆ −
 
 
 Overshoot = 333.0
96509890
98909970
=
−
−
=
b
a
 
 
 From Fig. 5.11, ζ ≈ 0.33 
 
 
 tp can be calculated by interpolating Fig. 5.8 
 
 For ζ ≈ 0.33 , tp ≈ 3.25 τ 
 
 Since tp is known to be 1.75 hr , τ = 0.54 
 
 
 2 2 2
80( )
2 1 0.29 0.36 1
KG s
s s s s
= =
τ + ζτ + + + 
 
 
Analytical approach 
 
The gain K doesn’t change: lb80 hr
psig
K = 
 To obtain the ζ and τ values, Eqs. 5-52 and 5-53 are used: 
 
 Overshoot = 333.0
96509890
98909970
=
−
−
=
b
a
 = exp(-ζpi/(1-ζ2)1/2) 
 
Resolving, ζ = 0.33 
 
7-13 
 
2
1.754 hence 0.527 hr
1
pt
piτ
= = τ =
− ζ
 
 2 2 2
80( )
2 1 0.278 0.35 1
KG s
s s s s
= =
τ + ζτ + + + 
 
 
c) Graphical approach 
 
From Fig. 5.8, ts/τ = 13 so ts = 2 hr (very crude estimation) 
 
 
Analytical approach 
 
 From settling time definition, 
 
 y = ± 5% KM so 9395.5 < y < 10384.5 
 
 
 (KM ± 5% KM) = KM[ 1-e(-0.633)[cos(1.793ts)+0.353sin(1.793ts)]] 
 
 
1 ± 0.05 = 1 – e(0.633 ts) cos(1.793 ts) + 0.353e (-0.633 ts) sin(1.7973 ts) 
 
 
 
 Solve by trial and error…………………… ts ≈ 6.9 hrs 
 
 
 
7.10 
 
 
a) 2 2
'( )
'( ) 2 1
T s K
W s s
=
τ + ζτ + 
 
o( ) (0) 156 140 C0.2
80 Kg/min
T TK
w
∞ − −
= = =
∆
 
 
From Eqs. 5-53 and 5-55, 
 
Overshoot = 344.0
140156
1565.161
=
−
−
=
b
a
 = exp(-ζpi (1-ζ2)1/2 
By either solving the previous equation or from Figure 5.11, ζ= 0.322 
(dimensionless) 
7-14 
There are two alternatives to find the time constant τ : 
 
 1.- From the time of the first peak, tp ≈ 33 min. 
 
One could find an expression for tp by differentiating Eq. 5-51 and 
solving for t at the first zero. However, a method that should work 
(within required engineering accuracy) is to interpolate a value of 
ζ=0.35 in Figure 5.8 and note that tp/τ ≈ 3 
 
 Hence τ ≈ min105.9
5.3
33
−≈ 
 
2.- From the plot of the output, 
 
Period =
2
2
1
P piτ=
− ζ
= 67 min and hence τ =10 min 
 
Therefore the transfer function is 
 
 
144.6100
2.0
)('
)('
)( 2 ++
==
sssW
sT
sG 
 
b) After an initial period of oscillation, the ramp input yields a ramp output 
with slope equal to KB. The MATLAB simulation is shown below: 
 
0 10 20 30 40 50 60 70 80 90 100
140
142
144
146
148
150
152
154
156
158
160
time
O
u
tp
u
t
 
 Figure S7.10. Process output for a ramp input 
 
7-15 
 We know the response will come from product of G(s) and Xramp = B/s2 
Then 2 2 2( ) ( 2 1)
KBY s
s s s
=
τ + ζτ + 
From the ramp response of a first-order system we know that the response 
will asymptotically approach a straight line with slope = KB. Need to find 
the intercept. By using partial fraction expansion: 
 
3 41 2
2 2 2 2 2 2( ) ( 2 1) 2 1
sKBY s
s s s s s s s
α + αα α
= = + +
τ + ζτ + τ + ζτ + 
 
Again by analogy to the first-order system, we need to find only α1 and 
α2. Multiply both sides by s2 and let s→ 0, α2 = KB (as expected) 
 
Can’t use Heaviside for α1, so equate coefficients 
 
2 2 2 2 3 2
1 2 3 4( 2 1) ( 2 1)KB s s s s s s s= α τ + ζτ + + α τ + ζτ + + α + α 
 
We can get an expression for α1 in terms of α2 by looking at terms 
containing s. 
 
s: 0 = α1+α22ζτ → α1 = -KB2ζτ 
 
and we see that the intercept with the time axis is at t = 2ζτ. Finally, 
presuming that there must be some oscillatory behavior in the response, 
we sketch the probable response (See Fig. S7.10) 
 
 
 
7.11 
 
 
a) Replacing τ by 5, and K by 6 in Eq. 7-34 
 
/ 5 /5( ) ( 1) [1 ]6 ( 1)t ty k e y k e u k−∆ −∆= − + − − 
 
b) Replacing τ by 5, and K by 6 in Eq. 7-32 
 
( ) (1 ) ( 1) 6 ( 1)
5 5
t ty k y k u k∆ ∆= − − + − 
 
In the integrated results tabulated below, the values for ∆t = 0.1 are shown 
only at integer values of t, for comparison. 
 
 
7-16 
t y(k) (exact) 
y(k) 
(�t=1) 
y(k) 
(�t=0.1) 
0 3 3 3 
1 2.456 2.400 2.451 
2 5.274 5.520 5.296 
3 6.493 6.816 6.522 
4 6.404 6.653 6.427 
5 5.243 5.322 5.251 
6 4.293 4.258 4.290 
7 3.514 3.408 3.505 
8 2.877 2.725 2.864 
9 2.356 2.180 2.340 
10 1.929 1.744 1.912 
 
Table S7.11. Integrated results for the first order differential equation 
 
Thus ∆t = 0.1 does improve the finite difference model bringing it closer 
to the exact model. 
 
 
 
7.12 
 
 To find 1a′ and 1b , use the given first order model to minimize 
 
 
10
2
1 1
1
( () ( 1) ( 1))
n
J y k a y k b x k
=
′= − − − −∑ 
 0)1())(1()1()((2 11
10
11
=−−−−−′−=
′∂
∂ ∑
=
kykxbkyaky
a
J
n
 
 
10
1 1
11
2( ( ) ( 1) ( 1))( ( 1)) 0
n
J y k a y k b x k x k
b
=
∂
′= − − − − − − =
∂ ∑ 
Solving simultaneously for 1a′ and 1b gives 
 
 
10 10
1
1 1
1 10
2
1
( ) ( 1) ( 1) ( 1)
( 1)
n n
n
y k y k b y k x k
a
y k
= =
=
− − − −
′ =
−
∑ ∑
∑
 
 
10 10 10 10
2
1 1 1 1
1 210 10 10
2 2
1 1 1
( 1) ( ) ( 1) ( 1) ( 1) ( 1) ( )
( 1) ( 1) ( 1) ( 1)
n n n n
n n n
x k y k y k y k x k y k y k
b
x k y k y k x k
= = = =
= = =
− − − − − −
=  
− − − − −  
∑ ∑ ∑ ∑
∑ ∑ ∑
 
 
7-17 
 Using the given data, 
 
 212.35)()1(
10
1
=−∑
=
kykx
n
 , 749.188)()1(
10
1
=−∑
=n
kyky 
 
 14)1(
10
1
2
=−∑
=n
kx , 112.198)1(
10
1
2
=−∑
=n
ky 
 
 409.24)1()1(
10
1
=−−∑
=n
kxky 
 
 Substituting into expressions for 1a′ and 1b gives 
 
 1a′ = 0.8187 , 1b = 1.0876 
 
Fitted model is )(0876.1)(8187.0)1( kxkyky +=+ 
 
or )1(0876.1)1(8187.0)( −+−= kxkyky (1) 
 
 Let the first-order continuous transfer function be 
 
 
( )
( ) 1
Y s K
X s s
=
τ +
 
 
From Eq. 7-34, the discrete model should be 
 
 
/ /( ) ( 1) [1 ] ( 1)t ty k e y k e Kx k−∆ τ −∆ τ= − + − − (2) 
 
 
Comparing Eqs. 1 and 2, for ∆t=1, gives 
 
 τ = 5 and K = 6 
 
 Hence the continuous transfer function is 6/(5s+1) 
7-18 
 
0 1 2 3 4 5 6 7 8 9 10
2
3
4
5
6
7
8
time,t
y(t
)
actual data
fitted model
 
 Figure S7.12. Response of the fitted model and the actual data 
 
 
 
7.13 
 
 
 To fit a first-order discrete model 
 
)1()1()( 11 −+−′= kxbkyaky 
 
Using the expressions for 1a′ and b1 from the solutions to Exercise 7.12, 
with the data in Table E7.12 gives 
 
 918.01 =′a , 133.01 =b 
 
Using the graphical (tangent) method of Fig.7.5 . 
 
1=K , 0.68θ = , and 6.8τ = 
 
The response to unit step change for the first-order model given by 
 
0.68
6.8 1
se
s
−
+
 is 8.6/)68.0(1)( −−−= tety 
7-19 
 
 Figure S7.13- Response of the fitted model, actual data and graphical method 
 
 
 
 
0
0,1
0,2
0,3
0,4
0,5
0,6
0,7
0,8
0,9
1
0 2 4 6 8 10time,t
y(t
)
actual data
fitted model
graphical method
8-1 
��������	�
 
 
 
 
 8.1 
 
 
a) For step response, 
 input is Mtu =′ )( , 
s
M
sU =′ )( 
 )(
1
1)( sU
s
s
KsY
D
D
ca
′


+ατ
+τ
=′ = 


+ατ
+τ
)1(
1
ss
s
MK
D
D
c 
 
 =′ )(sYa )1(1 +ατ++ατ
τ
ss
MK
s
MK
D
c
D
Dc
 
 
 Taking inverse Laplace transform 
 
 )1()( )/()/( DD tctca eMKe
MK
ty ατ−ατ− −+
α
=′ 
 
 As α →0 
 MKdtetMKty c
t
t
ca
D
+
α
δ=′ ∫∞
=
ατ−
0
)/(
)()( 
 MKtMKty cDca +τδ=′ )()( 
 
 Ideal response, 
 
 )()()( sUsGsY ii ′=′ = 

 +τ
s
s
MK Dc
1
 = KcMτD + 
s
MKc
 
 
MKtMKty cDci +δτ=′ )()( 
 
Hence )()( tyty ia ′→′ as 0→α 
 
 For ramp response, 
input is Mttu =′ )( , 2)( s
M
sU =′ 
 
Solution Manual for Process Dynamics and Control, 2nd edition, 
Copyright © 2004 by Dale E. Seborg, Thomas F. Edgar and Duncan A. Mellichamp. 
8-2 
 )(
1
1)( sU
s
s
KsY
D
D
ca
′


+ατ
+τ
=′ = 


+ατ
+τ
)1(
1
2 ss
s
MK
D
D
c 
 
 =′ )(sYa )1()1( 2 +ατ++ατ
τ
ss
MK
ss
MK
D
c
D
Dc
 
 
 


+ατ
ατ
++
ατ−
+


+ατ
ατ
−τ=
1
)(1
1
1 2
2 sss
MK
ss
MK
D
DD
c
D
D
Dc 
 
 Taking inverse Laplace transform 
 
 
[ ] [ ])1(1)( )/()/( −ατ++−τ=′ ατ−ατ− DD tDctDca etMKeMKty 
 
 As α →0 
 
 MtKMKty cDca +τ=′ )( 
 
 Ideal response, 
 
 =′ )(sYi 

 +τ
2
1
s
s
MK Dc = 2s
MK
s
MK cDc +τ 
 
MtKMKty cDci +=′ τ)( 
 
Hence )()( tyty ia ′→′ as 0→α 
 
b) It may be difficult to obtain an accurate estimate of the derivative for use 
in the ideal transfer function. 
 
c) Yes. The ideal transfer function amplifies the noise in the measurement by 
taking its derivative. The approximate transfer function reduces this 
amplification by filtering the measurement. 
 
 
 
 8.2 
 
a) 
11)(
)(
1
2121
2
1
1
+τ
+τ+
=+
+τ
=
′
s
KsKK
K
s
K
sE
sP
 








+τ
+
+
τ
+=
1
1
)(
1
21
12
21
s
s
KK
K
KK 
8-3 
 b) Kc = K1 + K2 → K2 = Kc − K1 
 
 Dατ=τ1 
 
 
21
2
21
12
KK
K
KK
K D
D +
ατ
=
+
τ
=τ 
 
or 
21
21
KK
K
+
α
= 
 
 α=+ 221 KKK 
 
 )1(2221 −α=−α= KKKK 
 
 Substituting, 
 
 111 )1()1()1)(( KKKKK cc −α−−α=−α−= 
 
 Then, 
 cKK 


α
−α
=
1
1 
 
c) If Kc = 3 , τD = 2 , α = 0.1 then, 
 
 273
1.0
9.0
1 −=×
−
=K 
 
 30)27(32 =−−=K 
 
 τ1 = 0.1 × 2 = 0.2 
 
Hence 
 
 K1 + K2 = -27 + 30 = 3 
 
 2
3
2.030
21
12
=
×
=
+
τ
KK
K
 
 
 


+
+
=
12.0
123)(
s
s
sGc 
 
 
8-4 
 8.3 
 
 
a) From Eq. 8-14, the parallel form of the PID controller is : 



τ′+
τ′
+′= s
s
KsG D
I
ci
11)( 
 
From Eq. 8-15, for α →0, the series form of the PID controller is: 
 
 
[ ]111)( +τ


τ
+= s
s
KsG D
I
ca 
 
 


τ+
τ
+
τ
τ
+= s
s
K D
II
D
c
11 
 
 














τ
τ
+
τ
+
τ



τ
τ
+
+



τ
τ
+=
I
D
D
I
I
DI
D
c
s
s
K
11
111 
 
 Comparing Ga(s) with Gi(s) 
 
 



τ
τ
+=′
I
D
cc KK 1 
 



τ
τ
+τ=τ′
I
D
II 1 
 
I
D
D
D
τ
τ
+
τ
=τ′
1
 
 
b) Since 



+
I
D
τ
τ1 ≥ 1 for all τD, τI, therefore 
 cc KK ′≤ , II τ′≤τ and DD τ′≥τ 
 
c) For Kc = 4, τI=10 min , τD =2 min 
 
8.4=′cK , min12=τ′I , min67.1=τ′D 
d) Considering only first-order effects, a non-zero α will dampen all 
responses, making them slower. 
8-5 
 8.4 
 
 
 Note that parts a), d), and e) require material from Chapter 9 to work. 
 
a) System I (air-to-open valve) : Kv is positive. 
 
System II (air-to-close valve) : Kv is negative. 
 
b) System I : Flowrate too high → need to close valve →decrease controller 
output → reverse acting 
 
System II: Flow rate too high →need to close valve → increase 
controller output → direct acting. 
 
c) System I : Kc is positive 
System II : Kc is negative 
 
 d) Kc Kv Kp Km 
 
 System I : + + + + 
System II : − − + + 
 
 Kc and Kv must have same signs 
 
e) Any negative gain must have a counterpart that "cancels" its effect. Thus, 
the rule: 
 
# of negative gains to have negative feedback = 0 , 2 or 4. 
# of negative gains to have positive feedback = 1 or 3. 
 
 
 
 8.5 
 
 
a) From Eqs. 8-1 and 8-2, 
 
[ ])()()( tytyKptp mspc −+=(1) 
 
The liquid-level transmitter characteristic is 
 
 ym(t) = KT h(t) (2) 
 
where h is the liquid level 
 KT > 0 is the gain of the direct acting transmitter. 
8-6 
The control-valve characteristic is 
 
 q(t) = Kvp(t) (3) 
 
where q is the manipulated flow rate 
 Kv is the gain of the control valve. 
 
From Eqs. 1, 2, and 3 
 
 
[ ] [ ])()()()( thKtyKKptpKqtq TspcVv −=−=− 
 
 )(
)(
thKy
qtqKK
Tsp
cV
−
−
= 
For inflow manipulation configuration, q(t)> q when ysp(t)>KTh(t). Hence 
KvKc > 0 
 
then for "air-to-open" valve (Kv>0), Kc>0 : reverse acting controller 
and for "air-to-close" valve (Kv<0), Kc<0 : direct acting controller 
 
For outflow manipulation configuration, KvKc <0 
 
then for "air-to-open" valve, Kc<0 : direct acting controller 
and for "air-to-close" valve, Kc>0 : reverse acting controller 
 
b) See part(a) above 
 
 
 
 8.6 
 
 
 For PI control 
 
 



τ
++= ∫t
I
c dtteteKptp
0
**)(1)()( 
 
 



τ
+=′ ∫t
I
c dtteteKtp
0
**)(1)()( 
 
 Since 
e(t) = ysp – ym and ym= 2 
 
 
8-7 
 Then 
 
 e(t)= -2 
 
 



τ
−−=



−
τ
+−=′ ∫ tKdtKtp
I
c
t
I
c
22*)2(12)(
0
 
 
 Initial response = − 2 Kc 
 
 Slope of early response = 
I
cK
τ
−
2
 
 
 − 2 Kc = 6 → Kc = -3 
 
 
I
cK
τ
−
2
= 1.2 min-1 → τI = 5 min 
 
 
 8.7 
 
 
a) To include a process noise filter within a PI controller, it would be placed 
in the feedback path 
b) 
 
f
1
� ��s




τ
+
s
K
I
c
11
 
 
c) The TF between controller output )(sP′ and feedback signal Ym(s) would 
be 
 
8-8 
)1(
)1(
)(
)(
+ττ
+τ−
=
′
ss
sK
sY
sP
fI
Ic
m
 Negative sign comes from comparator 
 
 For 
s
M
sYm =)( 
 
 



+τ
++
τ
−
=




+τ
+τ
τ
−
=′
1)1(
1)( 22 s
C
s
B
s
AMK
ss
sMK
sP
fI
c
f
I
I
c
 
 
The 
1+τ s
C
f
 term gives rise to an exponential. 
 
To see the details of the response, we need to obtain B (= τI - τf) and A(=1) 
by partial fraction expansion. 
 
The response, shown for a negative change in Ym, would be 
 
 
 
Note that as 0→
τ
τ
I
f
 , the two responses become the same. 
d) If the measured level signal is quite noisy, then these changes might still 
be large enough to cause the controller output to jump around even after 
filtering. 
 
One way to make the digital filter more effective is to filter the process 
output at a higher sampling rate (e.g., 0.1 sec) while implementing the 
controller algorithm at the slower rate (e.g., 1 sec). 
 
A well-designed digital computer system will do this, thus eliminating the 
need for analog (continuous) filtering. 
time
y
-KcM
-KcM(1-τf/τI)
"Ideal" PI
Filtered PI
Slope = KcM/τI
8-9 
 8.8 
 
a) From inspection of Eq. 8-26, the derivative kick = r
t
K Dc ∆∆
τ
 
b) Proportional kick = rK c∆ 
 
c) e1 = e2 = e3 = …. = ek-2 = ek-1 = 0 
 
ek = ek+1 = ek+2 = …= ∆r 
 
 ppk =−1 
 


∆
∆
τ
+∆
τ
∆
+∆+= r
t
r
t
rKpp D
I
ck 
 


∆
τ
∆
++∆+=+ r
tirKpp
I
cik )1( , i = 1, 2, … 
 
 
 
c) To eliminate derivative kick, replace (ek – ek-1) in Eq. 8-26 by (yk-yk-1). 
 
 
 
 
 
 
k k+1 k+2 k+3
r
t
K Dc ∆∆
τ
rKc∆ r
tK
I
c ∆
τ
∆
kp
p
k-1
8-10 
 8.9 
 
 
a) The digital velocity P algorithm is obtained by setting 1/τI = τD = 0 in Eq. 
8-28 as 
 
∆pk = Kc(ek – ek-1) 
 
 = ( )[ ]1)( −−−− kspkspc yyyyK 
 
 = [ ]kkc yyK −−1 
 
The digital velocity PD algorithm is obtained by setting 1/τI = 0 in Eq. 8-
28 as 
∆pk = Kc [(ek – ek-1) + 
t
D
∆
τ (ek – 2ek-1 + ek-2)] 
 = Kc [ (-yk + yk-1) + 
t
D
∆
τ (-yk – 2yk-1 + yk-2) ] 
 In both cases, ∆pk does not depend on spy . 
 
b) For both these algorithms ∆pk = 0 if yk-2 = yk-1 = yk. Hence steady state is 
reached with a value of y that is independent of the value of spy . Use of 
these algorithms is inadvisable if offset is a concern. 
 
c) If the integral mode is present, then ∆pk contains the term Kc )( ksp
I
yyt −
τ
∆
. 
Thus, at steady state when ∆pk = 0 and yk-2 = yk-1 = yk , yk = spy and the 
offset problem is eliminated. 
 
 
 
 8.10 
 
 
 a) 



+ατ
τ
+
τ
+=
′
1
11)(
)(
s
s
s
K
sE
sP
D
D
I
c 
 
 
( )( 1) 1
( 1)
I D D D I
c
I D
s s s s s
K
s s
τ ατ + + ατ + + τ τ
=
τ ατ +
 
 


+αττ
ττα++ατ+τ+
= )1(
)1()(1 2
ss
ss
K
DI
DIDI
c 
 
8-11 
 Cross- multiplying 
 
 ( ) )()1()(1)()( 22 sEssKsPss DIDIcIDI ττα++ατ+τ+=′τ+τατ 
 
 

ατ+τ+=
′
τ+
′
τατ
dt
tde
teK
dt
tpd
dt
tpd
DIcIDI
)()()()()(2
2


ττα++ 2
2 )()1(
dt
ted
DI 
 
 b) 



+ατ
τ




τ
+τ
=
′
1
1
)(
)(
s
s
s
s
K
sE
sP
D
D
I
I
c 
 
 Cross-multiplying 
 
 ( ) )()1)(1()()1(2 sEssKsPss DIcDI +τ+τ=′+αττ 
 
 

τ+τ+=
′
τ+
′
τατ
dt
tde
teK
dt
tpd
dt
tpd
DIcIDI
)()()()()(2
2


ττ+ 2
2 )(
dt
ted
DI 
 
c) We need to choose parameters in order to simulate: 
 
 e.g., 2=cK , 3=τ I , 5.0=τD , 1.0=α , M = 1 
 
 By using Simulink-MATLAB 
Step Response
Time 
0 2 4 6 8 10
2
4
6
8
10
12
14
16
18
20
22
Parallel PID with a derivative filter
Series PID with a derivative filter
p'(t) 
 
Figure S8.10. Step responses for both parallel and series PID controllers 
 with derivative filter. 
8-12 
 8.11 
 
 
 a) ( )11)(
)(
+τ



τ
+τ
=
′
s
s
s
K
sE
sP
D
I
I
c 
 
 ( ) )()1)(1()( sEssKsPs DIcI +τ+τ=′τ 
 
 
( ) ( )( ) ( )c I D
I
Kdp t de t
e t
dt dt
′ 
= + τ + ττ  

ττ+ 2
2 )(
dt
ted
DI 
 
 
 b) With the derivative mode active, an impulse response will occur at t = 0. 
 Afterwards, for a unit step change in e(t), the response will be a ramp 
 with slope = ( ) /c I D IK τ + τ τ and intercept = /c IK τ for 0>t . 
 
 
t
p '
slope =
Impulse at t=0
c
I
K
τ
( )c I D
I
K τ + τ
τ
 
 
 
9-1 
��������	�
 
 
 
 
 9.1 
 
 
a) Flowrate pneumatic transmitter: 
 
qm(psig)= 15 psig - 3 psig ( gpm - 0 gpm) 3 psig400 gpm-0 gpm q
 
+   
 = 
psig0.03 (gpm) 3 psig
gpm
q
 
+   
 
 Pressure current transmitter: 
 
Pm(mA)= 20 mA - 4 mA ( in.Hg 10 in.Hg) 4 mA30 in.Hg -10 in.Hg p
 
− +   
 = 
mA0.8 (in.Hg) 4 mA
in.Hg
p
 
−   
 
 Level voltage transmitter: 
 
hm(VDC)= 5 VDC -1 VDC ( (m) - 0.5m) 1 VDC20 m - 0.5 m h
 
+   
 = 
VDC0.205 (m) 0.897 VDC
m
h  +   
 
 Concentration transmitter: 
Cm(VDC)= 10 VDC -1 VDC ( (g/L)-2 g/L)+1 VDC20 g/L - 2 g/L C
    
 = 
VDC0.5 (g/L)
g/L
C    
 
b) The gains, zeros and spans are: 
 
 
 
Solution Manual for Process Dynamics and Control, 2nd edition, 
Copyright © 2004 by Dale E. Seborg, Thomas F. Edgar and Duncan A. Mellichamp 
9-2 
 PNEUMATIC CURRENT VOLTAGE VOLTAGE 
GAIN 0.03psig/gpm0.8mA/in.Hg 0.205 VDC/m 0.5VDC/g/L 
ZERO 0gal/min 10 in.Hg 0.5m 2g/L 
SPAN 400gal/min 20 in.Hg 19.5m 18g/L 
*The gain is a constant quantity 
 
 
 9.2 
 
 
a) The safest conditions are achieved by the lowest temperatures and 
pressures in the flash vessel. 
 
VALVE 1.- Fail close 
VALVE 2.- Fail open 
VALVE 3.- Fail open 
VALVE 4.- Fail open 
VALVE 5.- Fail close 
 
Setting valve 1 as fail close prevents more heat from going to flash drum 
and setting valve 3 as fail open to allow the steam chest to drain. Setting 
valve 3 as fail open prevents pressure build up in the vessel. Valve 4 
should be fail-open to evacuate the system and help keep pressure low. 
Valve 5 should be fail-close to prevent any additional pressure build-up. 
 
b) Vapor flow to downstream equipment can cause a hazardous situation 
 
VALVE 1.- Fail close 
VALVE 2.- Fail open 
VALVE 3.- Fail close 
VALVE 4.- Fail open 
VALVE 5.- Fail close 
 
Setting valve 1 as fail close prevents more heat from entering flash drum 
and minimizes future vapor production. Setting valve 2 as fail open will 
allow the steam chest to be evacuated, setting valve 3 as fail close prevents 
vapor from escaping the vessel. Setting valve 4 as fail open allows liquid 
to leave, preventing vapor build up. Setting valve 4 as fail-close prevents 
pressure buildup. 
 
c) Liquid flow to downstream equipment can cause a hazardous situation 
 
VALVE 1.- Fail close 
VALVE 2.- Fail open 
VALVE 3.- Fail open 
VALVE 4.- Fail close 
VALVE 5.- Fail close 
9-3 
Set valve 1 as fail close to prevent all the liquid from being vaporized 
(This would cause the flash drum to overheat). Setting valve 2 as fail open 
will allow the steam chest to be evacuated. Setting valve 3 as fail open 
prevents pressure buildup in drum. Setting valve 4 as fail close prevents 
liquid from escaping. Setting valve 5 as fail close prevents liquid build-up 
in drum 
 
 
 9.3 
 
 
a) Assume that the differential-pressure transmitter has the standard range of 
3 psig to 15 psig for flow rates of 0 gpm to qm(gpm). Then, the pressure 
signal of the transmitter is 
 
PT = 3 + 22
12 q
qm 





 
KT = 2
24T
m
dP q
dq q
 
=    
 
 2.4/qm , q = 10% of qm 
 
 12/qm , q = 50% of qm 
KT = 
 18/qm , q = 75% of qm 
 
 21.6/qm , q = 90% of qm 
 
 
b) Eq. 9-2 gives 
1/ 2
( ) ( )vv m
s
Pq C f q f
g
 ∆
= =  � � 
 
For a linear valve, 
 
 ( )f P= = α� � , where α is a constant. 
 
 KV = m
dq q
dP
= α 
 
Hence, linear valve gain is same for all flowrates 
 
 
 
9-4 
 For a square-root valve, 
 
 ( )f P= = α� � 
 KV = 
1 1
2 22
m m m
m
q q qdq q
dP qp
α α
= α = =
�
 
 
 5qmα , q = 10% of qm 
 
 qmα , q = 50% of qm 
KV = 
 0.67qmα , q = 75% of qm 
 
 0.56qmα , q = 90% of qm 
 
 For an equal-percentage valve, 
 
 
11)( −− == PRRf α�� 
 
 KV = 1 ln lnm m
m
dq qq R R q R
dP q
−
 
= α = α   
�
 
 
 0.1qmαlnR , q = 10% of qm 
 
 0.5qmαlnR , q = 50% of qm 
KV = 
 0.75qmαlnR , q = 75% of qm 
 
 0.9qmαlnR , q = 90% of qm 
 
c) The overall gain is 
 
KTV = KTKV 
 
 Using results in parts a) and b) 
 
 For a linear valve 
 
 2.4α , q = 10% of qm 
 
 12α , q = 50% of qm 
KTV = 
 18α , q = 75% of qm 
 
 21.6α , q = 90% of qm 
9-5 
 For a square-root valve 
 
 KTV = 12α for all values of q 
 
 For an equal-percentage valve 
 
 
 0.24αlnR , q = 10% of qm 
 
 6.0αlnR , q = 50% of qm 
KTV = 
 13.5αlnR , q = 75% of qm 
 
 19.4αlnR , q = 90% of qm 
 
 
The combination with a square-root valve gives linear characteristics over 
the full range of flow rate. For R = 50 and α = 0.067 values, a graphical 
comparison is shown in Fig. S9.3 
 
0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1
0
1
2
3
4
5
6
7
q/qm
Linear valve
Square valve
% valve
 K
TV 
 
 Fig. S9.3.- Graphical comparison of the gains for the three valves 
 
d) In a real situation, the square-root valve combination will not give an 
exactly linear form of the overall characteristics, but it will still be the 
combination that gives the most linear characteristics. 
 
9-6 
 9.4 
 
 
 Nominal pressure drop over the condenser is 30 psi 
 
 ∆Pc = Kq2 
 
 30 = K (200)2 , K = 2
3 psi
4000 gpm
 
 ∆Pc = 24000
3 q 
 
Let ∆Pv be the pressure drop across the valve and cv PP ∆∆ , be the 
nominal values of ∆Pv , ∆Pc, respectively. Then, 
 
 ∆Pv = ( )v cP P∆ + ∆ −∆Pc = ( ) 2330 4000vP q+ ∆ − (1) 
 
 Using Eq. 9-2 
 
 
2/1
)( 


 ∆
=
s
v
v g
PfCq � (2) 
 and 
 
 
1/ 2 1/ 2
200
0.5 1.11( )
v v
v
s
q P PC
gf l
− −   ∆ ∆
= =       (3) 
 
 Substituting for ∆Pv from(1) and Cv from(3) into (2) , 
 
 
1/ 2
21/ 2 330
4000400 ( )
1.11 1.11
v
v
P qPq f
−
−
 + ∆ −  ∆
=        
� (4) 
 
a) vP∆ = 5 
 
Linear valve: �� =)(f , and Eq. 4 becomes 
 
2/12
11.1
00075.035
5.188
−




−
=
qql 
 
9-7 
Equal % valve: 11 20)( −− == ��� Rf assuming R=20 
 
20ln
11.1
00075.035
5.188
ln
1
2/12







 −
+=
−
qq
l 
 
0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1
0
50
100
150
200
250
l (valve lift)
q(g
pm
)
Linear valve
Equal % valve
 
 Figure S9.4a. Control valve characteristics for vP∆ = 5 
 
 
b) vP∆ = 30 
 
Linear valve: �� =)(f , and Eq. 4 becomes 
 
2/12
11.1
00075.060
94.76
−




−
=
qql 
 
Equal % valve: 120)( −= ��f ; Eq. 4 gives 
 
20ln
11.1
00075.060
94.76
ln
1
2/12







 −
+=
−
qq
l 
 
 
 
9-8 
0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1
0
50
100
150
200
250
300
l (valve lift)
q 
(gp
m
)
Linear valve
Equal % valve
 
 Figure S9.4b. Control valve characteristics for vP∆ = 30 
 
 
 
c) vP∆ = 90 
 
Linear valve: �� =)(f , and Eq. 4 becomes 
 
2/12
11.1
00075.0120
42.44
−




−
=
qql 
 
Equal % valve: 120)( −= ��f ; Eq. 4 gives 
 
 
20ln
11.1
00075.0120
42.44
ln
1
2/12







 −
+=
−
qq
l 
 
 
 
 
9-9 
0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1
0
50
100
150
200
250
300
l (valve lift)
Eq
ua
l %
 
v
al
v
e
Linear valve
Equal % valve
 
 Figure S9.4c. Control valve characteristics for vP∆ = 90 
 
 
 Conclusions from the above plots: 
 
1) Linearity of the valve 
 
For vP∆ = 5, the linear valve is not linear and the equal % valve is 
linear over a narrow range. 
 
For vP∆ = 30, the linear valve is linear for very low � and equal 
% valve is linear over a wider range of � . 
 
For vP∆ = 90, the linear valve is linear for � <0.5 approx., equal % 
valve is linear for � >0.5 approx. 
 
2) Ability to handle flowrates greater than nominal increases as vP∆ 
increases, and is higher for the equal % valve compared to that for the 
linear valve for each vP∆ . 
 
3) The pumping costs are higher for larger vP∆ . This offsets the 
advantage of large vP∆ in part 1) and 2)9-10 
 9.5 
 
 
 Let ∆Pv/∆Ps = 0.33 at the nominal 320 gpmq = 
 
 ∆Ps = ∆PB + ∆Po = 40 + 1.953×10-4 q2 
 
 ∆Pv= PD - ∆Ps = (1 –2.44×10-6 q2)PDE – (40 + 1.953×10-4 q2) 
 
 33.0)32010 1.953 + (40
)32010 1.953 + (40 - )P32010 2.44- (1
2 4-
2 -4
DE
2 -6
=
××
××××
 
 
 PDE = 106.4 psi 
 
 Let qdes = 320 gpmq = 
 
For rated Cv, valve is completely open at 110% qdes i.e., at 352 gpm or the 
upper limit of 350 gpm 
 
2
1
−



 ∆
=
s
v
v q
p
qC 
 
2
1
2426
9.0
)35010953.140(4.106)3501044.21(350
−
−−


 ××+−××−
= 
 
Then using Eq. 9-11 
 
50ln
9.0
1055.44.66
6.101
ln
1
2/124







 ×−
+=
−
− qq
l 
 
 
 
 
9-11 
0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1
0
50
100
150
200
250
300
350
400
l (valve lift)
q 
(gp
m
)
Cv = 101.6
Cv = 133.5
 
 Figure S9.5. Control valve characteristics 
 
 
From the plot of valve characteristic for the rated Cv of 101.6, it is evident 
that the characteristic is reasonably linear in the operating region 250 ≤ q 
≤ 350. 
 
The pumping cost could be further reduced by lowering the PDE to a value 
that would make ∆Pv/∆Ps = 0.25 at 320=q gpm. Then PDE = 100.0 and 
for qdes = 320 gpm, the rated Cv = 133.5. However, as the plot shows, the 
valve characteristic for this design is more nonlinear in the operating 
region. Hence the selected valve is Cv = 101.6 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
9-12 
 9.6 
 
 
 a) 
 
 The "square" valve appears similar to the equal percentage valve in Fig. 9.8 
 
 
 b) 
Valve Gain ( �ddf / ) � =0 � =0.5 � =1 
Quick open �2/1 ∞ 0.707 0.5 
Linear 1 1 1 1 
Slow open �2 0 1 2 
 
 
The largest gain for quick opening is at � =0 (gain = ∞), while largest for 
slow opening is at � =1 (gain = 2). A linear valve has constant gain. 
 
c) 
s
v
v g
PfCq ∆= )(� 
 
For gs = 1 , ∆Pv = 64 , q = 1024 
 
 Cv is found when )(�f =1 (maximum flow): 
 
0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1
0 0.2 0.4 0.6 0.8 1
l
f
Linear
"Square root"
"Square"
9-13 
 =
∆
=
sv
v gP
qC 1/22
1024 gal/min 1024 gal.in
= =128
8 min.(lb)64 lb/in
 
 
d) � in terms of applied pressure 
 
� =0 when p = 3 psig 
� =1 when p = 15 psig 
 
 Then 25.0
12
1)3()315(
)01(
−=−
−
−
= pp� 
 
 e) q = 128 vP∆2� for slow opening ("square") valve 
 = 128
2
25.0
12
1 


−∆ pPv 
 = ( ) 22 )3(8889.03
144
128
−∆=−∆ pPpP vv 
 
 p = 3 , q = 0 for all ∆Pv 
 
 p =15 , q = 128 vP∆ 
 
 = 0 for ∆Pv = 0 
 
 = 1024 for ∆Pv = 64 
 
 
 looks O.K 
 
 
 
 
 9.7 
 
 
Because the system dynamic behavior would be described using deviation 
variables, all that is important are the terms involving x, dx/dt and d2x/dt2. 
Using the values for M, K and R and solving the homogeneous o.d.e: 
 
 03600000,153.0 2
2
=++ x
dt
dx
dt
xd
 
 
This yields a strongly overdamped solution, with ζ=228, which can be 
approximated by a first order model by ignoring the d2x/dt2 ter 
9-14 
 9.8 
 
 
A control system can incorporate valve sequencing for wide range along 
with compensation for the nonlinear curve (Shinskey, 1996). It features a 
small equal-percentage valve driven by a proportional pH controller. The 
output of the pH controller also operates a large linear valve through a 
proportional-plus-reset controller with a dead zone. The system is shown 
in Fig. E9.8 
 
 
 
 
 Figure S9.8. Schematic diagram for pH control 
 
 
Equal-percentage valves have an exponential characteristic, similar to the 
pH curve. As pH deviates from neutrality, the gain of the curve decreases; 
but increasing deviation will open the valve farther, increasing its gain in a 
compensating manner. As the output of the proportional controller drives 
the small valve to either of its limits, the dead zone of the two-mode 
controller is exceeded. The large valve is moved at a rate determined by 
the departure of the control signal from the dead zone and by the values of 
proportional and reset. When the control signal reenters the dead zone, the 
large valve is held in its last position. The large valve is of linear 
characteristic, because the process gain does not vary with flow, as some 
gains do. 
 
 
 
pHC
Percent
Linear
Influent
Reagent
9-15 
 9.9 Note: in the book’s second printing, the transient response in this problem will be 
 modified by adding 5 minutes to the time at which each temperature reading was taken. 
 
 We wish to find the model: 
 
 
( )
( ) 1
m m
m
T s K
T s s
′
=
′ τ +
 
 
where Tm is the measurement 
 T is liquid temperature 
 
From Eq. 9-1, 
 
 
o o o o
range of instrument output 20 mA - 4 mA 16 mA mA
= = =0.04
range of instrument input 400 C - 0 C 400 C Cm
K = 
 
From Fig. 5.5, τ can be found by plotting the thermometer reading vs. 
time and the transmitter reading vs. time and drawing a horizontal line 
between the two ramps to find the time constant. This is shown in Fig. 
S9.9. 
 
Hence, ∆τ = 1.33 min = 80 sec 
 
To get τ, add the time constant of the thermometer (20 sec) to ∆τ to get 
 
τ = 100 sec. 
2 2.5 3 3.5 4 4.5 5
106
108
110
112
114
116
118
120
122
time (min)
T 
(de
g 
F)
Thermometer
Transmitter
<Time constant>
 
 Figure S9.9. Data test from the Thermometer and the Transmitter 
9-16 
 9.10 
 
 
 
 precision = 0.1 psig 0.5%
20 psig
= of full scale 
 
 accuracy is unknown since the "true" pressure in the tank is unknown 
 
 resolution = 0.1 psig 0.5%
20 psig
= of full scale 
 
 repeatability = ±0.1 psig =±0.5%
20 psig
 of full scale 
 
 
 
9.11 
 
 
 Assume that the gain of the sensor/transmitter is unity. Then, 
 
 )11.0)(1(
1
)(
)(
++
=
′
′
sssT
sTm
 
 
 
 where T is the quantity being measured 
 Tm is the measured value 
 
 T ′ (t) = 0.1 t °C/s , T ′ (s) = 2
1.0
s
 
 2
1.0
)11.0)(1(
1)(
sss
sTm ×++
=′ 
 
 11.01.0111.00011.0)( 10 −++−=′ −− teetT ttm 
 
 Maximum error occurs as t→∞ and equals |0.1t − (0.1t − 0.11)| = 0.11 °C 
 
If the smaller time constant is neglected, the time domain response is a bit 
different for small values of time, although the maximum error (t→∞) 
doesn't change. 
 
 
9-17 
0 2 4 6 8 10 12 14 16 18 20
0
0.2
0.4
0.6
0.8
1
1.2
1.4
1.6
1.8
2
t(s)
T
',
 T
m
' 
(C
)
Tm'(t)
T'(t)
 
 Figure S9.11. Response for process temperature sensor/transmitter 
 
10-1 
��������	
�
 
 
 
10.1 
 
According to Guideline 6, the manipulated variable should have a large 
effect on the controlled variable. Clearly, it is easier to control a liquid 
level by manipulating a large exit stream, rather than a small stream. 
Because R/D>1, the reflux flow rate R is the preferred manipulated 
variable. 
 
 
10.2 
 
 
Exit flow rate w4 has no effect on x3 or x4 because it does not change the 
relative amounts of materials that are blended. The bypass fraction f has a 
dynamic effect on x4 but no steady-state effect because it also does not 
change the relative amounts of materials that are blended. Thus, w2 is the 
best choice.10.3 
 
 
Both the steady-state and dynamic behavior needs to be considered. From 
a steady-state perspective, the reflux stream temperature TR would be a 
poor choice because it is insensitive to changes in xD, due to the small 
nominal value of 5 ppm. For example, even a 100% change from 5 to 10 
ppm would result in a negligible change in TR. Similarly, the temperature 
of the top tray would be a poor choice. An intermediate tray temperature 
would be more sensitive to changes in the tray composition but may not be 
representative of xD. Ideally, the tray location should be selected to be the 
highest tray in the column that still has the desired degree of sensitivity to 
composition changes. 
 
The choice of an intermediate tray temperature offers the advantage of 
early detection of feed disturbances and disturbances that originate in the 
stripping (bottom) section of the column. However, it would be slow to 
respond to disturbances originating in the condenser or in the reflux drum. 
But on balance, an intermediate tray temperature is the best choice. 
 
Solution Manual for Process Dynamics and Control, 2nd edition, 
Copyright © 2004 by Dale E. Seborg, Thomas F. Edgar and Duncan A. Mellichamp 
Rev: 12-6-03 
10-2 
10.4 
 
 
For the flooded condenser in Fig. E10.4, the area available for heat 
transfer changes as the liquid level changes. Consequently, pressure 
control is easier when the liquid level is low and more difficult when the 
level is high. By contrast, for the conventional process design in Fig. 10.5, 
the liquid level has a very small effect on the pressure control loop. Thus, 
the flooded condenser is more difficult to control because the level and 
pressure control loops are more interacting than they are for the 
conventional process design in Fig. 10.5. 
 
 
10.5 
 
 
(a) The larger the tank, the more effective it will be in “damping out” 
disturbances in the reactor exit stream. A large tank capacity also provides 
a large feed inventory for the distillation column, which is desirable for 
periods where the reactor is shut down. Thus a large tank is preferred from 
a process control perspective. However a large tank has a high capital cost, 
so a small tank is appealing from a steady-state, design perspective. Thus, 
the choice of the storage tank size involves a tradeoff of control and 
design objectives. 
 
 
 (b) After a set-point change in reactor exit composition occurs, it would be 
desirable to have the exit compositions for both the reactor and the storage 
tank change to the new value as soon as possible. But the concentration in 
the storage tank will change gradually due to its liquid inventory. The time 
constant for the storage tank is proportional to the mass of liquid in the 
tank (cf. blending system models in Chapters 2 and 4). Thus, a large 
storage tank will result in sluggish responses in its exit composition, which 
is not desirable when frequent set-point changes are required. In this 
situation, the storage tank size should be smaller than for case (a). 
 
 
 
10.6 
 
 
 Variables : q1, q2,…. q6, h1, h2 NV = 8 
 
 Equations : 
 
3 flow-head relations: 3 1 1vq C h= 
10-3 
 5 2 2vq C h= 
 )( 214 hhKq −= 
 
2 mass balances: 
 
 
1
1 1 6 3 4� �� �dhA q q q qdt = + − − 
 
2
2 2 4 5� �� �dhA q q qdt = + − 
 
 
Thus NE = 5 
 
 
Degrees of freedom: NF = NV – NE = 8 − 5 = 3 
 
 
Disturbance variable : q6 ND = 1 
 
 NF = NFC + ND 
 NFC = 3 − 1 = 2 
 
 
 
10.7 
 
 
Consider the following energy balances assuming a reference temperature 
of Tref = 0 : 
 
Heat exchanger: 
 
 0 1 1 2(1 ) ( ) ( )c c C C h h h hC f w T T C w T T− − = − (1) 
 
Overall: 
 
 )()( 2112 hhhhCCcc TTwCTTwC −=− (2) 
 
Mixing point: 
 
 ccc fwwfw +−= )1( (3) 
 
Thus, 
 
10-4 
 NE =3 , NV = 8 ),,,,,,,( 21021 hhccchc TTTTTwwf 
 
 NF =NV − NE = 8 − 3 = 5 
 
 NFC =2 (f, wh) 
also 
 
 ND = NF − NFC = 3 (wc, Tc1, Tc2) 
 
The degrees of freedom analysis is identical for both cocurrent and 
countercurrent flow because the mass and energy balances are the same 
for both cases. 
 
 
 
10.8 
 
 
 The dynamic model consists of the following material balances: 
 
 Mass balance on the tank: 
 
 1 2 3� �� �dhA f w w wdt = − + − (1) 
 
 Component balance on the tank: 
 
 
3
1 1 2 2 3 3
( )� �� �d hxA f x w x w x w
dt
= − + − (2) 
 
 Mixing point balances: 
 
 w4 = w3 + fw1 (3) 
 
 x4w4 = x3w3 + fx1w1 (4) 
 
 Thus, 
 
 NE = 4 (Eqs.1-4) 
 
 NV = 10 1 2 3 4 1 2 3 4( , , , , , , , , , )h f w w w w x x x x 
 
 NF = NV − NE = 6 
 
Because two variables 2( and )w f can be independently adjusted, it 
would appear that there are two control degrees of freedom. However, the 
10-5 
fraction of bypass flow rate, f , has no steady-state effect on x4. To 
confirm this assertion, consider the overall steady-state component 
balance for the tank and the mixing point: 
 
 442211 wxwxwx =+ (5) 
 
This balance does not depend on the fraction bypassed, f, either directly or 
indirectly, 
 
 Conclusion : NFC = 1 (w2) 
 
 
 
10.9 
 
 
 
 
 Let Ci = concentration of N2 in the inlet stream = 100% 
 C = concentration in the vessel = exit concentration (perfect mixing) 
 
 Assumptions: 
 
 1. Perfect mixing 
 2. Initially, the vessel contains pure air, that is, C(0) = 79%. 
 
 N2 balance on the vessel: 
 
 )( CCq
dt
dCV i −= (1) 
 
 Take Laplace transforms and let τ=V/q: 
 
 �� � � � ��� � �iCsC s C t C s
s
− = = −
 
 
 Rearrange, 
 
Vessel
Ci
q
C
q
10-6 
 
( 0)( ) (� �� � �
iC C tC s
s s s
=
= +
+ +
 
 
 Take inverse Laplace transforms (cf. Chapter 3), 
 
 
/ � 	 �( ) (1 ) ( 0)t tiC t C e C t e− −= − + = (2) 
 
 Also, 
 
 
3
3
20,000 L 1 m� 
��
�
1000 L0.8m / min
V
q
  
= = =       
 
 Substitute for τ, Ci and C(0) into (2) and rearrange 
 
 
21%(25min) ln
100% ( )t C t
 
=  
−  (3) 
 
Let C(t) = 98% N2 (i.e., 2% O2). From (3), 
 
 t = 58.7 min 
 
 
 
10.10 
 
 
Define k as the number of sensors that are working properly. We are 
interested in calculating )2( ≥kP , when P(E) denotes the probability that 
an event, E, occurs. 
 
Because k = 2 and k = 3 are mutually exclusive events, 
 
 )3()2()2( =+==≥ kPkPkP (1) 
 
These probabilities can be calculated from the binomial distribution 1 
and the given probability of a sensor functioning properly (p = 0.99): 
 ( )1 23( 2) 0.01 (0.99) 0.0294
2
P k  = = =   
 
 ( )0 33( 3) 0.01 (0.99) 0.9703
3
P k  = = =   
 
10-7 
where the notation, 



r
n
, refers to the number of combinations of n objects 
taken r at a time, when the order of the r objects is not important. Thus 
3
2
3
=



 and 1
3
3
=



. From Eq.(1), 
 
 ( 2) 0.0294 0.9703 0.9997P k ≥ = + = 
 
1
 See any standard probability or statistics book, e.g., Montgomery D.C and G.C. Runger, 
Applied Statistics and Probability for Engineers, 3rd ed., John Wiley, NY (2003). 
 
 
 
 
10.11 
 
 
 Assumptions: 
 
1. Incompressible flow. 
 
2. Chlorine concentration does not affect the air sample density. 
 
3. T and P are approximately constant. 
 
The time tT that is required to detect a chlorine leak in the processing area 
is given by: 
 
tT =ttube + tA 
 
 where: 
 
ttube is the time that the air sample takes to travel through the tubing 
 
tA is the time that the analyzer takes to respond after chlorine first 
reaches it. 
 
The volumetric flow rate q is the product of the velocity v and the cross-
sectional area A: 
 
qq vA v
A
= =∴ 
 
 then: 
 
10-8 
 
( )22 2
3
2 2
3.14 6.35 0.762
24.5 mm
4 4
10 cm / 40.8 cm / s
24.5 10 cm
DA
s
v
pi
−
−
= = =
= =
×
 
 
 Thus, 
 
 
4000 cm 98.1 s
40.8 cm / stube
t = = 
 
 Finally, 
 
 tT = 98.1 + 5 = 103.1 s 
 
 Carbon monoxide (CO) is one of the most widely occurring toxic gases, 
especially in confined spaces. High concentrations of carbon monoxide 
can saturate a person’s blood in a matter of minutes and quickly lead to 
respiratory problems or even death. Therefore, this amount of time is not 
acceptable if the hazardous gas is CO. 
 
 
10.12 
 
 
 The key safety concerns include: 
 
 1. Early detection of any leaks to the surroundings 
 
 2. Over pressurizing the flash drum 
 
3. Maintain enough liquid level so that the pumps do not cavitate. 
 
 4. Avoid having liquid entrained in the gas. 
 
 
 These concerns can be addressed by the following instrumentation. 
 
1. Leak detection: sensors for hazardous gases should be located in 
the vicinity of the flash drum. 
 
2. Over pressurization: Use a high pressure switch (PSH) to shut 
off the feed when a high pressure occurs. 
 
3. Liquid inventory: Use a low level switch (LSL) to shut down 
the pump if a low level occurs. 
10-9 
 
4. Liquid entrainment: Use a high level alarm to shut off the feed 
if the liquid level becomes too high. 
 
 
This SIS system is shown below with conventional control loops for 
pressure and liquid level. 
 
 
 
 
Figure S10.12. 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
10-10 
10.13 
 
 
 The proposed alarm/SIS system is shown in Figure S10.13: 
 
 
 
 The solenoid-operated valves are normally open. If the column pressure 
exceeds a specified limit, the high pressure switch (PSH) shuts down both 
the feed stream and the steam flow to the reboiler. Both actions tend to 
reduce the pressure in the column. 
11-1 
��������		�
 
 
 
 
 11.1 
 
 
 
11.2 
 
 



τ
+=
s
KsG
I
cc
11)( 
The closed-loop transfer function for set-point changes is given by Eq. 11-
36 with Kc replaced by 



τ
+
s
K
I
c
11 , 
 
)1(
1111
)1(
111
)(
)(
+τ



τ
++
+τ



τ
+
=
′
′
ss
KKKK
ss
KKKK
sH
sH
I
mpvc
I
mpvc
sp
 
 
12
)1(
)(
)(
33
2
3 +τζ+τ
+τ
=
′
′
ss
s
sH
sH I
sp
 
 
where ζ3,τ3 are defined in Eqs. 11-62, 11-63 , Kp = R = 1.0 min/ft2 , 
and τ = RA = 3.0 min 
 
Solution Manual for Process Dynamics and Control, 2nd edition, 
Copyright © 2004 by Dale E. Seborg, Thomas F. Edgar and Duncan A. Mellichamp 
11-2 
 36.1
ft
psi7.1
ft
min0.1
psi
min/ft2.0)4( 2
3
=







== mpvcOL KKKKK 
 
 
22
3 min62.636.1
min)3min)(3(
==
ττ
=τ
OL
I
K
 
 
 2 ζ3 τ3 = min21.5336.1
36.21
=×=τ


 +
I
OL
OL
K
K
 
 
 
121.2
1
)121.2()10.3(
13
)(
)(
+
=
+++
+
=
′
′
sss
s
sH
sH
sp
 
 
 For 
ss
sH sp
1)23()( =−=′ 
 
 
21..2/1)( teth −−=′ 
 
 [ ])(1ln21.2 tht ′−−= 
 
 ft5.2)( =th ft5.0)( =′ th min53.1=t 
 
 ft0.3)( =th ft0.1)( =′ th ∞→t 
 
 Therefore, 
 
 ft5.2min)53.1( ==th 
 
 ft0.3)( =∞→th 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
11-3 
11.3 
 
 
 ma/ma5)( == cc KsG 
 
 Assume τm = 0, τv = 0, and K1 = 1, in Fig 11.7. 
 
a) Offset = FFFTTsp ��� 86.014.45)()( =−=∞′−∞′ 
b) 



+τ
+



+τ
=
′
′
1
1
1
)(
)(
2
2
s
K
KKKK
s
K
KKKK
sT
sT
vIPcm
vIPcm
sp
 
 
Using the standard current range of 4-20 ma, 
 
Fma/32.0
50
ma4ma20
�
�
=
−
=
F
K m 
 
2.1=vK , psi/ma75.0=IPK , τ =5 min , 
s
sTsp
5)( =′ 
 
)440.115(
20.7)(
2
2
Kss
K
sT
++
=′ 
 
 )440.11(
20.7)(lim)(
2
2
0 K
K
sTsT
s +
=′=∞′
→
 
 
 F14.4)( �=∞′T psi/F34.32 �=K 
 
c) From Fig. 11-7, since 0=′iT 
 
)()( 2 ∞′=∞′ TKKP vt , psi03.1)( =∞′tP 
 
and TKTKKP ivt =+′ 12 , psi74.3=tP 
 
 psi77.4)()( =∞′−=∞ ttt PPP 
 
 
 
 
 
11-4 
11.4 
 
 
 a) 
 
 
 
 b) smm meKsG θ−=)( assuming τm = 0 
 
ss
m eesG
2
3
2
3
sol/ftlb
ma67.2
ft
sollb)39(
ma)420()( −− 



=
−
−
= 




τ
+=
s
KsG
I
cc
11)( 
 
psi/ma3.0)( == IPIP KsG 
 
psi
USGPM67.1)612(
USGPM)2010()( −=
−
−
==
psi
KsG vv 
 
Overall material balance for the tank, 
 
hCqq
dt
dhA v−+=


213ft
USgallons481.7 (1) 
 
Component balance for the solute, 
 
32211
3 )()(481.7 chCcqcq
dt
hCdA v−+= (2) 
 
Linearizing (1) and (2) gives 
11-5 
h
h
C
q
dt
hdA v ′



−
′=
′
2
481.7 2 (3) 
 
( ) 33222233
2
481.7 chCh
h
C
ccqqc
dt
cdh
dt
hd
cA vv ′−′



−
′+′=

 ′
+
′
 
 
Subtracting (3) times 3c from the above equation gives 
 
)(481.7 323 ccdt
cdhA −=
′ ( ) 3222 chCcqq v ′−′+′ 
 
Taking Laplace transform and rearranging gives 
 
)(
1
)(
1
)( 22213 sC
s
K
sQ
s
K
sC ′
+τ
+′
+τ
=′ 
 
where 
 
 
USGPM
sol/ftlb08.0
3
32
1 =
−
=
hC
cc
K
v
 
 
 6.022 ==
hC
q
K
v
 
 
 min15481.7 ==τ
vC
hA
 
 
 since 22 ft6.124/ =pi= DA , and 
 
 ft
C
qq
C
qh
vv
4
2
21
2
3
=


 +
=



= 
 
 Therefore, 
 
 
115
08.0)(
+
=
s
sG p 
 
 
115
6.0)(
+
=
s
sGd 
 
11-6 
c) The closed-loop responses for disturbance changes and for setpoint 
 changes can be obtained using block diagram algebra for the block 
 diagram in part (a). Therefore, these responses will change only if any of 
 the transfer functions in the blocks of the diagram change. 
 
i. 2c changes. Then block transfer function )(sG p changes 
due to K1. Hence Gc(s) does need to be changed, and 
retuning is required. 
 
ii. Km changes. Block transfer functions do change. Hence 
 Gc(s) needs to be adjusted to compensate for changes in 
 block transfer functions. The PI controller should be 
 retuned. 
 
iii. Km remains unchanged. No block transfer function changes. 
 The controller does not need to be retuned. 
 
 
 
11.5 
 
 
 a) 
 
One example of a negative gain process that we have seen is the liquid 
level process with the outlet stream flow rate chosen as the manipulated 
variable 
 
With an "air-to-open" valve, w increases if p increases. However, h 
decreases as w increases. Thus Kp <0 since ∆h/∆w is negative. 
 
b) KcKp must be positive. If Kp is negative, so is Kc. See (c) below. 
 
c) If h decreases, p must also decrease. This is a direct acting controller 
whose gain is negative [ )()(()( thtrKtp c ′−′=′ ] 
h
ωωωω
LT
p
c
11-7 
11.6 
 
 
 For proportional controller, cc KsG =)( 
 
Assume thatthe level transmitter and the control valve have negligible 
dynamics. Then, 
 
mm KsG =)( 
 vv KsG =)( 
 
The block diagram for this control system is the same as in Fig.11.8. 
Hence Eqs. 11-26 and 11-29 can be used for closed-loop responses to 
setpoint and load changes, respectively. 
 
The transfer functions )(sG p and )(sGd are as given in Eqs. 11-66 and 
11-67, respectively. 
 
a) Substituting for Gc, Gm, Gv, and Gp into Eq. 11-26 gives 
 
1
1
11
1
+τ
=



−+



−
=
sK
As
KK
As
KKK
Y
Y
mvc
vcm
sp
 
where 
mvc KKK
A
−=τ (1) 
 
For a step change in the setpoint, sMsYsp /)( = 
 
M
s
sM
sssYtY
ss
=


+τ
==∞→
→→ 1
/lim)(lim)(
00
 
 
Offset = 0)()( =−=∞→−∞→ MMtYtYsp 
 
b) Substituting for Gc, Gm, Gv, Gp , and Gd into (11-29) gives 
1
1
11
1
)(
)(
+τ




−
=



−+



=
s
KKK
K
As
KK
As
sD
sY mvc
mvc
 
 
where τ is given by Eq. 1. 
11-8 
For a step change in the disturbance, sMsD /)( = 
 
mvc
mvc
ss KKK
M
ss
KKKM
sssYtY −=


+τ
−
==∞→
→→ )1(
)/(lim)(lim)(
00
 
 
Offset = 00)()( ≠



−
−=∞→−∞→
mvc
sp KKK
M
tYtY 
 
Hence, offset is not eliminated for a step change in disturbance. 
 
 
 
 
11.7 
 
 
 Using block diagram algebra 
 
 UGDGY pd += (1) 
 
 
( )[ ]UGYYGU pspc ~−−= (2) 
 
 From (2), 
pc
cspc
GG
YGYG
U
~1−
−
= 
 
Substituting for U in Eq. 1 
 [ ] spcppcdppc YGGDGGGYGGG +−=−+ )~1()~(1 
 
Therefore, 
 
 
)~(1 ppc
cp
sp GGG
GG
Y
Y
−+
= 
and 
 
(1 )
1 ( )
d c p
c p p
G G GY
D G G G
−
=
+ −
�
�
 
 
 
 
 
 
11-9 
11.8 
 
 
 The available information can be translated as follows 
 
1. The outlets of both the tanks have flow rate q0 at all times. 
 
2. 0)( =sTo 
 
3. Since an energy balance would indicate a first-order transfer function 
between T1 and Q0 , 
 
1/1)(
)( τ−
−=
∞′
′ te
T
tT
 or 1/121
3
2 τ−
−= e , τ1 = 10 min 
 
Therefore 
 
110
4
110
)75.0/(3
)(
)(
0
1
+
−=
+
−
=
ss
gpmF
sQ
sT �
 
 
1
67.2
1
)75.0/()35(
)(
)(
220
3
+
−=
+
−−
=
ss
gpmF
sQ
sT
ττ
�
 for T2(s) = 0 
 
4. 
110
4
110
)1012/()7078(
)(
)(
1
1
+
=
+
−−
=
ss
VF
sV
sT �
 
 
110
5.2
110
)1012/()8590(
)(
)(
2
3
+
=
+
−−
=
ss
VF
sV
sT �
 
 
5. 5τ2 =50 min or τ2 = 10 min 
 
Since inlet and outlet flow rates for tank 2 are q0 
 
 
10.10
1
1
/
)(
)(
2
00
2
3
+
=
+τ
=
ss
qq
sT
sT
 
 
6. 15.0)(
)(
3
3
=
sT
sV
 
7. )5.0(
60
30)( 112 −=


−= tTtTtT 
11-10 
 
se
sT
sT 5.0
1
2
)(
)(
−
= 
 
 Using these transfer functions, the block diagrams are as follows. 
 
 a) 
Gc-+
+
+0.15
T3sp V2
0.15
-2.67V1
V3sp
2.5
T3
T1
V3
1
10s+1
+
+e-0.5s
- 4
10s+1
+
+
T2
Q0
1
 
 
 
b) 
Gc-+ ++0.15
T3sp V1
0.15
-2.67
V2
V3sp T3T1
V3
1
10s+1
+
+
e-0.5s
- 4
10s+1
T2
Q0
+
+
-2.5
1
 
 
11-11 
c) The control configuration in part a) will provide the better control. As is 
evident from the block diagrams above, the feedback loop contains, in 
addition to Gc, only a first-order process in part a), but a second-order-
plus-time-delay process in part b). Hence the controlled variable responds 
faster to changes in the manipulated variable for part a). 
 
 
 
11.9 
 
 
 
The given block diagram is equivalent to 
 
 For the inner loop, let 
 
 
)1(~1 ~* sc
c
c
eGG
GG
E
P
θ−
−+
=′= 
 
 In the outer loop, we have 
 
 
GG
GG
D
Y
c
d
′+
=
1
 
 
 Substitute for cG ′ , 
 
 
)1(~1
1
~
* s
c
c
d
eGG
GG
GG
D
Y
θ−
−+
+
= 
 
 
( )
GGeGG
eGGGG
D
Y
c
s
c
s
cd
+−+
−+
=
θ−
θ−
)1(~1
)1(~1
~
*
~
*
 
 
 
11-12 
11.10 
 
 
a) Derive CLTF: 
 
3 2 3 2Y Y Y G Z G P= + = + 
 
 3 1 2 cY G ( D Y ) G K E= + + 
 
 EKGEKGGDGY cc 2133 ++= 
 
 EKGKGGDGY cc )( 2133 ++= YKE m−= 
 
 YKGGGKDGY mc )( 2133 +−= 
 
 
mc KGGGK
G
D
Y
)(1 213
3
++
= 
 
b) Characteristic Equation: 
 
0)(1 213 =++ mc KGGGK 
0
12
4
1
51 =


+
+
−
+
ss
K c 
 
0)12)(1(
)1(4)12(51 =


+−
−++
+
ss
ssK c 
 
 [ ] 0)1(4)12(5)12)(1( =−++++− ssKss c 
 
 0)44510(12 2 =−+++−− ssKss c 
 
 0)1()114(2 2 =−+−+ cc KsKs 
 
 Necessary conditions: 14/1>cK and 1>cK 
 
For a 2nd order characteristic equation, these conditions are also sufficient. 
 Therefore, 1>cK for closed-loop stability. 
 
 
 
 
11-13 
11.11 
 
 
 a) 
 
 
 
c) Transfer Line: 
 
Volume of transfer line = pi /4 (0.5 m)2(20m)= 3.93 m3 
 
Nominal flow rate in the line = min/m5.7 3=+ FA qq 
 
 Time delay in the line = min52.0
/minm7.5
m3.93
3
3
= 
 
 
s
TL esG
52.0)( −= 
 
 Composition Transmitter: 
 
 33 kg/m
ma0.08
kg/m0)(200
ma4)(20)( =
−
−
== mm KsG 
 
 Controller 
 
 From the ideal controller in Eq. 8.14 
 
 
[ ])()(~)(11)( sCsCsKsE
s
KsP mspDc
I
c
′
−
′τ+



τ
+=′ 
In the above equation, set 0)(~ =′ sCsp in order to get the derivative on the 
process output only. Then, 
11-14 
 



τ
+=
s
KsG
I
cPI
11)( 
 
 sKsG DcD τ−=)( 
 
with Kc >0 as the controller should be reverse-acting, since P(t) should 
increase when Cm(t) decreases. 
 
 I/P transducer 
 
 
ma
psig0.75
ma4)(20
psig3)(15
=
−
−
=IPK 
 
 Control valve 
 
 
1
)(
+τ
=
s
K
sG
v
v
v 
 
 15 =τv , min2.0=τv 
 
 
12
3
)20)(20)(ln12/1(03.0
−
=
==
v
vv
p
ppv
A
v dp
dqK 
 
 
12
3
)20(03.017.05.0
−
+==
vp
Aq 
 
 33.017.05.0)20(03.0 12
3
=−=
−vp
 
 
 
psig
/minm082.0)33.0)(20)(ln12/1(
3
==vK 
 
 
12.0
082.0)(
+
=
s
sGv 
 
 Process 
 
 Assume cA is constant for pure A. Material balance for A: 
 
 cqqcqcq
dt
dcV FAFFAA )( +−+= (1) 
 
11-15 
 Linearizing and writing in deviation variable form 
 
 AFAFFAA qccqqcqqcdt
cdV ′−′+−′+′=′ )( 
 
 Taking Laplace transform 
 
 [ ] )()()()()( sCqsQccsCqqVs FFAAFA ′+′−=′++ (2) 
 
 From Eq. 1 at steady state, 0/ =dtdc , 
 
 
3kg/m100)/()( =++= FAFFAA qqcqcqc 
 
 Substituting numerical values in Eq. 2, 
 
 [ ] )(7)(700)(5.75 sCsQsCs FA ′+′=′+ 
 
 [ ] )(93.0)(3.93)(167.0 sCsQsCs FA ′+′=′+ 
 
 
167.0
3.93)(
+
=
s
sG p 
 
167.0
93.0)(
+
=
s
sGd 
 
 
 
11.12 
 
 
 
The stability limits are obtained from the characteristic Eq. 11-83. Hence 
if an instrumentation change affects this equation, then the stability limits 
will change and vice-versa. 
 
a) The transmitter gain, Km, changes as the span changes. Thus Gm(s) 
changes and the characteristic equation is affected. Stability limits would 
be expected to change. 
 
b) The zero on the transmitter does not affect itsgain Km. Hence Gm(s) 
remains unchanged and stability limits do not change. 
 
c) Changing the control valve trim changes Gv(s). This affects the 
characteristic equation and the stability limits would be expected to 
change as a result. 
 
11-16 
11.13 
 
 
 
a) )1)(1()( ++τ= ss
KK
sG ca 
 
b) )1)(1(
)1()(
++ττ
+τ
=
sss
sKK
sG
I
Ic
b 
 
For a) 
 
pcc KKssKKsssNsD +++τ+τ=+++τ=+ 1)1()1)(1()()( 2 
 
Stability requirements: 
 
01 >+ pc KK or 1−>>∞ pc KK 
 
For b) 
 
)1()1)(1()()( +τ+++ττ=+ sKKsssNsD IcI 
 
 pcpcIII KKsKKss ++τ++ττ+ττ= )1()1( 23 
 
 Necessary condition: 0>pc KK 
 
 Sufficient conditions (Routh array): 
 
 ττ I )1( pcI KK+τ 
 
 )1( +ττ I pc KK 
 
 )1(
)1)(1(2
+ττ
ττ−++ττ
I
pcIpcI KKKK
 
 pc KK 
 
 Additional condition is: 
 
 0)()1)(1( >τ−++ττ pcpcI KKKK 
 
 (since Iτ and τare both positive) 
11-17 
 0)1()1( >τ−+ττ++ττ pcpcII KKKK 
 
 [ ] )1()1( +ττ−>τ−+ττ IpcI KK 
 
 Note that RHS is negative for all positive Iτ and τ 
 
(∴ RHS is always negative) 
 
 
Case 1: 
 
 If 0)1( >τ−+ττ I 


+τ
τ
>τ
1
.,. Iei 
 
 then KcKp > 0 > 


τ−+ττ
+ττ−
)1(
)1(
I
I
 
In other words, this condition is less restrictive than KcKp >0 and doesn't 
apply. 
 
 
 
Case 2: 
 
 
 If 0)1( <τ−+ττ I 


+
<
1
.,.
τ
τ
τ Iei 
 
 then KcKp < 


−+
+−
τττ
ττ
)1(
)1(
I
I 
 
In other words, there would be an upper limit on KcKp so the 
controller gain is bounded on both sides 
 
0 < KcKp < 
τ−+ττ
+ττ−
)1(
)1(
I
I 
 
c) Note that, in either case, the addition of the integral mode decreases the 
range of stable values of Kc. 
 
 
 
 
11-18 
11.14 
 
 
 From the block diagram, the characteristic equation is obtained as 
 
 
0
10
1
1
2
3
4)5.0(1
3
4)5.0(
1 =


+


−











+
+



+
+
ss
s
sK c 
 
that is, 
 
0
10
1
1
2
5
21 =


+


−



+
+
sss
K c 
 
 
Simplifying, 
 
 0)504(3514 23 =−+++ cKsss 
 
 
The Routh Array is 
 
 1 35 
 
14 4Kc-50 
 
14
)504(490 −− cK
 
4Kc – 50 
 
 
 For the system to be stable, 
 
0
14
)504(490
>
−− cK or Kc < 135 
 
 and 0504 >−cK or Kc > 12.5 
 
 Therefore 12.5 < Kc < 135 
 
 
 
11-19 
11.15 
 
 
 a) 
1
1
)1/(
1
1
1
1
)(
)(
+
+
τ
−
+
=
+τ−
=
τ−
+
τ−
=
s
KK
KKKK
KKs
KK
s
KK
s
KK
sY
sY
c
cc
c
c
c
c
sp
 
 
 For stability 0
1
>
+
τ
−
KK c
 
 
 Since τ is positive, the denominator must be negative, i.e., 
 
 01 <+ KK c 
 
 1−<KK c 
 
 KK c /1−< 
 Note that 
KK
KK
K
c
c
CL +
=
1
 
 
 b) If 1−<KK c and KK c+1 is negative, 
 
 then CL gain is positive. ∴ it has the proper sign. 
 
 
c) K = 10 and τ = 20 
 
 and we want 10
1
=
+
τ
−
KK c
 
 
 or cK)10)(10(1020 +=− 
 cK10030 =− 
 3.0−=cK 
 
 Offset: 5.1
2
3
)10)(3.0(1
)10)(3.0(
=
−
−
=
−+
−
=CLK 
 
 ∴ Offset = +1 − 1.5 = − 50% (Note this result implies overshoot) 
 
 
11-20 
d) 
KKss
KK
ss
KK
ss
KK
sY
sY
cm
c
m
c
m
c
sp ++ττ−
=
+ττ−
+
+ττ−
= )1)(1(
)1)(1(1
)1)(1(
)(
)(
 
 
KKss
KK
cmm
c
++τ−τ+ττ−
=
1)(2 
 
 
1
11
)1/(
2 +
+
τ−τ
+
+
ττ
−
+
=
s
KK
s
KK
KKKK
c
m
c
m
cc
 (standard form) 
 
 For stability, 
 
 (1) 0
1
>
+
ττ
−
KK c
m
 (2) 0
1
>
+
τ−τ
KK c
m
 
 
 From (1) Since 01 <+ KK c 
 1−<KK c 
 
K
K c
1
−< 
 
From (2) Since 01 <+ KK c 
 0<τ−τm 
 mτ−<τ− 
 mτ>τ 
 For K = 10 , τ = 20 , Kc = –0.3 , τm = 5 
 
 
 
15.250
5.1
1)31(
)205(
31
)5)(20(
5.1
)(
)(
2
2 ++
=
+
−
−
+
−
−
=
ss
ss
sY
sY
sp
 
 
 
 Underdamped but stable. 
 
 
 
 
 
11-21 
11.16 
 
 
 



τ
+=
s
KsG
I
cc
11)( 
 
 
1167.0
3.1
1)60/10()( +
−
=
+
=
ss
K
sG vv 
 
 
sAs
sG p 4.22
11)( −=−= since 
ft
gal4.22ft3 2 ==A 
 
 4)( == mm KsG 
 
 Characteristic equation is 
 
 



τ
++
s
K
I
c
111 0)4(
4.22
1
1167.0
3.1
=

 −


+
−
ss
 
 
 0)2.5()2.5()4.22()73.3( 23 =+τ+τ+τ cIcII KsKss 
 
 The Routh Array is 
 
 Iτ73.3 IcK τ2.5 
 
 Iτ4.22 cK2.5 
 
 cIc KK 867.02.5 −τ 
 
 cK2.5 
 
 For stable system, 
 0>τ I , 0867.02.5 >−τ cIc KK 0>cK 
 
 That is, 
 
 0>cK 
 min167.0>τ I 
 
 
 
 
11-22 
11.17 
 
 
 )(
)(
)110(
51)( 2 sD
sN
ss
s
KsG
I
I
cOL =



+



τ
+τ
= 
 
 0)1(5)120100()()( 2 =+τ+++τ=+ sKssssNsD IcI 
 
 05)51(20100 23 =+τ++τ+τ= cIcII KsKss 
 
a) Analyze characteristic equation for necessary and sufficient conditions 
 
Necessary conditions: 
 
 05 >cK → 0>cK 
 
 0)51( >τ+ IcK → 0>τ I and 5
1
−>cK 
 
Sufficient conditions obtained from Routh array 
 
Iτ100 IcK τ+ )51( 
 
 Iτ20 cK5 
 
 
I
cIcI KK
τ
τ−+τ
20
500)51(20 2
 
 
 cK5 
 
Then, 
 
0500)51(20 2 >τ−+τ cIcI KK 
ccI KK 25)51( >+τ or 
c
c
I K
K
51
25
+
>τ 
 
b) Sufficient condition is appropriate. Plot is shown below. 
11-23 
0 1 2 3 4 5 6 7
0
1
2
3
4
5
6
7
K
c
τ 
 I
 Stability region
 
c) Find Iτ as ∞→cK 
 
 5
5/1
25lim
51
25lim =


+
=


+ ∞→∞→ cKcc
c
Kc KK
K
 
 
∴ 5>τ I guarantees stability for any value of Kc. Appelpolscher is 
wrong yet again. 
 
 
 
11.18 
 
 
 cc KsG =)( 
 
1
)(
+τ
=
s
K
sG
V
V
V 
 
ma
lbm/sec106.0
4122
6.0
12
=
−
==
=p
s
v dp
dw
K 
 sec205 =τv sec4=τv 
 
110
5.2)(
+
=
−
s
e
sG
s
p 
 
FF
KsG mm
��
ma4.0)120160(
ma)420()( =
−
−
== 
 
 Characteristic equation is 
 
11-24 
 
( ) 04.0
110
5.2
14
106.0)(1 =



+



+
+
−
s
e
s
K
s
c (1) 
 
a) Substituting s=jω in (1) and using Euler's identity 
 
e
-jω
=cosω – j sin ω 
gives 
 
 -40ω2 +14jω + 1 + 0.106 Kc (cosω – jsinω)=0 
Thus 
 -40ω2 + 1 + 0.106 Kc cosω = 0 (2) 
 
and 14ω - 0.106Kc sinω =0 (3) 
 
From (2) and (3), 
 
140
14
tan 2
−ω
ω
=ω (4) 
 
Solving (4), ω = 0.579 by trial and error. 
 
Substituting for ω in (3) gives 
 
 Kc = 139.7 = Kcu 
 
Frequency of oscillation is 0.579 rad/sec 
 
b) Substituting the Pade approximation 
 
s
s
e s
5.01
5.01
+
−
≈
− 
 
into (1) gives 
 
0)106.01()053.05.14(4720 23 =++−++ cc KsKss 
 
 The Routh Array is 
 
20 14.5 –0.053 Kc 
 
47 1+ 0.106 Kc 
 
14.07 – 0.098 Kc 
 
1 + 0.106 Kc11-25 
 For stability, 
 
 0098.007.14 >− cK or Kc < 143.4 
 
 and 0106.01 >+ cK or Kc > -9.4 
 
Therefore, the maximum gain, Kcu = 143.4, is a satisfactory approximation 
of the true value of 139.7 in (a) above. 
 
 
 
11.19 
 
 
a) )12)(14)(125(
)51(4)(
+++
−
=
sss
s
sG 
 
cc KsG =)( 
 
0)51(4)12)(14)(125()()( =−++++=+ sKssssNsD c 
 
 129100 2 ++ ss 
 12 +s 
 sss 258200 23 ++ 
 129100 2 ++ ss 
 
0204131158200 23 =−++++ sKKsss cc 
041)2031(158200 23 =++−++ cc KsKss 
 
 Routh array: 
 
200 31-20 Kc 
 
158 1+4 Kc 
 
158(31-20Kc)-200(1+4Kc) 4898 –3160Kc –200 –800Kc 
 = 
158 158 
 
1+ 4 Kc 
 
∴ 4698 –3960 Kc > 0 or Kc < 1.2 
 
 
11-26 
b) 04)12)(14)(125( =++++ cKsss 
 
Routh array: 
 
0)41(31158200 23 =++++ cKsss 
 
200 31 
 
158 1 + 4 Kc 
 
158 (31) − 200(1+4Kc) = 4898 –200 –800Kc 
 
1+ 4 Kc 
 
∴ 4698 –800 Kc > 0 or Kc < 5.87 
 
 
c) Because Kc can be much higher without the RHP zero present, the process 
can be made to respond faster. 
 
 
 
 
11.20 
 
 
 The characteristic equation is 
 
 0
110
5.01
3
=
+
+
−
s
eK sc (1) 
 
a) Using the Pade approximation 
 
s
s
e s )2/3(1
)2/3(13
+
−
≈
− 
in (1) gives 
 
0)5.01()75.05.11(15 2 =++−+ cc KsKs 
 
For stability, 
 
 075.05.11 >− cK or 33.15<cK 
 
and 05.01 >+ cK or 2−>cK 
11-27 
Therefore 33.152 <<− cK 
 
b) Substituting s = jω in (1) and using Euler's identity. 
 
)3sin()3cos(3 ω−ω=ω− je j 
 
gives 
 
 [ ] 0)3sin()3cos(5.0110 =ω−ω++ω jKj c 
 
Then, 
 
 0)3cos(5.01 =ω+ cK (2) 
 
and 0)3sin(5.010 =ω−ω cK (3) 
 
 
 From (3), one solution is ω = 0, which gives Kc = -2 
 
 Thus, for stable operation Kc > -2 
 
 From (2) and (3) 
 
 tan(3ω) = -10ω 
 
Eq. 4 has infinite number of solutions. The solution for the range pi/2 < 3ω 
< 3pi/2 is found by trial and error to be ω = 0.5805. 
 
 Then from Eq. 2, Kc = 11.78 
 
The other solutions for the range 3ω > 3pi/2 occur at values of ω for which 
cos(3ω) is smaller than cos(3×5.805). Thus, for all other solutions of ω, 
Eq. 2 gives values of Kc that are larger than 11.78. Hence, stability is 
ensured when 
 
 -2 < Kc < 11.78 
 
 
 
 
 
 
 
 
11-28 
11.21 
 
 
 
a) To approximate GOL(s) by a FOPTD model, the Skogestad approximation 
technique in Chapter 6 is used. 
 
Initially, 
 
)12)(13)(15)(160(
3
)12)(13)(15)(160(
3)(
2)2.03.05.1(
++++
=
++++
=
−++−
ssss
eK
ssss
eK
sG
s
c
s
c
OL 
 
Skogestad approximation method to obtain a 1st -order model: 
 
 Time constant ≈ 60 + (5/2) 
 
 Time delay ≈ 2 +(5/2) + 3 + 2 =9.5 
 
Then 
 
 
15.62
3)(
5.9
+
≈
−
s
eK
sG
s
c
OL 
 
 
b) The only way to apply the Routh method to a FOPTD transfer function is 
to approximate the delay term. 
 
175.4
175.45.9
+
+−
≈
−
s
s
e s (1st order Pade-approximation) 
 
Then 
 
 )175.4)(15.62(
)175.4(3
)(
)()(
++
+−
≈≈
ss
sK
sD
sN
sG cOL 
 
The characteristic equation is: 
 
 )175.4(3)175.4)(15.62()()( +−+++=+ sKsssNsD c 
 
033.1413.67297 2 =+−++ cc KsKss 
 
0)31()3.143.67(297 2 =++−+ cc KsKs 
 
 
11-29 
 Necessary conditions: 
 
 03.143.67 >− cK 031 >+ cK 
 3.673.14 −>− cK 13 −>cK 
 71.4<cK 3/1−>cK 
 
Range of stability: 71.43/1 <<− cK 
 
c) Conditional stability occurs when 4.71c cuK K= = 
 
With this value the characteristic equation is: 
 
0)71.431()71.43.143.67(297 2 =×++×−+ ss 
 
013.15297 2 =+s 
 
297
13.152 −
=s 
 
 We can find ω by substituting jω → s 
 
 226.0=ω at the maximum gain. 
 
12-1 
Chapter 12 
 
 
 
 
 12.1 
 
 
For K = 1.0, τ1=10, τ2=5, the PID controller settings are obtained using 
Eq.(12-14): 
 
1 2τ τ1 15
τ τc c c
K
K
+= = , τI = τ1+τ2=15 , 
 1 2
1 2
τ ττ 3.33
τ τD
= =+ 
 
 The characteristic equation for the closed-loop system is 
 
 1 1.0 α1 1 τ 0
τ (10 1)(5 1)c DI
K s
s s s
    ++ + + =    + +   
 
 
 Substituting for Kc, τI, and τD, and simplifying gives 
 
 τ (1 α) 0cs + + = 
 
 Hence, for the closed-loop system to be stable, 
 
 τc > 0 
 
 and (1+α) > 0 or α > −1. 
 
(a) Closed-loop system is stable for α > −1 
 
(b) Choose τc > 0 
 
(c) The choice of τc does not affect the robustness of the system to changes in 
α. For τc ≤0, the system is unstable regardless of the value of α. For τc > 0, 
the system is stable in the range α > −1 regardless of the value of τc. 
 
 
 
Solution Manual for Process Dynamics and Control, 2nd edition, 
Copyright © 2004 by Dale E. Seborg, Thomas F. Edgar and Duncan A. Mellichamp 
Revised: 1-3-04
12-2 
 
 
 
 12.2 
 
 1.6(1 0.5 )
(3 1)v p m
sG G G G
s s
− −= = + 
 
The process transfer function contains a zero at s = +2. Because the 
controller in the Direct Synthesis method contains the inverse of the 
process model, the controller will contain an unstable pole. Thus, Eqs. 
(12-4) and (12-5) give: 
 
 ( )( )
3 11 1
τ 2τ 1 0.5c c c
s
G
G s s
+= = − − 
 
Modeling errors and the unstable controller pole at s = +2 would render 
the closed-loop system unstable. 
 
Modify the specification of Y/Ysp such that Gc will not contain the 
offending (1-0.5s) factor in the denominator. The obvious choice is 
 
 1 0.5
τ 1sp cd
Y s
Y s
  −=   + 
 
 
Then using Eq.(12-3b), 
 
 3 1
2τ 1c c
sG += − + 
 
which is not physically realizable because it requires ideal derivative 
action. Modify Y/Ysp, 
 
 2
1 0.5
(τ 1)
  −=   + sp cd
Y s
Y
 
 
Then Eq.(12-3b) gives 
 
 2
3 1
2τ 4τ 1
c
c c
sG
s
+= − + + 
 
 which is physically realizable. 
 
 
 
 
 
12-3 
 12.3 
 
 
 K = 2 , τ = 1, θ = 0.2 
 
(a) Using Eq.(12-11) for τc = 0.2 
 
Kc = 1.25 , τI = 1 
 
(b) Using Eq.(12-11) for τc = 1.0 
 
Kc = 0.42 , τI = 1 
 
(c) From Table 12.3 for a disturbance change 
 
KKc = 0.859(θ/τ)-0.977 or Kc = 2.07 
τ/τI = 0.674(θ/τ)-0.680 or τI = 0.49 
 
(d) From Table 12.3 for a setpoint change 
 
KKc = 0.586(θ/τ)-0.916 or Kc = 1.28 
τ/τI = 1.03 −0.165(θ/τ) or τI = 1.00 
 
(e) Conservative settings correspond to low values of Kc and high values of τI. 
Clearly, the Direct Synthesis method (τc = 1.0) of part (b) gives the most 
conservative settings; ITAE of part (c) gives the least conservative 
settings. 
 
(f) A comparison for a unit step disturbance is shown in Fig. S12.3. 
 
0 3 6 9 12 15
-0.2
0
0.2
0.4
0.6
0.8
1
1.2
time
y
Controller for (b)
Controller for (c)
 
 Fig S12.3. Comparison of part (e) PI controllers for unit step disturbance. 
12-4 
 
12.4 
 
 
 The process model is, 
 
 
θ
( )
sKeG s
s
−
=� (1) 
 
 Approximate the time delay by Eq. 12-24b, 
 
 θ 1 θse s− = − (2) 
 
 Substitute into (1): 
 
 (1 θ )( ) K sG s
s
−=� (3) 
 
 Factoring (3) gives ( ) 1 θG s s+ = −� and sKsG /)(~ =− . 
 
The DS and IMC design methods give identical controllers if, 
 
 fG
Y
Y
dsp
+=


 ~ (12-23) 
 
 For integrating process, f is specified by Eq. 12-32: 
 
 
0
θ
s
dGC
ds
+
=
= = −� (4) 
 
 2 2
(2τ ) 1(2τ θ) 1
(τ 1) (τ 1)
c c
c c
C s sf
s s
− + + += =+ + (5) 
 
 Substitute +G
~ and f into (12-23): 
 
 (1 θ )
sp d
Y s
Y
  = −    2
(2τ θ) 1
(τ 1)
c
c
s
s
 + + +  
 (6) 
 
 The Direct Synthesis design equation is: 
 
12-5 
 













−




=
dsp
dsp
c
Y
Y
Y
Y
G
G
1
~
1 (12-3b) 
 
 Substitute (3) and (6) into (12-3b): 
 
 
2
2
(2τ θ) 1(1 θ )
(τ 1)
(1 θ ) (2τ θ) 11 (1 θ )
(τ 1)
c
c
c
c
c
ss
ssG
K s ss
s
 + +−  +    =  −    + +− −  +  
 (7) 
 
 or 
 
 [ ]2
(2τ θ) 1
(τ 1) (1 θ ) (2τ θ) 1
c
c
c c
ssG
K s s s
+ += + − − + + (8) 
 
 Rearranging, 
 
 2 2 2
(2τ θ) 1 (2τ θ) 11 1
τ 2τ θ θ (τ θ)
c c
c
c c c
s sG
Ks Kss
+ + + += =+ + + (9) 
 
 The standard PI controller can be written as 
 
 τ 1
τ
I
c c
I
sG K
s
+= (10) 
 
 Comparing (9) and (10) gives: 
 
 τ 2τ θI c= + (11) 
 
 ( )2
1 1
τ τ θ
c
I c
K
K
=
+
 (12) 
 
 Substitute (11) into (12) and rearrange gives: 
 
 ( )2
2τ θ1
τ θ
c
c
c
K
K
+=
+
 (13) 
 
Controller M in Table 12.1 has the PI controller settings of Eqs. (11) and 
(13). 
12-6 
 12.5 
 
 
Assume that the process can be modeled adequately by a first-order-plus- 
time-delay model as in Eq. 12-10. Then using the given step response 
data, the model fitted graphically is shown in Fig. S12.5, 
Figure S12.5 Process data; first order model estimation. 
 
 
 This gives the following model parameters: 
 
 K = KIP Kv Kp Km = 
psi psi 16.9 12.0 mA0.75 0.9
mA psi 20 18 psi
   −       −      = 1.65 
 θ = 1.7 min 
 
 θ + τ = 7.2 min or τ = 5.5 min 
 
(a) Because θ/τ is greater than 0.25, a conservative choice of τc = τ / 2 is 
used. Thus τc = 2.75 min. 
 
Settling θc = θ and using the approximation e-θs ≈ 1 -θs, Eq. 12-11 gives 
 
 1 τ 0.75
θ τc c
K
K
= =+ , τI = τ = 5.5 min, τD = 0 
 
 (b) From Table 12.3 for PID settings for set-point change, 
 
KKc = 0.965(θ/τ)-0.85 or Kc = 1.58 
τ/τI = 0.796 − 0.1465 (θ/τ) or τI = 7.33 min 
τD/τ = 0.308 (θ/τ)0.929 or τD = 0.57 min 
 
12
13
14
15
16
17
18
0 2 4 6 8 10 12
Time (min)
Output
(mA)
12-7 
(c) From Table 12.3 for PID settings for disturbance input, 
 
KKc = 1.357(θ/τ)-0.947 or Kc = 2.50 
τ/τI = 0.842 (θ/τ)-0.738 or τI = 2.75 min 
τD/τ = 0.381 (θ/τ)0.995 or τD = 0.65 min 
 
 
 
12.6 
 
 
Let G be the open-loop unstable process. First, stabilize the process by 
using proportional-only feedback control, as shown below. 
 
Then, 
 
GG
GG
GK
GK
G
GK
GK
G
Y
Y
c
c
c
c
c
c
c
c
sp ′+
′=
++
+=
1
1
1
1
1
1
1
1
 
where 
GK
GK
G
c
c
1
1
1+=′ 
Then Gc is designed using the Direct Synthesis approach for the stabilized, 
modified process G′ . 
 
 
 
12.7 
 
 
(a.i) The model reduction approach of Skogestad gives the following 
approximate model: 
 
)122.0)(1(
)(
028.0
++=
−
ss
esG
s
 
G G-+
EYsp P
D
Y
-+c Kc1 +
+
12-8 
 Applying the controller settings of Table 12.5 (notice that τ1 ≥ 8θ) 
 
 Kc = 35.40 
 τI = 0.444 
 τD = 0.111 
 
 (a.ii) By using Simulink, the ultimate gain and ultimate period are found: 
 
 Kcu = 30.24 
 Pu = 0.565 
 
 From Table 12.6: 
 
 Kc = 0.45Kcu = 13.6 
 τI = 2.2Pu = 1.24 
 τD = Pu/6.3 = 0.089 
 
 (b) 
0 0.5 1 1.5 2 2.5 3 3.5 4
-0.01
0
0.01
0.02
0.03
0.04
0.05
0.06
0.07
0.08
time
y
Controller (i)
Controller (ii)
 
 Figure S12.7. Closed-loop responses to a unit step change in a disturbance. 
 
 
 
12.8 
 
 
 From Eq.12-39: 
 
0
1( ) ( ) ( ) ( *) * τ
τ
t m
c sp m c D
I
dyp t p K by t y t K e t dt
dt
  = + − + −    ∫ 
 
12-9 
 This control law can be implemented with Simulink as follows: 
 
 
Closed-loop responses are compared for b = 1, b = 0.7, b = 0.5 and 
b = 0.3: 
0 50 100 150 200 250 300
0
0.5
1
1.5
2
2.5
3
3.5
4
Time
y
b=1
b=0.7
b=0.5
b=0.3
 
 Figure S12.8. Closed-loop responses for different values of b. 
 
As shown in Figure E12.8, as b increases, the set-point response becomes 
faster but exhibits more overshoot. The value of b = 0.5 seems to be a 
good choice. The disturbance response is independent of the value of b. 
 
-+b
WEIGHTING FACTOR
++
KC
-+
INTEGRAL
ACTION
PROPORTIONAL
ACTION
SET POINT
CONTROLLER
CONTROLLER
INPUT
CONTROLLER
OUTPUT
12-10 
 
12.9 
 
 
In order to implement the series form using the standard Simulink form of 
PID control (the expanded form in Eq. 8-16), we first convert the series 
controller settings to the equivalent parallel settings. 
 
(a) From Table 12.2, the controller settings for series form are: 
 
τ1 0.971
τ
D
c c
I
K K
 ′′= + = ′ 
 
 
τ τ τ 26.52I I D′ ′= + = 
 
τ ττ 2.753
τ τ
I D
D
I D
′ ′= =′ ′+ 
 
 By using Simulink, closed-loop responses are shown in Fig. S12.9: 
0 50 100 150 200 250 300
0
0.5
1
1.5
2
2.5
3
Time
y
Parallel form
Series form
 
 
 Figure S12.9. Closed-loop responses for parallel and series form. 
12-11 
 
The closed-loop responses to the set-point change are significantly 
different. On the other hand, the responses to the disturbance are slightly 
closer. 
 
(b) By changing the derivative term in the controller block, Simulink shows 
that the system becomes more oscillatory as τD increases. For the parallel 
form, system becomes unstable for τD ≥5.4; for the series form, system 
becomes unstable for τD ≥4.5. 
 
 
 
12.10 
 
 
 
(a) 
 
 
 
 
(b) Process and disturbance transfer functions: 
 
 
Overall material balance: 021 =−+ www (1) 
Component material balance: 1 1 2 2 ρ
dxw x w x wx V
dt
+ − = (2) 
 
 
 Substituting (1) into (2) and introducing deviation variables: 
 
GC Gp-+
E ++GvKm
X'sp
Gm
Gd
X'P'
X'm
(mA) (mA)
X1'
(mA)
X'sp W'2
(Kg/min)(mA)
12-12 
 1 1 2 2 1 2 2 ρ
dxw x w x w x w x w x V
dt
′′ ′ ′ ′+ − − − = 
 
 Taking the Laplace transform, 
 
 1 1 2 2 1 2w X (s) (x x)W (s) (w w ρVs)X (s)′ ′ ′+ − = + + 
 
 Finally: 
 
2
2 1 2
2 1 2
( )( )
( ) ρ 1 τp
x x
x x w wX sG s
W s w w Vs s
−
′ − += = =′ + + + 
 
 
1
1 1 2
1 1 2
( )( )
( ) ρ 1 τd
w
w w wX sG s
X s w w Vs s
′ += = =′ + + + 
 
 where 
1 2
ρτ +�
V
w w
 
 
 Substituting numerical values: 
 
 
s
sGp 71.41
106.2)(
4
+
×=
−
 
 
 
s
sGd 71.41
65.0)( += 
 
 Composition measurement transfer function: 
 
 ssm eesG
−− =−= 32
5.0
420)( 
 
 Final control element transfer function: 
 
 
10833.0
5.187
10833.0
2.1/300
420
315)( +=+×−
−=
ss
sGv 
 
 Controller: 
 
 Let == mpv GGGG 10833.0
5.187
+s s71.41
106.2 4
+
× − se−32 
 
12-13 
 then 
)10833.0)(171.4(
56.1
++=
−
ss
eG
s
 
 
For a process with a dominant time constant, τ τ / 3c dom= is 
recommended. 
 
Hence τ 1.57.c = From Table 12.1, 
 
 Kc = 1.92 
 τI = 4.71 
 
(c) By using Simulink, 
0 5 10 15 20 25 30 35
0
0.005
0.01
0.015
0.02
0.025
0.03
0.035
0.04
Time
yFigure S12.10c. Closed-loop response for step disturbance. 
 
12-14 
 
(d) By using Simulink 
0 5 10 15 20
-0.12
-0.1
-0.08
-0.06
-0.04
-0.02
0
Time
y
 
 Figure S12.10d. Closed-loop response for a set-point change. 
 
The recommended value of τ 1.57c = gives very good results. 
 
(e) Improved control can be obtained by adding derivative action: τ 0.4D = . 
 
0 5 10 15 20
-0.12
-0.1
-0.08
-0.06
-0.04
-0.02
0
Time
y
 
 Figure S12.10e. Closed-loop response by adding derivative action. 
 
12-15 
(e) For θ =3 min, the closed-loop response becomes unstable. It's well known 
that the presence of a large process time delay limits the performance of a 
conventional feedback control system. In fact, a time delay adds phase lag 
to the feedback loop which adversely affects closed-loop stability (cf. Ch. 
14). Consequently, the controller gain must be reduced below the value 
that could be used if smaller time delay were present. 
 
0 5 10 15 20 25 30 35
-0.8
-0.6
-0.4
-0.2
0
0.2
0.4
0.6
Time
y
 
 Figure S12.10f. Closed-loop response for θ =3min. 
 
 
12.11 
 
 
The controller tuning is based on the characteristic equation for standard 
feedback control. 
 
 1 + GcGI/PGvGpGm = 0 
 
Thus, the PID controller will have to be retuned only if any of the transfer 
functions, GI/P, Gv, Gp or Gm, change. 
 
(a) Km changes. The controller may have to be retuned. 
 
(b) The zero does not affect Gm. Thus, the controller does not require retuning. 
 
(c) Kv changes. Retuning may be necessary. 
 
(d) Gp changes. Controller may have to be retuned. 
12-16 
 
12.12 
 
 
(a) Using Table 12.4, 
 
 0.14 0.28τ
θc
K
K K
= + 
 
τI = 6.8θ0.33θ 10θ+τ+ 
 
(b) Comparing to the Z-N settings, the H-A settings give much smaller Kc and 
slightly smaller τI, and are therefore more conservative. 
 
(c) The Simulink responses for the two controllers are compared in 
Fig. S12.12. The controller settings are: 
 
H-A: Kc = 0.49 , τI =1.90 
 Cohen-Coon: Kc = 1.39 , τI =1.98 
 
 
 
0 10 20 30 40 50 60
0
0.2
0.4
0.6
0.8
1
1.2
1.4
1.6
1.8
2
Time
y
Hagglund-Astrom
Cohen and Coon
 
 
 Fig. S12.12. Comparison of Häggland-Åström and Cohen-Coon 
 controller settings. 
 
 
12-17 
From Fig. S12.12, it is clear that the H-A parameters provide a better set-
point response, although they produce a more sluggish disturbance 
response. 
 
 
 
12.13 
 
 
From the solution to Exercise 12.5, the process reaction curve method 
yields 
 K = 1.65 
 θ = 1.7 min 
 τ = 5.5 min 
 
(a) Direct Synthesis method: 
 
From Table 12.1, Controller G: 
 
1 τ 1 5.5 0.94
τ θ 1.65 (5.5 / 3) 1.7c c
K
K
= = =+ + 
 
 τI = τ = 5.5 min 
 
(b) Ziegler-Nichols settings: 
 
 
1 71 65( )
5 5 1
. s. eG s
. s
−
= + 
 
In order to find the stability limits, consider the characteristic equation 
 
1 + GcG = 0 
 
Substituting the Padé approximation, 1 0 85
1 0 85
s . se
. s
− −≈ + , gives: 
2
1 65 (1 0 85 )1 1
4 675 6 35 1
c
c
. K . sG G
. s . s
−+ = + + + 
or 
 4.675s2 + (6.35 –1.403Kc)s + 1 + 1.65Kc = 0 
 
Substitute s = jωu and Kc = Kcu, 
 
− 4.675 ωu2 + j(6.35 − 1.403Kcu)ωu + 1 +1.65Kcu = 0 + j0 
 
Equating real and imaginary coefficients gives, 
12-18 
 
 (6.35 − 1.403Kcu)ωu = 0 , 1+ 1.65Kcu − 4.675 ωu2 = 0 
 
Ignoring ωu = 0, Kcu = 4.526 and ωu = 1.346 rad/min. Thus, 
 
 2 4 67 minu
u
P .π= =ω 
 
ThePI settings from Table 12.6 are: 
 
Kc τI (min) 
Ziegler-
Nichols 2.04 3.89 
 
 
The ultimate gain and ultimate period can also be obtained using 
Simulink. For this case, no Padé approximation is needed and the results 
are: 
 
 Kcu = 3.76 Pu = 5.9 min 
 
The PI settings from Table 12.6 are: 
 
Kc τI (min) 
Ziegler-
Nichols 1.69 4.92 
 
Compared to the Z-N settings, the Direct Synthesis settings result in 
smaller Kc and larger τI. Therefore, they are more conservative. 
 
 
 
12.14 
 
 2
5 1
s
v p m
eG G G
s
−
= + 
 
 To find stability limits, consider the characteristic equation: 
 
1 + GcGvGpGm = 0 
 or 
 2
2 (1 0.5 )
1 0
2.5 5.5 1
−+ =+ +
cK s
s s
 
 
12-19 
 Substituting a Padé approximation, 1 0.5
1 0.5
s se
s
− −≈ + , gives: 
 
 2.5s2 + (5.5 –Kc)s + 1 + 2Kc = 0 
 
 Substituting s = jωu and Kc = Kcu. 
 
 − 2.5 ωu2 + j(5.5 − Kcu)ωu + 1 +2Kcu = 0 + j0 
 
 Equating real and imaginary coefficients, 
 
 (5.5 − Kcu)ωu = 0 , 1+ 2Kcu − 2.5 ωu2 = 0 
 
 Ignoring ωu= 0, Kcu = 5.5 and ωu= 2.19. Thus, 
 
 2π 2.87
ωu u
P = = 
 
 Controller settings (for the Padé approximation): 
 
Kc τI τD 
Ziegler-Nichols 3.30 1.43 0.36 
Tyreus-Luyben 2.48 6.31 0.46 
 
 
The ultimate gain and ultimate period could also be found using Simulink. 
For this approach, no Padé approximation is needed and: 
 
Ku = 4.26 Pu = 3.7 
 
 Controller settings (exact method): 
 
Kc τI τD 
Ziegler-Nichols 2.56 1.85 0.46 
Tyreus-Luyben 1.92 8.14 0.59 
 
 
The set-point responses of the closed-loop systems for these controller 
settings are shown in Fig. S12.14. 
12-20 
0 10 20 30 40 50 60
0
0.2
0.4
0.6
0.8
1
1.2
1.4
1.6
1.8
2
Time
y
Hagglund-Astrom
Cohen and Coon
 
 
Figure S12.14. Closed-loop responses for a unit step change in the set point. 
 
 
12.15 
 
 
Eliminate the effect of the feedback control loop by opening the loop. That 
is, operate temporarily in open loop by switching the controller to the 
manual mode. This action provides a constant controller output signal. If 
oscillations persist, they must be due to external disturbances. If the 
oscillations vanish, they were caused by the feedback loop. 
 
 
 
12.16 
 
 
The sight glass observation confirms that the liquid level is actually rising. 
Since the controller output is saturated in response to the rising level, the 
controller is working properly. Thus, either the actual feed flow is higher 
than recorded, or the actual liquid flow is lower than recorded, or both. 
Because the flow transmitters consist of orifice plates and differential 
pressure transmitters, a plugged orifice plate could lead to a higher 
recorded flow. Thus, the liquid-flow-transmitter orifice plate would be the 
prime suspect. 
 
13-1 
��������	
�
 
 
 
 
13.1 
 
 )()(
)(3)(
32
1
ωω
ω
=ω= jGjG
jGjGAR 
 
 
14
13
1)2(
1)(3
2
2
2
2
+ωω
+ω
=
+ωω
+ω−
= 
 
From the statement, we know the period P of the input sinusoid is 0.5 min 
and, thus, 
 
 rad/min4
5.0
22
pi=
pi
=
pi
=ω
P
 
 
Substituting the numerical value of the frequency: 
 
 
�24.0212.02
1644
1163ˆ
2
2
=×=×
+pipi
+pi
=×= AARA 
Thus the amplitude of the resulting temperature oscillation is 0.24 degrees. 
 
 
 
13.2 
 
 
First approximate the exponential term as the first two terms in a truncated 
Taylor series 
 
se s θ−≈θ− 1 
 
Then ω−=ω jjG 1)( 
and 222 1)(1 θω+=ωθ−+=termtwoAR 
 
 )(tan)(tan 11 ωθ−=ωθ−=φ −−termtwo 
 
Solution Manual for Process Dynamics and Control, 2nd edition, 
Copyright © 2004 by Dale E. Seborg, Thomas F. Edgar and Duncan A. Mellichamp 
13-2 
For a first-order Pade approximation 
 
 
2
1
2
1
s
s
e s θ
+
θ
−
≈
θ−
 
from which we obtain 
 
 1=PadeAR 
 

 ωθ
−=φ −
2
tan2 1Pade 
Both approximations represent the original function well in the low 
frequency region.At higher frequencies, the Padé approximation matches 
the amplitude ratio of the time delay element exactly (ARPade = 1), while 
the two-term approximation introduces amplification (ARtwo term >1). For 
the phase angle, the high-frequency representations are: 
 
 �90−→φ termtwo 
 �180−→φPade 
 
Since the angle of ωθ− je is negative and becomes unbounded as ∞→ω , 
we see that the Pade representation also provides the better approximation 
to the time delay element's phase angle, matching φ of the pure time delay 
element to a higher frequency than the two-term representation. 
 
 
 
13.3 
 
 
 Nominal temperature F123
2
F119F127
�
��
=
+
=T 
 F4)F119F127(
2
1ˆ ���
=−=A 
 .,sec5.4=τ rad/s189.0sec)60/8.1(2 =pi=ω 
 
 Using Eq. 13-2 with K=1, 
 
 ( ) F25.51)5.4()189.0(41ˆ 2222 �=+=+τω= AA 
 
 Actual maximum air temperature = F25.128 �=+ AT 
 Actual minimum air temperature = F75.117 �=− AT 
13-3 
13.4 
 
 
 
12.0
1
)(
)(
+
=
′
′
ssT
sTm
 
 
 )()12.0()( sTssT m′+=′ 
 
 amplitude of T ′=3.464 467.31)2.0( 2 =+ω 
 
 phase angle of T ′= ϕ + tan-1(0.2ω) = ϕ + 0.04 
 
Since only the maximum error is required, set ϕ = 0 for the comparison of 
T ′ and mT ′ . Then 
 
Error = mT ′ − T ′=3.464 sin (0.2t) – 3.467sin(0.2t + 0.04) 
 = 3.464 sin(0.2t) –3.467[sin(0.2t) cos 0.04 + cos(0.2t)sin 0.04] 
 = 0.000 sin(0.2t) − 0.1386 cos(0.2t) 
 
Since the maximum absolute value of cos(0.2t) is 1, 
 
maximum absolute error = 0.1386 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
13-4 
13.5 
 
 
a) 
 
Bode Diagram
Frequency (rad/sec)
Ph
as
e 
(de
g)
M
ag
ni
tu
de
 
(ab
s)
10-2
10-1
100
10
-2
10
-1
10
0
10
1
-180
-135
-90
-45
0
 
ωωωω AR (absolute) ϕϕϕϕ 
0.1 4.44 -32.4° 
1 0.69 -124° 
10 0.005 -173° 
 b) 
 
Bode Diagram
Frequency (rad/sec)
Ph
as
e 
(de
g)
M
ag
ni
tu
de
 
(ab
s)
10-2
100
10
-2
10
-1
10
0
10
1
-270
-225
-180
-135
-90
-45
0
 
ωωωω AR (absolute) ϕϕϕϕ 
0.1 4.42 -38.2° 
1 0.49 -169° 
10 0.001 -257° 
 
13-5 
c) 
 
Bode Diagram
Frequency (rad/sec)
Ph
as
e 
(de
g)
M
ag
ni
tu
de
 
(ab
s)
10-1
100
10
-2
10
-1
10
0
10
1
10
2
-90
-45
0
 
ωωωω AR (absolute) ϕϕϕϕ 
0.1 4.48 -22.1° 
1 2.14 -44.9° 
10 0.003 -87.6° 
 
 
 d) 
 
Bode Diagram
Frequency (rad/sec)
Ph
as
e 
(de
g)
M
ag
n
itu
de
 
(ab
s)
10-1
10
0
10-2 10-1 100 101 102
-270
-225
-180
-135
-90
-45
0
 
ωωωω AR (absolute) ϕϕϕϕ 
0.1 4.48 -33.6° 
1 1.36 -136° 
10 0.04 -266° 
 
13-6 
e) 
 
Bode Diagram
Frequency (rad/sec)
Ph
as
e 
(de
g)
M
ag
n
itu
de
 
(ab
s)
10-1 100 101
-180
-150
-120
-90
10-2
10
-1
100
101
102
 
ωωωω AR (absolute) ϕϕϕϕ 
0.1 44.6 -117° 
1 0.97 -169° 
10 0.01 -179° 
 
 f) 
 
Bode Diagram
Frequency (rad/sec)
Ph
as
e
 
(de
g)
M
ag
ni
tu
de
 
(ab
s)
10-1 100 101
-180
-150
-120
-90
10-1
100
101
102
 
 
ωωωω AR (absolute) ϕϕϕϕ 
0.1 44.8 -112° 
1 1.36 -135° 
10 0.04 -158° 
 
 
 
 
13-7 
13.6 
 
 
a) Multiply the AR in Eq. 13-41a by 122 +τω a . Add to the value of ϕ in 
Eq. 13-41b the term + )(tan 1 aωτ− . 
 
KjG =ω)( 222222 )4.0()1(1 ωτ+τω−+τω a 



τω−
ωτ−
=ω∠ − 22
1
1
4.0
tan)( jG + )(tan 1 aωτ− . 
 
 b) 
Bode Diagram
ωτωτωτωτ
Ph
as
e 
(de
g)
N
or
m
ali
z
ed
 
Am
plt
u
de
 
ra
tio
 
(ab
s)
10-2 10-1 100 101 102
-180
-135
-90
-45
0
45
90
10-4
10-2
100
102
Ratio = 0
Ratio = 0.1
Ratio = 1
Ratio = 10
Ratio = 0
Ratio = 0.1
Ratio = 1
Ratio = 10
 
 Figure S13.6. Frequency responses for different ratios τa/τ 
 
 
 
 
 
 
 
 
 
13-8 
13.7 
 
 
 Using MATLAB 
Bode Diagram
Frequency (rad/sec)
Ph
as
e 
(de
g)
M
ag
n
itu
de
 
(ab
s)
10
-1
10
0
10
1
10
2
-270
-225
-180
-135
-90
-45
10-10
10-5
100
105
 
 Figure S13.7. Bode diagram of the third-order transfer function. 
 
 
The value of ω that yields a -180° phase angle and the value of AR at that 
frequency are: 
 
ω = 0.807 rad/sec 
AR = 0.202 
 
 
 
 
 
 
 
 
 
 
 
 
13-9 
13.8 
 
 
 Using MATLAB, 
Bode Diagram
Frequency (rad/sec)
Ph
as
e 
(de
g)
M
ag
n
itu
de
 
(ab
s)
10-2 10-1 100 101
-250
-200
-150
-100
-50
0
10-1
100
G(s)
G(s) w ith Pade approx.
 
 Figure S13.8. Bode diagram for G(s) and G(s) with Pade approximation. 
 
 
 
13.9 
 
 ω=2pif where f is in cycles/min 
 
 For the standard thermocouple, using Eq. 13-20b 
 
 ϕ1 = -tan-1(ωτ1) = tan-1(0.15ω) 
 
 Phase difference ∆ϕ = ϕ1 – ϕ2 
 
 Thus, the phase angle for the unknown unit is 
 
 ϕ2 = ϕ1 − ∆ϕ 
 
 and the time constant for the unknown unit is 
13-10 
 τ2 = )tan(1 2ϕ−
ω
 
 
using Eq. 13-20b . The results are tabulated below 
 
f ωωωω ϕϕϕϕ1 ∆∆∆∆ϕϕϕϕ ϕϕϕϕ2 ττττ2 
0.05 0.31 -2.7 4.5 -7.2 0.4023 
0.1 0.63 -5.4 8.7 -14.1 0.4000 
0.2 1.26 -10.7 16 -26.7 0.4004 
0.4 2.51 -26.6 24.5 -45.1 0.3995 
0.8 6.03 -37 26.5 -63.5 0.3992 
1 6.28 -43.3 25 -68.3 0.4001 
2 12.57 -62 16.7 -78.7 0.3984 
4 25.13 -75.1 9.2 -84.3 0.3988 
 
That the unknown unit is first order is indicated by the fact that ∆ϕ→0 as 
ω→∞, so that ϕ2→ϕ1→-90° and ϕ2→-90° for ω→∞ implies a first-order 
system. This is confirmed by the similar values of τ2 calculated for 
different values of ω, implying that a graph of tan(-ϕ2) versus ω is linear 
as expected for a first-order system. Then using linear regression or taking 
the average of above values, τ2 = 0.40 min. 
 
 
 
13.10 
 
 
 From the solution to Exercise 5-19, for the two-tank system 
 
 
1132.1
01.0
)(
/)(
1
max11
+τ
=
+
=
′
′′
s
K
ssQ
hsH
i
 
 22
1
max22
)1()132.1(
01.0
)(
/)(
+τ
=
+
=
′
′′
s
K
ssQ
hsH
i
 
 
 22
1
2
)1(
1337.0
)132.1(
1337.0
)(
)(
+τ
=
+
=
′
′
sssQ
sQ
i
 
 
 and for the one-tank system 
 
 
12164.2
01.0
)(
/)(
1
max
+τ
=
+
=
′
′′
s
K
ssQ
hsH
i
 
 
 
12
1337.0
164.2
1337.0
)(
)(
1 +τ
=
+
=
′
′
sssQ
sQ
i
 
13-11 
For a sinusoidal input ,sin)(1 tAtq i ω=′ the amplitudes of the heights and 
flow rates are 
 
 [ ] 14//ˆ 22max +τω=′′ KAhhA (1) 
 
 [ ] 14/1337.0ˆ 22 +τω=′ AqA (2) 
 
for the one-tank system, and 
 
 [ ] 1//ˆ 22
max11 +τω=′′ KAhhA (3) 
 
 [ ] 222
max22 )1(//
ˆ +τω=′′ KAhhA (4) 
 
 [ ] 2222 )1(/1337.0ˆ +τω=′ AqA (5) 
 
for the two-tank system. 
 
Comparing (1) and (3), for all ω 
 
 [ ] [ ]maxmax11 /ˆ/ˆ hhAhhA ′′≥′′ 
 
Hence, for all ω, the first tank of the two-tank system will overflow for a 
smaller value of A than will the one-tank system. Thus, from the overflow 
consideration, the one-tank system is better for all ω. However, if A is 
small enough so that overflow is not a concern, the two-tank system will 
provide a smaller amplitude in the output flow for those values of ω that 
satisfy 
 
 
[ ] [ ]qAqA ′≤′ ˆˆ 2or 
14
1337.0
)1(
1337.0
22222 +τω
≤
+τω
AA
 
 
or ω ≥ τ/2 = 1.07 
 
Therefore, the two-tank system provides better damping of a sinusoidal 
disturbance for ω ≥ 1.07 if and only if 
 
 [ ] 1/ˆ
max11 ≤′′ hhA , that is, 01.0
132.1 22 +≤ ωA 
 
13-12 
13.11 
 
 
 Using Eqs. 13-48 , 13-20, and 13-24, 
 
 AR=
141100
12
22
22
+ω+ω
+τω a
 
 
 φ = tan-1(ωτa) – tan-1(10ω) – tan-1(2ω) 
 
 The Bode plots shown below indicate that 
 
i) AR does not depend on the sign of the zero. 
ii) AR exhibits resonance for zeros close to origin. 
iii) All zeros lead to ultimate slope of –1 for AR. 
iv) A left-plane zero yields an ultimate φ of -90°. 
v) A right-plane zero yields an ultimate φ of -270°. 
vi) Left-plane zeros close to origin can give phase lead at low ω. 
vii) Left-plane zeros far from the origin lead to a greater lag (i.e., 
smaller phase angle) than the ultimate value. φu 90−= º with a left-
plane zero present. 
Bode Diagram
Frequency (rad/sec)
Ph
as
e 
(de
g)
M
ag
n
itu
de
 
(ab
s)
10-2 10-1 100 101
-270
-180
-90
0
90
10-2
10-1
100
101
Case i
Case ii(a)
Case ii(b)
Case iii
 
 Figure
 
S13.11. Bode plot for each of the four cases of numerator dynamics. 
13-13 
13.12 
 
 
a) From Eq. 8-14 with τI = 4τD 
 
s
sK
s
ssKsG
D
D
c
D
DD
cc
τ
+τ
=
τ
τ++τ
=
4
)12(
4
)414()(
222
 
ωτ
+ωτ
=
ωτ
 +ωτ
=ω
D
D
c
D
D
cc KKjG 4
14
4
14
)(
22
2
22
 
 
b) From Eq. 8-15 with τI = 4τD and α = 0.1 
 
( )
( )11.04
1)14()(
+ττ
+τ+τ
=
ss
ss
KsG
DD
DD
cc 
 
101.04
1116
)(
22
2222
+ωτωτ
 +ωτ +ωτ
=ω
DD
DD
cc KjG 
 
The differences are significant for 0.25 < ωτD < 1 by a maximum of 0.5 Kc 
at ωτD = 0.5, and for ωτD >10 by an amount increasing with ωτD . 
 
10-2 10-1 100 101 102
10-1
100
101
102
ωτωτωτωτ
AR
/K
c
Parallel controller (actual) 
Series controller (actual)
Series controller with filter (asymptote) 
Parallel controller (asymptote) 
D 
 
 Figure
 
S13.12. Nominal amplitude ratio for parallel and series controllers. 
 
13-14 
13.13 
 
 
MATLAB does not allow the addition of transfer functions with different 
time delays. Hence the denominator time delay needs to be approximated 
if a MATLAB program is used. However, the use of Mathematica or even 
Excel to evaluate derived expressions for the AR and angle, using various 
values of omega, and to make the plots will yield exact results: 
 
 MATLAB - Padé approximation: 
 
Substituting the 1/1 Padé approximation gives: 
 
42
)2(
1
2
2
)( 2 +τ+θτ
+θ
=
+


θ+
θ−
+τ
≈
ss
sK
s
s
s
K
sG (1) 
 
By using MATLAB, 
Bode Diagram
Frequency (rad/sec)
Ph
as
e 
(de
g)
M
ag
n
itu
de
 
(ab
s)
10-2 10-1 100 101 102
-90
-45
0
10-2
10-1
100
101
 
 Figure
 
S13.13. Bode plot by using Padé approximation. 
 
 
 
13-15 
13.14 
 
 
 
min
rad15080
cycle
radians2
rotation
cycles4
min
rotations600 =pi××=ω 
 
 psig2=A psig02.0ˆ =A 
 
 01.0/ˆAR == AA 
 
Volume of the pipe connecting the compressor to the reactor is 
 
32
2
ft982.0ft
12
3
4
ft20 =

pi
×=pipeV 
 
Two-tank surge system 
 
Using the figure and nomenclature in Exercise 2.5, the 0.02 psig variation 
in Aˆ refers to the pressure before the valve Rc, namely the pressure P2. 
Hence the transfer function )(/)(2 sPsP d′′ is required in order to use the 
value of AR. Mass balance for the tanks is (referring to the solution for 
Exercise 2.5. 
 
ba wwdt
dP
RT
MV
−=
1
1
1
 (1) 
 
cb wwdt
dP
RT
MV
−=
2
2
2
 (2) 
 
where the ideal-gas assumption has been used. For linear valves, 
 
a
d
a R
PP
w 1
−
= , 
b
b R
PP
w 21
−
= , 
c
f
c R
PP
w
−
=
2
 
 
 At nominal conditions, 
 
 psig200=dP 
 
 lb/min100lb/hr6000 ==== cba www 
 
 psig10
2
1.0
211 ==−=−
d
d
P
PPPP 
13-16 
 b
ba
d
a R
w
PP
w
PP
R =
−
===
−
=
211
lb/min
psig1.0
lb/min100
psig10
 
 
 Assume vac RRR == 
 Assume R792F30012 �� === TT 
 Given VVV == 21 
 
 Then equations (1) and (2) become 
 
 21211
1 2)( PPPPPPP
dt
dP
R
RT
VM
ddv +−=−−−=


 
 
 
 ffv PPPPPPPdt
dP
R
RT
VM
−−=−−−=


21221
2 2)( 
 
 
Taking deviation variables, Laplace transforming, and noting that fP′ is 
zero since fP is constant, gives 
 
)(
2
1)()(
2
1)( 211 sPsPsPsPs d ′+′−′=′τ (3) 
)()(
2
1)( 212 sPsPsPs ′−′=′τ (4) 
 
where 
 
 


=τ vRRT
VM
2
1
 
 
 
( ) ( )R792
Rmolelb
psigft731.10
lb/min
psig1.0
molelb
lb28ft
2
1 33 �
� 








= V 
 
 = min)10647.1( 4V−× 
 
From Eq. 3 
 
 )()1(2
1)()1(2
1)( 21 sP
s
sP
s
sP d ′+τ
+′
+τ
=′ 
 
13-17 
Substituting for )(1 sP′ into Eq. 4 
 
 )()1(4
1)()1(4
1)()1( 22 sP
s
sP
s
sPs d ′+τ
+′
+τ
=′+τ 
 
 
 or 
 
 
384
1
1)1(4
1
)(
)(
222
2
+τ+τ
=
−+τ
=
′
′
ssssP
sP
d
 
 
 
ωτ+τω−
=
ω′
ω′
8)43(
1
)(
)(
22
2
jjP
jP
d
 
 
 
 
94016
1
64)43(
1AR
224422222 +τω+τω
=
τω+τω−
= 
 
 Setting AR = 0.01 gives 
 
 1000094016 2244 =+τω+τω 
 
 099914016 2244 =−τω+τω 
 
 ( ) 77.2399911644040
162
1 222
=××++−
×
=τω 
 
 min10233.3875.477.23 4−×=
ω
=
ω
=τ 
 
 
3
4 ft963.110647.1
=
×
τ
=
−
V 
 
 Total surge volume 3ft926.32 == VVsurge 
 
Letting the connecting pipe provide part of this volume, the volume of 
each tank = 3ft472.1)(
2
1
=− pipesurge VV 
 
 
 
13-18 
Single-tank system 
 
In the figure for the two-tank system, remove the second tank and the 
valve before it (Rb). Now, Aˆ refers to P1 and AR refers to )(/)(1 sPsP d′′ . 
 
Mass balance for the tank is 
 
ca wwdt
dP
RT
MV
−=
1
1
1
 
 
where 
a
d
a R
PP
w 1
−
= , 
c
f
c R
PP
w
−
=
1
 
 
 At nominal conditions 
 
 psig201.01 ==− dd PPP 
 
 
lb/min
psig0.2
lb/min100
psig201
==
−
=
a
d
a
w
PP
R 
 
 Assume Rc = Ra = Rv 
 
 Then Eq. 1 becomes 
 
 71711
1
1
1 2)( PPPPPPP
dt
dP
R
RT
MV
ddv +−=−−−=



 
 
 Using deviation variables and taking the Laplace transform 
 
 
1
2/1
)(
)(1
+τ
=
′
′
ssP
sP
d
 
 
 where 
 
 =



=τ vRRT
MV
1
1
2
1
 min)10294.3( 14V−× 
 
AR = 0.01= 15.0 22 +τω , 310315.3 −×=τ min, 31 ft06.10=V 
 
 Volume of single tank = 331 ft472.14ft084.9)( ×>=− pipeVV 
 Hence, recommend two surge tanks, each with volume 3ft472.1 
13-19 
13.15 
 
By using MATLAB 
Nyquist Diagram
Real Axis
Im
ag
in
ar
y 
Ax
is
-1 0 1 2 3 4 5 6
-4
-3
-2
-1
0
1
2
3
4
 
 Figure
 
S13.15a. Nyquist diagram. 
 
Nyquist Diagram
Real Axis
Im
a
gin
ar
y 
Ax
is
-1 0 1 2 3 4 5 6
-4
-3
-2
-1
0
1
2
3
4
 
 Figure
 
S13.15b. Nyquist diagram by usingPade approximation. 
13-20 
The two plots are very different in appearance for large values of ω. The 
reason for this is the time delay. If the transfer function contains a time 
delay in addition to poles and zeros, there will be an infinite number of 
encirclements of the origin. This result is a consequence of the unbounded 
phase shift for the time delay. 
 
A subtle difference in the two plots, but an important one for the Nyquist 
design methods of Chapter 14, is that the plot in S13.5a “encircles” the -1, 
0 point while that in S13.5b passes through it exactly. 
 
 
 
 
13.16 
 
 
 By using MATLAB, 
10-2
10-1
10
0
101
102
M
ag
n
itu
de
 
(ab
s)
10-2 10-1 100
-270
-225
-180
-135
-90
-45
0
Ph
as
e 
(de
g)
Parallel
Series w ith filter
Parallel
Series w ith filter
Bode Diagram
Frequency (rad/sec)
 Figure
 
S3.16. Bode plot for Exercise 13.8 Transfer Function multiplied by PID 
 Controller Transfer Function. Two cases: a)Parallel b) Series with Deriv. 
 Filter (α=0.2). 
. 
 
13-21 
 Amplitude ratios: 
 
 Ideal PID controller: AR= 0.246 at ω = 0.80 
 Series PID controller: AR=0.294 at ω = 0.74 
 
 There is 19.5% difference in the AR between the two controllers. 
 
 
 
13.17 
 
 
a) Method discussed in Section 6.3: 
 
)12.2)(18(
12)(ˆ
3.0
1 ++
=
−
ss
e
sG
s
 
 
 Visual inspection of the frequency responses: 
 
 )185.2)(164.5(
12)(ˆ
4.0
2 ++
=
−
ss
e
sG
s
 
 
b) Comparison of three models: 
 
Bode plots: 
Bode Diagram
Frequency (rad/sec)
Ph
a
s
e 
(de
g)
M
a
gn
itu
de
 
(ab
s
)
10-10
10-5
100
105
G(s)
G1(s)
G2(s)
10-1 100 101 102
-720
-360
0
 
 Figure S13.17a. Bode plots for the exact and approximate models. 
13-22 
Impulse responses: 
Impulse Response
Time (sec)
Am
pli
tu
de
0 5 10 15 20 25 30 35 40 45
0
0.2
0.4
0.6
0.8
1
1.2
1.4
G(s)
G1(s)
G2(s)
 
 Figure S13.17b. Impulse responses for the exact and approximate models 
 
 
 
 
13.18 
 
 
 The original transfer function is 
 
 )1)(14)(120(
)12(10)(
2
+++
+
=
−
sss
es
sG
s
 
 
 The approximate transfer function obtained using Section 6.3 is: 
 
 
13-23 
Bode Diagram
Frequency (rad/sec)
Ph
as
e 
(de
g)
M
ag
n
itu
de
 
(ab
s)
10-2
10-1
100
101
10-2 10-1 100 101
-2880
-2520
-2160
-1800
-1440
-1080
-720
-360
0
G(s)
G'(s)
 
 Figure S13.18. Bode plots for the exact and approximate models. 
 
 
 As seen in Fig.S13.18, the approximation is good at low frequencies, but 
 not that good at higher frequencies. 
14-1 
Chapter 14 
 
 
 
14.1 
 
 
 Let GOL(jωc) = R + jI 
 
where ωc is the critical frequency. Then, according to the Bode stability 
criterion 
 
 | GOL(jωc)| = 1 = 22 IR + 
 ∠GOL(jωc) = -π = tan –1 (I/R) 
 
 Solving for R and I: R = -1 and I = 0 
 
Substituting s = jωc into the characteristic equation gives, 
 
 1 + GOL(jωc) = 0 
 
 I + R + jI = 0 or R = -1 , I = 0 
 
 Hence, the two approaches are equivalent. 
 
 
14.2 
 
 
 Because sustained oscillations occur at the critical frequency 
 
 12ω 0.628min
10 minc
π −= = 
(a) Using Eq. 14-7, 
 
1 = (Kc)(0.5)(1)(1.0) or Kc = 2 
 
(b) Using Eq. 14-8, 
 
 – π= 0 + 0 +(-θωc) + 0 
 or θ = 5 min
ωc
π = 
 
Solution Manual for Process Dynamics and Control, 2nd edition, 
Copyright © 2004 by Dale E. Seborg, Thomas F. Edgar and Duncan A. Mellichamp 
Revised: 1-3-04
14-2 
 
 
14.3 
 
 
(a) From inspection of the Bode diagrams in Tables 13.4 and 13.5, the 
transfer function is selected to be of the following form 
 
G(s) = 
)1)(1(
)1(
21 +τ+τ
+τ
sss
sK a 
 
where τa, τ1, τ2 correspond to frequencies of ω = 0.1, 2, 20 rad/min, 
respectively. 
 
 Therefore, τa = 1/0.1 = 10 min 
 
 τ1=1/2 = 0.5 min 
 
 τ2= 1/20 = 0.05 min 
 
 For low frequencies, AR ≈ |K/s| = K/ω 
 
 At ω = 0.01, AR = 3.2, so that K = (ω)(AR) = 0.032 
 
 Therefore, 
 
 G(s) = 
)105.0)(15.0(
)110(032.0
++
+
sss
s 
 
(b) Because the phase angle does not cross -180°, the concept of GM is 
meaningless. 
 
 
14.4 
 
 
The following process transfer can be derived in analogy with Eq. 6-71: 
 
 
2
1 1
1 1 1 2 2 1 1 2 1 2 2
( )
( ) ( ) ( ) 1
= + + + +
H s R
Q s A R A R s A R A R A R s
 
 
 For R1=0.5, R2 = 2, A1 = 10, A2 = 0.8: 
 
14-3 
 Gp(s) = 
178
5.0
2 ++ ss (1) 
 
For R2 = 0.5: Gp(s) = 
18.52
5.0
2 ++ ss (2) 
 
 (a) For R2 = 2 
 
∠Gp= tan-1 



ω−
ω−
281
7
c
c , |Gp| =
2 2 2
0.5
(1 8 ) (7 )− ω + ωc c
 
 
For Gv = Kv = 2.5, ϕv=0, |Gv| = 2.5 
 
For Gm = 
15.0
5.1
+s , ϕm= -tan
-1(0.5ω) , |Gm| = 
1)5.0(
5.1
2 +ωc
 
 
Kcu and ωc are obtained using Eqs. 14-7 and 14-8: 
 
 -180° = 0 + 0 + tan-1 



ω−
ω−
281
7
c
c − tan-1(0.5ωc) 
 
Solving, ωc = 1.369 rad/min. 
 
 



ω+ω−
=
222 )7()81(
5.0)5.2)((1
cc
cuK 



+ω 1)5.0(
5.1
2
c
 
 
 Substituting ωc = 1.369 rad/min, Kcu = 10.96, ωcKcu = 15.0 
 
For R2=0.5 
 
 ∠Gp = tan-1 



ω−
ω−
221
8.5
c
c , |Gp| = 



ω+ω− 222 )8.5()21(
5.0
cc
 
 
 -180° = 0 + 0 + tan-1 



ω−
ω−
221
8.5
c
c − tan-1(0.5ωc) 
 
Solving, ωc = 2.51 rad/min. 
 
 Substituting ωc = 2.51 rad/min, Kcu = 15.93, ωcKcu = 40.0 
 
14-4 
(a) From part (a), for R2=2, 
 
ωc = 1.369 rad/min, Kcu = 10.96 
Pu = 
cω
π2 = 4.59 min 
 
Using Table 12.6, the Ziegler-Nichols PI settings are 
 
Kc = 0.45 Kcu = 4.932 , τI= Pu/1.2 = 3.825 min 
 
Using Eqs. 13-63 and 13-62 , 
 
ϕc= -tan-1(-1/3.825ω) 
 
|Gc| = 4.932 1
825.3
1 2 +


ω 
 
Then, from Eq. 14-7 
 
-180° = tan-1 


ω
−
c825.3
1 + 0 + tan-1 



ω−
ω−
281
7
c
c − tan-1(0.5ωc) 
 
Solving, ωc = 1.086 rad/min. 
 
 Using Eq. 14-8, 
 
 Ac = AROL|ω=ωc = 
 
= 



ω+ω−







+



ω 222
2
)7()81(
5.0)5.2(1
825.3
1932.4
ccc
 



+ω 1)5.0(
5.1
2
c
 
 
= 0.7362 
 
Therefore, gain margin GM =1/Ac = 1.358. 
 
Solving Eq.(14-16) for ωg 
 
 AROL|ω=ωc = 1 at ωg = 0.925 
 
14-5 
 Substituting into Eq. 14-7 gives ϕg=ϕ|ω=ωg = −172.7°. 
 
 Therefore, phase margin PM = 180+ ϕg = 7.3°. 
 
 
 
14.5 
 
 
 (a) K=2 , τ = 1 , θ = 0.2 , τc=0.3 
 
 Using Eq. 12-11, the PI settings are 
 
11 =τ+θ
τ=
c
c K
K , τI = τ = 1 min, 
 
Using Eq. 14-8 , 
 
 -180° = tan-1 



ω
−
c
1 − 0.2ωc − tan-1(ωc) = -90° − 0.2ωc 
 
 or ωc = 
2.0
2/π = 7.85 rad/min 
 
 Using Eq. 14-7, 
 
 255.02
1
211AR
22
=ω=





+ω
+ω== ω=ω ccc
OLc
c
A 
 
 From Eq. 14-11, GM = 1/Ac = 3.93. 
 
 
(b) Using Eq. 14-12, 
 
 ϕg = PM − 180° = − 140 ° = tan-1(-1/0.5ωg) − 0.2ωg − tan-1(ωg) 
 
Solving by trial and error, ωg = 3.04 rad/min 
 








+ω
+



ω==ω=ω 1
21
5.0
11AR
2
2
gg
cOL K
g
 
 
Substituting for ωg gives Kc = 1.34. Then from Eq. 14-8 
14-6−180° = tan-1 



ω
−
c5.0
1 − 0.2ωc − tan-1(ωc) 
 
Solving by trial and error, ωc =7.19 rad/min. 
 
From Eq. 14-7, 
383.0
1
21
5.0
134.1AR
2
2
=



+ω
+



ω== ω=ω
cc
OLc
c
A 
 
From Eq. 14-11, GM = 1/Ac = 2.61 
 
 
(c) By using Simulink-MATLAB, these two control systems are compared for 
a unit step change in the set point. 
0 1 2 3 4 5
0
0.2
0.4
0.6
0.8
1
1.2
1.4
time
y
part (a)
part (b)
 
 Fig S14.5. Closed-loop response for a unit step change in set point. 
 
The controller designed in part a) (Direct Synthesis) provides better 
performance giving a first-order response. Part b) controller yields a large 
overshoot. 
 
14-7 
 
 
14.6 
 
 
(a) Using Eqs. 14-7 and 14-8, 
 
2
2 2 2
4 1 2 0.4AR (1.0)
0.01 1 0.25 1 25 1
m
OL c
sp
Y K
Y
ω
ω ω ω ω
   + = =      + + +   
 
 
ϕ= tan-1(2ω) − tan-1(0.1ω) − tan-1(0.5ω) – (π/2) − tan-1(5ω) 
Bode Diagram
Frequency (rad/sec)
Ph
as
e 
(d
eg
)
AR
/K
c
10
-4
10
-2
10
0
10
2
10
-2
10
-1
10
0
10
1
10
2
-270
-225
-180
-135
-90
 
 Figure S14.6a. Bode plot 
 
(b) Using Eq.14-12 
 
ϕg = PM – 180° = 30°− 180° = −150° 
 
From the plot of ϕ vs. ω: ϕg = -150° at ωg = 1.72 rad/min 
 
14-8 
From the plot of 
c
OL
K
AR
 vs ω: 
g
c
OL
K ω=ω
AR
= 0.144 
Because 
g
OL ω=ωAR = 1 , Kc = 144.0
1 = 6.94 
 
 (c) From the phase angle plot: 
 
 ϕ = -180° at ωc = 4.05 rad/min 
 
From the plot of 
c
OL
K
AR
 vs ω, 
c
c
OL
K ω=ω
AR
= 0.0326 
Ac = 
c
OL ω=ωAR = 0.326 
 
From Eq. 14-11, GM = 1/Ac = 3.07. 
Bode Diagram
Frequency (rad/sec)
Ph
as
e 
(d
eg
)
AR
/K
c
10
-4
10
-2
10
0
10
2
10
-2
10
-1
10
0
10
1
10
2
-270
-225
-180
-135
-90
 
 Figure S14.6b. Solution for part (b) using Bode plot. 
 
14-9 
Bode Diagram
Frequency (rad/sec)
Ph
as
e 
(d
eg
)
AR
/K
c
10
-4
10
-2
10
0
10
2
10
-2
10
-1
10
0
10
1
10
2
-270
-225
-180
-135
-90
 
 Figure S14.6c. Solution for part (c) using Bode plot. 
 
 
 
14.7 
 
 
(a) For a PI controller, the |Gc| and ∠ Gc from Eqs. 13.62 and 13.63 need to 
be included in the AR and ϕ given for GvGpGm to obtain AROL and ϕOL. 
The results are tabulated below 
 
ω AR |Gc|/Kc AROL/Kc ϕ ∠Gc ϕOL 
0.01 2.40 250 600 -3 -89.8 -92.8 
0.10 1.25 25.020 31.270 -12 -87.7 -99.7 
0.20 0.90 12.540 11.290 -22 -85.4 -107.4 
0.50 0.50 5.100 2.550 -41 -78.7 -119.7 
1.00 0.29 2.690 0.781 -60 -68.2 -128.2 
2.00 0.15 1.601 0.240 -82 -51.3 -133.3 
5.00 0.05 1.118 0.055 -122 -26.6 -148.6 
10.00 0.02 1.031 0.018 -173 -14.0 -187.0 
15.00 0.01 1.014 0.008 -230 -9.5 -239.5 
 
 
From Eq. 14-12, ϕg = PM – 180° = 45°− 180° = -135°. 
 
Interpolating the above table, ϕOL= -135° at ωg = 2.5 rad/min and 
14-10 
g
c
OL
K ω=ω
AR
= 0.165 
 
Because 
g
OL ω=ωAR = 1 , Kc = 165.0
1 = 6.06 
 
 
(b) From the table above, 
 ϕOL= -180° at ωc = 9.0 rad/min and 
c
c
OL
K ω=ω
AR
= 0.021 
 Ac = 
c
OL ω=ωAR = 0.021 Kc = 0.127 
 
 From Eq. 14-11, 
 
 GM = 1/Ac = 1/0.127 = 7.86 
 
 
(c) From the table in part (a), 
 
ϕOL= -180° at ωc = 10.5 rad/min and 
cω=ωAR = 0.016. 
 
Therefore, Pu = 
cω
π2 = 0.598 min and Kcu = 
c
AR ω=ω
1 = 62.5. 
 
Using Table 12.6, the Ziegler-Nichols PI settings are 
 
Kc = 0.45 Kcu = 28.1, τI = Pu/1.2 = 0.50 min 
 
Tabulating AROL and ϕOL as in part (a) and the corresponding values of M 
using Eq. 14-18 gives: 
 
ω |Gc| ∠Gc AROL ϕOL M 
0.01 5620 -89.7 13488 -92.7 1.00 
0.10 563.0 -87.1 703 -99.1 1.00 
0.20 282.0 -84.3 254 -106.3 1.00 
0.50 116.0 -76.0 57.9 -117.0 1.01 
1.00 62.8 -63.4 18.2 -123.4 1.03 
2.00 39.7 -45.0 5.96 -127.0 1.10 
5.00 30.3 -21.8 1.51 -143.8 1.64 
10.00 28.7 -11.3 0.487 -184.3 0.94 
15.00 28.3 -7.6 0.227 -237.6 0.25 
 
Therefore, the estimated value is Mp =1.64. 
14-11 
 
 
14.8 
 
 
Kcu and ωc are obtained using Eqs. 14-7 and 14-8. Including the filter GF 
into these equations gives 
 
-180° = 0 + [-0.2ωc − tan-1(ωc)]+[-tan-1(τFωc)] 
 
 Solving, 
 
 ωc = 8.443 for τF = 0 
 ωc = 5.985 for τF = 0.1 
 
Then from Eq. 14-8, 
 
( ) 



+ωτ





+ω
=
1
1
1
21
222
cFc
cuK 
 
Solving for Kcu gives, 
 
 Kcu = 4.251 for τF = 0 
 Kcu = 3.536 for τF = 0.1 
 
Therefore, 
 
 ωcKcu = 35.9 for τF = 0 
 ωc Kcu= 21.2 for τF = 0.1 
 
Because ωcKcu is lower for τF = 0.1, filtering the measurement results in 
worse control performance. 
 
 
14.9 
 
 
(a) Using Eqs. 14-7 and 14-8, 
 
)0.1(
1
1
1100
51
25
1AR
222 



+ω





+ω




 +ω= cOL K 
 
ϕ = tan-1(-1/5ω) + 0 + (-2ω − tan-1(10ω)) + (- tan-1(ω)) 
14-12 
Bode Diagram
Frequency (rad/sec)
Ph
as
e 
(d
eg
)
AR
/K
c
10
-2
10
-1
10
0
10
1
10
2
10
-2
10
-1
10
0
10
1
10
2
-350
-300
-250
-200
-150
-100
 
 Figure S14.9a. Bode plot 
 
 
(b) Set ϕ = 180° and solve for ω to obtain ωc = 0.4695. 
 
Then 
c
OL ω=ωAR = 1 = Kcu(1.025) 
 
Therefore, Kcu = 1/1.025 = 0.976 and the closed-loop system is stable for 
Kc ≤ 0.976. 
 
 
(c) For Kc = 0.2, set AROL = 1 and solve for ω to obtain ωg = 0.1404. 
 
Then ϕg = 
gω=ωϕ = -133.6° 
 
From Eq. 14-12, PM = 180° + ϕg = 46.4° 
 
 
(d) From Eq. 14-11 
14-13 
 
GM = 1.7 = 
cA
1 = 
c
OL ω=ωAR
1 
 
From part (b), 
c
OL ω=ωAR = 1.025 Kc 
 
Therefore, 1.025 Kc = 1/1.7 or Kc = 0.574 
Bode Diagram
Frequency (rad/sec)
Ph
as
e 
(d
eg
)
AR
/K
c
10
-2
10
-1
10
0
10
1
10
2
10
-2
10
-1
10
0
10
1
10
2
-350
-300
-250
-200
-180
-150
 
 Figure S14.9b. Solution for part b) using Bode plot. 
Bode Diagram
Frequency (rad/sec)
Ph
as
e 
(d
eg
)
AR
/K
c
10
-2
10
-1
10
0
10
1
10
2
10
-2
10
-1
10
0
10
1
10
2
-350
-300
-250
-200
-180
-150
 
 Figure S14.9c. Solution for part c) using Bode plot. 
 
 
 
14-14 
 
14.10 
 
(a) 
1083.0
264.5112
1083.0
047.0)( +=×+= sssGv 
 
)1017.0)(1432.0(
2)( ++= sssGp 
 
 0.12( )
0.024 1m
G s
s
= + 
 
 Using Eq. 14-8 
 
 -180° = 0 − tan-1(0.083ωc) − tan-1(0.432ωc) − tan-1(0.017ωc) 
− tan-1(0.024ωc) 
 
 Solving by trial and error, ωc = 18.19 rad/min. 
 
 Using Eq. 14-7, 




+ω+ω
⋅



+ω
=
1)017.0(1)432.0(
2
1)083.0(
624.5)(1
222
ccc
cuK 
 
2
0.12
(0.024 ) 1cω
  ×  + 
 
 Substituting ωc=18.19 rad/min, Kcu = 12.97. 
 
Pu = 2π/ωc = 0.345 min 
 
 Using Table 12.6, the Ziegler-Nichols PI settings are 
 
 Kc = 0.45 Kcu = 5.84 , τI=Pu/1.2 = 0.288 min 
 
 
(b) Using Eqs.13-62 and 13-63 
 
ϕc = ∠ Gc = tan-1(-1/0.288ω)= -(π/2) + tan-1(0.288ω) 
 
|Gc| = 5.84 1288.0
1 2 +


ω 
 
 Then, from Eq. 14-8, 
 
14-15 
 
-π = − (π/2) + tan-1(0.288ωc) − tan-1(0.083ωc) − tan-1(0.432ωc) 
 
 − tan-1(0.017ωc) − tan-1(0.024ωc) 
 
 
Solving by trial and error, ωc = 15.11 rad/min. 
 
Using Eq. 14-7, 
 




+ω
⋅


+



ω== ω=ω 1)083.0(
264.51
288.0
184.5AR
2
2
cc
cOLc
A 
2 2 2
2 0.12
(0.432 ) 1 (0.017 ) 1 (0.024 ) 1c c cω ω ω
      × ⋅   + + +   
 
 
 
= 0.651 
 
Using Eq. 14-11, GM = 1/Ac = 1.54. 
 
Solving Eq. 14-7 for ωg gives 
 
g
OL ω=ωAR = 1 at ωg = 11.78 rad/min 
 Substituting into Eq. 14-8 gives 
 
ϕg = 
gω=ωϕ = − (π/2) + tan-1(0.288ωg) − tan-1(0.083ωg) − 
tan-1(0.432ωg) − tan-1(0.017ωg) − tan-1(0.024ωg) = -166.8° 
 
Using Eq. 14-12, 
 
 PM = 180° + ϕg = 13.2 ° 
 
 
14.11 
 
 
 (a) 
2 2
10 1.5| | (1)
1 100 1
G
ω ω
  =     + +  
 
 
ϕ = − tan-1(ω) − tan-1(10ω) − 0.5ω 
14-16 
Bode Diagram
Frequency (rad/sec)
Ph
as
e 
(d
eg
)
AR
10
-1
10
0
10
1
10
2
10
-2
10
-1
10
0
10
1
-270
-180
-90
0
 
 Figure S14.11a. Bode plot for the transfer function G=GvGpGm. 
 
(b) From the plots in part (a) 
 
∠G = -180° at ωc = 1.4 and |G|ω=ωc = 0.62 
 
c
OL ω=ωAR = 1= (- Kcu) |G|ω=ωc 
 
Therefore, Kcu = -1/0.62 = -1.61 and 
 
Pu = 2π/ωc = 4.49 
 
Using Table 12.6, the Ziegler-Nichols PI-controller settings are: 
 
Kc = 0.45Kcu = -0.72 , τI = Pu/1.2 = 3.74 
 
Including the |Gc| and ∠Gc from Eqs. 13-62 and 13-63 into the results of 
part (a) gives 
 
 



+ω+ω
+


ω= 11001
151
74.3
172.0AR
22
2
OL 
14-17 
 
ω+ω+ω
+ω=
11001
10.1489.2
22
2
 
ϕ = tan-1(-1/3.74ω) − tan-1(ω) − tan-1(10ω) − 0.5ω 
 
Bode Diagram
Frequency (rad/sec)
Ph
as
e 
(d
eg
)
AR
10
-2
10
0
10
2
10
4
10
-2
10
-1
10
0
10
1
-360
-270
-180
-90
0
90
 
Figure S14.11b. Bode plot for the open-loop transfer function GOL=GcG. 
 
(c) From the graphs in part (b), 
 
ϕ = -180° at ωc=1.15 
 
c
OL ω=ωAR = 0.63 < 1 
 
Hence, the closed-loop system is stable. 
14-18 
 
Bode Diagram
Frequency (rad/sec)
Ph
as
e 
(d
eg
)
AR
10
-2
10
0
10
2
10
4
10
-2
10
-1
10
0
10
1
-270
-180
-90
` 
 Figure S14.11c. Solution for part (c) using Bode plot. 
 
 
(d) From the graph in part b), 
 
5.0
AR =ωOL = 2.14 = 
amplitude of ( )
amplitude of ( )
m
sp
y t
y t
 
Therefore, the amplitude of ym(t) = 5.114.2 × = 3.21. 
 
(e) From the graphs in part (b), 
5.0
AR =ωOL = 2.14 and 5.0=ωϕ =-147.7°. 
 
Substituting into Eq. 14-18 gives M = 1.528. Therefore, the amplitude of 
y(t) = 1.528×1.5 = 2.29 which is the same as the amplitude of ym(t) 
because Gm is a time delay. 
 
(f) The closed-loop system produces a slightly smaller amplitude for ω = 0.5. 
As ω approaches zero, the amplitude approaches one due to the integral 
control action. 
14-19 
 
 
14.12 
 
 
(a) Schematic diagram: 
 
 
 Block diagram: 
 
 
(b) GvGpGm = Km = 6 mA/mA 
 
GTL = e-8s 
 
 GOL = GvGpGmGTL = 6e-8s 
 
 
If GOL = 6e-8s, 
 
 | GOL(jω) | = 6 
 
 ∠ GOL (jω) = -8ω rad 
 
 
 Find ωc: Crossover frequency generates − 180° phase angle = − π radians 
 
 -8ωc = -π or ωc = π/8 rad/s 
 
Hot fluid
TT
TC
Cold fluid
Mixing Point Sensor
14-20 
 Find Pu: Pu = 
2π 2π 16s
ω π / 8c
= = 
 Find Kcu: Kcu = 167.0
6
1
|)(|
1 ==ωcp jG
 
 
 
 Ziegler-Nichols ¼ decay ratio settings: 
 
 PI controller: 
 
Kc = 0.45 Kcu = (0.45)(0.167) = 0.075 
 τI = Pu/1.2 = 16/1.2 = 13.33 sec 
 
 PID controller: 
 
Kc = 0.6 Kcu = (0.6)(0.167) = 0.100 
 τI = Pu/2 = 16/2 = 8 s 
 τD = Pu/8 = 16/8 = 2 s 
 
 (c) 
0 30 60 90 120 150
0
0.2
0.4
0.6
0.8
1
1.2
1.4
PID control
PI control
y 
t 
 
 Fig. S14.12. Set-point responses for PI and PID control. 
 
14-21 
(d) Derivative control action reduces the settling time but results in a more 
oscillatory response. 
 
 
 
 
14.13 
 
 
(a) From Exercise 14.10, 
 
1083.0
264.5)( += ssGv 
)1017.0)(1432.0(
2)( ++= sssGp 
 
)1024.0(
12.0)( += ssGm 
 The PI controller is 

 +=
s
sGc 3.0
115)( 
 Hence the open-loop transfer function is 
 
 mpvcOL GGGGG = 
 
 Rearranging, 
 
 
sssss
sGOL ++++×
+= − 23455 556.005738.000168.01046.1
06.21317.6 
 
14-22 
 By using MATLAB, the Nyquist diagram for this open-loop system is 
 
Nyquist Diagram
Real Axis
Im
ag
in
ar
y 
Ax
is
-3 -2.5 -2 -1.5 -1 -0.5 0
-4
-3.5
-3
-2.5
-2
-1.5
-1
-0.5
0
0.5
1
 
 Figure S14.13a. The Nyquist diagram for the open-loop system. 
 
(b) Gain margin = GM = 
cAR
1 
 
where ARc is the value of the open-loop amplitude ratio at the critical 
frequency ωc. By using the Nyquist plot, 
Nyquist Diagram
Real Axis
Im
ag
in
ar
y 
Ax
is
-3 -2.5 -2 -1.5 -1 -0.5 0
-4
-3.5
-3
-2.5
-2
-1.5
-1
-0.5
0
0.5
1
 
 Figure S14.13b. Graphical solution for part (b). 
14-23 
 θ = -180 ⇒ ARc = | G(jωc)| = 0.5 
 
 Therefore the gain margin is GM = 1/0.5 = 2. 
 
 
14.14 
 
To determine 
p
m M
e 1||max <
ω
, we must calculate Mp based on the CLTF 
with IMC controller design. In order to determine a reference Mp, we 
assume a perfect process model (i.e. GG ~− = 0 ) for the IMC controller 
design. 
 
GG
R
C
c
*=∴ 
 
Factoring, 
 
 −+= GGG ~~~ 
 
12
10~,~ +== −
−
+ s
GeG s 
 fsGc 10
12* +=∴ 
 
 Filter Design: Because τ = 2 s, let τc = τ/3 = 2/3 s. 
 
 ⇒ 
132
1
+= sf 
 
 
10
12* +=∴ sGc 10320
12
132
1
+
+=+ s
s
s
 
 
 
10320
10
12
10
10320
12*
+=



+



+
+==∴
−−
s
e
s
e
s
sGG
R
C ss
c 
 
 1=∴ pM 
 
 The relative model error with K as the actual process gain is: 
 
14-24 
10
10
12
10
12
10
12
~
~ −=
+



+−


+=−=∴ −
−−
K
s
e
s
e
s
Ke
G
GGe s
ss
m 
Since Mp = 1, 110
10||max <−=
ω
Kem 
 
 ⇒ 1
10
10 <−K ⇒ K < 20 
 
 1
10
10 −>−K ⇒ K > 0 
 
 
 ∴ 200 << K for guaranteed closed-loop stability. 
 
 
 
14.15 
 
 
Denote the process model as, 
 
1
2~ 2.0
+=
−
s
eG
s
 
 
 and the actual process as: 
 
 
1
2 2.0
+τ=
−
s
eG
s
 
 
 The relative model error is: 
 
 
1
)1(
)(~
)(~)()( +τ
τ−=−=∆∴
s
s
sG
sGsGs 
 
 Let s = jω. Then, 
 
 
 
|1|
|)1(|
1
)1(
+ωτ
ωτ−=+ωτ
ωτ−=∆∴
jj
j (1) 
 
 
14-25 
 or 
 
 
2 2
| (1 ) |
1
τ ω
τ ω
−∆ =
+
 
 
 Because | ∆ | in (1) increases monotonically with ω, 
 
 τ
τ−=∆=∆ ∞→ωω
|1|||lim||max (2) 
 
 Substituting (2) and Mp = 1.25 into Eq. 14-34 gives: 
 
 8.0|1| <τ
τ− 
 
 This inequality implies that 
 
 8.01 <τ
τ−
 ⇒ 1 < 1.8τ ⇒ τ > 0.556 
 and 
 
 8.01 <τ
−τ
 ⇒ 0.2τ < 1 ⇒ τ < 5 
 
 Thus, closed-loop stability is guaranteed if 
 
0.556 < τ < 5 
 
 
 
15-1 
Chapter 15 
 
 
 
 
15.1 
 
 
 For Ra=d/u 
 
2u
d
u
R
K ap −=∂
∂= 
 
which can vary more than Kp in Eq. 15-2, because the new Kp depends on 
both d and u. 
 
 
 
15.2 
 
 
 By definition, the ratio station sets 
 
 (um – um0) = KR (dm – dm0) 
 
 Thus 
2
1
2
2
1
2
2
0
0 

==−
−=
d
u
K
K
dK
uK
dd
uu
K
mm
mm
R (1) 
 
For constant gain KR, the valuesof u and d in Eq. 1 are taken to be at the 
desired steady state so that u/d=Rd, the desired ratio. Moreover, the 
transmitter gains are 
 
21
mA)420(
dS
K −= , 22 mA)420(
uS
K −= 
 
Substituting for K1, K2 and u/d into (1) gives: 
 
2
2
2
2



==
u
d
dd
d
u
R S
SRR
S
SK 
 
 
Solution Manual for Process Dynamics and Control, 2nd edition, 
Copyright © 2004 by Dale E. Seborg, Thomas F. Edgar and Duncan A. Mellichamp 
 
Revised: 1-3-04
 
15-2 
15.3 
 
 
(a) The block diagram is the same as in Fig. 15.11 where Y ≡ H2, Ym ≡ H2m, 
Ysp ≡ H2sp, D ≡ Q1, Dm ≡ Q1m, and U ≡ Q3. 
 
 
b) (A steady-state mass balance on both tanks gives 
 
0 = q1 – q3 or Q1 = Q3 (in deviation variables) (1) 
 
From the block diagram, at steady state: 
 
 Q3 = Kv Kf Kt Q1 
 
From (1) and (2), Kf = 
1
v tK K
 (2) 
 
c) (No, because Eq. 1 above does not involve q2. 
 
 
15.4 
 
 
 
 
 
 
 
(b) From the block diagram, exact feedforward compensation for Q1 would 
result when 
 
Q1 + Q2 = 0 
 
 
 Substituting Q2 = KV Gf Kt Q1, 
 
Gf = − 1
v tK K
 
 
15-3 
 
(c) Same as part (b), because the feedforward loop does not have any dynamic 
elements. 
 
 
 (d) 
 
 
 
 For exact feedforward compensation 
 
 Q4 + Q3 = 0 (1) 
 
 From the block diagram, Q2 = KV Gf Kt Q4 (2) 
 
Using steady-state analysis, a mass balance on tank 1 for no variation in q1 
gives 
 
 Q2 − Q3 = 0 (3) 
 
 Substituting for Q3 from (3) and (2) into (1) gives 
 
 Q4 + KV Gf Kt Q4 = 0 
 
or Gf = − 1
v tK K
 
 
For dynamic analysis, find Gp1 from a mass balance on tank 1, 
 
 
 11 1 2 1 1
dhA q q C h
dt
= + − 
 
 
 
 
15-4 
Linearizing (4), noting that 1q′ = 0, and taking Laplace transforms: 
 
 11 2 1
12
CdhA q h
dt h
′ ′ ′= − 
or 1 11
2 1 1 1
(2 / )( )
( ) (2 / ) 1
h CH s
Q s A h C s
′ =′ +
 (5) 
 
 Since 
3 1 1
1
3 1
12
q C h
Cq h
h
=
′ ′= or 
3 1
1 1
( )
( ) 2
Q s C
H s h
′ =′ (6) 
 
 From (5) and (6), 
 
 
1
3
2 1 1 1
( ) 1
( ) (2 / ) 1
P
Q s G
Q s A h C s
′ = =′ + (7) 
 
 
Substituting for Q3 from (7) and (2) into (1) gives 
 
 4 4
1 1 1
1( ) 0
(2 / ) 1
v f tQ K G K Q
A h C s
+ =
+ 
 
 or 1 1 1
1 [(2 / ) 1]f
v t
G A h C s
K K
= − + 
 
 
 
15.5 
 
 
(a) For a steady-state analysis: 
 
Gp=1, Gd=2, Gv = Gm = Gt =1 
 
From Eq.15-21, 
 
Gf = 2
)1)(1)(1(
2 −=−=−
ptv
d
GGG
G
 
 
(b) Using Eq. 15-21, 
 
15-5 
 Gf = 
15
2
1
1)1)(1(
)15)(1(
2
+
−=



+
++
−
=−
s
s
ss
GGG
G
ptv
d 
 
(c) Using Eq. 12-19, 
 
 −+=+== GGsGGGG mpv
~~
1
1~
 
where 
11,
1
G G
s+ −= = +
� � 
 
For τc=2, and r=1, Eq. 12-21 gives 
 
 f =
12
1
+s 
 
From Eq. 12-20 
 
* 1 1 1( 1)
2 1 2 1
−
−
+ = = + = + + 
�
c
sG G f s
s s
 
 
From Eq. 12-16 
 
s
s
s
s
s
GG
G
G
c
c
c 2
1
12
11
12
1
~1
+=
+−
+
+
=−= ∗
∗
 
 
(d) For feedforward control only, Gc=0. For a unit step change in disturbance, 
D(s) = 1/s. 
 
Substituting into Eq. 15-20 gives 
 
 Y(s) = (Gd+GtGfGvGp) s
1 
 
For the controller of part (a) 
 
 Y(s) = 
ssss
1
1
1)1)(2)(1(
)15)(1(
2 

 


+−+++ 
 
 
15-6 
 Y(s) = 
5/1
5.2
1
5.2
15
2/25
1
2/5
)15)(1(
10
+
−−+=+
−++=++
−
ssssss
 
 
or y(t) = 2.5 (e-t – e- t/5) 
 
 
For the controller of part (b) 
 
 Y(s) = 01
1
1)1(
15
2)1(
)15)(1(
2 =

 


+


+
−+++ sssss 
 
 or y(t) = 0 
 
 The plots are shown in Fig. S15.5a below. 
 
 
 
Figure S15.5a. Closed-loop response using feedforward control only. 
 
 
(e) Using Eq. 15-20: 
 
For the controller of parts (a) and (c), 
 
 Y(s) = 
2 1(1)( 2)(1)
1( 1)(5 1) 1
1 11 (1) (1)
2 1
s s s
s s
s s
  + −   + + +  +    +     +    
 
 
15-7 
 or Y(s) = =+++
−
)12)(15)(1(
20
sss
s
12
3/40
15
3/25
1
5
+
−++++ sss 
 =
5/1
3/5
2/1
3/20
1
5
+++−+ sss 
 
or y(t) = 5e-t –
3
20 e-t/2 + 
3
5 e- t/5 
 
and for controllers of parts (b) and (c) 
 
Y(s) = 
s
ss
s
ssss 1
)1(
1
1)1(
2
11
1
1)1(
15
2)1(
)15)(1(
2











+

 ++



+


+
−+++ = 0 
 
 or y(t) = 0 
 
 The plots of the closed-loop responses are shown in Fig. S15.5b. 
 
 
 Figure S15.5b. Closed-loop response for feedforward-feedback control. 
 
 
 
 
15.6 
 
 
(a) The steady-state energy balance for both tanks takes the form 
 
0 = w1 C T1 + w2 C T2 − w C T4 + Q 
 
15-8 
 
where Q is the power input of the heater 
 C is the specific heat of the fluid. 
 
Solving for Q and replacing unmeasured temperatures and flow rates by 
their nominal values, 
 
 Q = C ( )42211 TwTwTw −+ (1) 
 
Neglecting heater and transmitter dynamics, 
 
 Q = Kh p (2) 
 
 T1m = T1m0 + KT(T1-T10) (3) 
 
 wm = wm0 + Kw(w-w0) (4) 
 
 
 Substituting into (1) for Q, T1, and w from (2), (3), and (4), gives 
 
0 0 0 0
1 1 1 1 2 2 4
1 1[ ( ( )) ( ( ))]m m m m
h T w
Cp w T T T w T T w w w
K K K
= + − + − + −
 
 
(b) Dynamic compensation is desirable because the process transfer function 
Gp= T4(s)/P(s) is different from each of the disturbance transfer functions, 
Gd1= T4(s)/T1(s), and Gd2= T4(s)/w(s); this is more so for Gd1 which has a 
higher order. 
 
 
15.7 
 
 
(a) 
 
 
 
(b) A steady-state material balance for both tanks gives, 
 
15-9 
 
0 = q1 + q2 + q4 − q5 
 
Because 2q′ = 4q′ = 0, the above equation gives 
 
 0 = 1q′ – 5q′ or 0 = Q1 – Q5 (1) 
 
From the block diagram, 
 
 Q5 = Kv Gf Kt Q1 
 
 Substituting for Q5 into (1) gives 
 
 0 = Q1 − Kv Gf Kt Q1 or Gf = 
tv KK
1 
 
 (c) To find Gd and Gp, the mass balance on tank 1 is 
 
 112111 hCqqdt
dhA −+= 
 
 where A1 is the cross-sectional area of tank 1. 
 
 Linearizing and setting 2q′ = 0 leads to 
 
 1 11 1 1
12
dh CA q h
dt h
′ ′ ′= − 
 
 Taking the Laplace transform, 
 
 1 1
1 1 1
( )
( ) 1
H s R
Q s A R s
′ =′ + where 1
1
1
2
C
h
R ≡ (2) 
 
 Linearizing q3 = C1 1h gives 
 
 3 1
1
1q h
R
′ ′= or 3
1 1
( ) 1
( )
Q s
H s R
′ =′ (3) 
 
 Mass balance on tank 2 is 
 
 54322 qqqdt
dhA −+= 
 
15-10 
 
 Using deviation variables, setting 4q′ = 0, and taking Laplace transform 
 
 ( ) ( ) ( )2 2 3 5A sH s Q s Q s′ ′ ′= − 
 
 2
3 2
( ) 1
( )
H s
Q s A s
′ =′ (4) 
 and 
 
 2
5 2
( ) 1 ( )
( ) p
H s G s
Q s A s
′ = − =′ 
 
 32 2 1
1 3 1 1 2 1 1
( )( ) ( ) ( ) 1( )
( ) ( ) ( ) ( ) ( 1)d
Q sH s H s H sG s
Q s Q s H s Q s A s A R s
′′ ′ ′= = =′ ′ ′ ′ + 
 
 upon substitution from (2), (3), and (4). 
 
Using Eq. 15-21, 
 
)/1(
)1(
1
2
112
sAKK
sRAsA
GGGGG
vtpvt
d
f −
+−=−= 
 
1
11
11 +
+=
sRAKK tv
 
 
15.8 
 
 
For the process model in Eq. 15-22 and the feedforward controller in Eq. 
15-29, the correct values of τ1 and τ2 are given by Eq. 15-42 and (15-43). 
 
 Therefore, 
 
 τ1 − τ2 = τp − τL (1) 
 
for a unit step change in d, and no feedback controller, set D(s)=1/s, and 
Gc= 0 in Eq. 15-20 to obtain 
 
 Y(s) = [ ]
s
GGGGG pvftd
1+ 
 Setting Gt = Gv = 1, and using Eqs. 15-22 and 15-29, 
 
 
15-11 
 1
2
/ (τ 1) 1( ) (1) (1)
τ 1 τ 1 τ 1
pd d P
d p
KK K K sY s
s s s s
   − += +     + + +     
 
 
 12 1 2
2 2 2
(τ τ )ττ τ (τ τ )1 1 1 1
τ 1 (τ τ ) τ 1 τ τ τ 1
p pd
d
d p p p
K
s s s s s
 −−= − − − − + − + − +  
 
 
 or y(t) 2 1 / τ/ τ/ τ 1 2
2 2
τ τ(τ τ )
τ τ τ τ
−−− −−= − − − − −  
pp ttt
d
p p
K e e e 
 
 
0 0
( ) ( )e t dt y t dt
∞ ∞=∫ ∫ 12 1 2
2 2
τ (τ τ )τ (τ τ )τ
τ τ τ τ
p p
d d
p p
K
 −−= − + + − −  
 
 
 2 22 2 1 2 1 2 2
2
τ τ τ τ τ τ τ τ τ τ (τ τ τ τ )
τ τ
d
d d p p p p p
p
K−  = − + − − + + − − 
 
 1 2(τ τ ) (τ τ )d p dK  = − − − −  
 
 0= when (1) holds. 
 
 
 
 
 
 
15.9 
 
 
(a) For steady-state conditions 
 
Gp=1, Gd=2, Gv = Gm = Gt =1 
 
Using Eq. 15-21, 
 
Gf = 2
)1)(1)(1(
2 −=−=−
ptv
d
GGG
G
 
 
(b) Using Eq. 15-21, 
 
 
15-12 
 Gf = 
15
2
1
1)1)(1(
)15)(1(
2
+
−=



+
++
−
=−
−
−
se
s
ss
e
GGG
G
s
s
ptv
d 
 
(c) Using Eq. 12-19, 
 
−+
−
=+== GGs
eGGGG
s
mpv
~~
1
~ 
where 
1
1~ , ~ +== −
−
+ s
GeG s 
 
For τc=2, and r = 1, Eq. 12-21gives 
 
 f =
12
1
+s 
 
From Eq. 12-20 
 
12
1
12
1)1(~
1* +
+=++== − s
s
s
sf
G
Gc 
 
From Eq. 12-16 
 
 
s
s
s
s
s
GG
G
G
c
c
c 2
1
12
11
12
1
~*1
* +=
+−
+
+
=−= 
 
(d) For feedforward control only, Gc=0. For a unit step disturbance, 
D(s) = 1/s. 
 
Substituting into Eq. 15-20 gives 
 
 Y(s) = (Gd+GtGfGvGp) s
1 
 
For the controller of part (a) 
 
 Y(s) = 
ss
e
ss
e ss 1
1
)1)(2)(1(
)15)(1(
2 






+−+++
−−
 
 
15-13 
 = 
)15)(1(
10
++
− −
ss
e s 
 
or y(t) = 2.5 (e-(t-1) – e-( t-1)/5)S(t-1) 
 
 
For the controller of part (b) 
 
 Y(s) = 01
1
)1(
15
2)1(
)15)(1(
2 =






+


+
−+++
−−
ss
e
sss
e ss 
 
 or y(t) = 0 
 
 The plots are shown in Fig. S15.9a below. 
 
 
Figure S15.9a. Closed-loop response using feedforward control only. 
 
 
(e) Using Eq. 15-20: 
 
For the controllers of parts (a) and (c), 
 
 Y(s) = 
s
s
e
s
s
s
e
ss
e
s
ss
1
)1(
1
)1(
2
11
1
)1)(2)(1(
)15)(1(
2














+

 ++




+−+++
−
−−
 
 
15-14 
 
and for the controllers of parts (b) and (c), 
 
Y(s) = 
s
ss
s
ssss 1
)1(
1
1)1(
2
11
1
1)1(
15
2)1(
)15)(1(
2











+

 ++



+


+
−+++ = 0 
 
 or y(t) = 0 
 
 The plots of the closed-loop responses are shown in Fig. S15.9b. 
 
 
 Figure S15.9b. Closed-loop response for the feedforward-feedback control. 
 
 
 
15.10 
 
 
 (a) For steady-state conditions 
 
Gp=Kp, Gd=Kd, Gv = Gm = Gt =1 
 
Using Eq. 15-21, 
 
 
15-15 
Gf = 25.0
)2)(1)(1(
5.0 −=−=−
ptv
d
GGG
G
 
 
(b) Using Eq. 15-21, 
 
 Gf = ss
s
ptv
d e
s
s
s
e
s
e
GGG
G 10
20
30
)160(
)195(25.0
195
2)1)(1(
160
5.0
−
−
−
+
+−=




+
+
−
=− 
(c) Using Table 12.1, a PI controller is obtained from equation G, 
 
95.0
)2030(
95
2
11 =+=θ+τ
τ=
cp
c K
K 
τ τ 95I = = 
 
(d) As shown in Fig.S15.10a, the dynamic controller provides significant 
improvement. 
(e) 
0 100 200 300 400 500
-0.04
-0.02
0
0.02
0.04
0.06
0.08
time
y(t)
Controller of part (a)
Controller of part (b)
 
 
 Figure S15.10a. Closed-loop response using feedforward control only. 
 
 
15-16 
 
0 100 200 300 400 500
-0.06
-0.04
-0.02
0
0.02
0.04
0.06
time
y(t)
Controllers of part (a) and (c)
Controllers of part (b) and (c)
 
Figure S15.10b. Closed-loop response for feedforward-feedback control. 
 
 
f) As shown in Fig. S15.10b, the feedforward configuration with the 
dynamic controller provides the best control. 
 
 
 
15.11 
 
 
 Energy Balance: 
 
 )()()1()( aLLcci TTAUTTAqUTTwCdt
dTVC −−−+−−=ρ (1) 
 
 Expanding the right hand side, 
 
 )()( ci TTUATTwCdt
dTVC −−−=ρ 
 )( aLLccc TTAUTUAqTUAq −−+− (2) 
 
 Linearizing, 
 
 cccc qTTqTqTq ′+′+≈ (3) 
 
Substituting (3) into (2), subtracting the steady-state equation, and 
introducing deviation variables, 
 
 
15-17 
 TqUAqTUATUATTwC
dt
TdVC cci ′−′−′−′−′=′ρ )( 
 TAUqUAT LLcc ′−′+ (4) 
 
 Taking the Laplace transform and assuming steady-state at t = 0 gives, 
 
 ( ) ( ) ( ) ( )i c cVCsT s wCT s UA T T Q sρ −′ ′ ′= + − 
 
 )()( sTAUqUAUAwC LLc ′+++− (5) 
 
 Rearranging, 
 
 ( ) ( ) ( ) ( ) ( )L i p cT s G s T s G s Q s′ ′ ′= + (6) 
 
 where: 
 
 ( )
τ 1
L
d
KG s
s
= + 
 
1
)( +τ= s
K
sG pp 
 d
wCK
K
= (7) 
 
K
TTUAK cp
)( −= 
 
K
VCρ=τ 
 LLc AUqUAUAwCK +++= 
 
 The ideal FF controller design equation is given by, 
 
 dF
t v p
GG
G G G
−= (17-27) 
 
 But, stt eKG
θ−= and Gv=Kv (8) 
 
 Substituting (7) and (8) gives, 
 
 
)( TTUAKK
wCeG
cvt
s
F −
−=
θ+
 (9) 
 
 In order to have a physically realizable controller, ignore the e+θs term, 
 
15-18 
 
 
)( TTUAKK
wCG
cvt
F −
−= (10) 
 
 
15.12 
 
 
a) A component balance in A gives: 
 
A
Ai A A
dcV qc qc Vkc
dt
= − − (1) 
 
At steady state, 
 
 
 0 Ai A Aq c q c Vkc= − − (2) 
 
 Solve for ,q 
 
 
AAi
A
CC
CkVq −= (3) 
 
 For an ideal FF controller, replace AiC by AiC , q by q1 and AC by AspC : 
 
 Asp
Ai Asp
kVC
q
C C
= − 
 
b) Linearize (1): 
 
A
iA Ai Ai A A A A
dcV q c qc c q q c qc c q Vkc
dt
′ ′ ′ ′= + + − − − − 
 
Subtract (2), 
 
A
iA Ai A A A
dcV qc c q qc c q Vkc
dt
′ ′ ′ ′ ′ ′= + − − − 
 
Take the Laplace transform, 
 
( ) ( ) ( ) ( ) ( ) ( )A iA Ai A A AsVc s qc s c Q s qc s c Q s Vkc s′ ′ ′ ′ ′ ′= + − − − 
 
Rearrange, 
 
15-19 
 
 ( )( ) ( )Ai AA Ai c cqC s C s Q ssV q Vk sV q Vk
−′ ′ ′= ++ + + + (6) 
or 
 
 ( ) ( ) ( ) ( ) ( )A d iA pC s G s C s G s Q s′ ′ ′= + (7) 
 
 The ideal FF controller design equation is, 
 
 ( )( )
( ) ( ) ( )
d
F
v p t
G sG s
G s G s G s
= − (8) 
 
 Substitute from (6) and (7) with Gv(s)=Kv and Gt(s)=Kt : 
 
 ( )
( )F v Ai A t
qG s
K c c K
= − − (9) 
 
 . 
Note: )(/)()( sCsPsG mAiF ′′= where P is the controller output and cAim 
is the measured value of cAi. 
 
 
15.13 
 
 
 (a) Steady-state balances: 
 
 3150 qqq −+= (1) 
 
 4230 qqq −+= (2) 
 0 
 3311550 qxqxqx −+= (3) 
 
 4422330 qxqxqx −+= (4)Solve (4) for 33qx and substitute into (3), 
 
 4422550 qxqxqx −+= (5) 
 
 Rearrange, 
 
 
15-20 
 
2
5544
2 x
qxqxq −= (6) 
 
In order to derive the feedforward control law, let 
 
4 4 ,spx x→ 2 2 ( ),x x t→ 5 5( ),x x t→ and )(22 tqq → 
 
Thus, 
 
 
2
5544
2
)()(
)(
x
tqtxqx
tq sp
−= (7) 
 
Substitute numerical values: 
 
 
990.0
)()()3400(
)( 5542
tqtxx
tq sp
−= (8) 
 or 
 
 )()(01.13434)( 5542 tqtxxtq sp −= (9) 
 
Note: If transmitter and control valve gains are available, then an 
expression relating the feedforward controller output signal, p(t), to the 
measurements , x5m(t) and q5m(t), can be developed. 
 
(b) Dynamic compensation: It will be required because of the extra 
dynamic lag preceding the tank on the left hand side. The stream 5 
disturbance affects x3 while q3 does not. 
 
 
 
 
 
16-1 
��������	
�
 
 
 
16.1 
 
The difference between systems A and B lies in the dynamic lag in the 
measurement elements Gm1 (primary loop) and Gm2(secondary loop). With 
a faster measurement device in A, better control action is achieved. In 
addition, for a cascade control system to function properly, the response of 
the secondary control loop should be faster than the primary loop. Hence 
System A should be faster and yield better closed-loop performance than 
B. 
 
Because Gm2 in system B has an appreciable lag, cascade control has the 
potential to improve the overall closed-loop performance more than for 
system A. Little improvement in system A can be achieved by cascade 
control versus conventional feedback. 
 
Comparisons are shown in Figs. S16.1a/b. PI controllers are used in the 
outer loop. The PI controllers for both System A and System B are 
designed based on Table 12.1 (τc = 3). P controllers are used in the inner 
loops. Because of different dynamics the proportional controller gain of 
System B is about one-fourth as large as the controller gain of System A 
 
System A: Kc2 = 1 Kc1=0.5 τI=15 
System B: Kc2 = 0.25 Kc1=2.5 τI=15 
 
0 10 20 30 40 50 60 70 80 90 100
0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
time
O
ut
pu
t
Cascade
Standard feedback
 
Figure S16.1a. System A. Comparison of D2 responses (D2=1/s) for cascade 
control and conventional PI control. 
 
Solution Manual for Process Dynamics and Control, 2nd edition, 
Copyright © 2004 by Dale E. Seborg, Thomas F. Edgar and Duncan A. Mellichamp 
16-2 
In comparing the two figures, it appears that the standard feedback results 
are essentially the same, but the cascade response for system A is much 
faster and has much less absolute error than for the cascade control of B 
 
 
0 10 20 30 40 50 60 70 80 90 100
0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
time
O
ut
pu
t
Cascade
Standard feedback
 
Figure S16.1b. System B .Comparison of D2 responses (D2=1/s) for cascade 
control and conventional PI control. 
 
 
 
 
 
Figure S16.1c. Block diagram for System A 
 
 
16-3 
 
 
 
 
 Figure S16.1d. Block diagram for System B 
 
 
 
16.2 
 
 
a) The transfer function between Y1 and D1 is 
 
11
1 2
1 1
2 2
1
1
d
c v
c p m
c v m
GY
D G GG G G
G G G
=  
+  
+ 
 
 
and that between Y1 and D2 is 
 
21
2 2 2 2 1 11
p d
c v m c v m c p
G GY
D G G G G G G G G
=
+ +
 
 
using 
1
5
+
=
s
Gv , 2 1dG = , 1
1
3 1d
G
s
=
+
 , 
 
 )14)(12(
4
++
=
ss
G p , 05.01 =mG , 2.02 =mG 
16-4 
For Gc1 = Kc1 and Gc2 = Kc2, we obtain 
 
3 2
2 2 21
4 3 2
1 2 2 1 2 2 1
8 (14 8 ) (7 6 ) 1
24 (50 24 ) [10 (9 3 )] (35 26 ) (1 ) 1
c c c
c c c c c c
s K s K s KY
D s K s K K s K s K K
+ + + + + +
=
+ + + + + + + + + +
 
 
1
3 2
2 2 2 2 1
4( 1)
8 (14 8 ) (7 6 ) (1 ) 1c c c c
Y s
D s K s K s K K
+
=
+ + + + + + +
 
 
 
The figures below show the step load responses for Kc1=43.3 and for 
Kc2=25. Note that both responses are stable. You should recall that the 
critical gain for Kc2=5 is Kc1=43.3. Increasing Kc2 stabilizes the controller, 
as is predicted. 
 
0 5 10 15 20 25 30
-0.8
-0.6
-0.4
-0.2
0
0.2
0.4
0.6
0.8
1
time
O
ut
pu
t
 
0 5 10 15 20 25 30
-2
0
2
4
6
8
10
12
x 10-3
time
O
ut
pu
t
 
Figure S16.2a. Responses for unit load change in D1 (left) and D2 (right) 
 
 
b) The characteristic equation for this system is 
 
1+Gc2GvGm2+Gc2GvGm1Gc1Gp = 0 (1) 
 
Let Gc1=Kc2 and Gc2=Kc2. Then, substituting all the transfer functions into 
(1), we obtain 
 
1)1()67()814(8 122223 +++++++ cccc KKsKsKs (2) 
 
Now we can use the Routh stability criterion. The Routh array is 
 
Row 1 8 267 cK+ 
Row 2 2814 cK+ )1(1 12 cc KK ++ 
Row 3 
2
212
2
2
47
4456624
c
cccc
K
KKKK
+
−++
 0 
Row 4 )1(1 12 cc KK ++ 
 
16-5 
0.00
20.00
40.00
60.00
80.00
100.00
120.00
140.00
160.00
0 5 10 15 20
Kc2
K c
1,
 
u
lti
m
a
te
Kc2 Kc1,u 
1 33.75 
2 34.13 
3 38.25 
4 43.31 
5 48.75 
6 54.38 
7 60.11 
8 65.91 
9 71.75 
10 77.63 
11 83.52 
12 89.44 
13 95.37 
14 101.30 
15 107.25 
16 113.20 
17 119.16 
18 125.13 
19 131.09 
20 137.06 
 
For 1 ≤ Kc2≤ 20, there is no impact on stability by the term 14+8Kc2 in the 
second row. The critical Kc1 is found by varying Kc2 from 1 to 20, and 
using 
 
04456624 212
2
2 ≥−++ cccc KKKK (3) 
0)1(1 12 ≥++ cc KK (4) 
 
Rearranging (3) and (4), we obtain 
 
2
2
2
2
1 4
456624
c
cc
c K
KK
K
++
≤ (5) 



 +
−≥
2
2
1
1
c
c
c K
K
K (6) 
 
Hence, for normal (positive) values of Kc1 and Kc2, 
 
2
2 2
1,
2
24 66 45
4
c c
c u
c
K KK
K
+ +
= 
The results are shown in the table and figure below. Note the nearly linear 
variation of Kc1 ultimate with Kc2. This is because the right hand side is 
very nearly 6 Kc2+16.5. For larger values of Kc2, the stability margin on 
Kc1 is higher. There don’t appear to be any nonlinear effects of Kc2 on Kc1, 
especially at high Kc2. 
 
There is no theoretical upper limit for Kc2, except that large values may 
cause the valve to saturate for small set-point or load changes. 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
Figure S16.2b. Effect of Kc2 on the critical gain of Kc1 
16-6 
c) With integral action in the inner loop, 
 
11 cc KG = 



+=
s
Gc 5
1152 
 
Substitution of all the transfer functions into the characteristic equation 
yields 
 
 1)05.0(1
5
5
115)2.0(
1
5
5
1151 cK
ssss +



++
+



++ 
 
 0)12)(14(
4
=
++ ss
 
 
 Rearrangement gives 
 
 01)512(45548 11234 =++++++ cc KsKsss 
 
The Routh array is: 
 
Row 1 8 45 11 cK+ 
 
Row 2 54 1512 cK+ 0 
Row 3 
27
201167 1cK−
 11 cK+ 
Row 4 
1
1
2
1
201167
125464137100
c
cc
K
KK
−
++−
 0 
 
Row 5 11 cK+ 
 
Using the Routh array analysis 
 
Row 3: 0201167 1 >− cK ∴ 35.581 <cK 
 01 1 >+ cK ∴ 11 −>cK 
 
Row 4: Since 1201167 cK− is already positive, 
 0125464137100 1
2
1 >++− cc KK 
 Solving for the positive root, we get 3.431 <cK 
 
16-7 
The ultimate 1cK is 43.3, which is the same resultas for proportional only 
control of the secondary loop. 
 
With integral action in the outer loop only, 
 



+=
s
KG cc 5
1111 
52 =cG 
 
Substituting the transfer functions into the characteristic equation. 
 
0)12)(14(
4
5
11)05.0(
1
55)2.0(
1
551 1 =
++



+
+
+
+
+
sss
K
ss
c 
 
∴ 0)56(37548 11234 =+++++ cc KsKsss 
 
The Routh array is 
 
Row 1 8 37 1cK 
 
Row 2 54 156 cK+ 0 
Row 3 
27
20975 1cK−
 1cK 
Row 4 
1
1
2
1
20975
58503297100
c
cc
K
KK
−
++−
 0 
 
Row 5 1cK 
 
Using the Routh array analysis, 
 
Row 3: 020975 >− ∴ 75.481 <cK 
 01 >cK 
 
Row 4: Since 120975 cK− is already positive, 
 058503297100 1
2
1 >++− cc KK 
 Solving for the positive root, we get 66.341 <cK 
 
Hence, Kc1<34.66 is the limiting constraint. Note that due to integral 
action in the primary loop, the ultimate controller gain is reduced. 
 
 
16-8 
Calculation of offset: 
 
For 1 1
1
11c c
I
G K
s
 
= + 
τ  , 22 cc KG = , 2( )Iτ = ∞ 
 
 
1 2 21
1
2 2 2 1 1
1
(1 )
11 1
d c v m
c v m c v m c p
I
G K G GY
D
K G G K G G K G
s
+
=  
+ + + 
τ 
 
 
1
1
( 0) 0Y s
D
= = 
 
Since Gc1 contains integral action, a step-change in D1 does not produce an 
offset in Y1. 
 
21
2
2 2 2 1 1
1
11 1
p d
c v m c v m c p
I
G GY
D
K G G K G G K G
s
=  
+ + + 
τ 
 
1
2
( 0) 0Y s
D
= = 
 
Thus, for the same reason as before, a step-change in D2 does not produce 
an offset in Y1. 
 
For 11 cc KG = (ie. 1 )Iτ = ∞ , 2 2
2
11c c
I
G K
s
 
= + 
τ  
 
1 2 2
21
1
2 2 2 1 1
2 2
1(1 1 )
1 11 1 1
d c v m
I
c v m c v m c p
I I
G K G G
sY
D
K G G K G G K G
s s
 
+ + 
τ 
=    
+ + + +   
τ τ   
 
1
1
( 0) 0Y s
D
= ≠ 
 
Therefore, when there is no integral action in the outer loop, a primary 
disturbance produces an offset. 
 
Thus, there is no offset for a step-change in the secondary disturbance. 
. 
 
 
 
16-9 
 
21
2
2 2 2 1 1
2 2
1 11 1 1
p d
c v m c v m c p
I I
G GY
D
K G G K G G K G
s s
=    
+ + + +   
τ τ   
 
 
1
2
( 0) 0Y s
D
= = 
 
 Thus, there is no offset for a step-change in the secondary disturbance. 
 
 
16.3 
 
 
a) Tuning the slave loop: 
 
The open-loop transfer function is 
 )1)(15)(12()(
2
+++
=
sss
K
sG c 
Since a proportional controller is used, a high Kc2 reduces the steady-state 
offset. The highest Kc2 which satisfies the bounds on the gain and phase 
margins is 5.3. For this Kc2, the gain margin is 2.38, and the phase margin 
is 30.7°. 
 
By using MATLAB, the Bode plot of G(s) with Kc2 = 5.3 is shown below. 
Bode Diagram
Frequency (rad/sec)
Ph
as
e 
(de
g)
M
ag
n
itu
de
 
(ab
s)
0
1
2
3
4
5
6
10-2 10-1 100 101
-270
-225
-180
-135
-90
-45
0
Gain margin graphical solution 
Phase margin graphical solution 
 
 Figure S16.3a. Bode plot for the inner open-loop; gain and phase margins. 
16-10 
b) Tuning the master loop: 
 
 The input-output transfer function of the inner loop is 
 
 
 
3.681710
)1(3.5)( 23 +++
+
=
sss
s
sG
ni (with Kc2 = 5.3) 
 
The ultimate gain Kc1,u can be found by simulation. In doing so, 
 
 Kc1,u = 3.2491 
 
The corresponding period of oscillation is 
 
Pu = 2pi/ω = 8.98 time units. 
 
 
 The Ziegler-Nichols tuning criteria for a PI-controller yield 
 
 Kc1 = Kc1,u / 2.2 = 1.48 
 
 48.72.1/1 == uI Pτ 
 
The closed-loop response with these tuning constant values (Kc1=1.48, 
1Iτ = 7.48 , Kc2 = 5.3) is shown below. 
0 10 20 30 40 50 60 70 80 90 100
0
0.2
0.4
0.6
0.8
1
1.2
1.4
Time
O
u
tp
u
t
 
 Fig S16.3b. Closed-loop response for a unit step set-point change. 
16-11 
16.4 
 
 
 For the inner controller (Slave controller), IMC tuning rules are used 
 
 2 3
2 2
1 (2 1)(5 1)( 1)
* ( 1)c c
s s sG
G s−
+ + +
= =
τ +
 
 
Closed-loop responses for different values of τc2 are shown below. A τc2 value of 
3 yields a good response. 
 
 
For the Master controller, 
 
 
−
=
1
1
1
*
G
Gc where 1 3
1
(2 1)(5 1)( 1) 1
( 1) (10 1)c
s s sG
s s
−
+ + +
=
τ + +
 
 
This higher-order transfer function is approximated by first order plus time delay 
using a step test: 
0 10 20 30 40 50 60
0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1
time
O
u
tp
u
t
 
Figure S16.4. Reaction curve for the higher order transfer function 
 
 Hence )132.15(
38.0
1 +
≈
−
−
s
eG
s
 
 
From Table 12.1: (PI controller, Case G): 
1
15.32
0.38c c
K =
τ +
 and 15.32iτ = 
Closed-loop responses are shown for different values of τc1. A τc1 value of 7 
yields a good response. 
16-12 
 
0 10 20 30 40 50 60
0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
time
O
u
tp
u
t
Tauc2=0.5
Tauc2=3
Tauc2=7
Tauc2=5
0 10 20 30 40 50 60 70 80 90 100
0
0.2
0.4
0.6
0.8
1
1.2
1.4
time
O
u
tp
u
t
Tauc1=3
Tauc1=5
Tauc1=7
 
 Figure S16.4b. Closed-loop response for τc2 Figure S16.4c. Closed-loop response for τc1 
 
Hence for the master controller, Kc = 2.07 and τI = 15.32 
 
 
16.5 
 
 
 
a) The T2 controller (TC-2) adjusts the set-point, T1sp, of the T1 controller (TC-
1). Its output signal is added to the output of the feedforward controller. 
 
 
 
 Figure S16.5a. Schematic diagram for the control system 
 
b) This is a cascade control system with a feedforward controller being used 
to help control T1. Note that T1 is an intermediate variable rather than a 
disturbance variable since it is affected by V1. 
16-13 
c) Block diagram: 
 
 
 Figure S16.5b. Block diagram for the control system in Exercise 16.5. 
 
 
 
16.6 
 
 
 a) For the inner loop, the characteristic equation reduces to: 
 
 0
3
11 =
−
+
+
s
sK inner � s − 3 + Kinner s + Kinner = 0 
 � s(1+ Kinner) –3 + Kinner = 0 
 Hence, 
inner
inner
K
K
s
+
−
=
1
3
 
The inner loop will be stable if this root is negative. Thus, we conclude 
that this loop will be stable if either Kinner>3 or Kinner<−1. 
b) The servo transfer function for the outer loop is: 
( ) ( )( )
( ) 1 ( ) ( ) ( )
c inner p
sp inner p c inner p
G s K G sY s
Y s K G s G s K G s
=
+ +
 
16-14 
The complex closed-loop poles arise when the characteristic polynomial 
is factored. This polynomial is 
 (s2 + s + 0.313) = (s + 0.5 + 0.25 i) (s + 0.5 −0.25i) 
11 11 6 6 0
3 3
I
c
I
ss sK
s s s
 τ ++ +
+ + = 
− τ −  
( ) 26 6I I c IK sτ + τ + τ� 
 ( 3 6 6 6 )I I I c cK K s+ − τ + τ + τ + 
 06 =+ cK 
The poles are also the roots of the characteristic equation: 
Hence, the PI controller parameters can be found easily: 
 052.0=cK 
 0.137Iτ = 
 
 
16.7 
 
 
Using MATLAB-Simulink, the block diagram for the closed-loop system 
is shown below. 
 
 
 
 Figure S16.7a. Block diagram for Smith predictor 
16-15 
where the blockrepresents the time-delay term e-θs. 
 
 
The closed-loop response for unit set-point and disturbance changes are 
shown below. Consider a PI controller designed by using Table 12.1(Case 
A) with τc = 3 and set Gd = Gp. Note that no offset occurs, 
0 5 10 15 20 25 30
0
0.2
0.4
0.6
0.8
1
1.2
time
O
ut
pu
t
Servo response
Regulatory response
 
 Figure S16.7b. Closed-loop response for setpoint and disturbance changes. 
 
 
 
16.8 
 
 
 The block diagram for the closed-loop system is 
 
 
 
Figure S6.8. Block diagram for the closed-loop system 
 
 where 1
1 1
s
pI
c c ps
I
K esG K and G
s e s
−θ
−θ
 + τ
= = 
+ τ − + τ  
16-16 
 a) 
 
1
1 1
1 11
1 1
s
I
c p s
c p I
s
sp c p I
c p s
I
s eK K
G G s e sY
Y G G s eK K
s e s
−θ
−θ
−θ
−θ
 + τ + τ − + τ 
= =
+  + τ
+  + τ − + τ 
 
Since 
p
c K
K 1= and τI = τ 
 
1
1
1
1
s
s s
I
s ss
sp I
s
I
e
s eY e
Y s e ee
s e
−θ
−θ
−θ
−θ −θ
−θ
−θ
  + τ − 
= =
+ τ − + 
+  + τ − 
 
 
 Hence dead-time is eliminated from characteristic equation: 
 
 
1
s
sp I
Y e
Y s
−θ
=
+ τ
 
 
 
b) The closed-loop response will not exhibit overshoot, because it is a first 
order plus dead-time transfer function. 
 
 
16.9 
 
For a first-order process with time delay, use of a Smith predictor and 
proportional control should make the process behave like a first-order 
system, i.e., no oscillation. In fact, if the model parameters are accurately 
known, the controller gain can be as large as we want, and no oscillations 
will occur. 
Appelpolscher has verified that the process is linear, however it may not 
be truly first-order. If it were second-order (plus time delay), proportional 
control would yield oscillations for a well-tuned system. Similarly, if there 
are errors in the model parameters used to design the controller even when 
the actual process is first-order, oscillations can occur. 
 
 
 
 
16-17 
16.10 
 
 
a) Analyzing the block diagram of the Smith predictor 
1 (1 )
s
c p
s s
sp c p c p
G G eY
Y G G e G G e
−θ
−θ −θ
′
=
′ ′+ − +
�
�
 
 
1
s
c p
s s
c p c p c p
G G e
G G G G e G G e
−θ
−θ −θ
′
=
′ ′ ′+ + −
�
� �
 
Note that the last two terms of the denominator can when pp GG ′=′
~
 and 
θ = θ� 
The characteristic equation is 
1 0s sc p c p c pG G G G e G G e
−θ −θ
′ ′ ′= + + − =
�
� �
 
 
b) The closed-loop responses to step set-point changes are shown below for 
 the various cases. 
 
 
 
Figure S16.10a. Simulink diagram block; base case 
 
16-18 
0 5 10 15 20 25 30 35 40 45 50
0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1
time
O
u
tp
u
t
 
0 5 10 15 20 25 30 35 40 45 50
0
0.2
0.4
0.6
0.8
1
1.2
1.4
time
O
u
tp
ut
 
Figure S16.10b. Base case Figure S16.10c. Kp = 2.4 
 
0 5 10 15 20 25 30 35 40 45 50
0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1
time
O
u
tp
u
t
 
0 5 10 15 20 25 30 35 40 45 50
0
0.2
0.4
0.6
0.8
1
1.2
1.4
time
O
ut
pu
t
 
Figure S16.10d. Kp = 1.6 Figure S16.10e. τ = 6 
 
0 5 10 15 20 25 30 35 40 45 50
0
0.2
0.4
0.6
0.8
1
1.2
1.4
time
O
u
tp
u
t
 
0 5 10 15 20 25 30 35 40 45 50
-6
-4
-2
0
2
4
6
time
O
ut
pu
t
 
Figure S16.10f. τ = 4 Figure S16.10g. θ=2.4 
16-19 
 
0 5 10 15 20 25 30 35 40 45 50
-25
-20
-15
-10
-5
0
5
10
15
20
25
time
O
ut
pu
t
 
 Figure S16.10h. θ = 1.6 
It is immediately evident that errors in time-delay estimation are the most 
serious. This is because the terms in the characteristic equation which 
contain dead-time do not cancel, and cause instability at high controller 
gains. 
When the actual process time constant is smaller than the model time 
constant, the closed–loop system may become unstable. In our case, the 
error is not large enough to cause instability, but the response is more 
oscillatory than for the base (perfect model) case. The same is true if the 
actual process gain is larger than that of the model. If the actual process 
has a larger time constant, or smaller gain than the model, there is no 
significant degradation in closed loop performance (for the magnitude of 
the error, ± 20% considered here). Note that in all the above simulations, 
the model is considered to be 
15
2 2
+
−
s
e
s
 and the actual process parameters 
have been assumed to vary by ± 20% of the model parameter values. 
 
c) The proportional controller was tuned so as to obtain a gain margin of 2.0. 
This resulted in Kc = 2.3. The responses for the various cases are shown 
below 
0 5 10 15 20 25 30 35 40 45 50
0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
time
O
ut
pu
t
 
0 5 10 15 20 25 30 35 40 45 50
0
0.2
0.4
0.6
0.8
1
1.2
1.4
time
O
ut
pu
t
 
 Base case Kp = 3 
16-20 
0 5 10 15 20 25 30 35 40 45 50
0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
time
O
ut
pu
t
 
0 5 10 15 20 25 30 35 40 45 50
0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
time
O
ut
pu
t
 
 Kp = 1 τ = 1 
 
0 5 10 15 20 25 30 35 40 45 50
0
0.2
0.4
0.6
0.8
1
1.2
1.4
time
O
ut
pu
t
 
0 5 10 15 20 25 30 35 40 45 50
0
0.2
0.4
0.6
0.8
1
1.2
1.4
time
O
ut
pu
t
 
 τ = 2.5 θ = 3 
0 5 10 15 20 25 30 35 40 45 50
0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
 
 θ = 1 
 
 Nyquist plots were prepared for different values of Kp, τ and θ, and 
 checked to see if the stability criterion was satisfied. The stability regions 
 when the three parameters are varied one to time are. 
Kp ≤ 4.1 (τ = 5 , θ = 2) 
τ ≥ 2.4 (Kp=2 , θ = 2) 
θ ≤ 0.1 and 1.8 ≤ θ ≤ 2.2 (Kp = 2 , τ = 5) 
16-21 
16.11 
 
 
 From Eq. 16-24, 
 
 
( )( )1 1
1
s
d c
c
G G G eY
D G G
∗ −θ
∗
+ −
=
+
 
 
 that is, 
 
( )3 32 21 1
21
s sc c I
I
c c I
I
K K s
e e
s s sY
K K sD
s s
− −
 + τ
+ − 
τ 
=
+ τ
+
τ
 
 
 Using the final value theorem for a step change in D: 
 
 )(lim)(lim
0
ssYty
st →∞→
= 
 
 then 
 
 
00
lim)(lim
→→
=
ss
ssY
( )3 32 21 1
1
21
s sc c I
I
c c I
I
K K s
e e
s s s
s K K s s
s s
− −
 + τ
+ − 
τ 
+ τ
+
τ
 
 
 
0
lim
→
=
s
( )3 32 2( ) 1
2( )
s s
I c c I
I c c I
e s K K s e
s s
s K K s
s
− −
 
τ + + τ −  
τ + + τ
 
 
 Multiplying both numerator and denominator by s2, 
 
 
0
lim
→
=
s
( )( )3 2 3
3
2 ( )2 1
( )2
s s
I c c I
I c c I
e s K K s e
s K K s s
− −τ + + τ −
τ + + τ
 
 
 Applying L'Hopital's rule: 
 
 
0
lim
→
=
s
 
( )( )3 2 3
2
6 ( )2 1
3 2( 2 )
s s
I c c I
I c c I
e s K K s e
s K K s
− −
− τ + + τ −
τ + + τ
 
+ 
3 3 3 3
2
2 (2 6 2 2 6 )
3 2( 2 )
s s s s
I c c I c I c I
I c c I
e s K e K K e K se
s K K s
− − − −τ + + τ −τ + τ
τ + + τ
 = 6 
16-22 
Therefore 
 
)(lim)(lim
0
ssYty
st →∞→
= = 6 
 
 and the PI control will not eliminate offset. 
 
 
16.12 
 
 
 For a Smith predictor, we have the following system 
 Figure S16.12. Smith Predictor diagram block 
where the process model is Gp(s) = Q(s) e-θs 
For this system, 
pc
pc
sp GG
GG
Y
Y
′+
′
=
1
 
where Gc’ is the transfer function for the system in the dotted box. 
1 (1 )
c
c s
c
GG
G Q e−θ′ = + − 
1 (1 )
1
1 (1 )
c p
s
c
c psp
s
c
G G
G Q eY
G GY
G Q e
−θ
−θ
+ −
∴ =
+
+ −
 
Simplification gives 
16-23 
( )
1
s
sc
sp c
G QeY P s e
Y G Q
−θ
−θ
= =
+
 
where ( )
1
c
c
G QP s
G Q= + 
If P(s) is the desired system performance (after the time delay has elapsed) 
under feedback control, then we can solve for Gc in terms of P(s). 
 
 ))(1)((
)(
sPsQ
sPGc
−
= 
 
 The IMC controller requires that we define 
 
sG e−θ+ =� 
 )(~ sQG =
−
 (the invertible part of Gp) 
 Let the filter for the controller be f(s) = 1
1F sτ +
 
 Therefore, the controller is 
 )(
)()(~ 1
sQ
sf
sfGGc == −− 
The closed-loop transfer function is 
 
 
1
s
c p
sp F
Y eG G G f
Y s
−θ
+= = =+ τ
�
 
 
Note that this is the same closed-loop form as analyzed in part (a), which 
led to a Smith Predictor type of controller. Hence, the IMC design also 
provides time-delay compensation. 
 
 
 
 
16-24 
16.13 
 
Referring to Example 4.8, if q, the flowrate, and Ti, the inlet temperature, 
are know and are constant, then the Laplace transform models in (4-79) 
and (4-80) are 
 )()()( 1211 sTasCas A ′=′− (4-79) 
 )()()()( 21222 sTbsCasTas sA ′+′=′− (4-80) 
where ( )sT s′ is the coolant temperature. Using Eq. 4-86, we can directly 
compute concentration from the temperature signal, i.e., 
 )()(
11
12 sT
as
a
sCA ′
−
=′ 
which is a first-order filter operating on ( )T s′ 
So inferential control of concentration using temperature would be 
feasible in this case. If q and Ti varied, a more general expression for the 
linearized model would be necessary, but there would still be a direct way 
to infer CA from T. 
 
 
16.14 
 
One possible solution would be to use a split range valve to handle the 
100≤ p≤ 200 and higher pressure ranges. Moreover, a high-gain controller 
with set-point = 200 psi can be used for the vent valve. This valve would 
not open while the pressure is less than 200 psi, which is similar to how a 
selector operates. 
Stephanopoulos (Chemical Process Control, Prentice-Hall, 1989) has 
described many applications for this so-called split-range control. A 
typical configuration consists of 1 controller and 2 final control elements 
or valves. 
 
 
 
 
16-25 
 
 Figure S16.14. Process instrumentation diagram 
 
 
16.15 
The amounts of air and fuel are changed in response to the steam pressure. 
If the steam pressure is too low, a signal is sent to increase both air and 
fuel flowrates, which in turn increases the heat transfer to the steam. 
Selectors are used to prevent the possibility of explosions (low air-fuel 
ratio). If the air flowrate is too low, the low selector uses that 
measurement as the set-point for the fuel flow rate controller. If the fuel 
flowrate is too high, its measurement is selected by the high selector as the 
set-point for the air flow controller. This also protects against dynamic 
lags in the set-point response. 
 
 
16.16 
 
 Figure S16.16. Control condensate temperature in a reflux drum 
CONDENSATE
FC
L
 T T
 T T
TC
TC
COOLING
WATER
REACTOR
PT
INLET OUTLET
VENT
SPLIT RANGE
CONTROLLER
16-26 
16.17 
 
 
 Supposing a first-order plus dead time process, the closed-loop transfer 
 function is 
 ( )
1
c p
CL
c p
G G
G s
G G
=
+
 ∴ 
11
( 1)( )
11
1 ( 1)
s
D
I
c p
p
CL
s
D
I
c p
p
s e
s
K K
s
G s
s e
s
K K
s
−θ
−θ
 
+ + τ 
τ 
τ +
=  
+ + τ 
τ +
τ +
 
Notice that Kc and Kp always appear together as a product. Hence, if we 
want the process to maintain a specified performance (stability, decay 
ratio specification, etc.), we should adjust Kc such that it changes inversely 
with Kp; as a result, the product KcKp is kept constant. Also note, that since 
there is a time delay, we should adjust Kc based upon the future estimate 
of Kp: 
 ( )
ˆ ( )
ˆ ( )
c p c p
c
p
K K K K
K t bK t a
M t
= =
+ θ +
+ θ
 
 where ˆ ( )pK t + θ is an estimate of Kp θ time units into the future. 
 
 
16.18 
 
 
This is an application where self-tuning control would be beneficial. In 
order to regulate the exit composition, the manipulated variable (flowrate) 
must be adjusted. Therefore, a transfer function model relating flowrate to 
exit composition is needed. The model parameters will change as the 
catalyst deactivates, so some method of updating the model (e.g., periodic 
step tests) will have to be derived. The average temperature can be 
monitored to determine a significant change in activation has occurred, 
thus indicating the need to update the model. 
 
 
 
16-27 
16.19 
 
 a) 1
1 1
c p
c p c
G G
G G s
=
+ τ +
 
1
1 1 1
11
1
c
c
p c
p
c
sG
G s
G
s
τ +
∴ = =
τ 
− 
τ + 
 
 Substituting for Gp 
 
2
1 2 1 2( ) 11( )c
c p
s sG s
s K
τ τ + τ + τ +
=
τ
 1 2 1 2
1 1( )
p c
s
K s
 
= τ + τ + τ τ + τ   
Thus, the PID controller tuning constants are 
1 2( )
c
p c
K
K
τ + τ
=
τ
 
1 2Iτ = τ + τ 
1 2
1 2
D
τ τ
τ =
τ + τ
 
 (See Eq. 12-14 for verification) 
b) For τ1 = 3 and τ2 = 5 and τc = 1.5, we have 
 Kc = 5.333 τI = 8.0 and τD = 1.875 
Using this PID controller, the closed-loop response will be first order 
when the process model is known accurately. The closed-loop response to 
a unit step-change in the set-point when the model is known exactly is 
shown above. It is assumed that τc was chosen such that the closed loop 
response is reasonable, and the manipulated variable does not violate any 
bounds that are imposed. An approximate derivative action is used by 
Simulink-MATLAB, namely 
1
Ds
s
τ
+ β when β=0.01 
 
16-28 
 
 
 Figure S16.19a. Simulink block diagram. 
 
0 5 10 15 20 25 30 35 40
0
0.2
0.4
0.6
0.8
1
1.2
1.4
time
O
ut
pu
t
 
0 5 10 15 20 25 30 35 40
-200
0
200
400
600
800
1000
1200
time
M
an
ip
ul
at
ed
 
v
ar
ia
bl
e
 
 Figure S16.19b. Output (no model error) Figure S16.19c. Manipulated variable (no 
model error) 
0 5 10 15 20 25 30 35 40
0
0.2
0.4
0.6
0.8
1
1.2
1.4
time
O
ut
pu
t
 
0 5 10 15 20 25 30 35 40
-200
0
200
400
600
800
1000
1200
time
M
an
ip
ul
at
ed
 
v
ar
ia
bl
e
 
 Figure S16.19d. Output (Kp = 2) Figure S16.19e. Manipulated variable 
 (Kp = 2) 
16-29 
0 5 10 15 20 25 30 35 40
0
0.2
0.4
0.6
0.8
1
1.2
1.4
time
O
ut
pu
t
 
0 5 10 15 20 25 30 35 40
0
200
400
600
800
1000
1200
time
M
an
ip
u
la
te
d 
v
ar
ia
bl
e
 
Figure S16.19f. Output (Kp = 0.5) Figure S16.19g. Manipulated variable 
(Kp =0.5) 
0 5 10 15 20 25 30 35 40
0
0.20.4
0.6
0.8
1
1.2
1.4
time
O
ut
pu
t
 
0 5 10 15 20 25 30 35 40
0
200
400
600
800
1000
1200
time
M
an
ipu
la
te
d 
v
ar
ia
bl
e
 
 Figure S16.19h. Output (τ2 = 10) Figure S16.19i. Manipulated variable 
 (τ2 = 10) 
0 5 10 15 20 25 30 35 40
0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1
time
O
ut
pu
t
 
0 5 10 15 20 25 30 35 40
-200
0
200
400
600
800
1000
1200
time
M
s
an
ip
ul
at
e
d 
va
ria
bl
e
 
 Figure E16.9 j.- Output (τ2 = 1) Figure E16.9 k.- Manipulated variable 
(τ2 = 1) 
 
(1) The closed-loop response when the actual Kp is 2.0 is shown above. The 
controlled variable reaches its set-point much faster than for the base case 
(exact model), but the manipulated variable assumes values that are more 
negative (for some period of time) than the base case. This may violate 
some bounds. 
16-30 
(2) When Kp = 0.5, the response is much slower. In fact, the closed-loop time 
constant seems to be about 3.0 instead of 1.5. There do not seem to be any 
problems with the manipulated variable. 
 
 
(3) If (τ2 = 10), the closed-loop response is no longer first-order. The settling 
time is much longer than for the base case. The manipulated variable does 
not seem to violate any bounds. 
 
(4) Both the drawbacks seen above are observed when τ2 = 1. The settling 
time is much longer than for the base case. Also the rapid initial increase 
in the controlled variable means that the manipulated variable drops off 
sharply, and is in danger of violating a lower bound. 
 
 
 
 
16.20 
 
Based on discussions in Chapter 12, increasing the gain of a controller 
makes it more oscillatory, increasing the overshoot (peak error) as well as 
the decay ratio. Therefore, if the quarter-decay ratio is a goal for the 
closed-loop response (e.g., Ziegler-Nichols tuning), then the rule proposed 
by Appelpolscher should be satisfactory from a qualitative point of view. 
However, if the controller gain is increased, the settling time is also 
decreased, as is the period of oscillation. Integral action influences the 
response characteristics as well. In general, a decrease in τI gives 
comparable results to an increase in Kc. So, Kc can be used to influence the 
peak error or decay ratio, while τI can be used to speed up the settling time 
(a decrease in τI decreases the settling time). See Chapter 8 for typical 
response for varying Kc and τI. 
 
 
16.21 
 
 SELECTIVE CONTROL 
Selectors are quite often used in forced draft combustion control system to 
prevent an imbalance between air flow and fuel flow, which could result 
in unsafe operating conditions. 
For this case, a flow controller adjusts the air flowrate in the heater. Its set-
point is determined by the High Selector, which chooses the higher of the 
two input signals: 
 .- Signal from the fuel gas flowrate transmitter (when this is too high) 
16-31 
 .- Signal from the outlet temperature control system. 
Similarly, if the air flow rate is too low, its measurement is selected by the 
low selector as the set-point for the fuel-flow rate. 
CASCADE CONTROLLER 
The outlet temperature control system can be considered the master 
controller that adjusts the set-point of the fuel/air control system (slave 
controller). If a disturbance in fuel or air flow rate exists, the slave control 
system will act very quickly to hold them at their set-points. 
FEED-FORWARD CONTROL 
The feedforward control scheme in the heater provides better control of 
the heater outlet temperature. The feed flowrate and temperature are 
measured and sent to the feedback control system in the outflow. Hence 
corrective action is taken before they upset the process. The outputs of the 
feedforward and feedback controller are added together and the combined 
signal is sent to the fuel/air control system. 
 
 
16.22 
 
 ALTERNATIVE A.- 
 
Since the control valves are "air to close", each Kv is positive (cf. Chapter 
9). Consequently, each controller must be reverse acting (Kc>0) for the 
flow control loop to function properly. 
 
Two alternative control strategies are considered: 
 
Method 1: use a default feed flowrate when Pcc > 80% 
 
 Let : Pcc = output signal from the composition controller (%) 
 =spF
~ (internal) set point for the feed flow controller (%) 
 
 Control strategy: 
 
 If Pcc > 80% , =spF
~
lowspF ,
~
 
 
 where lowspF ,
~
 is a specified default flow rate that is lower than the normal 
 value, nomspF
~
. 
16-32 
 Method 2: Reduce the feed flow when Pcc > 80% 
 
 Control strategy: 
 
 If Pcc < 80%, =spF
~
nomspF
~
 − K(Pcc – 80%) 
 
 where K is a tuning parameter (K > 0) 
 
 Implementation: 
 Note: A check should be made to ensure that 0 ≤ spF
~
 ≤ 100% 
 ALTERNATIVE B.- 
 A selective control system is proposed: 
 
 Figure S16.22. Proposed selective control system 
Both control valves are A-O and transmitters are “direct acting”, so the 
controller have to be “reverse acting”. 
When the output concentration decreases, the controller output increases. 
Hence this signal cannot be sent directly to the feed valve (it would open 
V--1
CT
CC
V--2FT
> FC
HS
Pcc
80 % Fnom
-
+ K -+
80 %
~
Fsp
~
16-33 
the valve). Using a high selector that chooses the higher of these signals 
can solve the problem 
 
 .- Flow transmitter 
 .- Output concentration controller 
Therefore when the signal from the output controller exceeds 80%, the 
selector holds it and sends it to the flow controller, so that feed flow rate is 
reduced. 
 
 
16.23 
 
ALTERNATIVE A.- 
 
Time delay.- Use time delay compensation, e.g., Smith Predictor 
 
Variable waste concentration.- Tank pH changes occurs due to this 
unpredictable changes. Process gain changes also (c,f. literature curve for 
strong acid-strong base) 
 
Variable waste flow rate.- Use FF control or ratio qbase to qwaste. 
 
Measure qbase .- This suggests you may want to use cascade control to 
compensate for upstream pressure changes, etc 
ALTERNATIVE B.- 
Several advanced control strategies could provide improved process 
control. A selective control system is commonly used to control pH in 
wastewater treatment .The proposed system is shown below (pH T = pH 
sensor; pH C = pH controller) 
 Figure S16.23. Proposed selective control system. 
`
V--3V--4
T-1
pH CpH T
SFC
FTFT
FCS
16-34 
where S represents a selector ( < or >, to be determined) 
In this scheme, several manipulated variables are used to control a single 
process variable. When the pH is too high or too low, a signal is sent to the 
selectors in either the waste stream or the base stream flowrate controllers. 
The exactly configuration of the system depends on the transmitter, 
controller and valve gains. 
In addition, a Smith Predictor for the pH controller is proposed due to the 
large time delay. There would be other possibilities for this process such 
as an adaptive control system or a cascade control system. However the 
scheme above may be good enough 
 Necessary information: 
.- Descriptions of measurement devices, valves and controllers; direct 
action or reverse action. 
.- Model of the process in order to implement the Smith Predictor 
 
 
16.24 
 
 
For setpoint change, the closed-loop transfer function with an integral 
controller and steady state process (Gp = Kp) is: 
 
 
1
1
11 1 1
P
C P I P
sp IC P I PP
I P
KG G s KY
Y G G s KK ss K
τ
= = ==
τ+ τ ++ +τ
 
Hence a first order response is obtained and satisfactory control can be 
achieved. 
 
 
For disturbance change (Gd = Gp): 
 
( )
11 1 1
d P P I I
IC P I PP
I P
G K K s sY
D G G s KK ss K
τ τ
= = = =
τ+ τ ++ +τ
 
 
Therefore a first order response is also obtained for disturbance change. 
 
 
17-1 
��������	
�
 
 
 
 
 17.1 
 
 
 Using Eq. 17-9, the filtered values of xD are shown in Table S17.1 
 
 
time(min) α = 1α = 1α = 1α = 1 α = α = α = α = 0.8 α = α = α = α = 0.5 
0 0 0 0 
1 0.495 0.396 0.248 
2 0.815 0.731 0.531 
3 1.374 1.245 0.953 
4 0.681 0.794 0.817 
5 1.889 1.670 1.353 
6 2.078 1.996 1.715 
7 2.668 2.534 2.192 
8 2.533 2.533 2.362 
9 2.908 2.833 2.635 
10 3.351 3.247 2.993 
11 3.336 3.318 3.165 
12 3.564 3.515 3.364 
13 3.419 3.438 3.392 
14 3.917 3.821 3.654 
15 3.884 3.871 3.769 
16 3.871 3.871 3.820 
17 3.924 3.913 3.872 
18 4.300 4.223 4.086 
19 4.252 4.246 4.169 
20 4.409 4.376 4.289 
 
 Table S17.1. Unfiltered and filtered data. 
To obtain the analytical solution for xD, set 
s
sF 1)( = in the given transfer 
function, so that 
 
5 5 1 1( ) ( ) 5
10 1 (10 1) 1 10DX s F ss s s s s
 
= = = − 
+ + +  
Taking inverse Laplace transform 
 
 xD(t) = 5 (1 − e-t/10) 
 
A graphical comparison is shown in Fig. S17.1 
 
Solution Manual for Process Dynamics and Control, 2nd edition, 
Copyright © 2004 by Dale E. Seborg, Thomas F. Edgar and Duncan A. Mellichamp 
17-2 
0 2 4 6 8 10 12 14 16 18 20
0
0.5
1
1.5
2
2.5
3
3.5
4
4.5
time (min)
noisy data
alpha = 0.5
alpha = 0.8
analytical solution
 X D 
 
Fig S17.1. Graphical comparison for noisy data, filtered data and analytical 
 solution. 
 
As α decreases, the filtered data give a smoother curve compared to the 
no-filter (α=1) case, but this noise reduction is traded off with an increase 
in the deviation of the curve from the analytical solution. 
 
 
 17.2 
 
 
 The exponential filter output in Eq. 17-9 is 
 
 ( ) ( ) (1 ) ( 1)F m Fy k y k y k= α + − α − (1) 
 
 Replacing k by k-1 in Eq. 1 gives 
 
 ( 1) ( 1) (1 ) ( 2)F m Fy k y k y k− = α − + − α − (2) 
 
 Substituting for ( 1)Fy k − from (2) into (1) gives 
 
 
2( ) ( ) (1 ) ( 1) (1 ) ( 2)F m m Fy k y k y k y k= α + − α α − + − α − 
 
 Successive substitution of ( 2)Fy k − , ( 3)Fy k − ,… gives the final form 
 
1
0
( ) (1 ) ( ) (1 ) (0)
k
i k
F m F
i
y k y k i y
−
=
= − α α − + − α∑ 
17-3 
 17.3 
 
 
Table S17.3 lists the unfiltered output and, from Eq. 17-9, the filtered data 
for sampling periods of 1.0 and 0.1. Notice that for sampling period of 0.1, 
the unfiltered and filtered outputs were obtained at 0.1 time increments, 
but they are reported only at intervals of 1.0 to preserve conciseness and 
facilitate comparison. 
 
The results show that for each value of ∆t, the data become smoother as α 
decreases, but at the expense of lagging behind the mean output y(t)=t. 
Moreover, lower sampling period improves filtering by giving smoother 
data and less lagg for the same value of α. 
 
 
 
 ∆∆∆∆t=1 ∆∆∆∆t=0.1 
t αααα=1 αααα=0.8 αααα=0.5 αααα=0.2 αααα=0.8 αααα=0.5 αααα=0.2 
0 0 0 0 0 0 0 0 
1 1.421 1.137 0.710 0.284 1.381 1.261 0.877 
2 1.622 1.525 1.166 0.552 1.636 1.678 1.647 
3 3.206 2.870 2.186 1.083 3.227 3.200 2.779 
4 3.856 3.659 3.021 1.637 3.916 3.973 3.684 
5 4.934 4.679 3.977 2.297 4.836 4.716 4.503 
6 5.504 5.339 4.741 2.938 5.574 5.688 5.544 
7 6.523 6.286 5.632 3.655 6.571 6.664 6.523 
8 8.460 8.025 7.046 4.616 8.297 8.044 7.637 
9 8.685 8.553 7.866 5.430 8.688 8.717 8.533 
10 9.747 9.508 8.806 6.293 9.741 9.749 9.544 
11 11.499 11.101 10.153 7.334 11.328 11.078 10.658 
12 11.754 11.624 10.954 8.218 11.770 11.778 11.556 
13 12.699 12.484 11.826 9.115 12.747 12.773 12.555 
14 14.470 14.073 13.148 10.186 14.284 14.051 13.649 
15 14.535 14.442 13.841 11.055 14.662 14.742 14.547 
16 15.500 15.289 14.671 11.944 15.642 15.773 15.544 
17 16.987 16.647 15.829 12.953 16.980 16.910 16.605 
18 17.798 17.568 16.813 13.922 17.816 17.808 17.567 
19 19.140 18.825 17.977 14.965 19.036 18.912 18.600 
20 19.575 19.425 18.776 15.887 19.655 19.726 19.540 
 
 Table S17.3. Unfiltered and filtered output for sampling periods of 1.0 and 0.1 
 
 
 
 
 
 
 
 
 
17-4 
 Graphical comparison: 
 
0 2 4 6 8 10 12 14 16 18 20
0
2
4
6
8
10
12
14
16
18
20
time, t
y(t
)
α =1
α = 0.8
α = 0.5
α = 0.2
 
 Figure S17.3a. Graphical comparison for ∆t = 1.0 
 
 
 
0 2 4 6 8 10 12 14 16 18 20
0
2
4
6
8
10
12
14
16
18
20
time, t
y(t
)
α=1
α=0.8
α=0.5
α=0.2
 
 Figure S17.3b. Graphical comparison for ∆t = 0.1 
 
 
17-5 
 17.4 
 
 
Using Eq. 17-9 for α = 0.2 and α = 0.5, Eq. 17-18 for N* = 4, and Eq. 17-
19 for ∆y=0.5, the results are tabulated and plotted below. 
 
 
 (a) (a) (b) (c) 
t αααα=1 αααα=0.2 αααα=0.5 N*=4 ∆∆∆∆y=0.5 
0 0 0 0 0 0 
1 1.50 0.30 0.75 0.38 0.50 
2 0.30 0.30 0.53 0.45 0.30 
3 1.60 0.56 1.06 0.85 0.80 
4 0.40 0.53 0.73 0.95 0.40 
5 1.70 0.76 1.22 1.00 0.90 
6 1.50 0.91 1.36 1.30 1.40 
7 2.00 1.13 1.68 1.40 1.90 
8 1.50 1.20 1.59 1.68 1.50 
 
 
 Table S17.4. Unfiltered and filtered data. 
 
 
0 1 2 3 4 5 6 7 8
0
0.5
1
1.5
2
2.5
time, t
y(t
)
α=1 
α=0.2 
α=0.5 
 N*=4
∆ y=0.5
 
 Figure S17.4. Graphical comparison for filtered data and the raw data. 
 
 
 
 
 
17-6 
 17.5 
 
 
 Let C denote the controlled output. Then 
 
 
( )
( ) 1
d
c v p m F
GC s
d s K G G G G
=
+
 , 
1
1)( 2 += ssd 
 
 For τF = 0, GF = 1 and 
 
 )15/(11
)15/(1)(
++
+
=
s
s
sC 2 2
1 1
1 (5 2)( 1)s s s=+ + + 
 
 For τF = 3, GF = 1/(3s+1) and 
 
 [ ][ ]
1/(5 1)( )
1 1/(5 1) 1/(3 1)
sC s
s s
+
=
+ + + )1)(2815(
13
1
1
222 +++
+
=
+ sss
s
s
 
 
 
 By using Simulink-MATLAB, 
 
0 2 4 6 8 10 12 14 16 18 20
-0.4
-0.3
-0.2
-0.1
0
0.1
0.2
0.3
time. t
y(t
)
No filtering
Exponetial filtering
 
Figure S17.5. Closed-loop response comparison for no filtering and for 
an exponential filter (τF = 3 min) 
17-7 
 17.6 
 
1 1 1( ) ( )
1 1
Y s X s
s s s
= =
+ +
 , then y(t) = 1 − e-t 
 
For noise level of ± 0.05 units, several different values of α are tried in 
Eq. 17-9 as shown in Fig. S17.6a. While the filtered output for α = 0.7 is 
still quite noisy, that for α = 0.3 is too sluggish. Thus α = 0.4 seems to 
offer a good compromise between noise reduction and lag addition. 
Therefore, the designed first-order filter for noise level ± 0.05 units is α = 
0.4, which corresponds to τF = 1.5 according to Eq. 17-8a. 
 
 Noise level = ± 0.05 
0 2 4 6 8 10 12 14 16 18 20
0
0.2
0.4
0.6
0.8
1
1.2
1.4
t
y(t
)
α=1.0
α=0.7
α=0.4
α=0.3
 
 Figure S17.6a. Digital filters for noise level = ± 0.05 
 
 Noise level = ± 0.1 
0 5 10 15 20
0
0.2
0.4
0.6
0.8
1
1.2
1.4
t
y(t
)
α=1
α=0.3
α=0.2
α=0.15
 
 Figure S17.6b. Digital filters for noise level = ± 0.1 
17-8 
Noise level = ± 0.01 
0 5 10 15 20
0
0.2
0.4
0.6
0.8
1
1.2
1.4
t
y(t
)
α=1
 
Figure S17.6c. Response for noise level = ± 0.01; no filter needed. 
 
Similarly, for noise level of ± 0.1 units, a good compromise isα =0.2 or τF 
= 4.0 as shown in Fig. S17.6b. However, for noise level of ±0.01 units, no 
filter is necessary as shown in Fig. S17.6c. thus α=1.0, τF = 0 
 
 
17.7 
 
 y(k) = y(k-1) − 0.21 y(k-2) + u(k-2) 
 
k u(k) u(k-1) u(k-2) y(k) 
0 1 0 0 0 
1 0 1 0 0 
2 0 0 1 1.00 
3 0 0 0 1.00 
4 0 0 0 0.79 
5 0 0 0 0.58 
6 0 0 0 0.41 
7 0 0 0 0.29 
8 0 0 0 0.21 
9 0 0 0 0.14 
10 0 0 0 0.10 
11 0 0 0 0.07 
12 0 0 0 0.05 
13 0 0 0 0.03 
14 0 0 0 0.02 
15 0 0 0 0.02 
16 0 0 0 0.01 
17 0 0 0 0.01 
18 0 0 0 0.01 
19 0 0 0 0.00 
17-9 
 Plotting this results 
 
0 2 4 6 8 10 12 14 16 18 20
0
0.2
0.4
0.6
0.8
1
1.2
k 
y 
 
 Fig S17.7. Graphical simulation of the difference equation 
 
 The steady state value of y is zero. 
 
 
17.8 
 
 
a) By using Simulink and STEM routine to convert the continuous signal to a 
series of pulses, 
0 5 10 15 20 25 30 35 40 45 50
0
2
4
6
8
10
12
time
T
m
'(
t)
 
 Figure S17.8. Discrete time response for the temperature change. 
 
Hence the maximum value of the logged temperature is 80.7° C. 
This maximum point is reached at t = 12 min. 
17-10 
17.9 
 
 
 a) 
 
1 1
2 2
( ) 2.7 ( 3) 2.7 8.1
( ) 0.5 0.06 0.5 0.06
Y z z z z
U z z z z z
− −+ +
= =
− + − +
 
 
 Dividing both numerator and denominator by z2 
 
 
2 3
1 2
( ) 2.7 8.1
( ) 1 0.5 0.06
Y z z z
U z z z
− −
− −
+
=
− +
 
 
 Then 1 2 2 3( )(1 0.5 0.06 ) ( )(2.7 8.1 )Y z z z U z z z− − − −− + = + 
 
or y(k) = 0.5y(k-1) − 0.06y(k-2) + 2.7u(k-2) + 8.1u(k-3) 
 
 
The simulation of the difference equation yields 
 
 
k u(k) u(k-2) u(k-3) y(k) 
0 1 0 0 0 
1 1 0 0 0 
2 1 1 0 2.70 
3 1 1 1 12.15 
4 1 1 1 16.71 
5 1 1 1 18.43 
6 1 1 1 19.01 
7 1 1 1 19.20 
8 1 1 1 19.26 
9 1 1 1 19.28 
10 1 1 1 19.28 
11 1 1 1 19.28 
12 1 1 1 19.29 
13 1 1 1 19.29 
14 1 1 1 19.29 
15 1 1 1 19.29 
16 1 1 1 19.29 
17 1 1 1 19.29 
18 1 1 1 19.29 
19 1 1 1 19.29 
20 1 1 1 19.29 
 
 
 
 
 
17-11 
 b) 
 
0 2 4 6 8 10 12 14 16 18 20
0
2
4
6
8
10
12
14
16
18
20
∆
Difference equation
Simulink
k t 
y 
 
 Fig S17.9. Simulink response to a unit step change in u 
 
c) The steady state value of y can be found be setting z =1. In doing so, 
 
y =19.29 
 
This result is in agreement with data above. 
 
 
 
17.10 
 
1( ) 2 1
8c
G s
s
 
= +   
 
 Substituting s ≅ (1-z-1)/∆t and accounting for ∆t=1 
 
 
1
1 1
1 2.25 2( ) 2 1
8(1 ) (1 )c
zG z
z z
−
− −
  −
= + = 
− −  
 
By using Simulink-MATLAB, the simulation for a unit step change in the 
controller error signal e(t) is shown in Fig. S17.10 
 
17-12 
0 5 10 15 20 25 30
0
10
20
30
40
50
60
70
k
b(k
)
 
 Fig S17.10. Open-loop response for a unit step change 
 
 
 
17.11 
 
 
a) 2
( ) 5( 0.6)
( ) 0.41
Y z z
U z z z
+
=
− +
 
 
Dividing both numerator and denominator by z2 
 
 
1 2
1 2
( ) 5 3
( ) 1 0.41
Y z z z
U z z z
− −
− −
+
=
− +
 
 
Then 1 2 1 2( )(1 0.41 ) ( )(5 3 )Y z z z U z z z− − − −− + = + 
or y(k) = y(k-1) − 0.41y(k-2) + 5u(k-1) + 3u(k-2) 
 
 
 
b) The simulation of the difference equation yields 
 
 
 
 
 
17-13 
k u(k) u(k-1) u(k-2) y(k) 
1 1 1 0 5 
2 1 1 1 13.00 
3 1 1 1 18.95 
4 1 1 1 21.62 
5 1 1 1 21.85 
6 1 1 1 20.99 
7 1 1 1 20.03 
8 1 1 1 19.42 
9 1 1 1 19.21 
10 1 1 1 19.25 
11 1 1 1 19.37 
12 1 1 1 19.48 
13 1 1 1 19.54 
14 1 1 1 19.55 
15 1 1 1 19.54 
16 1 1 1 19.52 
17 1 1 1 19.51 
18 1 1 1 19.51 
19 1 1 1 19.51 
 
 
c) By using Simulink-MATLAB, the simulation for a unit step change in u 
yields 
 
0 2 4 6 8 10 12 14 16 18 20
0
5
10
15
20
25
∆
Difference equation
Simulink
k t 
y
 
 Fig S17.11. Simulink response to a unit step change in u 
 
d) The steady state value of y can be found be setting z =1. In doing so, 
 
y =19.51 
 
 This result is in agreement with data above. 
17-14 
17.12 
 
a) 11
1
−
− z
 
 
0 1 2 3 4 5
0
1
2
3
4
5
6
7
Time
O
ut
pu
t
 
 
 b) 17.01
1
−+ z
 
 
0 1 2 3 4 5
0
0.2
0.4
0.6
0.8
1
Time
O
u
tp
u
t
 
 
c) 17.01
1
−
− z
 
0 1 2 3 4 5
0
0.5
1
1.5
2
2.5
3
Time
O
u
tp
u
t
 
 
17-15 
d) )3.01)(7.01(
1
11 −−
−+ zz
 
 
0 1 2 3 4 5
0
0.2
0.4
0.6
0.8
1
Time
O
ut
pu
t
 
 
e) )3.01)(7.01(
5.01
11
1
−−
−
−+
−
zz
z
 
 
 
0 1 2 3 4 5
0
0.2
0.4
0.6
0.8
1
Time
O
u
tp
u
t
 
 
f) )3.01)(6.01(
2.01
11
1
−−
−
−+
−
zz
z
 
 
0 1 2 3 4 5
0
0.2
0.4
0.6
0.8
1
 
17-16 
 Conclusions: 
 
 .- A pole at z = 1 causes instability. 
 
 .- Poles only on positive real axis give oscillation free response. 
 
 .- Poles on the negative real axis give oscillatory response. 
 
 .- Poles on the positive real axis dampen oscillatory responses. 
 
 ..- Zeroes on the positive real axis increase oscillations. 
 
 .- Zeroes closer to z = 0 contribute less to the increase in oscillations. 
 
 
 
17.13 
 
 
By using Simulink, the response to a unit set-point change is shown in Fig. 
S17.13a 
0 5 10 15 20 25 30 35 40
0
0.2
0.4
0.6
0.8
1
1.2
1.4
1.6
1.8
Time
O
u
tp
u
t
 
Fig S17.13a. Closed-loop response to a unit set-point change (Kc = 1) 
 
 
Therefore the controlled system is stable. 
 
The ultimate controller gain for this process is found by trial and error 
 
17-17 
0 5 10 15 20 25 30 35 40
0
1
2
3
4
5
6
7
8
Time
O
u
tp
ut
 
Fig S17.13b. Closed-loop response to a unit set-point change (Kc =21.3) 
 
Then Kcu = 21.3 
 
 
 
17.14 
 
 
 By using Simulink-MATLAB, these ultimate gains are found: 
 
 
 ∆∆∆∆t = 0.01 
 
 
0 1 2 3 4 5 6
0
0.2
0.4
0.6
0.8
1
1.2
1.4
1.6
1.8
2
Time
O
ut
pu
t
 
Fig S17.14a. Closed-loop response to a unit set-point change (Kc =1202) 
 
17-18 
 ∆∆∆∆t = 0.1 
 
 
0 5 10 15
0
0.2
0.4
0.6
0.8
1
1.2
1.4
1.6
1.8
2
Time
O
u
tp
ut
 
Fig S17.14b. Closed-loop response to a unit set-point change (Kc =122.5) 
 
 
 ∆∆∆∆t = 0.5 
 
0 5 10 15
0
0.2
0.4
0.6
0.8
1
1.2
1.4
1.6
1.8
2
Time
O
u
tp
ut
 
Fig S17.14c. Closed-loop response to a unit set-point change (Kc =26.7) 
 
 Hence 
 ∆t = 0.01 Kcu = 1202 
 ∆t = 0.1 Kcu = 122.5 
 ∆t = 0.5 Kcu = 26.7 
 
As noted above, decreasing the sampling time makes the allowable 
controller gain increases. For small values of ∆t, the ultimate gain is large 
enough to guarantee wide stability range. 
17-19 
17.15 
 
 
By using Simulink-MATLAB 
 
 Kc = 1 
 
 
0 5 10 15 20 25 30 35 40 45 50
0
0.2
0.4
0.6
0.8
1
1.2
1.4
time
O
ut
pu
t
 
Fig S17.15a. Closed-loop response to a unit set-point change (Kc =1) 
 
 
Kc = 10 
 
0 5 10 15 20 25 30 35 40 45 50
0
0.2
0.4
0.6
0.8
1
1.2
1.4
1.6
1.8
time
O
ut
pu
t
 
 Fig S17.15b. Closed-loopresponse to a unit set-point change (Kc =10) 
 
17-20 
Kc = 17 
 
 
0 5 10 15 20 25 30 35 40 45 50
-0.5
0
0.5
1
1.5
2
2.5
time
O
ut
pu
t
 
 Fig S17.15c. Closed-loop response to a unit set-point change (Kc =17) 
 
 
 Thus the maximum controller gain is 
 
 Kcm = 17 
 
 
 
17.16 
 
 
 Gv(s) = Kv = 0.1 ft3 / (min)(ma) 
 
 Gm(s) = 15.0
4
+s
 
 
 In order to obtain Gp(s), write the mass balance for the tank as 
 
 321 qqqdt
dhA −+= 
 
 Using deviation variables and taking Laplace transform 
 
 )()()()( 321 sQsQsQsHAs ′−′+′=′ 
 
Therefore, 
17-21 
 
3
( ) 1 1( ) ( ) 12.6p
H sG s Q s As s
′ − −
= = =
′
 
 
 By using Simulink-MATLAB, 
 
 
 Kc = -10 
 
 
0 5 10 15 20 25 30 35 40 45 50
0
0.2
0.4
0.6
0.8
1
1.2
1.4
time
y(t
)
 
Fig S17.16a. Closed-loop response to a unit set-point change (Kc = -10) 
 
 
Kc = -50 
 
0 5 10 15 20 25 30 35 40 45 50
0
0.2
0.4
0.6
0.8
1
1.2
1.4
1.6
1.8
time
y(t
)
 
Fig S17.16b. Closed-loop response to a unit set-point change (Kc = -50) 
17-22 
Kc = -92 
 
0 5 10 15 20 25 30 35 40 45 50
-1.5
-1
-0.5
0
0.5
1
1.5
2
2.5
3
3.5
time
y(t
)
 
Fig S17.16c. Closed-loop response to a unit set-point change (Kc = -92) 
 
Hence the closed loop system is stable for 
 
 -92 < Kc < 0 
 
 As noted above, offset occurs after a change in the setpoint. 
 
 
 
17.17 
 
 
a) The closed-loop response for set-point changes is 
 
( )( )
( ) 1 ( )
c
sp c
G G sY s
Y s G G s
=
+
 then 
( / )1( )
1 ( / )
sp
c
sp
Y Y
G z
G Y Y
=
−
 
 
We want the closed-loop system exhibits a first order plus dead time 
response, 
 
 ( / )
1
hs
sp
eY Y
s
−
=
λ +
 or 
1
1
(1 )( / )
1
N
sp
A zY Y
Az
− −
−
−
=
−
 where A = e-∆t/λ 
 
 Moreover, 
 
17-23 
 
13
)(
2
+
=
−
s
e
sG
s
 or 1
3
716.01
284.0)(
−
−
−
=
z
z
zG 
 
 Thus, the resulting digital controller is the Dahlin's controller Eq. 17-66. 
 
 
1
1 1
(1 ) 1 0.716( )
1 (1 ) 0.284c N
A zG z
Az A z
−
− − −
− −
=
− − −
 (1) 
 
 If a value of λ=1 is considered, then A = 0.368 and Eq. 1 is 
 
 
1
1 3
0.632 1 0.716( )
1 0.368 0.632 0.284c
zG z
z z
−
− −
−
=
− −
 (2) 
 
b) (1-z-1) is a factor of the denominator in Eq. 2, indicating the presence of 
integral action. Then no offset occurs. 
 
c) From Eq. 2, the denominator of Gc(z) contains a non-zero z-0 term. Hence 
the controller is physically realizable. 
 
d) First adjust the process time delay for the zero-order hold by adding ∆t/2 
to obtain a time delay of 2 + 0.5 = 2.5 min. Then obtain the continuos PID 
controller tuning based on the ITAE (setpoint) tuning relation in Table 
12.3 with K = 1, τ=3, θ = 2.5. Thus 
 
KKc = 0.965(2.5/3) − 0.85 , Kc = 1.13 
 
τ/τI = 0.796 + (-0.1465)(2.5/3) , , τI = 4.45 
 
τD/τ = 0.308(2.5/3)
0.929
 , τD = 0.78 
 
Using the position form of the PID control law (Eq. 8-26 or 17-55) 
 
1
1
1( ) 1.13 1 0.225 0.78(1 )
1c
G z z
z
−
−
  
= + + −  
−   
 
 
1 2
1
2.27 2.89 0.88
1
z z
z
− −
−
− +
=
−
 
 
 
By using Simulink-MATLAB, the controller performance is examined: 
 
17-24 
0 5 10 15 20 25 30 35 40 45 50
0
0.2
0.4
0.6
0.8
1
1.2
1.4
Time
y(t
)
 
Fig S17.17. Closed-loop response for a unit step change in set point. 
 
 
 Hence performance shows 21% overshoot and also oscillates. 
 
 
 
17.18 
 
 
 a) 
 
 
 The transfer functions in the various blocks are as follows. 
 
 Km = 2.5 ma / (mol solute/ft3) 
 
 Gm(s) = 2.5e-s 
 
17-25 
 H(s)=
s
e s−−1
 
 
 Gv(s) = Kv = 0.1 ft3/min.ma 
 
 To obtain Gp(s) and Gd(s), write the solute balance for the tank as 
 
 
3
1 1 2 2 3 3( ) ( ) ( )
dcV q c q t c t q c t
dt
= + − 
 
 Linearizing and using deviation variables 
 
 332222
3 cqqccq
dt
cdV ′−′+′=
′
 
 
 Taking Laplace transform and substituting numerical values 
 
 )(3)(1.0)(5.1)(30 3223 sCsCsQsCs ′−′+′=′ 
 
 Therefore, 
 
 
110
5.0
330
5.1
)(
)()(
2
3
+
=
+
=
′
′
=
sssQ
sC
sG p 
 
 
3
2
( ) 0.1 0.033( ) ( ) 30 3 10 1d
C sG s
C s s s
′
= = =
′ + +
 
 
 
b) 1
2
3
9.01
05.0
)(
)()(
−
−
==
zzQ
zC
zG p 
 
 
A proportional-integral controller gives a first order exponential response 
to a unit step change in the disturbance C2. This controller will also give a 
first order response to setpoint changes. Therefore, the desired response 
could be specified as 
 
 
1( / )
1sp
Y Y
s
=
λ +
 
 
 
 
 
17-26 
17.19 
 
 
 
( ) ( )
1 ( ) ( )
p m c
sp p m c
HG z K G zY
Y HG G z G z
=
+
 
 
 Solving for Gc(z) 
 
 ( )
( ) ( )
sp
c
p m p m
sp
Y
Y
G z YHG z K HG G z
Y
=
−
 (1) 
 
 Since the process has no time delay, N = 0. Hence 
 
 
1
1
(1 )
1sp d
Y A z
Y Az
−
−
 
−
=  
−  
 
 Moreover 
 
 1
1
1
)(
−
−
−
=
z
z
zHG p 
 
 1
2
1
)(
−
−
−
=
z
z
zGHG mp 
 
 Km = 1 
 
 Substituting into (1) gives 
 
 
1
1
1 2 1
1 1 1
(1 )
1( ) (1 )
1 1 1
c
A z
AzG z
z z A z
z z Az
−
−
− − −
− − −
−
−
=
−
−
− − −
 
 
 Rearranging, 
 
 21
1
)1(1
)1()1()(
−−
−
−−−
−−−
=
zAAz
zAA
zGc 
 
By using Simulink-MATLAB, the closed-loop response is shown for 
different values of A (actually different values of λ) : 
17-27 
 λ = 3 A = 0.716 
 λ = 1 A = 0.368 
 λ = 0.5 A = 0.135 
 
0 5 10 15 20 25 30 35 40 45 50
0
0.5
1
1.5
2
2.5
3
3.5
4
4.5
5
Time
y(t
)
λ=3
λ=1
λ=0.5
 
 Fig S17.19. Closed-loop response for a unit step change in disturbance. 
 
 
 
17.20 
 
 
 The closed-loop response for a setpoint change is 
 
 
( ) ( )
1 ( ) ( ) ( )
v c
sp v m c
HG z K G zY
Y HG z K K z G z
=
+
 
 
 Hence 
 
1( ) ( )
sp
c
v v m
sp
Y
Y
G z YHG z K K K
Y
=
−
 
 
 The process transfer function is 
 
17-28 
 
110
5.2)(
+
=
s
sG or 1
1
819.01
453.0)(
−
−
−
=
z
z
zHG (θ = 0 so N = 0) 
 
 Minimal prototype controller implies λ = 0 (i.e., A )0→ . Then, 1
sp
Y
z
Y
−
= 
 
 Therefore the controller is 
 
1
1
1
1
)25.0)(2.0(2.0453.0
819.01)(
−
−
−
−
−
−
=
z
z
z
z
zGc 
Simplifying, 
 
1
1
21
21
023.0091.0
819.01
023.0091.0
819.0)(
−
−
−−
−−
−
−
=
−
−
=
z
z
zz
zz
zGc 
 
 
 
17.21 
 
 
a) From Eq. 17-71, the Vogel-Edgar controller is 
 
11
21
1
21
2
2
1
1
))(1()1)((
)1)(1(
−−−−
−−
+−−−+
−++
= NVE zzbbAAzbb
AzazaG 
 
where A = e-∆t/λ = e –1/5 = 0.819 
 
Using z-transforms, the discrete-time version of the second-order transfer 
function yields 
 
 a1 = -1.625 
a2 = 0.659 
 b1 = 0.0182 
 b2 = 0.0158 
 
 Therefore111
21
)0158.00182.0(181.0)819.01)(0158.00182.0(
181.0)659.0625.11(
−−−
−−
+−−+
+−
=
zzz
zzGVE 
 
 21
21
003.0031.0034.0
119.0294.0181.0
−−
−−
−−
+−
=
zz
zz
 
 
17-29 
By using Simulink-MATLAB, the controlled variable y(k) and the 
controller output p(k) are shown for a unit step change in ysp. 
 
 
Controlled variable y(k): 
 
0 5 10 15 20 25
0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1
k
y(k
)
 
Figure S17.21a. Controlled variable y(k) for a unit step change in ysp. 
 
 
 
Controller output p(k): 
0 5 10 15 20 25
0.5
1
1.5
2
2.5
3
3.5
4
4.5
5
5.5
k
p(k
)
 
Figure S17.21b. Controlled output p(k) for a unit step change in ysp. 
 
 
17-30 
17.22 
 
 
Dahlin's controller 
 
From Eq. 17-66 with a1 = e
-1/10
=0.9, N=1, and A=e-1/1 = 0.37, the Dahlin 
controller is 
 
)9.01(2
9.01
)37.01(37.01
)37.01()(
1
21
−
−
−−−
−
=
−
−−
z
zz
zGDC 
 )63.01)(1(
84.215.3
63.037.01
84.215.3
11
1
21
1
−−
−
−−
−
+−
−
=
−−
−
=
zz
z
zz
z
 
 
By using Simulink, controller output and controlled variable are shown 
below: 
 
0 5 10 15 20 25 30 35 40 45 50
0
0.5
1
1.5
2
2.5
3
3.5
time
p(t
)
 
 Fig S17.22a. Controller output for Dahlin controller. 
 
 
0 5 10 15 20 25 30 35 40 45 50
0
0.2
0.4
0.6
0.8
1
1.2
1.4
time
O
u
tp
u
t
 
 Fig S17.22b. Closed-loop response for Dahlin controller. 
17-31 
Thus, there is no ringing (this is expected for a first-order system) and no 
adjustment for ringing is required. 
 
 
 PID (ITAE setpoint) 
 
For this controller, adjust the process time delay for the zero-order hold by 
adding ∆t/2, and K=2, τ=10, θ=1.5 obtain the continuous PID controller 
tunings from Table 12.3 as 
 
KKc = 0.965(1.5/10) − 0.85 , Kc = 2.42 
 
τ/τI = 0.796 + (-0.1465)(1.5/10) , , τI = 12.92 
τD/τ = 0.308(1.5/10)0.929 , τD = 0.529 
 
Using the position form of the PID control law (Eq. 8-26 or 17-55) 
 
1
1
1 1( ) 2.42 1 0.529(1 )
12.92 1c
G z z
z
−
−
  
= + + −  
−   
 
 1
21
1
28.198.489.3
−
−−
−
+−
=
z
zz
 
 
 
 By using Simulink, 
 
 
0 5 10 15 20 25 30 35 40 45 50
-1
-0.5
0
0.5
1
1.5
2
2.5
3
3.5
4
time
p(t
)
 
 Fig S17.22c. Controller output for PID (ITAE) controller 
17-32 
 
0 5 10 15 20 25 30 35 40 45 50
0
0.2
0.4
0.6
0.8
1
1.2
1.4
Time
O
u
tp
u
t
 
 Fig S17.22d. Closed-loop response for PID (ITAE) controller. 
 
 
Dahlin's controller gives better closed-loop performance than PID because 
it includes time-delay compensation. 
 
 
 
 
17.23 
 
 
From Eq. 17-66 with a1 = e
-1/5
=0.819, N=5, and A=e-1/1 = 0.37, the Dahlin 
controller is 
 
)819.01(25.1
819.01
)37.01(37.01
)37.01()(
1
61
−
−
−−−
−
=
−
−−
z
zz
zGDC 
 
 )63.037.01(
28.278.2
61
1
−−
−
−−
−
=
zz
z
 
 
 
By using Simulink-MATLAB, the controller output is shown in Fig. 
S17.23 
 
17-33 
0 5 10 15 20 25
0.5
1
1.5
2
2.5
3
k
p(k
)
 
Figure S17.23. Controller output for Dahlin controller. 
 
 
As noted in Fig.S17.23, ringing does not occur. This is expected for a 
first-order system. 
 
 
 
 
17.24 
 
 
 Dahlin controller 
 
 Using Table 17.1 with K=0.5 , r =1.0, p =0.5, 
 
 
1 2
1 2
0.1548 0.0939( )
1 0.9744 0.2231
z zG z
z z
− −
− −
+
=
− +
 
 
 From Eq. 17-64, with λ = ∆t = 1, Dahlin's controller is 
 
 1
1
21
21
1
632.0
0939.01548.0
)2231.09744.01()(
−
−
−−
−−
−+
+−
=
z
z
zz
zz
zGDC 
 
 )0939.01548.0)(1(
141.0616.0632.0
11
21
−−
−−
+−
+−
=
zz
zz
 
 
From Eq. 17-63, 
 
17-34 
1
1
( ) 0.632
( ) 1 0.368sp
Y z z
Y z z
−
−
=
−
 
 
y(k) = 0.368 y(k-1) + 0.632 ysp(k-1) 
 
 Since this is first order, no overshoot occurs. 
 
 By using Simulink-MATLAB, the controller output is shown: 
 
0 5 10 15 20 25
-1
0
1
2
3
4
5
k
p(k
)
 
 Figure S17.24a. Controller output for Dahlin controller. 
 
As noted in Fig. S17.24 a, ringing occurs for Dahlin's controller. 
 
 Vogel-Edgar controller 
 
 From Eq. 17-71, the Vogel-Edgar controller is 
 
 21
21
239.0761.01
567.0476.2541.2)(
−−
−−
−−
+−
=
zz
zz
zGVE 
 
Using Eq. 17-70 and simplifying, 
 
1 2
1
( ) (0.393 0.239 )
( ) 1 0.368sp
Y z z z
Y z z
− −
−
+
=
−
 
 
 y(k) = 0.368 y(k-1) + 0.393 ysp(k-1) + 0.239 ysp (k-2) 
 
 Again no overshoot occurs since y(z)/ysp(z) is first order. 
 
 By using Simulink-MATLAB, the controller output is shown below: 
17-35 
 
0 5 10 15 20 25
0.8
1
1.2
1.4
1.6
1.8
2
2.2
2.4
2.6
k
p(k
)
 
 Figure S17.24b. Controller output for Vogel-Edgar controller. 
 
 
 As noted in Fig. S17.24 b, the V-E controller does not ring. 
 
 
 
 
17.25 
 
 
 
a) Material Balance for the tanks, 
 
1
1 1 2 1 2
1 ( )dhA q q h h
dt R
= − − − 
 
2
2 1 2
1 ( )dhA h h
dt R
= − 
 
 where A1 = A2 = pi/4(2.5)2=4.91 ft2 
 
 Using deviation variables and taking Laplace transform 
 
 1 1 1 2 1 2
1 1( ) ( ) ( ) ( ) ( )A sH s Q s Q s H s H s
R R
′ ′ ′ ′ ′= − − + (1) 
 2 2 1 2
1 1( ) ( ) ( )A sH s H s H s
R R
′ ′ ′= − (2) 
 
 
 
17-36 
 From (2) 
 
 2 1
2
1( ) ( )
1
H s H s
A Rs
′ ′=
+
 
 
 Substituting into (1) and simplifying 
 
 [ ] [ ]21 2 1 2 1 2 1 2( ) ( ) ( ) 1 ( ) ( )A A R s A A s H s A Rs Q s Q s′ ′ ′ + + = + −  
 
 
1 2
2
2 1 2 1 2
( ) ( 1) 0.204( 0.102)( ) ( ) ( ) ( ) ( 0.204)p
H s A Rs sG s Q s A A R s A A s s s
′
− + − +
= = =
′ + + +
 
 
 
1 2
2
1 1 2 1 2
( ) 1 0.204( 0.102)( ) ( ) ( ) ( ) ( 0.204)d
H s A Rs sG s Q s A A R s A A s s s
′ + +
= = =
′ + + +
 
 
Using Eq. 17-64, with N =0, A=e-∆t/λ and HG(z) = KtKvHGp(z), Dahlin's 
controller is 
 
1
1
1 (1 )( ) (1 )DC
A zG z
HG z
−
−
−
=
−
 
 
Using z-transforms, 
 
HG(z)=KtKvHGp(z)=
1 2
1 1
0.202 0.192
(1 )(1 0.9 )
z z
z z
− −
− −
− +
− −
 
 
Then, 
 
1 1
1 2
(1 )(1 0.9 )( ) ( 0.202 0.192 )DC
z zG z
z z
− −
− −
− −
= ⋅
− +
1
1
(1 )
(1 )
A z
z
−
−
−
−
 
 
 
1
1
(1 )(1 0.9 )
0.202 0.192
A z
z
−
−
− −
=
− +
 
 
 
b) 
1
1
(1 )(1 0.9 )
0.202 0.192DC
A zG
z
−
−
− −
=
− +
 
 
 
By using Simulink-MATLAB, 
17-37 
0 5 10 15 20 25 30 35 40 45 50
-2
-1.8
-1.6
-1.4
-1.2
-1
-0.8
-0.6
-0.4
-0.2
0
time
p(k
)
 
 Figure S17.25. Controller output for Dahlin's controller. 
 
 As noted in Fig. S17.25, the controller output doesn't oscillate. 
 
c) This controller is physically realizable since the z-0 coefficient in the 
denominator is non-zero. Thus, controller is physically realizable for all 
value of λ. 
 
d) λ is the time constant of the desired closed-loop transfer function. From 
the expression for Gp(s) the open-loop dominant time constant is 1/0.204 = 
4.9 min. 
 
A conservative initial guess for λ would beequal to the open-loop time 
constant, i.e., λ = 4.9 min. If the model accuracy is reliable, a more bold 
guess would involve a smaller λ, say 1/3 rd of the open-loop time constant. 
In that case, the initial guess would be λ = (1/3)×4.9 =1.5 min. 
 
 
 
17.26 
 
 1
2
( 1) ( )( )
1 ( )f
K s P sG s
s E s
τ +
= =
τ +
 
 
 Substituting s )1( 1−−≅ z / ∆t into equation above: 
 
 
1 1 1
1 1 1 1
1 1 1
2 2 2 2
(1 ) / 1 (1 ) ( )( ) (1 ) / 1 (1 ) ( )f
z t z t t zG z K K K
z t z t t z
− − −
− − −
τ − ∆ + τ − + ∆ τ + ∆ − τ
= = =
τ − ∆ + τ − + ∆ τ + ∆ − τ
 
17-38 
 Then, 
 
1
1 2
1
1
( )( )
1 ( )f
b b z P zG z
a z E z
−
−
+
= =
+
 
 
 where 11
2
( )K tb
t
τ + ∆
=
τ + ∆
 , 
1
2
2
Kb
t
− τ
=
τ + ∆
 and 21
2
a
t
−τ
=
τ + ∆
 
 
 Therefore, 
 
 
1 1
1 1 2(1 ) ( ) ( ) ( )a z P z b b z E z− −+ = + 
 
 
 Converting the controller transfer function into a difference equation form: 
 
 1 1 2( ) ( 1) ( ) ( 1)p k a p k b e k b e k= − − + + − 
 
 
Using Simulink-MATLAB, discrete and continuous responses are 
compared : ( Note that b1=0.5 , b2 = −0.333 and a1= −0.833) 
 
0 5 10 15 20 25 30 35 40 45 50
0.4
0.5
0.6
0.7
0.8
0.9
1
Time
O
ut
pu
t
Continuos response
Discrete response
 
Figure S17.26. Comparison between discrete and continuous controllers. 
 
 
 
 
 
 
17-39 
17.27 
 
 
 Using Table 17.1 with K= -1/3, r = 1/3, p = 1/5, 
 
 
1 5
1 1
( 0.131 0.124 )( ) ( )(1 0.716 )(1 0.819 )
z zG z G z
z z
− −
− −
− −
= ≡
− −
 
 
 Since the zero is at z =-0.95, it should be included in )z(G~+ . Therefore 
 
 )z(G~+
1 5
5 6( 0.131 0.124 ) 0.514 0.486( 0.131 0.124)
z z
z z
− −
− −
− −
= = +
− −
 
 
 =
−
)z(G~ 1 1
( 0.131 0.124)
(1 0.716 )(1 0.819 )z z− −
− −
− −
 
 
 For deadbeat filter, F(z) = 1 
 
 Using Eq. 17-77, the IMC controller is 
 
 
1
*( ) ( ) ( )cG z G z F z−−= =�
1 1(1 0.716 )(1 0.819 )
( 0.131 0.124)
z z− −− −
− −
 
 
By using Simulink-MATLAB, the IMC response for a unit step in load at 
t=10 is shown in Fig. S17.27 
0 5 10 15 20 25 30 35 40 45
0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
Time
O
ut
pu
t
 
Fig. S17.27. IMC close-loop response for a unit step change in load at t=10. 
19-1 
��������	
�
 
 
 
 
 19.1 
 
 
 From definition of xc, 0 ≤ xc ≤ 1 
 
 f(x) = 5.3 x e (-3.6x +2.7) 
 
Let three initial points in [0,1] be 0.25, 0.5 and 0.75. Calculate x4 using 
Eq. 19-8,. 
 
x1 f1 x2 f2 x3 f3 x4 
0.25 8.02 0.5 6.52 0.75 3.98 0.0167 
 
For next iteration, select x4, and x1 and x2 since f1 and f2 are the largest 
among f1, f2, f3. Thus successive iterations are 
 
x1 f1 x2 f2 x3 f3 x4 
0.25 8.02 0.5 6.52 0.017 1.24 0.334 
0.25 8.02 0.5 6.52 0.334 7.92 0.271 
0.25 8.02 0.334 7.92 0.271 8.06 0.280 
0.25 8.02 0.271 8.06 0.280 8.06 not needed 
 
 x
opt
 = 0.2799 7 function evaluations 
 
 
 19.2 
 
 
As shown in the drawing, there is both a minimum and maximum value of 
the air/fuel ratio such that the thermal efficiency is non- zero. If the ratio is 
too low, there will not be sufficient air to sustain combustion. On the other 
hand, problems in combustion will appear when too much air is used. 
 
The maximum thermal efficiency is obtained when the air/fuel ratio is 
stoichiometric. If the amount of air is in excess, relatively more heat will 
be “absorbed” by the air (mostly nitrogen). However if the air is not 
sufficient to sustain the total combustion, the thermal efficiency will 
decrease as well. 
 
Solution Manual for Process Dynamics and Control, 2nd edition, 
Copyright © 2004 by Dale E. Seborg, Thomas F. Edgar and Duncan A. Mellichamp 
19-2 
19.3 
 
 
By using Excel-Solver, this optimization problem is quickly solved. The 
selected starting point is (1,1): 
 
 
 X1 X2 
Initial values 1 1 
Final values 0.776344 0.669679 
 
 
max Y= 0.55419 
 
Constraints 
 
0 ≤≤≤≤ X1 ≤≤≤≤ 2 
0 ≤≤≤≤ X2 ≤≤≤≤2 
 
 Table S19.3. Excel solution 
 
 Hence the optimum point is ( X1*, X2* ) =(0.776, 0.700) 
 
 and the maximum value of Y is Ymax = 0.554 
 
 
 19.4 
 
 
 Let N be the number of batches/year. Then NP ≥ 300,000 
Since the objective is to minimize the cost of annual production, only the 
required amount should be produced annually and no more. That is, 
 
 NP = 300,000 (1) 
 
a) Minimize the total annual cost, 
 
 
min TC = 400,000 $
batch
    + 2 P
0.4 hr
batch
    50
$
hr
    N
batch
yr
    
 + 800 P0.7 



yr
$
 
 
 
 Substituting for N from (1) gives 
 
 min TC = 400,000 + 3x107 P–0.6 + 800 P0.7 
 
19-3 
b) There are three constraints on P 
 
i) P ≥ 0 
ii) N is integer. That is, 
 
(300,000/P) = 0, 1, 2,… 
 
iii) Total production time is 320 x 24 hr/yr 
 
 (2 P0.4 + 14) hr
batch
    × N
batch
yr
    ≤ 7680 
 
Substituting for N from (1) and simplifying 
 
 6×105P-0.6 + 4.2×106P-1 ≤ 7680 
 
 
 c) 
7 1.6 0.3( ) 0 3 10 ( 0.6) 800(0.7)d TC P P
dP
− −
= = × − + 
 
1/1.373 10 ( 0.6) lb2931
800(0.7) batch
optP
 × −
= = 
− 
 
 
 
2
7 2.6 1.3
2
( ) 3 10 ( 0.6)( 1.6) 800(0.7)( 0.3)d TC P P
dP
− −
= × − − + − 
 
2
2
2
( ) 2.26 10 0
optP P
d TC
dP
−
=
= × 〉 hence minimum 
 
 Nopt = 300,000/Popt = 102.35 not an integer. 
 
Hence check for Nopt = 102 and Nopt = 103 
 
For Nopt = 102, Popt = 2941.2, and TC = 863207 
 
For Nopt = 103, Popt = 2912.6, and TC = 863209 
 
Hence optimum is 102 batches of 2941.2 lb/batch. 
 
Time constraint is 
 
5 0.6 6 16 10 4.2 10 6405.8 7680P P− −× + × = ≤ , satisfied 
 
 
 
 
19-4 
 19.5 
 
 
 Let x1 be the daily feed rate of Crude No.1 in bbl/day 
 x2 be the daily feed rate of Crude No.2 in bbl/day 
 
 Objective is to maximize profit 
 
 max P = 2.00 x1 + 1.40 x2 
 
 Subject to constraints 
 
 gasoline : 0.70 x1 + 0.31 x2 ≤ 6000 
 kerosene: 0.06 x1 + 0.09 x2 ≤ 2400 
fuel oil: 0.24 x1 + 0.60 x2 ≤ 12,000 
 
 By using Excel-Solver, 
 
 x1 x2 
Initial values 1 1 
Final values 0 19354.84 
 
 
max P = 27096.77 
 
 
Constraints 
 
0.70 x1 + 0.31 x2 6000 
0.06 x1 + 0.09 x2 1741.935 
0.24 x1 + 0.60 x2 11612.9 
 
 Table S19.5. Excel solution 
 
 Hence the optimum point is (0, 19354.8) 
 
 Crude No.1 = 0 bbl/day Crude No.2 = 19354.8 bbl/day 
 
 
 
 
 
 
 
 
 
 
 
 
 
19-5 
 19.6 
 
 
 Objective function is to maximize the revenue, 
 
 max R = -40x1 +50x3 +70x4 +40x5 –2x1-2x2 (1) 
 
*Balance on column 2 
 
 x2 = x4 + x5 (2) 
 
* From column 1, 
 x1 = )(667.160.0
0.1
542 xxx += (3) 
 x3 = )(667.060.0
4.0
542 xxx += (4) 
 
Inequality constraints are 
 
x4 ≥ 200 (5) 
x4 ≤ 400 (6) 
x1 ≤ 2000 (7) 
x4 ≥ 0 x5 ≥ 0 (8) 
 
The restricted operating range for column 2 imposes additional inequality 
constraints. Medium solvent is 50 to 70% of the bottoms; that is 
 
 0.5 ≤ 4
2
x
x
≤ 0.7 or 0.5 ≤ 4
4 5
x
x x+
≤ 0.7 
 Simplifying, 
 
 x4 –x5 ≥ 0 (9) 
 0.3 x4 –0.7x5 ≤ 0 (10) 
 
No additional constraint is needed for the heavy solvent. Thatthe heavy 
solvent will be 30 to 50% of the bottoms is ensured by the restriction on 
the medium solvent and the overall balance on column 2. 
 
 
 By using Excel-Solver, 
 
 
 
 
 
 
19-6 
 
 x1 x2 x3 x4 x5 
Initial values 1 1 1 1 1 
Final values 1333.6 800 533.6 400 400 
 
 
max R = 13068.8 
 
 
 
 
Constraints 
x2 - x4 - x5 0 
x1 - 1.667x2 7.467E-10 
x3 - 0.667x2 -1.402E-10 
x4 400 
x4 400 
x1 - 1.667x2 1333.6 
x4 - x5 0 
0.3x4 - 0.7x5 -160 
 
 Table S19.6. Excel solution 
 
 Thus the optimum point is x1 =1333.6, x2 =800; x3=533.6, x4 = 400 and 
 x5 = 400. 
 
Substituting into (5), the maximum revenue is 13,068 $/day, and the 
percentage of output streams in column 2 is 50 % for each stream. 
 
 
 19.7 
 
 
The objective is to minimize the sum of the squares of the errors for the 
material balance, that is, 
 
 min E = (wA + 11.1 – 92.4)2 + (wA +10.8 –94.3)2 + (wA + 11.4 –93.8)2 
 
 Subject to wA ≥ 0 
 
 Solve analytically, 
 
 == 0
Adw
dE
2 (wA + 11.1 – 92.4) + 2(wA +10.8 –94.3) 
 +2(wA + 11.4 –93.8) 
 
 Solving for wA… wA 
opt 
= 82.4 Kg/hr 
 
 Check for minimum, 
 062222
2
>=++=
Adw
Ed
, hence minimum 
19-7 
 19.8 
 
 
a) Income = 50 (0.1 +0.3xA + 0.0001S – 0.0001 xAS) 
 
Costs = 2.0 + 10xA + 20 xA
2
 + 1.0 + 0.003 S + 2.0x10-6S2 
 
f = 2.0 +5xA + 0.002S – 20xA2 – 2.0x10-6S2 – 0.005xAS 
 
b) Using analytical method 
 
Sx
x
f
A
A
005.04050 −−==
∂
∂
 
 AxS
f 005.0100.4002.00 6 −×−==
∂
∂
− 
 
Solving simultaneously, xA = 0.074 , S = 407 which satisfy the given 
constraints. 
 
 
 
 19.9 
 
 
 By using Excel-Solver 
 
 
 ττττ1111 ττττ2222 
 Initial values 1 0.5 
 Final values 2.991562 1.9195904 
 
 
TIME EQUATION DATA SQUARE ERROR 
0 0.000 0.000 0.00000000 
1 0.066 0.058 0.00005711 
2 0.202 0.217 0.00022699 
3 0.351 0.360 0.00007268 
4 0.490 0.488 0.00000403 
5 0.608 0.600 0.00006008 
6 0.703 0.692 0.00012252 
7 0.778 0.772 0.00003428 
8 0.835 0.833 0.00000521 
9 0.879 0.888 0.00008640 
10 0.911 0.925 0.00019150 
 SUM= 0.00086080 
 
 
19-8 
Hence the optimum values are τ1=3 and τ2=1.92. The obtained model is 
compared with that obtained using MATLAB. 
 
0 5 10 15 20 25 30
0
0.2
0.4
0.6
0.8
1
time
Y/
K
MATLAB
data
equation
 
 Figure S19.9. Comparison between the obtained model with that obtained 
 using MATLAB 
 
 
19.10 
 
 
 Let x1 be gallons of suds blended 
 x2 be gallons of premium blended 
 x3 be gallons of water blended 
 
 Objective is to minimize cost 
 
 min C = 0.25x1 + 0.40x2 (1) 
 
 Subject to 
 
 x1 + x2 + x3 = 10,000 (2) 
 
 0.035 x1 + 0.050 x2 = 0.040×10,000 (3) 
 
x1 ≥ 2000 (4) 
 
x1 ≤ 9000 (5) 
 
19-9 
x2 ≥ 0 (6) 
 
x3 ≥ 0 (7) 
 
 
The problem given by Eqs. 1, 2, 3, 4, 5, 6, and 7 is optimized using Excel-
Solver, 
 
 x1 x2 x3 
Initial values 1 1 1 
Final values 6666.667 3333.333 0 
 
 
min C = 3000 
 
 
Constraints 
 
x1+x2+x3-10000 0 
0.035x1+0.050x2- 400 0.0E+00 
x1- 2000 4666.667 
x1- 9000 -2333.333 
x2 3333.333 
x3 0 
 
Table S19.10. Excel solution 
 
Thus the optimum point is x1 = 6667 , x2 = 3333 and x3 = 0. 
The minimum cost is $3000 
 
 
19.11 
 
 
 Let xA be bbl/day of A produced 
 xB be bbl/day of B produced 
 
 Objective is to maximize profit 
 
 max P = 10xA + 14xB (1) 
 
 Subject to 
 
 Raw material constraint: 120xA+ 100xB ≤ 9,000 (2) 
 
 Warehouse space constraint: 0.5 xA + 0.5 xB ≤ 40 (3) 
 
Production time constraint: (1/20)xA + (1/10)xB ≤ 7 (4) 
 
 
19-10 
 
 xA xB 
Initial values 1 1 
Final values 20 60 
 
 
max P = 1040 
 
 
Constraints 
 
120xA+ 100xB 8400 
0.5 xA + 0.5 xB 40 
(1/20)xA + (1/10)xB 7 
 
 Table S19.11. Excel solution 
 
Thus the optimum point is xA = 20 and xB = 60 
The maximum profit = $1040/day 
 
 
19.12 
 
 
PID controller parameters are usually obtained by using either process 
model, process data or computer simulation. These parameters are kept 
constant in many cases, but when operating conditions vary, supervisory 
control could involve the optimization of these tuning parameters. For 
instance, using process data, Kc ,τI and τD can be automatically calculated 
so that they maximize profits. Overall analysis of the process is needed in 
order to achieve this type of optimum control. 
 
Supervisory and regulatory control are complementary. Of course, 
supervisory control may be used to adjust the parameters of either an 
analog or digital controller, but feedback control is needed to keep the 
controlled variable at or near the set-point. 
 
 
 
19.13 
 
 
 Assuming steady state behavior, the optimization problem is, 
 
 max f = D e 
 
 Subject to 
 
 0.063 c –D e = 0 (1) 
 0.9 s e – 0.9 s c – 0.7 c – D c = 0 (2) 
19-11 
 -0.9 s e + 0.9 s c + 10D – D s = 0 (3) 
 D, e, s, c ≥ 0 
 
where f = f(D, e, c, s) 
 
 Excel-Solver is used to solve this problem, 
 
 c D e s 
Initial values 1 1 1 1 
Final values 0.479031 0.045063 0.669707 2.079784 
 
 
max f = 0.030179 
Constraints 
 
0.063 c –D e 2.08E-09 
0.9 s e – 0.9 s c – 0.7 c – Dc -3.1E-07 
-0.9 s e + 0.9 s c + 10D – Ds 2.88E-07 
 
 Table S19.13. Excel solution 
 
Thus the optimum value of D is equal to 0.045 h-1 
 
 
19.14 
 
 
 
 Material balance: 
 
 Overall : FA + FB = F 
 
 Component B: FB CBF + VK1CA – VK2CB = F CB 
 
 Component A: FA CAF + VK2CB – VK1CA = FCA 
 
 Thus the optimization problem is: 
 
 max (150 + FB) CB 
 
 Subject to: 
 
 0.3 FB + 400CA − 300CB = (150 + FB)CB 
 
 45 + 300 CB – 400 CA = (150 + FB) CA 
 
 FB ≤ 200 
 
19-12 
 CA, CB, FB ≥ 0 
 
 By using Excel- Solver, the optimum values are 
 
 FB = 200 l/hr 
 
 CA = 0.129 mol A/l 
 
 CB = 0.171 mol B/l 
 
 
 
19.15 
 
 
 Material balance: 
 
 Overall : FA + FB = F 
 
 Component B: FB CBF + VK1CA – VK2CB = F CB 
 
 Component A: FA CAF + VK2CB – VK1CA = FCA 
 
 Thus the optimization problem is: 
 
 max (150 + FB) CB 
 
 Subject to: 
 
 0.3 FB + 3×106e(-5000/T)CA V − 6×106e(-5500/T)CB V = (150 + FB)CB 
 
 45 + 6×106e(-5500/T)CB V – 3×106e(-5000/T) CA V = (150 + FB) CA 
 
 FB ≤ 200 
 300 ≤ T ≤ 500 
 CA, CB, FB ≥ 0 
 
 By using Excel- Solver, the optimum values are 
 
 FB = 200 l/hr 
 
 CA = 0.104 molA/l 
 
 CB = 0.177 mol B/l 
 
 T = 311.3 K 
20-1 
��������	
�
 
 
 
 
 20.1 
 
 
a) The unit step response is 



+
−
+
+=





++
==
−
−
110
20
15
5121)15)(110(
2)()()(
sss
e
sss
e
sUsGsY s
s
p 
Therefore, 
 [ ]10/)1(5/)1( 21)1(2)( −−−− −+−= tt eetSty 
 
For ∆t = 1.0, 
 
{ }...3096.0,2174.0,1344.0,06572.0,01811.0,0)()( ==∆= iytiyS i 
 
b) From the expression for y(t) in part (a) above 
 
y(t) = 0.95 (2) at t =37.8, by trial and error. 
 
Hence N = 38, for 95% complete response. 
 
 
20.2 
 
 
 Note that )()()()( sGsGsGsG mpv= . From Figure 12.2,