[Solucionário - Capítulo 10] Reações Químicas e Reatores Químicas George W. Roberts (Wiley, 2009)

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Copyright© - George W. Roberts 1 May 8, 2008
Solution to Problem 10-1 (NIR-11)
Since the reaction takes place in the liquid phase, constant density can be
assumed. Therefore,
CA = CA0(1 - xA)
CB = CA0xA
-rA = kCA0(1 – xA)/(1 + KCA0xA)
Question 1:
For an ideal PFR,
 
dV
FA0
=
dxA
�rA
=
(1 + KCA0xA )dxA
kCA0 (1� xA )
kCA0dV
FA0
=
kdV
�
= kd� = (1 + KCA0xA )dxA
(1� xA )
Integrating from 0 to �, 0 to xA,
 
k� = (1+ KCA0xA )dxA
(1� xA )
=
dxA
(1� xA )
+ KCA0
xAdxA
(1� xA )0
xA
�
0
xA
�
0
xA
�
 k� = �[1 + KCA0 ] * ln(1� xA ) �KCA0xA (1)
Substituting numerical values:
k� = 4.08(min-1)*[200(L)/20(L/min)] = 40.8
KCA0 = 10(L/mol)*1.0(mol/L) = 10
40.8 = -11*ln(1 – xA) – 10xA
ln(1 – xA) + (10/11)xA + (40.8/11) = 0
Use GOALSEEK in EXCEL to find the value of xA that satisfies this equation. See
the spreadsheet “Solution to Problem NIR-11 (Solution to Question 1)”.
xPFR = 0.99
Question 2:
For a macrofluid,
 
x A = xA (t)E(t)dt
0
�
� (2)
The solution of the design equation for an isothermal, ideal batch reactor is the same as
Eqn.(1), except that the real time, t, is substituted for the space time, �.
 kt = �[1 + KCA0 ] * ln(1� xA ) �KCA0xA (3)
Since this equation is implicit in xA, the integration in Eqn.(2) will be performed
numerically.
 
x A = xA (t)(�2 /2t3 )dt
�/2
�
� = �
2
2
xA (t)dt
t3�/2
�
�
Numerical values:
� = V/� = 200(L)/20(L/min) = 10 min; �/2 = 5 min; �2/2 = 50 min2
 
x A = 50
xA (t)dt
t35
�
� (4)
Copyright© - George W. Roberts 2 May 8, 2008
Let’s calculate values of xA(t) at every minute between t = 5 and t = 25 by
applying GOALSEEK to Eqn.(3) (as we did in Question 1). A numerical integration of
Eqn.(4) will then be performed. The results of the calculations of xA(t) and a portion of
the numerical integration are shown in the spreadsheet “Solution to Problem NIR-11
(Solution to Question 2)”. Note that the numerical integration has been truncated at t =
25 min., rather than being carried out to t = �.
If the numerical integration is truncated at 25 min., by Simpson’s 1/3 rule,
 x A = 50(min
2 ) * (1(min)/3) * 0.0558(min�3 ) = 0.930
(The value 0.0558 (min-3) is the sum of the last column of the spreadsheet.)
Is it sufficiently accurate to terminate the integration at t = 25 min., or must the
integration be carried to longer times? Note that xA(25 min) = 0.99996. Let’s assume
that xA = 1.0 for all times longer that 25 min. and calculate a “correction” to the
numerical integration based on that assumption. This correction will be an over-
estimate since xA(actual) < 1.0.
 
"Correction"= 50
xA (t)dt
t325
�
� � 50 (1.0)dt
t325
�
� = 50 �1
2t2
� 
� 	 
� 
� 
 25
�
= 25 �0 + 1
252
� 
� 	 
� 
� 
 
= 0.040
Answer:
 xLFTR = 0.97
A final question is whether the step size (1 min.) that was used for the numerical
integration is sufficient for good accuracy. Let’s halve the step size to 0.5 min. and try
again. See Spreadsheet NIR-11(Check):
 x A = 50(min
2 ) * (0.50(min)/3) * 0.1116(min�3 ) = 0.930 OK
Question 3:
Rewrite the rate equation as:
 
�rA =
kCA
1+ K(CA0 �CA )
=
kCA
(1 + KCA0 ) �KCA
As CA increases, -rA increases faster than CA since the numerator increases in
direct proportion to CA and the denominator decreases as CA increases. The “effective
order” of the reaction is > 1. Therefore, the calculated conversion is an upper bound.
Copyright© - George W. Roberts 3 May 8, 2008
Solution to Problem NIR -11
SOLUTION TO QUESTION 1:
x k*Tau Residue
0.99003667 40.8 0.0002802
SOLUTION TO QUESTION 2:
Time, t k*Time Conversion, x Residue x(t)/t^3 Factor Product
5 20.4 0.93298 -3.5595E-05 0.00746383 1 0.00746383
6 24.48 0.95464 0.00019351 0.00441963 4 0.01767851
7 28.56 0.96911 -3.4801E-05 0.0028254 2 0.0056508
8 32.64 0.97887 -9.8012E-05 0.00191186 4 0.00764745
9 36.72 0.98551 -7.106E-05 0.00135186 2 0.00270373
10 40.8 0.99004 5.9474E-05 0.00099004 4 0.00396016
11 44.88 0.99315 2.2333E-05 0.00074616 2 0.00149233
12 48.96 0.99528 -1.8571E-05 0.00057597 4 0.00230389
13 53.04 0.99674 0.00046305 0.00045368 2 0.00090737
14 57.12 0.99776 -0.00014815 0.00036361 4 0.00145446
15 61.2 0.99845 5.2014E-05 0.00029584 2 0.00059168
16 65.28 0.99893 0.00089679 0.00024388 4 0.00097552
17 69.36 0.99926 5.7905E-05 0.00020339 2 0.00040678
18 73.44 0.99949 -0.00016093 0.00017138 4 0.00068552
19 77.52 0.99965 0.00028001 0.00014574 2 0.00029149
20 81.6 0.99976 1.0595E-05 0.00012497 4 0.00049988
21 85.68 0.99983 9.4631E-05 0.00010796 2 0.00021592
22 89.76 0.99988 5.4556E-05 9.3904E-05 4 0.00037561
23 93.84 0.99992 0.00060802 8.2183E-05 2 0.00016437
24 97.92 0.99995 2.206E-05 7.2334E-05 4 0.00028934
25 102 0.99996 1.2111E-05 6.3998E-05 1 6.3998E-05
Sum= 0.05582262
NOTES: "Residue" is value of k*t+(1+KCA0)*ln(1-x)+KCA0*x; "Factor" is weighting factor in 
 Simpson's 1/3 rule; "Product" is "Factor"*x(t)/t^3
Copyright© - George W. Roberts 4 May 8, 2008
Solution to Problem NIR –11 (Check)
SOLUTION TO QUESTION 2:
Time, t k*Time Conversion, x Residue x(t)/t^3 Factor Product
5 20.4 0.93298 -3.5595E-05 0.00746383 1 0.00746383
5.5 22.44 0.94494 -0.00025604 0.00567957 4 0.02271826
6 24.48 0.95464 0.00019351 0.00441963 2 0.00883926
6.5 26.52 0.96260 -1.3001E-05 0.00350513 4 0.01402053
7 28.56 0.96911 -3.4801E-05 0.0028254 2 0.0056508
7.5 30.6 0.97444 0.00083871 0.00230979 4 0.00923916
8 32.64 0.97887 -9.8012E-05 0.00191186 2 0.00382373
8.5 34.68 0.98250 7.8285E-05 0.00159985 4 0.00639938
9 36.72 0.98551 -7.106E-05 0.00135186 2 0.00270373
9.5 38.76 0.98799 9.8515E-06 0.00115234 4 0.00460936
10 40.8 0.99004 5.9474E-05 0.00099004 2 0.00198008
10.5 42.84 0.99174 -5.7807E-05 0.0008567 4 0.00342681
11 44.88 0.99315 2.2333E-05 0.00074616 2 0.00149233
11.5 46.92 0.99431 3.0511E-05 0.00065378 4 0.0026151
12 48.96 0.99528 -1.8571E-05 0.00057597 2 0.00115194
12.5 51 0.99608 0.00024014 0.00050999 4 0.00203997
13 53.04 0.99674 0.00046305 0.00045368 2 0.00090737
13.5 55.08 0.99730 0.00017364 0.00040534 4 0.00162138
14 57.12 0.99776 -0.00014815 0.00036361 2 0.00072723
14.5 59.16 0.99814 0.00019067 0.00032741 4 0.00130962
15 61.2 0.99845 5.2014E-05 0.00029584 2 0.00059168
15.5 63.24 0.99871 0.00024643 0.00026819 4 0.00107277
16 65.28 0.99893 0.00089679 0.00024388 2 0.00048776
16.5 67.32 0.99911 2.0147E-05 0.00022241 4 0.00088966
17 69.36 0.99926 5.7905E-05 0.00020339 2 0.00040678
17.5 71.4 0.99939 3.0511E-05 0.00018647 4 0.0007459
18 73.44 0.99949 -0.00016093 0.00017138 2 0.00034276
18.5 75.48 0.99958 2.3038E-05 0.00015787 4 0.00063148
19 77.52 0.99965 0.00028001 0.00014574 2 0.00029149
19.5 79.56 0.99971 1.1583E-05 0.00013482 4 0.0005393
20 81.6 0.99976 1.0595E-05 0.00012497 2 0.00024994
20.5 83.64 0.99980 2.5562E-05 0.00011605 4 0.00046421
21 85.68 0.99983 9.4631E-05 0.00010796 2 0.00021592
21.5 87.72 0.99986 0.0004162 0.00010061 4 0.00040242
22 89.76 0.99988 5.4556E-05 9.3904E-05 2 0.00018781
22.5 91.8 0.99990 2.8237E-05 8.7783E-05 4 0.00035113
23 93.84 0.99992 0.00060802 8.2183E-05 2 0.00016437
23.5 95.88 0.99993 0.0009342 7.7049E-05 4 0.0003082
24 97.92 0.99995 2.206E-05 7.2334E-05 2 0.00014467
24.5 99.96 0.99995 6.6172E-05 6.7996E-05 4 0.00027198
25 102 0.99996 1.2111E-05 6.3998E-05 1 6.3998E-05
Sum= 0.11156408
NOTES: "Residue" is value of k*t+(1+KCA0)*ln(1-x)+KCA0*x; "Factor" is weighting factor in 
 Simpson's 1/3 rule; "Product" is "Factor"*x(t)/t^3
Copyright� - George W. Roberts 1 May 29, 2008
Solution to Problem 10-2(NIR-13)
Question 1:
Let N0 = mmol of tracer injected and C(t) = measured tracer concentration in
outlet stream at time, t.
 
N0 = � C(t)dt
0
�
�Since the data stops at t = 30 min, the integral will be evaluated in two parts,
numerically from t = 0 to t = 30 min. and analytically from t = 30 min. to t = �. To
perform the analytical integration, we will fit an exponential to the data at longer times.
The data for 22 � t � 30 min. is plotted in Figure 1.
1
10
20 22 24 26 28 30 32
Fit of Exponential to "Late Time" Data
(NIR-13)
y = 215.63 * e^(-0.15609x) R= 0.99948 
T
ra
ce
r 
C
on
c(
m
m
ol
/L
)
Time(min)
An exponential fits the “late time” data quite well. The “best fit” function is:
C(t) = 215.63*exp(-0.15609t) (mmol/L)
The numerical integration was carried out using Simpson’s 1/3 Rule between 6
and 30 min. and the Trapezoid Rule between 4 and 6 min. See the attached spreadsheet,
Part A (Numerical Integrations).
Copyright� - George W. Roberts 2 May 29, 2008
 
C(t)dt
0
30 min
� = 360(mmol �min/L)
 
C(t)dt
30 min
�
� = (215.63)exp(�0.15609t)dt
30 min
�
� = 215.63�0.15609
� 
� � 
� 
� 	 
e�0.15609t ]
30
�
=
215.63
0.15609
� 
� � 
� 
� 	 
e�0.15609 *30
 
C(t)dt
30 min
�
� = 12.8(mmol �min/L)
 
C(t)dt
0
�
� = (360 + 13)(mmol �min/L) = 373(mmol �min/L)
Returning to Eqn.1:
N0 = 150(L/min)*373(mmol-min/L) = 55,950 mmol
N0 = 56,000 mmol = 56 mol
Question 2:
For a closed vessel:
 
� = V/� = t =
tC(t)dt
0
�
�
C(t)dt
0
�
�
tC(t)dt
0
�
� = tC(t)dt
0
30 min
� + tC(t)dt
30 min
�
�
From the attached spreadsheet, Part A (Numerical Integrations),
 
tC(t)dt = 5214(mmol �min2/L)
0
30 min
�
 
tC(t)dt = t(215.63)exp(�0.15609t)dt
30 min
�
�
30 min
�
� = 465(mmol �min2/L)
 
tC(t)dt = (5214 + 465)(mmol�min2/L)
0
�
� = 5679(mmol �min2/L)
 t = 5679(mmol�min2/L)/373(mmol �min/L) = 15.2min
 V = �t = 15.2(min) * 150(L/min)
V=2280 L
Question 3:
To calculate the conversion, we must first estimate the Dispersion Number,
D/uL, for the reactor. For the tracer experiment,
Copyright� - George W. Roberts 3 May 29, 2008
 
�2 =
t2C(t)dt
0
�
�
C(t)dt
0
�
�
� (t )2 (1)
From the attached spreadsheet, Part A (Numerical Integrations):
 
t2C(t)dt
0
30 min
� = 83,954(mmol �min3/L)
 
t2C(t)dt
30 min
�
� = t2 (215.63)exp(�0.15609t)dt = 17,429
30 min
�
�
From Eqn.(1),
 
�2 = (83,954 + 17,429)(mmol �min
3/L)
373(mmol �min/L)
� (15.2)2 (min)2 = 40.8min2
For a closed vessel with injection and detection on the boundaries,
 
�2
(t )2
= 2
D
uL
� 
� � 
� 
� � 
� 2 D
uL
� 
� � 
� 
� � 
2
(1� exp(�uL/D)) (2)
As an approximation,
 
D
uL
� 
� � 
� 
� � 
� �
2
2(t )2
=
40.8min2
2(15.2)2 min2
= 0.088
Using GOALSEEK on Eqn.(2) (see attached spreadsheet, Part B),
 
D
uL
= 0.0975
What will happen to the value of D/uL when the flowrate is reduced by a factor
of 3? In general, the intensity of dispersion (D�/ulc) will depend on the Reynolds
Number and therefore on u. However, since the dimensions of the experimental
packing are not given, the value of lc and the Reynolds Number cannot be calculated.
Fortunately, Figure 10-7 shows that the intensity of dispersion is essentially constant
over a wide range of Re, and a relatively weak function of Re at high values of Re.
Therefore, it is reasonable to assume that
 
D
udp
� constant(� f(u))
With this assumption, D1/u1 = D2/u2 and D/uL does not change.
The fractional conversion is given by:
 
(1� xA ) =
4a * exp(uL/2D)
(1+ a)2 * exp(auL/2D) � (1� a)2 * exp(�auL/2D)
(3)
where a = 1+ 4k�(D/uL) .
Since the volumetric flowrate is a factor of 3 lower than in the tracer experiment,
 � = 3t = 3(15.2) = 45.6min
where � is the space time for the reactor and t is the mean retention time in the tracer
experiment.
k� = 0.040(min-1)*45.6(min) = 1.824
 a = 1+ 4 * 1.824 * 0.0975 = 1.308
From Eqn.(3), the value of xA is:
Copyright� - George W. Roberts 4 May 29, 2008
xA = 0.80
Question 4:
For ideal plug flow,
 
� V 
FAo
=
dxA
�rA
=
0
xA
� dxAkCA0 (1� xA )0
xA
�
 
� V kCA0
FA0
= k � � = � ln(1� xA )
(The prime designates values for the ideal PFR.)
 
k � � = � ln(1� 0.80) = 1.61
k � � 
k�
=
1.61
1.82
=
k � V /�
kV/�
=
� V 
V
= 0.88
The ideal PFR requires about 12% less volume than the actual reactor.
Copyright� - George W. Roberts 5 May 29, 2008
CALCULATIONS FOR PROBLEM NIR-13
1) NUMERICAL INTEGRATIONS
Time(min.) C(t) Factor(f) f*C(t) f*t*C(t) f*t^2*C(t)
0 0
2 0
4 0
6 2 1 2 12 72
8 12 4 48 384 3072
10 37 2 74 740 7400
12 35 4 140 1680 20160
14 28 2 56 784 10976
16 22 4 88 1408 22528
18 16 2 32 576 10368
20 11 4 44 880 17600
22 7 2 14 308 6776
24 5 4 20 480 11520
26 3.8 2 7.6 197.6 5137.6
28 2.7 4 10.8 302.4 8467.2
30 2 1 2 60 1800
Sum 538 7812 125877
Simpson's 1/3 Rule between 6 and 30 min. 359 5208 83918
Trapezoid Rule between 4 and 6 min. 1 6 36
Value of Integral between 4 and 30 min. 360 5214 83954
(mmol,min/L) (mmol,min^2/L) (mmol,min^3/L)
Including the analytical integrations from t=30 min to t=�:
t(average)= 15.2 min
sigma^2= 40.8 min^2
CALCULATIONS FOR PROBLEM NIR-13 (CONTINUED)
B) GOAL SEEK FOR VALUE OF D/uL:
D/uL= 0.097566 -0.000498373 (=Residual)
C) CALCULATION OF FRACTIONAL CONVERSION
k*Tau= 1.824
a= 1.30837362
xA= 0.79776818
Copyright© - George W. Roberts 1 May 8, 2008
Solution to Problem 10-3 (NIR-4)
The next-to-last sentence suggests that the conversion in the LFTR will be the
same as that in a PFR, if the reaction is zero-order. This is not generally true. The
conversion in the PFR will exceed that in an LFTR operating at the same temperature,
feed concentration, and space time, except at very long space times, where the
conversion is 1.0 in both types of reactor.
For a zero-order reaction, the reactant will disappear completely, and the
reaction rate will be zero, for fluid elements that are in the reactor for longer that t =
CA0/k. Therefore, in a PFR, the conversion will be 1.0 if the space time , �, is greater
than CA0/k. However, since the velocity at the center of the LFTR is twice the average
velocity, there will be unconverted reactant at and near the center of the LFTR until � =
2CA0/k. Therefore, the conversions in the PFR and the LFTR will be the same (1.0) only
when � � 2CA0/k. For values of � between CA0/k and 2CA0/k, the conversion in the
LFTR will be lower than that in the PFR.
When � � CA0/k, the conversion in both types of reactor will be < 1.0. For this
case, the reaction rate every point in the PFR will be finite (-rA = k) since there will be
unconverted reactant at every point in the reactor. However, at the wall of the LFTR,
the fluid velocity approaches zero and the residence time of a fluid element approaches
infinity. Near the wall of the LFTR, the concentration of reactant will be zero and the
reaction rate will be zero. Therefore, the average rate in the LFTR will be lower than in
the PFR. Consequently, the final average conversion will be lower in the LFTR than in a
PFR.
Copyright© - George W. Roberts 1 June 6, 2008
Solution 10-5 (NIR-14)
Question 1:
Let: � = volumetric feed rate to reactor (L/h)
�z = volumetric exchange rate between V1 and V2 (L/h)
I° = initiator (AIBN) concentration in feed (mol/L)
I1, I2 = initiator concentrations in Regions 1 and 2, respectively (mol/L)
kI = rate constant for initiator decomposition (s
-1)
Initiator Balance – Region 1:
Rate In – Rate Out = Rate of Disappearance
�I° + �zI2 – �zI1 – �I1 = V1kII1 (1)
Initiator Balance – Region 2:
�zI1 – �zI2 – �I1 = V2kII2
I2 = [�z/(�z + V2kI)]I1 (2)
Substituting Eqn.(2)into Eqn.(1):
 
I1 =
�
�z + � + V1kI � [�z2 /(�z + V2kI)]
� 
� 
� 
� 
� 
� 
I0 (3)
Substituting Numbers:
V1kI = 700(L)*9.25(s
-1)*3600(s/h) = 23.3x106(L/h)
V2kI = 300(L)*9.25(s
-1)*3600(s/h) = 9.99x106(L/h)
[�z/(�z + V2kI) = 100(L/h)/(100(L/h) + 9.99x106(L/h)) = 10.0x10-6
[�z2/(�z + V2kI)] = 100(L/h)*10.0*10.0x10-6 = 10-3(L/h)
 
I1 =
1500(L/h)
100(L/h) + 1500(L/h) + 23.3x106 (L/h) � 10�3 (L/h)
� 
� 
� 
� 
� 
� 
x0.010(mol/L)
I1 = 6.44x10
-7 (mol/L)
From Eqn.(2),
I2 = 10.0x10
-6x6.44x10-7 mol/L
I2 = 6.44x10
-12 mole/L
Question 2:
Let: S° = styrene concentration in feed (mol/L)
S1, S2 = concentrations of styrene in Regions 1 and 2, respectively (mol/L)
Styrene Balance – Region 2:
Rate In – Rate Out = Rate of Disappearance
�zS1 – �zS2 = V2kS2 I2
1/2
 
S2 =
�z
�z + V2kI21/2
� 
� 
� 
� 
� 
� S1 (4)
Styrene Balance – Region 1:
 �S
0
+ �zS2 � �S1 � �zS1 = V1kI11/2S1
Substituting Eqn.(4):
 
�S0 + �z
2
�z + V2kI21/2
� 
� 
� 
� 
� 
� S1 = S1 (� + �z + V1kI11/2 )
Copyright© - George W. Roberts 2 June 6, 2008
 
S1
S0
= (1� xS ) =
�
� + �z + V1kI11/2 �
�z2
�z + V2kI21/2
� 
� 
� 
� 
� 
� 
� 
� 
 
 
	 
 
 
� 
� 
 
 
 
 
 
Here, xS is the fractional conversion of styrene monomer.
Numbers:
V1k I1
1/2 = 700(L)x0.925(L1/2/mol1/2-s)x3600(s/h)x(64.4x10-8)1/2(mol/L)1/2
V1k I1
1/2 = 1871 (L/h)
V2k I2
1/2 = 300(L)x0.925(L1/2/mol1/2-s)x3600(s/h)x(6.44x10-12)1/2(mol/L)1/2
V2k I2
1/2 = 2.54 (L/h)
�z2/(�z + V2k I2
1/2 ) = 1002(L/h)2/[100(L/h) + 2.54(L/h)] = 97.5 (L/h)
 
S1
S0
= (1� xM) =
1500(L/h)
1500(L/h) + 100(L/h) + 1871(L/h) � 97.5(L/h)
� 
� 
� 
� 
� 
� 
 = 0.445
xS = 0.555
Question 3:
Let n = average number of monomer units (styrene molecules) per molecule of
polymer
 
n =
rate of styrene consumption
2 x rate of initiator decomposition
The factor of 2 in the above equation results from the fact that each initiator molecule
starts 2 polymer chains, plus the fact that there is 1/2 of an initiator molecule in each
molecule of “dead” polymer.
 
n =
�(S0 � S1 )
2�(I0 � I1 )
=
(S0 � S1 )
2(I0 � I1 )
S1 = 0.445xS
0 =0.445x4.0(mol/L) = 1.78 (mol/L)
 
n =
(4.0 � 1.78)(mol/L)
2(0.010 � 6.44x10�7 )(mol/L)
 n = 111 styrene molecules per polymer molecule
This corresponds to a number-average polymer molecular weight of approximately 111
x 104 = 11,544 (104 is the molecular weight of styrene; the presence of the initiator
fragment in the polymer molecule has been neglected).
Copyright� - George W. Roberts 1 July 4, 2008
Solution 10-8 (NIR-16)
1) The fluid that flows at a given radius, r, will remain in the reactor for a time, t,
given by:
 
t =
L
u(r)
=
�R2L
�
� 
� 
� 
� 
� 
� /2 1�
r
R
� 
� � 
� 
� � 
2� 
	 
� 
 � 
� 
 
� 
� � 
(1)
The fraction of the total flow that leaves the reactor between t and (t +dt) is:
 
E(t)dt =
u(r)2�rdr
�
=
(L/t)2�rdr
�
(2)
Let � = V/� = �R2L/�. From Eqn.(1),
 
t = �/2 1� r
R
� 
� � 
� 
� � 
2� 
� 
 
	 
 
� 
� 
 
 
 
,
 
�
2t
= 1� r
R
� 
� � 
� 
� � 
2� 
� 
 
	 
 
� 
� 
 
 
 
,
 
r
R
� 
� � 
� 
� � 
= 1� �
2t
 
dr
R
� 
� � 
� 
� � 
=
1
1� �
2t
��
2
� 
� � 
� 
� � 
�1
t2
� 
� � 
� 
� � 
dt =
�dt
4t2 (r/R)
 
rdr =
R2�dt
4t2
Substituting this result into Eqn.(2),
 
E(t)dt =
(L/t)2� R
2�dt
4t2
� 
� 
� 
� 
� 
	 
�
=
�2dt
2t3
 
E(t) =
�2
2t3
=
(�R2L/�)2
2t3
This result is valid when t � �/2. For t < �/2, the velocity at r = 0 (where the
velocity is highest) is not high enough for the fluid to have traveled a distance, L.
Therefore,
E(t) = 0, t < �/2
E(t) = �2/2t3 = (�R2L/�)2/2t3, t � �/2
2) The mean residence time is given by:
 
t = tE(t)dt
0
�
�
For the laminar-flow reactor,
 
t = t
�2
2t3
dt
�/2
�
� = �
2
2
dt
t2�/2
�
� = �
2
2
�1
t
� 
� 	 
� 
� 
 �/2
�
= �
 t = � =�R2L/�
3) The second moment of E(t) about t , i.e., the variance is given by:
Copyright� - George W. Roberts 2 July 4, 2008
 
�2 = (t � t )2 E(t)dt
0
�
� = (t2 � 2tt + t 2 )E(t)dt
0
�
�
�2 = t2E(t)dt � 2t 
0
�
� tE(t)dt
0
�
� + t 2 E(t)dt
0
�
� = t2E(t)dt � t 2
0
�
�
�2 + t 2 = t2E(t)dt
0
�
�
For the laminar-flow reactor,
 
�2 + t 2 = t2 �
2
2t3
� 
� 
 
� 
	 
� dt
0
�
� = �
2
2
ln t[ ]�/2
�
=
�2
2
ln�� ln(�/2)[ ]
�2 = �
4) For a macrofluid,
 
x A = xA (t)E(t)dt
0
�
� (3)
For a second-order reaction with no volume change in an isothermal batch reactor:
 
�dCA
dt
= kCA = CA0
dxA
dt
= kCA0
2 (1� xA )2
dxA
(1� xA )20
xA
� = kCA0 dt
0
t
�
1
(1� xA ) 0
xA
= kCA0t
xA (t) =
kCA0t
(1+ kCA0t)
From Eqn.(3),
 
x A =
kCA0tE(t)dt
(1 + kCA0t)0
�
�
For the laminar-flow reactor,
 
x A =
kCA0t(�2 /2t3 )dt
(1 + kCA0t)�/2
�
� = kCA0�
2
2
dt
t2 (1+ kCA0t)�/2
�
�
 
x A =
kCA0�2
2
� 1
t
+ kCA0 ln
1+ kCA0t
t
� 
	 � 
 
� � 
� 
� 
 
� 
� 
� 
�/2
�
 
x A = kCA0� +
(kCA0�)2
2
ln
kCA0�
2 + kCA0�
� 
� 
� 
� 
� 
� 
Copyright - George W. Roberts 1 July 18, 2008
Problem 10-9 (NIR-19)
1) Amount of tracer injected (N):
 
€ 
N = υ C(t)dt
0
∞
∫
υ = volumetric flowrate = 0.10 L/min
C(t) = concentration of tracer in effluent from vessel (mmol/L)
The integral in the above equation can be evaluated numerically using Simpson’s rule
(see attached Excel spreadsheet).
 
€ 
C(t)dt ≅ 151(mmol −min/L)
0
∞
∫
N = 0.10 (L/min)x151(mmol-min/L)
N = 15.1 mmol
2) Volume
τ = V/υ = 
€ 
t = 
 
€ 
tC(t)dt/ C(t)dt
0
∞
∫
0
∞
∫
The values of the two integrals in the above equation are calculated in the attached
spreadsheet.
 
€ 
t = V/0.10 (L/min) = 1349 (mmol-min2/L)/151(mmol-min/L) = 8.93 min
V = 0.893 L
3) Variance
 
€ 
σ2 = (t − t )2 C(t)dt/ C(t)dt
0
∞
∫
0
∞
∫ =
t2C(t)dt − 2t tC(t)dt + t 2 C(t)dt
0
∞
∫
0
∞
∫
0
∞
∫
C(t)dt
0
∞
∫
 
€ 
σ2 =
t2C(t)dt
0
∞
∫
C(t)dt
0
∞
∫
− 2t 2 + t 2 =
t2C(t)dt
0
∞
∫
C(t)dt
0
∞
∫
− t 2
Substituting values of integrals from spreadsheet, and value of 
€ 
t from Part 2:
 
€ 
σ2 =
13189(mmol−min3/L)
151(mmol−min/L) − (8.93)
2 min2
σ2 = 7.60 min2
Copyright - George W. Roberts 2 July 18, 2008
PROBLEM NIR-19
Simpson's 1/3 Rule: Value of Integral =(h/3)*[f(0)+4f(1)+2f(2)+4f(3)+2f(4)+…+4f(N-1)+f(N)]
Tracer
Time Concentration
(min) (mmol/L) Factor C*Factor t*C*Factor t^2*C*Factor
0 0
1 0
2 0
3 1 1 1 3 9
4 2 4 8 32 128
5 8 2 16 80 400
6 18 4 72 432 2592
7 25 2 50 350 2450
8 22 4 88 704 5632
9 19 2 38 342 3078
10 16 4 64 640 6400
11 13 2 26 286 3146
12 10 4 40 480 5760
13 7 2 14 182 2366
14 5 4 20 280 3920
15 3 2 6 90 1350
16 2 4 8 128 2048
17 1 1 1 17 289
18 0
Sums= 452 4046 39568
Value of Integral= 151 1349 13189
min-mmol/L min^2-mmol/ min^3-mmol/
L L
	Solution 10-1(NIR-11).pdf
	Solution 10-2(NIR-13).pdf
	Solution 10-3 (NIR-4).pdf
	Solution 10-4(NIR-22).pdf
	Solution 10-5(NIR-14).pdf
	Solution 10-6(NIR-17).pdf
	Solution 10-7(NIR-18).pdf
	Solution 10-8.pdf
	Solution 10-9.pdf
	Solution 10-10(NIR-20).pdf
	Solution 10-11.pdf
	Solution 10-12(NIR-2).pdf