Assignment_2_Solution

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ASSIGNMENT 2 CHE 3473 
Solution 
#Problem 1: 3.3 
a) One kilogram of steam contained in a horizontal frictionless piston and cylinder is heated at 
constant pressure of 1.013 bar from 125 C to such a temperature that its volume doubles. 
Calculate the amount of heat that must be added to accomplish this change, the final 
temperature of the steam, the internal energy and enthalpy changes of the steam for this 
process. 
b) Repeat for heating at constant volume to a pressure that is twice the initial pressure 
c) Repeat part a assuming ideal gas with heat capacity of 34.4 J/mol K 
d) Repeat part b assuming ideal gas 
Choose system to be the contents of the piston and cylinder 
Stage 1: system at the initial state 
 2: system at the final state 
a) Mass balance equation for closed system 
M2 – M1 = 0  M1 = M2 = M 
Energy balance equation for closed system, no shaft work with negligible kinetic and potential 
energy 
𝑈2 − 𝑈1 = 𝑄 − ∫ 𝑃𝑑𝑉 = 𝑄 − 𝑃(𝑉2 − 𝑉1) 
𝑄 = 𝐻2 − 𝐻1 
At initial state 
P1 = 1.013 bar ≈ 0.1 MPa 
T1 = 125oC 
Interpolation from data table in appendix AIII, we get 
�̂�1 = 2726.3
𝑘𝐽
𝑘𝑔
 
�̂�1 = 1.8161
𝑚3
𝑘𝑔
 
�̂�1 = 2544.8
𝑘𝐽
𝑘𝑔
 
At final state 
P2 = 1.013 bar ≈ 0.1 MPa 
V2 = 2V1  �̂�2 = 2�̂�1 = 3.6322
𝑚3
𝑘𝑔
 
With known values of P2 and �̂�2, we could interpolate other values as below 
T2 = 514.5oC 
�̂�2 = 3519.5
𝑘𝐽
𝑘𝑔
 
�̂�2 = 3156.3
𝑘𝐽
𝑘𝑔
 
 Q = H2 – H1 = M(�̂�2 − �̂�1) = 793.2 kJ 
W = – P(V2 – V1) = −105
𝑁
𝑚2
∗ 1.8161
𝑚3
𝑘𝑔
∗ 1𝑘𝑔 = −181.61 𝑘𝐽 
∆U = U2 – U1 = M(�̂�2 − �̂�1) = 611.5 kJ 
∆H = Q = 793.2 kJ 
b) Mass balance equation for closed system 
M2 – M1 = 0  M1 = M2 = M 
Energy balance equation for closed system, no shaft work with negligible kinetic and potential 
energy, no change of volume of system 
U2 – U1 = Q 
At final state 
P2 = 0.2 MPa ; �̂�2 = 1.8161
𝑚3
𝑘𝑔
 (because V1 = V2; M1 = M2) 
Interpolation gives us 
T2 = 514.98oC 
�̂�2 = 3156.4
𝑘𝐽
𝑘𝑔
 
�̂�2 = 3519.6
𝑘𝐽
𝑘𝑔
 
Then we have 
Q = U2 – U1 = M(�̂�2 − �̂�1) = 611.6 kJ 
W = 0 
∆U = Q = 611.6 kJ 
∆H = M(�̂�2 − �̂�1) = 793.3 kJ 
c) Mass balance equation for closed system 
M2 – M1 = 0  M1 = M2 = M  N1 = N2 = N 
𝑃1𝑉1
𝑅𝑇1
=
𝑃2𝑉2
𝑅𝑇2
 
Since P1 = P2; V2 = 2V1 
 T2 = 2T1 = 2(125 + 273.15) = 796.3 K 
Q = ∆H = NCp∆T = 
1000
18
(𝑚𝑜𝑙) ∗ 34.4
𝐽
𝑚𝑜𝑙𝐾
∗ (796.3 − 398.15)𝐾 = 760.9𝑘𝐽 
𝑊 = −𝑃∆𝑉 = −𝑃 (
𝑁2𝑅𝑇2
𝑃2
−
𝑁1𝑅𝑇1
𝑃1
) = −𝑁𝑅(𝑇2 − 𝑇1) = −183.9𝑘𝐽 
∆U = Q + W = 577 kJ 
d) Mass balance equation for closed system 
M2 – M1 = 0  M1 = M2 = M  N1 = N2 = N 
𝑃1𝑉1
𝑅𝑇1
=
𝑃2𝑉2
𝑅𝑇2
 
Since 
P2 = 2P1; V2 = V1 
 T2 = 2T1 = 2(125 + 273.15) = 796.3 K 
Q = ∆U = NCv∆T = 
1000
18
(𝑚𝑜𝑙) ∗ (34.4 − 8.314)
𝐽
𝑚𝑜𝑙𝐾
∗ (796.3 − 398.15)𝐾 
= 577 kJ 
W = 0 due to ∆V = 0 
∆H = NCp∆T = 760.9 kJ 
#Problem 2 3.4 
In Joule’s experiments, the slow lowering of a weight turned a stirrer in an insulated container of 
water. As a result of viscosity, the kinetic energy transferred from the stirrer to the water 
eventually dissipated. In this process the potential energy of the water was first converted to 
kinetic energy of the stirrer and the water, and then as a result of viscous forces, the kinetic 
energy of the water was converted to thermal energy apparent as a rise in temperature. 
Assuming no friction in the pulleys and no heat losses, how large a temperature rise would be 
found in 1 kg of water as a result of a 1-kg weight being lowered 1 m? 
𝑚𝑔∆ℎ = 𝑚𝐶𝑝∆𝑇 
∆𝑇 =
𝑔∆ℎ
𝐶𝑝
 
∆𝑇 =
9.816 𝑚
2
𝑠2
⁄
4184 𝑚
2
𝑠2𝐾⁄
 
∆𝑇 = 0.002346 𝐾 
 
#Problem 3 3.5 
Steam at 500 bar and 600 C is to undergo a Joule-Thompson expansion to atmospheric pressure. 
What will be the temperature of the steam after the expansion? What would be the downstream 
temperature if the steam were replaced by an ideal gas? 
Steam undergoes a Joule-Thomson expansion from 500bar, 600oC to atmospheric pressure (≈ 
1bar) would have 
𝐻1(500𝑏𝑎𝑟, 600
𝑜𝐶) = 𝐻2(1𝑏𝑎𝑟, 𝑇2) 
From appendix A.III we have 
𝐻1(500𝑏𝑎𝑟, 600
𝑜𝐶) = 3247.6 (
𝑘𝐽
𝑘𝑔
) = 𝐻2(1𝑏𝑎𝑟, 𝑇2) 
With known values of P2, 𝐻2 we could obtain T2 = 385
oC from the steam table 
 
If the steam were replaced by an ideal gas, since enthalpy of ideal gas is a function of temperature 
only, we easily obtain T2 = T1 = 600oC 
#Problem 4 3.6 
Water in an open metal drum is to be heated from room Temperature (25 C) to 80 C by adding 
steam slowly enough that all the steam condenses. The drum initially contains 100 kg of water, 
and steam is supplied at 3.0 bar and 300 C. How many kilograms of steam should be added so 
that the final temp of the water is exactly 80 C. Neglect all heat losses. 
Choose system to be the contents inside the drum 
Mass balance equation for an open system 
M2 – M1 = Msteam 
In which M1: mass in the drum at initial time 
M2: mass in the drum at final time 
Msteam: mass of steam into the drum 
Energy balance equation with neglect of kinetic energy and potential energy change 
U2 – U1 = Msteam�̂�𝑠𝑡𝑒𝑎𝑚 + Q + Ws – ∫ 𝑃𝑑𝑉 
with 
Q = 0; Ws = 0, 
Liquids are incompressible, so 
∫ 𝑃𝑑𝑉 = 𝑃∆𝑉 ≈ 0 
 �̂�𝑠𝑡𝑒𝑎𝑚 = 3069.3 (𝑘𝐽/𝑘𝑔) (from appendix AIII) 
 We assume that internal energy of liquid water is independent of pressure, so that we have 
�̂�1(25 
𝑜𝐶, 1.013𝑏𝑎𝑟) = �̂�1(25 
𝑜𝐶, 3.169𝑘𝑃𝑎) = 104.88 (𝑘𝐽/𝑘𝑔) 
�̂�2(80 
𝑜𝐶, 1.013𝑏𝑎𝑟) = �̂�2(80 
𝑜𝐶, 47.39𝑘𝑃𝑎) = 334.86 (𝑘𝐽/𝑘𝑔) 
Substitute into energy balance equation, we have 
𝑀2�̂�2 − 𝑀1�̂�1 = 𝑀𝑠𝑡𝑒𝑎𝑚�̂�𝑠𝑡𝑒𝑎𝑚 
(100 + 𝑀𝑠𝑡𝑒𝑎𝑚) ∗ 334.86 − 100 ∗ 104.88 = 𝑀𝑠𝑡𝑒𝑎𝑚 ∗ 3069.3 
 Msteam = 8.41 kg 
 
 
#Problem 5 3.7 
Consider the following statement: “The adiabatic work necessary to cause a given change of state 
in a closed system is independent of the path by which that change occurs. 
a) Consider a process that changes system from state 1 to state 2 (state 1 and state 2 are fixed) 
Energy balance equation for closed system 
{𝑈 + 𝑀(
𝑣2
2
+ 𝜑)}
2
− {𝑈 + 𝑀 (
𝑣2
2
+ 𝜑)}
1
= 𝑄 + 𝑊 
State 1 and state 2 are fixed 
 Left hand side of the above equation = constant 
 Q + W = constant 
But Q = 0 for any adiabatic path 
 W = constant for any adiabatic path 
 
In illustration 3.4-6, we found that (Q + W) is the same for all path. This statement must also be 
hold in any adiabatic path. However, in adiabatic path, Q + W = W, thus W is the same for all 
adiabatic path. Because of this, this conclusion is not in contradict with illustration 3.4-16 
 
b) Consider a system receives work Wa to change from state 1 to state 2 in an adiabatic path. Then 
it releases work Wb to get back to state 1 in another adiabatic path 
For these two different paths, energy balance equations for closed system, no heat exchange, with 
neglect of kinetic and potential energy change give us (assume that U2 > U1) 
U2 – U1 = Wa (Wa > 0) 
U1 – U2 = Wb (Wb < 0) 
 Wa + Wb = 0 or |𝑊𝑏| = |𝑊𝑎| 
 
If |𝑊𝑏| > |𝑊𝑎| then Wa + Wb < 0: we have generated more energy than we put in. 
If |𝑊𝑏| < |𝑊𝑎| then Wa + Wb > 0: we just add energy to the system that goes nowhere. 
In the first case, if we add the work Wa to the system to change from state 1 to state 2, and then 
get back to state 1 on the path generating work Wb, then we would obtain a machine that generates 
energy. 
 
 
#Problem 6 3.8 
A nonconducting tank of negligible heat capacity and 1 m3 volume is connected to a pipeline 
containing steam at 5 bar and 370 C, filled with steam to a pressure of 5 bar, and disconnected 
from the pipeline. 
Mass balance equation for open system 
M2 – M1 = Msteam 
Energy balance equation for open system with neglect of change in kinetic and potential energy, 
no shaft work, constant volume 
𝑈2 − 𝑈1 = 𝑀𝑠𝑡𝑒𝑎𝑚�̂�𝑖𝑛 + 𝑄 
Since tank is nonconductingand has negligible heat capacity  Q = 0 
𝑈2 − 𝑈1 = 𝑀𝑠𝑡𝑒𝑎𝑚�̂�𝑖𝑛 
a) If the tank is initially evacuated: M1 = 0 
 M2 = Msteam 
Energy equation balance becomes 
𝑀2�̂�2 = 𝑀𝑠𝑡𝑒𝑎𝑚�̂�𝑖𝑛 
But 
M2 = Msteam, thus 
�̂�2 = �̂�𝑖𝑛 = 3209.38 (
𝑘𝐽
𝑘𝑔
) 𝑎𝑡 370𝑜𝐶 𝑎𝑛𝑑 5𝑏𝑎𝑟 
With known values of P2 and �̂�2 we can interpolate to get other data and obtain 
T2 = 547.3oC ; �̂�2 = 0.755 
𝑚3
𝑘𝑔
 
𝑀2 =
𝑉2
�̂�2
=
1
0.755
= 1.325(𝑘𝑔) 
 Msteam = M2 = 1.325 kg 
 
 
 
 
 
 
 
 
b) Now we have 
T1 = 150oC; P1 = 1bar 
 �̂�1= 1.9364 m
3/kg ; �̂�1= 2582.8 kJ/kg 
thus 
𝑀1 =
𝑉1
�̂�1
= 0.5164 (𝑘𝑔) 
Mass balance equation becomes 
M2 – M1 = Msteam  M2 = Msteam + M1 
Energy balance equation is 
(𝑀𝑠𝑡𝑒𝑎𝑚 + 𝑀1)�̂�2 − 𝑀1�̂�1 = 𝑀𝑠𝑡𝑒𝑎𝑚�̂�𝑖𝑛 
(𝑀𝑠𝑡𝑒𝑎𝑚 + 0.5164)�̂�2 − 0.5164 ∗ 2582.8 = 𝑀𝑠𝑡𝑒𝑎𝑚 ∗ 3209.38 
𝑀𝑠𝑡𝑒𝑎𝑚(�̂�2 − 3209.38) ∗ 2582.8 = 1333.76 − 0.5164�̂�2 
 
Due to lack of equations, we have to guess an initial value of T2, then infer values of �̂�2, �̂�2. Plug 
back value of �̂�2 into the above equation, we can calculate Msteam, and then M2 
With values of V2 and M2 we calculate �̂�2
′ =
𝑉2
𝑀2
. This value should be close enough to �̂�2 
(difference should be less than 3%) 
Finally we obtain T2 = 425oC 
Check: with T2 = 425oC; P2 = 5bar, we get 
�̂�2 = 3004.5
𝑘𝐽
𝑘𝑔
 
�̂�2 = 0.6407 
𝑚3
𝑘𝑔
 
 Msteam = 1.06288 kg 
 M2 = 1.57928 kg 
 �̂�2
′ =
𝑉2
𝑀2
= 0.633 
𝑚3
𝑘𝑔
 
(∆𝑉 =
�̂�2 − �̂�2′
�̂�2
= 1.2%) 
 
 
 
#Problem 7 3.11 
A frictionless piston-and-cylinder system shown here is subjected to 1.013 bar external pressure. 
The piston’s mass is 200 kg, it has an area of 0.15 m2 and the initial volume of the entrapped ideal 
gas is 0.12 m3. The piston and cylinder do not conduct heat, but heat can be added to the gas by 
a heating coil. The gas has a constant-volume heat capacity of 30.1 J/(mol K) and an initial 
temperature of 298 K, and 10.5 kJ of energy are to be supplied to the gas through the heating 
coil. 
a) Initial pressure of the system (contents in piston-cylinder) 
𝑃1 = 1.013 ∗ 10
5𝑃𝑎 +
200𝑘𝑔 ∗ 9.81
𝑚
𝑠2
0.15𝑚2
= 114380 𝑃𝑎 
For ideal gas we have 
𝑁1 =
𝑃1𝑉1
𝑅𝑇1
=
114380 ∗ 0.12
8.314 ∗ 298
= 5.54 𝑚𝑜𝑙 
Mass balance equation is 
M1 – M2 = 0  M1 = M2  N1 = N2 
Energy balance equation is 
𝑈2 − 𝑈1 = 𝑄 + 𝑊𝑠 − ∫ 𝑃𝑑𝑉 
Ws = 0 
∆V = 0 
 ∆U = Q = 10.5 kJ 
Thus 
∆𝑇 =
∆𝑈
𝑁𝐶𝑣
=
10500𝐽
5.54𝑚𝑜𝑙 ∗ 30.1 
𝐽
𝑚𝑜𝑙𝐾
= 62.97𝐾 
 
 T2 = T1 + ∆T = 360.97 K 
 
𝑃2 =
𝑁2𝑅𝑇2
𝑉2
=
5.54𝑚𝑜𝑙 ∗ 8.314
𝐽
𝑚𝑜𝑙𝐾
∗ 360.97𝐾
0.12𝑚3
= 138550 𝑃𝑎 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
b) Piston is allowed to move freely 
 P2 = P1 = 114380 Pa 
 
From mass balance equation 
N1 = N2 = 5.54 mol 
 
Energy balance equation for closed system with neglect of change in kinetic and potential 
energy is 
U2 – U1 = Q – P(V2 – V1) 
 Q = H2 – H1 = NCp(T2 – T1) 
With 
Cp = Cv + R = 30.1 + 8.314 = 38.414 J/molK 
Thus 
𝑇2 = 𝑇1 +
𝑄
𝑁𝐶𝑝
= 298 + 
10500𝐽
5.54𝑚𝑜𝑙 ∗ 38.414
𝐽
𝑚𝑜𝑙𝐾
= 347.34𝐾 
𝑉2 =
𝑁2𝑅𝑇2
𝑃2
=
5.54𝑚𝑜𝑙 ∗ 8.314
𝐽
𝑚𝑜𝑙𝐾
∗ 347.34𝐾
114380𝑃𝑎
= 0.1399𝑚3 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
#Problem 8 3.15 
An isolated chamber with rigid walls is divided into two equal compartments, one containing gas 
and the other evacuated. The partition between the compartments ruptures. After the passage 
of a sufficiently long period of time, the temperature and pressure are found to be uniform 
throughout the chamber. 
a) If the filled compartment initially contains an ideal gas of constant heat capacity at 1MPa and 
500 K. what are the final temperature and pressure in the chamber? 
Choose a system to be contents of both two compartments, named a and b. 
Mass balance equation for this system is 
M2 – M1 = 0  M1 = M2  N1 = N2 
With 
N1 = N1a + N1b = N1a  N1a = N2 
Energy balance equation is 
U2 – U1 = 0  U2 = U1 
With 
U1 = U1a + U1b = U1a  U1a = U2  T1a = T2 = 500K 
Since ideal gas is only a function of temperature. 
We have 
P1aV1a = N1aRT1a 
P2V2 = N2RT2 
With 
N1a = N2; T1a = T2; V2 = 2V1a 
 P2 = 0.5P1a = 0.5 MPa 
b) If the filled compartment initially contains steam at 1 MPa and 500 K, what are the final 
temperature and pressure in the compartment? 
For steam, we still have the following results 
N2 = N1a; M2 = M1a 
U2 = U1a 
V2 = 2V1a 
Thus 
�̂�2 = �̂�1𝑎 
�̂�2 = 2�̂�1𝑎 
Initially, at 1MPa and 500K 
�̂�1𝑎 = 2669.1 𝑘𝐽/𝑘𝑔 
�̂�1𝑎 = 0.2203 
𝑚3
𝑘𝑔
 
Thus 
�̂�2 = 2669.1 𝑘𝐽/𝑘𝑔 
�̂�2 = 0.4406 
𝑚3
𝑘𝑔
 
From these two intensive properties, state 2 is totally defined 
From steam table  T2 ≈ 216.1oC; P2 ≈ 0.5 MPa 
c) Repeat part (a) if the second compartment initially contains an ideal gas, but at half the pressure 
and 100 K higher temperature. 
Mass balance equation for this system is 
M2 – M1 = 0  M2 = M1  N2 = N1  N2 = N1a + N1b 
so 
𝑃2𝑉2
𝑇2
=
𝑃1𝑎𝑉1𝑎
𝑇1𝑎
+
𝑃1𝑏𝑉1𝑏
𝑇1𝑏
 
With 
V2 = 2V1a = 2V1b 
 
2𝑃2
𝑇2
=
𝑃1𝑎
𝑇1𝑎
+
𝑃1𝑏
𝑇1𝑏
=
1
500
+
0.5
600
= 2.833 ∗ 10−3 
Energy balance equation is 
U2 – U1 = 0 
 U2 – U1a – U1b = 0 
For ideal gas, this equation becomes 
𝑃1𝑎𝑉1𝑎
𝑅𝑇1𝑎
𝐶𝑣(𝑇2 − 𝑇1𝑎) +
𝑃1𝑏𝑉1𝑏
𝑅𝑇1𝑏
𝐶𝑣(𝑇2 − 𝑇1𝑏) = 0 
1
500
(𝑇2 − 500) +
0.5
600
(𝑇2 − 600) = 0 
 T2 = 529.4K 
 P2 = 529.4*2.833*10-3/2 = 0.75 MPa 
d) Repeat part (b) if the second compartment initially contains steam, but at half the pressure 
and 100 K higher temperature. 
From mass balance equation, we have 
𝑀2 = 𝑀1𝑎 + 𝑀1𝑏 
𝑉2
�̂�2
=
𝑉1𝑎
�̂�1𝑎
+
𝑉1𝑏
�̂�1𝑏
 
With 
V2 = 2V1a = 2V1b 
�̂�1𝑎 = 0.2203 
𝑚3
𝑘𝑔
 𝑎𝑡 1𝑀𝑃𝑎, 500𝐾 
�̂�1𝑏 = 0.5481 
𝑚3
𝑘𝑔
 𝑎𝑡 0.5𝑀𝑃𝑎, 600𝐾 
Plug back into the above equation and solve for �̂�2, we obtain 
�̂�2 = 0.3143
𝑚3
𝑘𝑔
 
From energy balance equation, we have 
U2 – U1 = 0  U2 = U1 
 
2𝑉2
�̂�2
�̂�2 =
𝑉1𝑎
�̂�1𝑎
�̂�1𝑎 +
𝑉1𝑏
�̂�1𝑏
�̂�1𝑏 
 
2�̂�2
0.3143
=
2669.1
0.2203
+
2845.7
0.5481
 
 
�̂�2 = 2719.9
𝑘𝐽
𝑘𝑔
 
From two values of �̂�2 and �̂�2, from steam data table, we obtain 
T2 = 284oC; P2 = 0.8 MPa 
 
Another estimate results in T2=253C and P2=0.76MPa. Use your judgment. 
#Problem 9 3.22 
Nitrogen gas is being withdrawn at the rate of 4.5 g/s from a 0.15-m3 cylinder, initially containing 
the gas at a pressure of 10 bar and 320 K. The cylinder does not conduct heat, nor does its 
temperature change during the emptying process. What will be the temperature and pressure of 
the gas in the cylinder after 5 minutes? What will be the rate of change of the gas temperature 
at this time? Nitrogen can be considered to be an ideal gas with C*p = 30 J/(mol K) 
Choose a system as contents inside the cylinder 
From Mass balance equation, we have 
𝑑𝑁
𝑑𝑡
= �̇� 
 
𝑉
𝑅
𝑑(
𝑃
𝑇
)
𝑑𝑡
= �̇� 
 
𝑑(
𝑃
𝑇
)
𝑑𝑡
= �̇�
𝑅
𝑉
=
−
4.5
28
𝑚𝑜𝑙
𝑠
∗ 8.314
𝐽
𝑚𝑜𝑙𝐾
0.15𝑚3
= −8.908 ∗ 10−5
𝑏𝑎𝑟
𝐾. 𝑠
 
Integrate this equation, we obtain 
(
𝑃
𝑇
)𝑓𝑖𝑛𝑎𝑙 = (
𝑃
𝑇
)𝑖𝑛𝑖𝑡𝑖𝑎𝑙 − 8.908 ∗ 10
−5𝑡 =
10
320
− 8.908 ∗ 10−5𝑡 (1) 
From illustration 3.4-5 in textbook, we have 
(
𝑇𝑓𝑖𝑛𝑎𝑙
𝑇𝑖𝑛𝑖𝑡𝑖𝑎𝑙
)
𝐶𝑝
𝑅 =
𝑃𝑓𝑖𝑛𝑎𝑙
𝑃𝑖𝑛𝑖𝑡𝑖𝑎𝑙
 
𝑇
𝑓𝑖𝑛𝑎𝑙
30
8.314
𝑃𝑓𝑖𝑛𝑎𝑙
=
𝑇
𝑖𝑛𝑖𝑡𝑖𝑎𝑙
30
8.314
𝑃𝑖𝑛𝑖𝑡𝑖𝑎𝑙
=
320
30
8.314
10
 (2) 
From (1)&(2)  Tfinal = 152.57K; Pfinal = 0.6907bar 
We have 
𝐶𝑣
𝑅
𝑑𝑙𝑛𝑇
𝑑𝑡
=
𝑑𝑙𝑛(
𝑃
𝑇
)
𝑑𝑡
=
𝑑𝑙𝑛𝑃
𝑑𝑡
−
𝑑𝑙𝑛𝑇
𝑑𝑡
=
𝑑𝑃
𝑃𝑑𝑡
−
𝑑𝑙𝑛𝑇
𝑑𝑡
 
𝐶𝑣
𝑅
𝑑𝑙𝑛𝑇
𝑑𝑡
=
𝑇
𝑃
[
1
𝑇
(
𝑑𝑃
𝑑𝑡
−
𝑃𝑑𝑙𝑛𝑇
𝑑𝑡
)] =
𝑇
𝑃
𝑑(
𝑃
𝑇
)
𝑑𝑡
=
�̇�𝑅𝑇
𝑃𝑉
 
 
𝐶𝑣
𝑅𝑇
𝑑𝑇
𝑑𝑡
=
�̇�𝑅𝑇
𝑃𝑉
 
 
𝑑𝑇
𝑑𝑡
=
𝑁(̇ 𝑅𝑇)2
𝐶𝑣𝑃𝑉
 
At t = 5 mins 
P = 0.6907 bar 
T = 152.57 K 
Cv = Cp – R = 30 – 8.314 = 21.686 J/molK 
Plub back into the above equation,we get 
𝑑𝑇
𝑑𝑡
= −1.151 
𝐾
𝑠
 
 
 
 
 
Alternatively, here is a little different way of looking at the problem: 
 
Open system, adiabatic, no shaft work, and no expansion work. So, the gas leaving the cylinder 
must have the same properties as the gas entering the valve. From the energy balance, we get 
the following: 
𝒅(𝑵𝑼)
𝒅𝒕
= −𝑵�̇� 
( )
( ) ( )
( )
d NU
NH
dt
d U d N dN
N U H
dt dt dt
NdU UdN HdN
NdU UdN HdN
NdU H U dN

 
 
 
 
 
 
Now, we can use definitions of enthalpy and internal energy for ideal gas: 
 
( ) ( )( ) ( )
( )( ) ( )
R R R R
R R R R
H Cp T T H Cv R T T U Cv T T RT
H U Cv R T T Cv T T RT RT
       
        
 
Now, plugging into equation above, and cancelling out terms, we get: 
)(RTdNNCvdT 
and finally, using separation of variables, we get: 
N
dN
T
dT
R
Cv
 
integrating, we get: 
𝐶𝑣
𝑅
ln (
𝑇2
𝑇1
) = ln (
𝑁2
𝑁1
) 
Now, we rearrange and the final number of moles can be computed by multiplying the rate at 
which nitrogen is being withdrawn, times 5 minutes. 
𝑇2 = 𝑇1 (
𝑁2
𝑁1
)
(
𝑅
𝐶𝑣)
 
𝑁2 = 𝑁1 −
𝑑𝑁
𝑑𝑡
∗ 𝑡 
𝑇2 = 𝑇1 (
𝑁2
𝑁1
)
(
𝑅
𝐶𝑣)
= 𝟏𝟓𝟐. 𝟓𝟗𝑲 
using the ideal gas law, we get that: P=0.691bar 
For the change, since T is a function of time, the above equation also can be written 
𝑇(𝑡) = 𝑇1 (
𝑁(𝑡)
𝑁1
)
(
𝑅
𝐶𝑣)
 
So to find dT/dt, we take the derivative of both sides with respect to t. Everything but N(t) is a 
constant, so we use our excellent Calculus I skills to find that 
𝑑𝑇
𝑑𝑡
= 𝑇1(
𝑅
𝐶𝑣
) (
1
𝑁1
)
(
𝑅
𝐶𝑣)
𝑁(𝑡)(
𝑅
𝐶𝑣−1)
𝑑𝑁
𝑑𝑡
 
We know all values except dT/dt, so we can find that 
𝒅𝑻
𝒅𝒕
= −𝟏. 𝟏𝟓
𝑲
𝒔
 
 
 
 
 
#Problem 10 3.32 
Nitrogen can be liquefied using a Joule-Thomson expansion process. This is done by rapidly and 
adiabatically expanding cold nitrogen gas from high pressure to a low pressure. If Nitrogen at 135 
K and 20 MPa undergoes a Joule-Thomson expansion to 0.4 MPa: 
a) Extimate the fraction of vapor and liquid present after the expansion and the temperature 
of this mixture using the pressure-enthalpy diagram for nitrogen 
 
As nitrogen goes through a Joule-Thomson process 
�̂�1 = �̂�2 
With 
T1 = 135K 
P1 = 20MPa 
 �̂�1= 153 kJ/kg 
 �̂�1 = �̂�2 = 153 𝑘𝐽/𝑘𝑔 
 
We also have 
P2 = 0.4MPa 
 T2 = 90K 
From Pressure-Enthalpy diagram for nitrogen, we find that 55% N2 is in vapor phase, 45% is in 
liquid phase 
b) Repeat the calculation assuming nitrogen to be an ideal gas with Cp*=29.3 J/(mol K) 
As nitrogen is an ideal gas, we have 
∆H = NCp∆T 
With ∆H = 0  ∆T = 0  T2 = T1 = 135K: 100% N2 is in vapor phase 
 
 
 
 
 
 
#Problem 11 
A piston-cylinder assembly contains 6kg of steam at a pressure of 100 bar and a temperature of 
400oC. It undergoes a process whereby it expands against a constant pressure of 20 bar, until the 
forces balance. During the process, the piston generates 1497480 J of work. Steam is not an ideal 
gas under these conditions. Determine the final temperature and the heat transferred during the 
process. 
Choose system as the steam inside cylinder 
Mass balance equation for this closed system is 
M2 = M1 = M 
Energy balance for this closed system with neglect of change of kinetic and potential energy, 
with no shaft work 
𝑈2 − 𝑈1 = 𝑄 + 𝑊 = 𝑄 − ∫ 𝑃𝑑𝑉 = 𝑄 − 𝑃(𝑉2 − 𝑉1) 
From appendix AIII 
�̂�1 = 0.02641
𝑚3
𝑘𝑔
 (𝑎𝑡 100 𝑏𝑎𝑟, 400𝑜𝐶) 
𝑉1 = 𝑀1�̂�1 = 6 ∗ 0.02641 = 0.15846 (𝑚
3) 
Thus 
𝑉2 = 𝑉1 −
𝑊
𝑃
= 0.15846 −
−1497480
20 ∗ 105
= 0.9072 𝑚3 
�̂�2 =
𝑉2
𝑀2
= 0.1512 (𝑚3/𝑘𝑔) 
With known values of �̂�2 and P2, we can get T2 = 400
oC (20 bar, 0.1512 m3/kg) 
Also from appendix AIII 
 
�̂�2 = 2945.2
𝑘𝐽
𝑘𝑔
 
�̂�1 = 2832.4
𝑘𝐽
𝑘𝑔
 
∆𝑈 = 𝑀(�̂�2 − �̂�1) = 6 ∗ (2945.2 − 2832.4) = 676.8 (𝑘𝐽) 
Q = ∆U – W = 676.8 + 1497.48 = 2174.28 kJ 
#Problem 12 3.20 
A clever chemical engineer has devised the thermally operated elevator shown in the 
accompanying diagram. The elevator compartment is made to rise by electrically heating the air 
contained in the piston-and-cylinder drive mechanism, and the elevator is lowered by opening a 
valve at the side of the cylinder, allowing the air in the cylinder to slowly escape. Once the 
elevator compartment is back to the lower level, a small pump forces out the air remaining in the 
cylinder and replaces it with air at 20 C and a pressure just sufficient to support the elevator 
compartment. This cycle can then be repeated. There is no heat transfer between the piston, 
cylinder, and the gas; the weight of the piston, elevator, and the elevator contents is 4000 kg; 
the piston has a surface area of 2.5 m2; and the volume contained in the cylinder when the 
elevator is at its lowest level is 25 m3. There is no friction between the piston and the cylinder, 
and the air in the cylinder is assumed to be an ideal gas with Cp*=30 J/(mol K) 
a) What is the pressure in the cylinder throughout the process? 
𝑃 =
𝐹
𝐴
 
𝑃𝑡𝑜𝑡 = 𝑃𝑎𝑡𝑚 + 𝑃𝑝𝑖𝑠𝑡𝑜𝑛 
 
𝑃𝑡𝑜𝑡 = 101,325 + 
4000 𝑘𝑔 ∗ 9.816 𝑚 𝑠2⁄
2.5𝑚2
= 117.03 𝑘𝑃𝑎 
b) How much heat must be added to the air during the process of raising the elevator 3 m, and 
what is the final temperature of the gas? 
We can use the ideal gas law to calculate the number of moles. 
𝑁 =
𝑃𝑉
𝑅𝑇
 
𝑁 =
117031 𝑃𝑎 ∗ 25 𝑚3
8.314 𝑚
3𝑃𝑎
𝐾 𝑚𝑜𝑙⁄ ∗ 293.15 𝐾
= 1200.44 𝑚𝑜𝑙 
We can also use the ideal gas law to calculate the final temperature of the gas. 
∆𝑉 = 2.5 ∗ 3 = 7.5𝑚3 
 
𝑉𝑓 = 25𝑚
3 + 2.5𝑚2 ∗ 3𝑚 = 32.5𝑚3 
𝑇𝑓 =
𝑃𝑉
𝑁𝑅
 
𝑇𝑓 =
117031 𝑃𝑎 ∗ 32.5 𝑚3
1200.44 𝑚𝑜𝑙 ∗ 8.314 𝑚
3𝑃𝑎
𝐾 𝑚𝑜𝑙⁄
= 381.1 𝐾 
Now the gas does work both against the atmosphere and the piston 
𝑊𝑎𝑡𝑚 = −𝑃𝑎𝑡𝑚 ∗ ∆𝑉 = −877.7325 𝑘𝐽 
𝑊𝑝𝑖𝑠𝑡𝑜𝑛 = −𝑚 ∗ 𝑔 ∗ ∆ℎ = −117.792 𝑘𝐽 
𝑊𝑇 = 𝑊𝑎𝑡𝑚 + 𝑊𝑝𝑖𝑠𝑡𝑜𝑛 = −995.9245 𝑘𝐽 
Using a simplified version of the first law, 
𝑄 + 𝑊 = 𝑁𝐶𝑝(𝑇𝑓 − 𝑇𝑖) 
𝑄 = 𝑁𝐶𝑝∆𝑇 − 𝑊 
𝑄 = 1200.44 𝑚𝑜𝑙 ∗ 30 
𝐽
𝑚𝑜𝑙 𝐾⁄ ∗ (381.1 𝐾 − 293.15 𝐾) + 995924.5 𝐽 = 4160.8 𝑘𝐽 
 
 
c) What fraction of the heat added is used in doing work, and what fraction is used in raising 
the temperature of the gas? 
%𝑄𝑊 =
995.9245
4160.8
= 24% 
 
%𝑄∆𝑇 =
1200.44 ∗ 30 ∗ 87.95
4160800
= 76% 
d) How many moles of air must be allowed to escape in order for the elevator to return to the 
lowest level? 
Using the ideal gas law 
𝑁 =
𝑃𝑉
𝑅𝑇
 
𝑁𝑓 =
117031 𝑃𝑎 ∗ 25 𝑚3
8.314 𝑚
3𝑃𝑎
𝐾 𝑚𝑜𝑙⁄ ∗ 381.1 𝐾
= 923.4 𝑚𝑜𝑙 
∆𝑁 = 𝑁𝑓 − 𝑁𝑖 = 1200.44 𝑚𝑜𝑙 − 923.4 𝑚𝑜𝑙 = 277.04 𝑚𝑜𝑙 
 
 
 
 
 
 
 
 
##Problem 13 
 An air compressor is designed to compress atmospheric air (assumed to be at 100 kPa, 20oC) to a 
pressure of 1 MPa. The heat transfer rate to the environment is anticipated to be about equal to 
10% of the power input to the compressor. The air enters at 50 m/s where the inlet area is 9x10-3 
m2 and leaves at 120m/s through an area 5x10-4m2. Determine the exit-air temperature and the 
power input to the compressor. The compressor is working at steady state, and air could be 
assumed to be an ideal gas. 
 
 
 
 
 
Solution 
 
Choose system to be the control volume inside the compressor 
 
Mass balance equation for this opened system is 
𝑑𝑀
𝑑𝑡
= �̇�𝑖𝑛 − �̇�𝑜𝑢𝑡 = 0 
�̇�𝑖𝑛 = �̇�𝑜𝑢𝑡 = �̇� 
 
�̇� = �̇�𝑖𝑛 =
𝐴𝑖𝑛𝑣𝑖𝑛
�̂�𝑖𝑛
=
𝐴𝑖𝑛𝑣𝑖𝑛𝑃𝑖𝑛𝑀𝑎𝑖𝑟
𝑅𝑇𝑖𝑛
 
�̇� = �̇�𝑖𝑛 =
9 ∗ 10−3 ∗ 50 ∗ 100 ∗ 28.97
8.314 ∗ 293
= 0.535
𝑘𝑔
𝑠
 
 
 
We have 
�̇�𝑖𝑛 = �̇�𝑜𝑢𝑡 = �̇� 
which is equivalent to 
𝐴𝑖𝑛𝑣𝑖𝑛𝑃𝑖𝑛𝑀𝑎𝑖𝑟
𝑅𝑇𝑖𝑛
=
𝐴𝑜𝑢𝑡𝑣𝑜𝑢𝑡𝑃𝑜𝑢𝑡𝑀𝑎𝑖𝑟
𝑅𝑇𝑜𝑢𝑡
 
𝑇𝑜𝑢𝑡 = 𝑇𝑖𝑛
𝐴𝑜𝑢𝑡𝑣𝑜𝑢𝑡𝑃𝑜𝑢𝑡
𝐴𝑖𝑛𝑣𝑖𝑛𝑃𝑖𝑛
 
 
𝑇𝑜𝑢𝑡 = 293 ∗ (
5 ∗ 10−4
9 ∗ 10−3
) (
120
50
) (
1000
100
) = 391 𝐾 
Energy balance equation for this system in steady state is 
0 = �̇�𝑖𝑛(�̂� +
𝑣2
2
+ 𝜑)𝑖𝑛− �̇�𝑜𝑢𝑡(�̂� +
𝑣2
2
+ 𝜑)𝑜𝑢𝑡 + �̇� + �̇� 
With �̇� = −0.1�̇�(< 0) 
Neglect potential energy change, we have 
0.9�̇� + �̇� (�̂�𝑖𝑛 − �̂�𝑜𝑢𝑡 +
𝑣𝑖𝑛
2 − 𝑣𝑜𝑢𝑡
2
2
) = 0 
�̇� = −
�̇�
0.9
(�̂�𝑖𝑛 − �̂�𝑜𝑢𝑡 +
𝑣𝑖𝑛
2 − 𝑣𝑜𝑢𝑡
2
2
) 
With 
�̂�𝑖𝑛 = 293.3
𝑘𝐽
𝑘𝑔
 
�̂�𝑜𝑢𝑡 = 392.1
𝑘𝐽
𝑘𝑔
 
 
Solving the above equation, we obtain 
 
�̇� = 62.27 𝑘𝑊 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
#Problem 14 
Steam enters a turbine with a pressure and temperature of 15MPa and 600oC and leaves at 100kPa 
as a saturated vapor. The flow area at the turbine inlet is 0.045m2 and at the exit it is 0.31 m2. The 
steam flows steadily through the turbine at a mass flow rate of 30kg/s. Calculate the power that 
can be produced by the turbine, assuming negligible heat transfer from the system. Specific volume 
and specific enthalpy of inlet air are 0.0249 m3/kg and 3581.5 kJ/kg. 
 
 
 
 
 
Choose system as a control volume inside the turbine 
 
Mass balance equation for this system is 
𝑑𝑀
𝑑𝑡
= �̇�𝑖𝑛 − �̇�𝑜𝑢𝑡 = 0 
 
�̇�𝑖𝑛 = �̇�𝑜𝑢𝑡 = �̇� 
 
�̇� =
𝐴𝑖𝑛𝑣𝑖𝑛
�̂�𝑖𝑛
=
𝐴𝑜𝑢𝑡𝑣𝑜𝑢𝑡
�̂�𝑜𝑢𝑡
 
 
�̂�𝑖𝑛 = 0.0249 
𝑚3
𝑘𝑔
 (𝑎𝑡 15 𝑀𝑃𝑎, 600𝑜𝐶) 
�̂�𝑜𝑢𝑡 = 1.694 
𝑚3
𝑘𝑔
 (100𝑘𝑃𝑎, 𝑠𝑎𝑡𝑢𝑟𝑎𝑡𝑒𝑑 𝑣𝑎𝑝𝑜𝑟) 
�̇� = 30
𝑘𝑔
𝑠
 
Thus we have 
𝑣𝑖𝑛 =
30 ∗ 0.0249
0.045
= 16.6
𝑚
𝑠
 
𝑣𝑜𝑢𝑡 =
30 ∗ 1.694
0.31
= 163.9
𝑚
𝑠
 
Energy balance equation for this system is 
0 = �̇�(�̂� +
𝑣2
2
+ 𝜑)𝑖𝑛 − �̇�(�̂� +
𝑣2
2
+ 𝜑)𝑜𝑢𝑡 + �̇� + �̇� 
If heat transfer is negligible, and neglect change in potential energy, we have 
�̇� = �̇�(�̂�𝑜𝑢𝑡 − �̂�𝑖𝑛 +
𝑣𝑜𝑢𝑡
2 − 𝑣𝑖𝑛
2
2
) 
With 
�̂�𝑖𝑛 = 3581.5
𝑘𝐽
𝑘𝑔
 
�̂�𝑜𝑢𝑡 = 2675.1
𝑘𝐽
𝑘𝑔
 
Thus 
 
�̇� = 26793 𝑘𝑊 
 
 
 
 
 
 
 
 
 
 
 
 
 
#Problem 15 
An insulated vessel has two compartments separated by a membrane. On one side is 2kg of steam 
at 500oC and 200 bar. The other side is evacuated. The membrane ruptures, filling the entire 
volume. The final pressure is 100 bar. Determine the final temperature of the steam and the volume 
of the vessel. 
 
 
Choose system as a whole volume inside the vessel 
 
Mass balance equation for this closed system is 
M2 = M1 = M1a + M1b = M1a (b is the evacuated compartment at beginning) 
Energy balance equation with neglect of change of kinetic and potential energy, with no heat 
transfer and work done for this closed system is 
U2 – U1a – U1b = U2 – U1a = 0 
thus 
�̂�2 = �̂�1𝑎 
From appendix AIII 
 
�̂�1𝑎 = 2942.9
𝑘𝐽
𝑘𝑔
 (500𝑜𝐶, 200 𝑏𝑎𝑟) 
Thus 
�̂�2 = 2942.9
𝑘𝐽
𝑘𝑔
 
We also have 
P2 = 100 bar 
Thus we can determine 
T2 = 449.7oC 
�̂�2 = 0.02973
𝑚3
𝑘𝑔
 
 
𝑉𝑣𝑒𝑠𝑠𝑒𝑙 = 𝑀2�̂�2 = 2 ∗ 0.02973 = 0.05946 (𝑚
3) 
#Problem 16 
Consider a piston-cylinder assembly containing 10kg of steam. Initially the gas has a pressure of 
20 bar and occupies a volume of 1.0 m3. Under these conditions, steam does not behave as an ideal 
gas. 
A) The system now undergoes a compression process in which it is compressed to 100 bar, the 
external pressure is slightly larger than and could be assumed to be equal to internal pressure. The 
pressure-volume relationship is given by 𝑃𝑉1.5 = 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡. What is the final temperature and 
internal energy of the system? Calculate the work done during this process. How much heat was 
exchanged? 
B) Consider a different process by which the system get to the same final state in part (A). In this 
case, a large block is placed on the piston, forcing it to compress. Calculate the work done during 
this process. How much heat was exchanged? 
a) Choose system as the steam inside the assembly. 
Mass balance equation for this closed system is 
M2 – M1 = 0  M1 = M2 
Energy balance equation for this closed system with neglect of change of kinetic and potential 
energy, no shaft work is 
𝑈2 − 𝑈1 = 𝑄 − ∫ 𝑃𝑑𝑉
2
1
 
We have 
𝑃1�̂�1
1.5 = 𝑃2�̂�2
1.5 
P1 = 20 bar, P2 = 100 bar 
and 
�̂�1 =
1
10
= 0.1 
𝑚3
𝑘𝑔
 
we can obtain 
�̂�2 = 0.0342
𝑚3
𝑘𝑔
 
�̂�1 = 2602.8
𝑘𝐽
𝑘𝑔
 
With know values of P2 and �̂�2, from appendix AIII we get 
 
T2 = 524.7 K 
�̂�2 = 3094.6
𝑘𝐽
𝑘𝑔
 
𝑈2 = 𝑀2�̂�2 = 10 ∗ 3094.6 = 30946 𝑘𝐽 
 
𝑊 = − ∫ 𝑃𝑑𝑉
2
1
= −𝑀 ∫
𝑃1�̂�1
1.5
�̂�1.5
𝑑�̂�
2
1
= 2839 𝑘𝐽 
 
Q = ∆U – W = U2 – U1 – W = 10(3094.6 – 2602.8) – 2839 = 2079 (kJ) 
b) 
𝑊 = − ∫ 𝑃𝑑𝑉
2
1
= −𝑀𝑃2 ∫ 𝑑�̂�
2
1
= −10 ∗ 100 ∗ (0.0342 − 0.1) 
W = 6580 kJ 
 
Q = ∆U – W = – 1662 kJ 
 
#Problem 17 
You wish to measure the temperature of steam flowing in a pipe at pressure of 9MPa. To do this 
task you connect a well-insulated tank of volume 0.4m3 to this pipe through a valve. This tank 
initially is at vacuum. The valve is opened, and the tank fills with steam until the pressure is 9 
MPa. At this point the pressure of the pipe and tank are equal, and no more steam flows through 
the valve. The valve is then closed. The temperature right after the valve is closed is measured to 
be 800oC. The process takes place adiabatically. Determine the temperature of the system flowing 
in the pipe. You may assume the steam in the pipe stays at the same temperature and pressure 
throughout this process. 
 
 
 
Mass balance equation for this open system with only one inlet stream is 
𝑑𝑀
𝑑𝑡
= �̇�𝑖𝑛 − �̇�𝑜𝑢𝑡 = �̇�𝑖𝑛 
∫ 𝑑𝑀
𝑀2
𝑀1
= ∫ �̇�𝑖𝑛 𝑑𝑡
𝑡
0
= 𝑀2 − 𝑀1 = 𝑀2 
Energy balance equation for this system with neglect of change of kinetic and potential energy is 
𝑑𝑈
𝑑𝑡
= �̇�𝑖𝑛 �̂�𝑖𝑛 − �̇�𝑜𝑢𝑡 �̂�𝑜𝑢𝑡 + �̇� + �̇� 
Since we have no outlet stream, adiabatic process, constant volume with no shaft work 
𝑑𝑈
𝑑𝑡
= �̇�𝑖𝑛 �̂�𝑖𝑛 
∫ 𝑑𝑈
2
1
= ∫ �̇�𝑖𝑛 �̂�𝑖𝑛𝑑𝑡
𝑡
0
= �̂�𝑖𝑛 ∫ �̇�𝑖𝑛 𝑑𝑡
𝑡
0
 
We can infer 
𝑈2 = 𝑀2�̂�𝑖𝑛 
𝑏𝑢𝑡 𝑈2 = 𝑀2�̂�2 
𝑡ℎ𝑢𝑠 �̂�2 = �̂�𝑖𝑛 
From appendix AIII we get 
�̂�2 = 3632.5
𝑘𝐽
𝑘𝑔
 (𝑎𝑡 9𝑀𝑃𝑎, 800𝑜𝐶) = �̂�𝑖𝑛 
Now we have values of �̂�𝑖𝑛 and Pin we can obtain Tin = 600
oC 
 
 
#Problem 18 
Refrigerant 12 flows steadily through a 40 mm diameter horizontal pipe. At a point where the 
velocity is 40m/s the temperature and pressure of the refrigerant are 40oC and 300kPa, 
respectively. As a result of heat transfer from the surroundings, the temperature at a point 
downstream reaches 50oC. 
A) Assuming a negligible pressure drop, determine the heat transfer rate to the refrigerant 12. 
Specific volumes of inlet and outlet streams are 0.06821m3/kg and 0.07077 m3/kg. Specific 
enthalpy of inlet and outlet streams are 214.31 kJ/kg and 220.77 kJ/kg. 
B) Do part (a) with consideration of 50kPa of pressure drop. Specific volume and specific enthalpy 
of outlet stream now are 0.0856 m3/kg and 221.33 kJ/kg. Mass flow rate in pipe is constant. 
 
Choose system as a control volume inside the pipe. 
a) Mass balance equation: 
𝑑𝑀
𝑑𝑡
= �̇�𝑖𝑛 − �̇�𝑜𝑢𝑡 = 0 
�̇�𝑖𝑛 = �̇�𝑜𝑢𝑡 = �̇� 
 
Energy balance equation with neglect of change of potential energy because pipe is horizontal is 
0 = �̇�𝑖𝑛(�̂� +
𝑣2
2
)𝑖𝑛 − �̇�𝑜𝑢𝑡(�̂� +
𝑣2
2
)𝑜𝑢𝑡 + �̇� + �̇� 
Since no work was done 
�̇� = �̇�[(�̂� +
𝑣2
2
)𝑜𝑢𝑡 − (�̂� +
𝑣2
2
)𝑖𝑛] 
Since �̇�𝑖𝑛 = �̇�𝑜𝑢𝑡 
𝐴𝑜𝑢𝑡𝑣𝑜𝑢𝑡
�̂�𝑜𝑢𝑡
=
𝐴𝑖𝑛𝑣𝑖𝑛
�̂�𝑖𝑛
 
Pipe has constant cross section  Aout = Ain 
We now have 
𝑣𝑜𝑢𝑡
�̂�𝑜𝑢𝑡
=
𝑣𝑖𝑛
�̂�𝑖𝑛
 
With 
�̂�𝑖𝑛 = 0.06821 
𝑚3
𝑘𝑔
 
�̂�𝑜𝑢𝑡 = 0.07077 
𝑚3
𝑘𝑔
 
 
Substitute into the above equation, we obtain 
vout = 1.037vin 
thus we may neglect change in kinetic energy in the energy balance equation 
�̇� = �̇�(�̂�𝑜𝑢𝑡 − �̂�𝑖𝑛) = 40 ∗ 𝜋 ∗
0.042
4
∗
1
0.06821
∗ (220.77 − 214.31) 
 
�̇� = 4.76 𝑘𝑊 
 
b) The exit velocity now is 
𝑣𝑜𝑢𝑡 =
�̂�𝑜𝑢𝑡𝑣𝑖𝑛
�̂�𝑖𝑛
= 50
𝑚
𝑠
 
Since there is big difference between vin and vout, we cannot neglectthe change in kinetic energy. 
Energy balance equation now becomes 
 
�̇� = �̇�[(�̂� +
𝑣2
2
)𝑜𝑢𝑡 − (�̂� +
𝑣2
2
)𝑖𝑛] 
 
�̇� = 40 ∗ 𝜋 ∗
0.042
4
∗
1
0.06821
(221.33 − 214.31 +
502 − 402
2 ∗ 1000
) 
 
 
�̇� = 5.505 𝑘𝑊 
#Problem 19 
Air at 1 atm and 20oC occupies an initial volume of 1000 cm3 in a cylinder. The air is confined by 
a piston, which has a constant restraining force so the gas pressure always remains constant. Heat 
is added to the air until its temperature reaches 260oC. Calculate the heat added, the work the gas 
does on the piston, and the change in internal energy of the gas. Air could be assumed to be ideal 
gas with Cp* = 1.005 x 103 J/kgoC 
Choose system as the air inside the cylinder 
 
Heat added to system is calculated as 
Q = mcp∆T 
Since air is an ideal gas, we have 
𝑚 = 
𝑀𝑃𝑉
𝑅𝑇
=
28.97 ∗ 105 ∗ 10−3
8.314 ∗ 293
= 1.189 𝑔 
 
Q = 1.189*10-3 * 1.005 * 103 * (260 – 20) = 287 J 
Mass balance equation for this system is 
m1 = m2 
thus 
N1 = N2 
𝑃1𝑉1
𝑇1
=
𝑃2𝑉2
𝑇2
 
𝑉2 = 𝑉1 (
𝑃1
𝑃2
) (
𝑇2
𝑇1
) = 10−3 ∗ 1 ∗ (
533
293
) = 1.819 ∗ 10−3 (𝑚3) 
𝑊 = − ∫ 𝑃𝑑𝑉
2
1
= −𝑃(𝑉2 − 𝑉1) = −10
5 ∗ (1.819 − 1) ∗ 10−3 = −81.9𝐽 
The change in internal energy is 
∆U = Q + W = 287 – 81.9 = 205.1 (J) 
 
 
 
 
 
 
 
 
 
 
#Problem 20 
Water is to be heated from its pipeline temperature of 20oC to 90oC using superheated steam at 
450oC and 2.5 MPa in a steady-state process to produce 10kg/s of heated water. In each of the 
processes below, assume there is no heat loss. 
A) The heating is to be done in a mixing tank by direct injection of the system, all of which 
condenses. Determine the two inlet mass flows needed to meet the desired hot water flow rate. 
B) Instead of direct mixing, a heat exchanger will be used in which the water to be heated will 
flow inside copper tubes and the steam will partially condense on the outside of the tubes. In this 
case heat will flow from the steam to the water, but the two streams are not mixed. Calculate 
the steam flow rate if the steam leaves the heat exchanger at 50 percent quality at 100oC 
A) define the system as the water and steam 
𝑀 = 𝑀𝑤𝑎𝑡𝑒𝑟 + 𝑀𝑠𝑡𝑒𝑎𝑚 
 
Because the system is a closed adiabatic system, 
10 ∗ 𝑈𝑓(90
𝑜𝐶, 2.5𝑀𝑃𝑎) = (10 − 𝑀𝑠𝑡𝑚) ∗ 𝑈(20
𝑜𝐶, 2.5𝑀𝑃𝑎) + 𝑀𝑠𝑡𝑚 ∗ 𝑈(450
𝑜𝐶, 2.5 𝑀𝑃𝑎) 
 
Plugging in given values from NIST or appendix A.III 
10 
𝑘𝑔
𝑠⁄ ∗ 376.33 
𝑘𝐽
𝑘𝑔⁄ = (10
𝑘𝑔
𝑠⁄ − 𝑀𝑠𝑡𝑚) ∗ 86.76
𝑘𝐽
𝑘𝑔⁄ + 𝑀𝑠𝑡𝑚 ∗ 3026.2 
𝑘𝐽
𝑘𝑔⁄ 
 
Using some simple algebra and solving for 𝑀𝑠𝑡𝑚 
𝑀𝑠𝑡𝑚 = 0.985
𝑘𝑔
𝑠⁄ 
 
B) Once again, the system is defined as both the water and the steam 
𝑀 = 𝑀𝑤𝑎𝑡𝑒𝑟 + 𝑀𝑠𝑡𝑒𝑎𝑚 
Since the system is still a closed adiabatic system, 
 
10 ∗ 𝑈𝑓(90
𝑜𝐶, 2.5𝑀𝑃𝑎) + 0.5𝑀𝑠𝑡𝑚
∗ (𝑈𝑔𝑎𝑠(100
𝑜𝐶, 0.10135 𝑀𝑃𝑎) + 𝑈𝑙𝑖𝑞(100
𝑜𝐶, 0.10135 𝑀𝑃𝑎))
= 10 ∗ 𝑈𝑖(20
𝑜𝐶, 2.5𝑀𝑃𝑎) + 𝑀𝑠𝑡𝑚 ∗ 𝑈𝑖(450
𝑜𝐶, 2.5 𝑀𝑃𝑎) 
Plugging in values from NIST or appendix A.III 
 
10 
𝑘𝑔
𝑠⁄ ∗ 376.33 
𝑘𝐽
𝑘𝑔⁄ + 0.5𝑀𝑠𝑡𝑚 ∗ (2505.5
𝑘𝐽
𝑘𝑔⁄ + 418.94
𝑘𝐽
𝑘𝑔⁄ )
= 10
𝑘𝑔
𝑠⁄ ∗ 86.76 
𝑘𝐽
𝑘𝑔⁄ + 𝑀𝑠𝑡𝑚 ∗ 3026.2 
𝑘𝐽
𝑘𝑔⁄ 
Using some simple algebra and solving for 𝑀𝑠𝑡𝑚 
 
𝑀𝑠𝑡𝑚 = 1.85 
𝑘𝑔
𝑠⁄