Thermodynamics Hw#4

1. Chemical and Mechanical Engineering 2300 / Thermodynamics I
Solution to Homework Assignment 4 (Lectures 1 - 9)
Prof. Geoff Silcox
Chemical Engineering
University of Utah
Due Monday, 2014 September 29, by 17:00
To ensure that you receive full credit for your solutions, write out all equations in
symbolic form, give numerical values for all variables and constants in the equations, and
write answers to definitions or conceptual problems in complete sentences. Approximate
answers are rounded to one significant figure. Your answers need to be reported with
three.
Problem 1
The properties of compressed liquids are commonly approximated by the equation,
written here for specific volume,
v(T ,P )  vf (T )
Enthalpy (h = u + Pv) is one property where this approximation can lead to significant
errors because pressure appears explicitly in its definition. Two possible approximations
for the enthalpy of compressed water are

h(T ,P ) h (T )
 
f
  
h(T,P) h (T ) v (T ) P P (T )
f f sat
The latter is a more accurate approximation than the former. Calculate h at 20C and 100
bar using these two approximations and compare your answers to that from Table A-7.
Solution
From Table A-4 at 20C, h = hf = 83.915 kJ/kg. From the more accurate approximation,
kJ m3 kJ 83 915 0 001002 10000 2 3392 kPa 93 92
   
f f sat kg kg kg h  h  v P P  .  .  .  .
From Table A-7 at 20C and 10 MPa, h = 93.28 kJ/kg. We conclude that the second
approximation is quite accurate while the first is low by more than 10%.
Problem 2
Complete the following table for water.

2. T, C P, kPa v, m3/kg Phase
description
140 0.05
550 saturated liquid
125 750
500 0.140
Solution
T, C P, kPa v, m3/kg Phase
description
140 361.53 0.05 sat. mixture
Tsat = 155.46 550 0.001097 saturated liquid
125 750 0.001065a compres. liq.
500 2500 0.140 superheated
Notes
v = vf at 125C
Problem 3
Complete the following table for water.
T, C P, kPa u, kJ/kg Phase
description
400 1450
220 saturated vapor
190 2500
4000 3040
Solution
T, C P, kPa u, kJ/kg Phase
description
143.61 400 1450 sat. mixture
220 2319.6 2601.3 sat. vapor
190 2500 uf = 806.00 compres. liq.
466.2a 4000 3040 superheated
Notes
a) Linear interpolation. 450 500 450 3040 3011 466.2
3100.3 3011.0
T

   


3. Problem 4
A 0.5-m3 rigid tank contains refrigerant-134a initially at 160 kPa and 40 percent quality.
Heat is now transferred to the refrigerant until the pressure reaches 700 kPa. Determine
(a) the mass of the refrigerant in the tank and (b) the amount of heat transferred. Sketch
the process on a P-v diagram with respect to saturation lines. Approximate answers: the
mass is roughly 10 kg and Qin is about 3000 kJ.
Solution
a) From Table A-12, the constant specific volume for this process is
m3 1 0.6 0.0007437 0.4 0.12348 0.04984
     
f g kg v   x v  xv   
The mass of the system is
3
0.5 m 10.03 kg
0.04984m
  
3
kg
m V
v
Because we will need it later, u1 is given by
m3 1 0.6 31.09 0.4 221.35 107.2
     
f g kg u   x u  xu   
b) Because v2 = 0.04984 m3/kg and P2 = 700 kPa, the final state is superheated vapor. To
determine the heat added, an energy balance on the system gives
  2 1 and in out in in U Q W Q Q  m u  u
Interpolation in Table A-13, slightly outside the bounds, gives u2 = 376.8 kJ/kg and
finally
    2 1 10.03 376.8 107.2 2705 in Q  m u  u    kJ
The process is sketched below on a P-v diagram.

4. P
2
Problem 5
The Rankine cycle is sketched below. We will discuss this cycle in Lesson 22 of the
online notes. The cycle is primarily used to generate electricity (the generator is attached
to the turbine and is not shown here). The working fluid for fossil-fuel-fired and nuclear
powered systems is water. For lower temperature cycles, the working fluid is commonly
propane. In the figure below, state 1 is a superheated vapor, state 2 is typically a
superheated vapor or saturated mixture of high quality, state 3 is a saturated liquid, and
state 4 is a compressed liquid.
If the temperature and pressure of the propane at point 1 in the cycle are 220C and 110
bar(a), calculate the specific volume (m3/kg) by the following techniques.
a) The ideal gas equation of state (EOS).
b) The compressibility factor obtained from the Nelson-Obert generalized
compressibility chart, Figure A-15 of the text.
c) The Redlich-Kwong EOS, where
v
1
Boiler Turbine
Pump Condenser
out , T W
out Q
in Q
in , P W
Boundary
of system
2
3
4
1

5. 3 2  2 
2
2 25
0 5
2 2
or 0
where
0 42748
0 08664
.
C
.
C
C
C
Z Pv
RT
P RT a Z Z A B B Z AB
v b v bv
R T
a .
P T
RT
b .
P
A aP
R T
B bP
RT

       
 




Include a plot of the cubic equation, g(Z )  Z3  Z2  A  B  B2 Z  AB , as part
of your answer. Also include your values of a, b, A, and B, with units as appropriate.
d) The NIST website, http://webbook.nist.gov/chemistry/fluid/.
Approximate answers: (a) 0.008 m3/kg, (b) 0.006 m3/kg.
Solution
(a) For the ideal gas EOS,
3
8 31446 kPa-m 493 15 K kmol-K 008453m
3
0 44 097 kg 11000 kPa kg
. . v  RT  
.
kmol
P .
(b) For the compressibility chart,
493 15 K 1 334 and 110 bar(a) 2 589
T  T  .  . P  P  
.
R T 369 . 8 K R P 42 .
49 bar(a)
c c
and
 3

 
3
0 68 8 31446 kPa-m 493 15 K
. . .
kmol-K m 0 68 and 0 005748
Z  . v  ZRT    
.
44 097 kg 11000 kPa kg
kmol
P .

6. (c) From the Redlich-Kwong EOS, the following values are obtained
Variable Value
a 8.236x10-4 kJ-m3/mol
b 6.269x10-5 m3
A 0.5389
B 0.1682
Z 0.6950
v 0.005875 m3/kg
The value of Z = 0.6950 agrees with the plot of g(Z) versus Z.
0.50
0.40
0.30
0.20
0.10
0.00
‐0.10
‐0.20
0.00 0.20 0.40 0.60 0.80 1.00 1.20
g(Z)
Compressibility Factor, Z
The plot was drawn in Excel. The roots of the equation for Z were obtained with the
MATLAB function “roots” and verified by an iterative solution in Excel using Newtons
method.
(d) From the NIST website at 493.15 K and 110 bar(a), v = 0.005996 m3/kg. The NIST
properties are calculated with a much more sophisticated model than the Redlich-Quong
EOS. References to the models used are available at
http://webbook.nist.gov/chemistry/fluid/. In summary, the four methods give the
following values for v.

7. Method Specific volume, m3/kg
Ideal gas EOS 0.00845
Chart 0.00575
Redlich-Quong EOS 0.00588
NIST property calculator 0.00600
Those from NIST can be considered the most accurate.
Problem 6
Steam in a piston-cylinder device expands adiabatically from 700C and 2.0 MPa to
300C and 0.5 MPa. How much work is done by the steam? Approximate answer: 700
kJ/kg
Solution