College of Engineering and Computer Science
Mechanical Engineering Department
Mechanical Engineering 370
Thermodynamics
Fall 2010 Number: 14319 Instructor: Larry Caretto
Solution to Third Quiz – Heat, Internal Energy and the First Law
1
A mass of 0.1 kg of water fills a piston-cylinder apparatus with an initial volume of V1 =
0.01 m3. The water then undergoes a three step path where the initial and final states
are the same:
a. an expansion to a volume of V2 = 0.02 m3 following the path equation P = a + b/V
where a = 3 MPa and b = –0.02 MPa·m3.
b. a constant volume decrease in pressure P3 = P1, the initial pressure.
c. a constant pressure decrease in volume to V4 = V1, the initial volume.
Find the heat transfer for (i) part a of the path and (ii) the entire path.
We can use the path equation to find the work for part a as pathPdV.
2
V 
b

PdV

 a  dV  aV2  V1   b ln  2 


V
 V1 
path a
V1 
V
Wa 
 0.02 m 3 
 V2 
3
3
3

Wa  aV2  V1   b ln    3 MPa  0.02 m  0.01 m   0.02 MPa  m ln 
3 
 V1 
 0.01 m 
 0.016137 MPa  m 3  0.016137 MJ  16.137 kJ

 

The heat transfer along this path can be found from the first law: Qa = Ua + W a = (u2 – u1) + W a.
We can find the internal energy, u1 at the initial state v1 = V1/m = (0.01 m3)/(0.1 kg) = 0.1 m3/kg
and P1 = a + b/V1 = (3 MPa) + (-0.02 MPa·m3) / (0.01 m3) = 1 MPa. From the saturation table at
P1 = 1 MPa we see that this specific volume is between vf and vg, so we must be in the mixed
region. We can find the internal energy from the quality:
x1 
v1  v f ( P1  1 MPa)
v fg ( P1  1 MPa)
 0.1 m 3  0.001127 m 3 
kg
kg 

 0.5115
3
m
0.1933
kg
u1  u f ( P1  1 MPa)  x1u fg ( P1  1 MPa)  761.37 kJ
kg
 (0.5115)1821.8 kJ   1693.2 kJ
kg 
kg

Since the tables do not have internal energy, we have to compute it from the enthalpy. At the
final state, P2 = a + b/V2 = 3 MPa + (–0.02 MPa·m3)/(0.2 m3) = 2 MPa, and v2 = V2/m = (0.02
m3)/(0.1 kg) = 0.2 m3/kg. At this pressure and specific volume we see that the temperature is
very close to 600oC. If we used interpolation we would find the final enthalpy as follows:
3802.7 kJ 3689.7 kJ

3689.7 kJ
kg
kg
h2 

3
kg
0.21144 m
0.19960 m 3

kg
kg
Jacaranda (Engineering) 3519
E-mail: lcaretto@csun.edu
 0.2 m 3 0.19960 m 3  3693.5 kJ

 

kg
kg
 kg

Mail Code
8348
Phone: 818.677.6448
Fax: 818.677.7062
Solutions to third quiz
ME 370, L. S. Caretto, Fall 2010
Page 2
We can now find the internal energy at the end of the first path, u2.
u 2  h2  P2 v2 
3689.7 kJ
0.2 m 3 1000 kJ
3293.5 kJ
 2 MPa 

3
kg
kg MPa  m
kg
We can now find the heat transfer for step a from the first law.
 3293.5 kJ 1693.2 kJ 
  16.137 kJ
Qa  mu 2  u1   Wa  0.1 kg 

kg
kg


Qa = 176.2 kJ
For the entire path the initial and final states are the same so the internal energy change is zero.
(U depends only on the state; it the initial and final states are the same, the value of u at both
states is the same so DU = 0) This gives Q = W = W a + W b + W c for the overall path. We have
found W a = 16.137 kJ and we know that W b = 0 for the constant volume step in part b of the path.
The third part of the path, at constant pressure P3 = P4 = P1 = 1 MPa is found as follows: W c =
Pconst (V4 – V3) = (1 MPa)(0.01 m 3 – 0.02 m3) = –0.01 MPa·m3 = –0.01 MJ = –10 kJ. The heat
transfer for the overall path is then found to be Q = W = W a + W b + W c = 16.137 kJ + 0 + (–10 kJ)
or Q = 6.137 kJ