College of Engineering and Computer Science
Mechanical Engineering Department
Mechanical Engineering 370
Thermodynamics
Fall 2010 Course Number: 14319 Instructor: Larry Caretto
Unit Twelve Homework Solutions, December 7, 2010
1
A refrigerator uses R-134a as the working fluid and operates on an ideal vaporcompression refrigeration cycle between 0.12 and 0.7 MPa. The mass flow rate of the
refrigerant is 0.05 kg/s. Show the cycle on a T-s diagram with respect to the saturation
lines. Determine (a) the rate of heat removal from the refrigerated space and the power
input to the compressor, (b) the rate of heat rejection to the environment, and (c) the
coefficient of performance.
3
2
Condenser
Throttling
Valv e
Compressor
Ev aporator
4
1
The state points for the cycle are shown on the
diagram. For the ideal cycle there is no
pressure drop or work in a heat transfer device
so P1 = P4 = 0.12 MPa and P2 = P3 = 0.7 MPa.
The compressor is assumed to be isentropic so
s2 = s1. State 1 is a saturated vapor and state
3 is a saturated liquid.
We apply the first law for open systems
assuming steady flow and negligible changes
in kinetic and potential energies; each device
has one inlet and one outlet. Thus our first law
 hout  hin 
for each device is Q  W  m
We can find the enthalpy values from tables A-11 to A-13.
h1 = hg(Pevap = 120 kPa) = 236.97 kJ/kg
s1 = sg(Pevap = 120 kPa) = 0.94779 kJ/kg∙K
h2 = h(P2 = Pcond = 700 kPa, s2 = s1= 0.94779 kJ/kg∙K) = 273.54 kJ/kg∙K by interpolation.
h3 = hf(Pcond = 700 kPa) = 88.82 kJ/kg
For the throttling valve, h4 = h3 so, h4 = hf(Pcond = 700 kPa) = 88.82 kJ/kg
Since there is no useful work in the evaporator, we can find the heat removed from the
refrigerated space (the evaporator) using the first law as follows.
0.05 kg  236.97 kJ 88.82 kJ  kW  s


Q evap  m h1  h4  

s 
kg
kg  1 kJ
Q evap  7.41 kW
Since there is no heat transfer to the compressor, we can find the compressor power input using
the first law as follows.
0.05 kg  236.97 kJ 273.54 kJ  kW  s


W comp  m h1  h2  

s 
kg
kg
 1 kJ
Jacaranda (Engineering) 3519
E-mail: lcaretto@csun.edu
Mail Code
8348
Wcomp  1.83 kW
Phone: 818.677.6448
Fax: 818.677.7062
Unit twelve homework solutions
ME 370, L. S. Caretto, Fall 2010
Page 2
The heat rejected to the environment is the evaporator heat transfer. We can find this high
temperature heat transfer from the general cycle relationship: Q H  Q L  W . For the
refrigeration cycle, this becomes.
Q H  Q cond  Q L  W  Q cond  W comp  7.41 kW  1.83 kW
Q H  9.24 kW
Finally, we find the coefficient of performance the values computed above for work and heat.
cop 
2
Q evap
W comp

7.41 kW
1.83 kW
cop = 4.05
If the throttling valve in Prob. 1 is replaced by an isentropic turbine, determine the
percentage increase in the COP and in the rate of heat removal from the refrigerated
space.
Refer to the diagram for problem 1. If the throttling valve is replaced by an isentropic turbine, only
the value of h4 will change. The values of h1, h2, and h3, will remain the same as they were in
problem 1. For an isentropic process from point 3 to point 4, s 4 = s3 = sf(Pcond = 0.7 MPa) =
0.33230 kJ/kg∙K. At the evaporator pressure of 120 kMPa, this entropy value is in the mixed
region. Here we compute the enthalpy, h4 from the quality in the usual fashion.
x4 
s 4  s f ( P4  120kPa)
s fg ( P4  120kPa)
0.33230 kJ 0.09725 kJ

kg  K
kg  K

 0.2749
0.85503 kJ
kg  K
h4  h f ( P4  120kPa)  x4 h fg ( P4  120kPa) 
22.49 kJ
214.48 kJ 81.45 kJ
 (0.2749)

kg
kg
kg
With this value of h4, (and the value of h1 from problem 10-11) the evaporator heat transfer
becomes
0.05 kg  236.97 kJ 81.45 kJ  kW  s


Q evap  m h4  h1  

 7.78 kW
s 
kg
kg  1 kJ
This is an increase of
7.78  7.41
= 4.97% for the heat removed form the refrigerated space
7.41
above the value in problem 1.
The isentropic turbine work can be used to reduce the work input to the compressor. The
available work from the turbine is found as follows.
0.05 kg  88.82 kJ 81.454 kJ  kW  s


Wturb  m h3  h4  

s  kg
kg
 1 kJ
Wturb  0.368 kW
The net cycle work is just the difference between the compressor input and the turbine output:
Unit twelve homework solutions
ME 370, L. S. Caretto, Fall 2010
Page 3
W net  1.46 kW
W net  W comp  W turb  1.828 kW  0.368 kW
For this cycle, the coefficient of performance is the ratio of Q evap to W net ,
cop 
Q evap
W net

7.78 kW
1.46 kW
cop = 5.32
The percent increase in the cop is (5.32 – 4.05)/4.05 = 31.5% .
3
A 300 kJ/min refrigeration system operates on a vapor-compression refrigeration cycle
with refrigerant 134a as the working fluid and an isentropic efficiency of 85 percent for the
compressor. The refrigerant enters the compressor as a saturated vapor at 140 kPa and is
compressed to 800 kPa. Determine (a) the quality of the refrigerant at the end of the
throttling process, (b) the coefficient of performance and (c) the power input to the
compressor.
3
2
Condenser
Throttling
Valv e
Compressor
Ev aporator
1
4
The state points for the cycle are shown on the
diagram. For the ideal cycle there is no
pressure drop or work in a heat transfer device
so P1 = P4 = 140 kPa and P2 = P3 = 800 kPa.
The ideal compressor is assumed to be
isentropic so s2,s = s1. State 1 is a saturated
vapor and state 3 is a saturated liquid.
We apply the first law for open systems
assuming steady flow and negligible changes
in kinetic and potential energies; each device
has one inlet and one outlet. Thus our first law
 hout  hin 
for each device is Q  W  m
We can find the enthalpy values from tables A-11 to A-13.
h1 = hg(Pevap = 140 kPa) = 239.16 kJ/kg
s1 = sg(Pevap = 140 kPa) = 0.94456 kJ/kg∙K
h2,s = h(P2 = Pcond = 800 kPa, s2,s = s1= 0.9322 kJ/kg∙K) = 275.39 kJ/kg∙K by interpolation.
Since the compressor is a work input device, we can find the actual value of h 2 by the equation h2
= h1 – (h1 – h2,s)/s = 239.16 kJ/kg – (239.16 kJ/kg – 275.39 kJ/kg∙K)/(85%) = 278.40 kJ/kg.
h3 = hf(Pcond = 800 kPa) = 93.42 kJ/kg
For the throttling valve, h4 = h3 so, h4 = hf(Pcond = 800 kPa) = 95.47 kJ/kg
We can find the quality at the throttling valve exit as follows.
x4 
h4  h f ( P4  140 kPa)
h fg ( P4  140 kPa)
95.47 kJ 27.08 kJ

kg
kg

212.08 kJ
kg
x4 = 32.2% .
Unit twelve homework solutions
ME 370, L. S. Caretto, Fall 2010
Page 4
We find the COP from the usual equation in terms of the cycle enthalpy values.
COP 
qevap
wcomp
239.16 kJ 95.47 kJ

h1  h4
kg
kg


h2  h1 281.78 kJ 239.16 kJ

kg
kg
COP = 3.37 .
The required power input to the compressor is found from the refrigerant heat load and the cop.
W comp
4
300 kJ min kW  s
min 60 s 1 kJ


cop
3.37
Q evap
W comp  1.48 kW
An ideal Otto cycle has a compression ratio of 8. At the beginning of the compression
process, air is at 95 kPa and 27oC, and 750 kJ/kg of heat is transferred to air during the
constant-volume heat addition process. Taking into account the variation of specific
heats with temperature, determine (a) the pressure and temperature at the end of the heataddition process, (b) the net work output, (c) the thermal efficiency, and (d) the mean
effective pressure for the cycle.
To handle the variable heat capacity, use the ideal gas tables for air (Table A-17, starting on page
849). At the initial temperature of 27oC = 300.15 K we find u1 ≈ u(300 K) = 214.07 kJ/kg and
vr(300 K) = 621.2. The isothermal compression process from point one to point two is computed
as follows using the vr values from Table A-17.
v2 v r (T2 )

v1 v r (T1 )
 v r (T2 )  v r (T1 )
v2
1
491.2 kJ
 621.2  77.65  T2  673.1 K and u2 
v1
8
kg
The values of T2 and u2 are found by interpolation. For the constant volume heat addition, we
apply the first law to get u3 = u2 + qH = 491.2 kJ/kg + 750 kJ/kg = 1241.2 kJ/kg. At this value of
internal energy, we find T3 = 1539 and vr(T3) = 6.588. The pressure at point 3 can be found from
the ideal gas law.
P3 
RT3 RT3 RT3 v1 RT3
P (CR)T3 (95 kPa)(8)(1539 K )



(CR)  1

 3,898 kPa
v3
v2
v1 v 2 RT1
T1
(300.15 K )
P1
The isentropic expansion from point 3 to point 4 is computed in the same way as the isentropic
compression from point 1 to point 2.
v 4 v r (T4 )

v3 v r (T3 )
 v r (T4 )  v r (T3 )
v4
571.69 kJ
 (6.588)(8)  52.70  T4  774.5 K and u4 
v3
kg
The low temperature heat rejection is found from the first law for closed-system, constant-volume
processes: |qL| = |u1 – u4| = |214.07 kJ/kg – 571.69 kJ/kg| = 357.62 kJ/kg. We can then find the
new work as the difference between the two cycle heat transfers:
|w| = |qH| - |qL| = 750 kJ/kg – 357.62 kJ/kg or |w| = 392.38 kJ/kg .
Unit twelve homework solutions
ME 370, L. S. Caretto, Fall 2010
Page 5
The cycle efficiency is found as from the usual equation.
 = |w|/|qH| = (392.38 kJ/kg) / (750 kJ/kg) or  = 52.3% .
The mean effective pressure is defined by the equation mep = |w|/(vmax – vmin). For the Otto
cycle, vmax = v1 and vmin = v2 = v1/CR. Thus, we can find our mean effective pressure as follows.
mep 
5
w
1 

v1 1 

 CR 

w
RT1 
1 
1 

P1  CR 

392.38 kJ
kg
0.287 kJ
(300.15 K )
kg  K
 1
1  
95 kPa
 8
mep = 495 kPa .
Air enters the compressor of a gas-turbine engine at 300 k and 100 kPa, where it is
compressed to 700 kPA and 580 K. Heat is transferred to air in the amount of 950 kJ/kg
before it enters the turbine. For a turbine efficiency of 86% determine (a) the fraction of
the turbine work ouput used to drive the compressor and (b) the thermal efficiency.
Assume variable specific heats for air.
For each component in the gas turbine cycle we assume that the component is steady flow with
negligible kinetic and potential energy changes. Each component has one inlet and one outlet so
that the general first law for each component in this system is q = w u + hout – hin. We assume that
there is no work input in the heat addition step and that the compressor and turbine are adiabatic.
We also assume that there is no pressure loss during the heat addition step so that the
compressor exit pressure of 700 kPa is the same as the turbine entrance pressure. We will use
the air tables (Table A-17, starting on page 849).to compute the air properties with variable heat
capacity.
For the compressor, the first law, with q = 0, becomes w = hin – hout. = h(300 K) – h(580 K) =
300.19 kJ/kg – 586.04 kJ/kg = -285.85 kJ/kg.
For the ideal, isentropic turbine, we find the end state from the ideal gas table relation for
isentropic processes: Pr(Tout,s) = Pr(Tin)(Pout/Pin) where Pr(Tin) = 474.11 and Pout/Pin = (100
kPa)/(700 kPa) so that Pr(Tout,s) = 474.11/7 = 67.73. Interpolation in the ideal gas tables gives
hout,s = 905.83 kJ/kg. The actual turbine work then is found as w = s(hin – hout,s). The inlet
enthalpy to the turbine is found from the first law for the heat addition step: q H = hturb,in – hcomp,out,
so that hturb,in = qH + hcomp,out,= 950 kJ/kg + 586.04 kJ/kg = 1536.04 kJ/kg. We can then find the
turbine work.
W turb = s(hin – hout,s) =(86%)(536.04 kJ/k - 905.83 kJ/kg) = 542.0 kJ/kg
The compressor work input as a fraction of the turbine work output is now found.
|wcomp|/|wturb| = |-285.85 kJ/kg| / (542.0 kJ/kg)
|wcomp|/|wturb| = 52.7% .
The efficiency is found from the usual equation.
 = |wnet|/|qH| = (542.0 kJ/kg - 285.85 kJ/kg) / (950 kJ/kg) or  = 27.0% .