Chapter 5
The maximum coefficient of performance β for the refrigeration cycle is given by
β =
QC
QC
TC
=
=
QH − QC TH − TC
Wcycle
The maximum coefficient of performance γfor the heat pump cycle is given by
γ =
QH
TH
QH
=
=
QH − QC TH − TC
Wcycle
Example 5.4-23. ---------------------------------------------------------------------------------By steadily circulating a refrigerant at low temperature through passages in the walls of the
freezer compartment, a refrigerator maintains the freezer compartment at −5°C when the air
surrounding the refrigerator is at 22°C. The rate of heat transfer from the freezer
compartment to the refrigerant is 8000 kJ/h and the power input required to operate the
refrigerator is 3200 kJ/h. Determine the coefficient of performance of the refrigerator and
compare with the coefficient of performance of a reversible refrigeration cycle operating
between reservoirs at the same two temperatures.
Solution -----------------------------------------------------------------------------------------High T = 295 K
Hot reservoir
QH
System
Wcycle = 3200 kJ/h
QC
Qc = 8000 kJ/h
Low T = 268K
Cold reservoir
Refrigeration cycle
The coefficient of performance β for the refrigeration cycle is given by
β =
QC
QC
8,000 kJ/h
=
=
= 2.5
3,200 kJ/h
Wcycle
Wcycle
The maximum coefficient of performance β for the refrigeration cycle is given by
3
Moran, M. J. and Shapiro H. N., Fundamentals of Engineering Thermodynamics, Wiley, 2008, pg. 236
5-9
β max =
QC
QC
TC
268 K
=
=
=
= 9.93
QH − QC TH − TC
295 K-268 K
Wcycle
Example 5.4-34. ---------------------------------------------------------------------------------A dwelling requires 6×105 Btu per day to maintain its temperature at 70°F when the outside
temperature is 32°F. (a) If an electric heat pump is used to supply this energy, determine the
minimum theoretical work input for one day of operation, in Btu/day. (b) Evaluating
electricity at 8 cents per kW⋅h, determine the minimum theoretical cost to operate the heat
pump, in $/day.
Solution -----------------------------------------------------------------------------------------High T = 530 R
Hot reservoir
5
QH = 6x10 Btu/h
QH
System
Wcycle = ?
QC
Low T = 492 R
Cold reservoir
Heat Pump cycle
The maximum coefficient of performance γ for the heat pump cycle is given by
γmax =
QH
TH
QH
=
=
QH − QC TH − TC
Wcycle
The minimum theoretical work input is then: Wmin =
Wmin = QH
QH
γ max
= QH
TH − TC
TH
( 530 - 492 ) R = 4.3×104 Btu/day
TH − TC
= (6×105 Btu/day)
TH
530 R
The minimum cost is
Cost = (4.3×104 Btu/day)
4
1 kW ⋅ h
3413 Btu
0.08 $
kW ⋅ h
= 1.01 $/day
Moran, M. J. and Shapiro H. N., Fundamentals of Engineering Thermodynamics, Wiley, 2008, pg. 238
5-10
5.5 Carnot Cycles
A Carnot cycle is the most efficient type of cycle we can possibly have. Figure 5.5-1 shows
an ideal gas in a piston-cylinder assembly undergoing a Carnot cycle. In this cycle, the gas
goes through four reversible processes through which it returns to its initial state.
Constant TH
Isothermal
expansion
QH
Adiabatic
compression
2
Adiabatic
expansion
3
Isothermal
compression
2 1
QC
Constant TC
1
3
4
4
Figure 5.5-1 Carnot power cycle for an ideal gas.
The four processes of the cycle are
Process 1-2: The gas is compress adiabatically to state 2, where the temperature is TH.
Process 2–3: The gas expands isothermally while receiving energy QH from the hot reservoir
at TH by heat transfer.
Process 3–4: The gas is allowed to continue to expand adiabatically until the temperature
drops to TC.
Process 4–1: The gas is compressed isothermally to its initial state while it discharges energy
QC to the cold reservoir at TC by heat transfer.
The net work obtained in a Carnot cycle is given by the sum of the work obtained in all
processes:
− Wcycle = W23 + W34 − W41 − W12
5-11
Figure 5.5-2 pv diagram for a Carnot cycle.
The sign of Wcycle is negative since the overall effect of the Carnot cycle shown in Figure 5.51 is to deliver work from the system to the surroundings. The Carnot power cycle can be
operated in the opposite direction to act as a refrigeration or heat pump cycle.
Example 5.5-15. ---------------------------------------------------------------------------------Consider 1 mole of an ideal gas in a piston-cylinder assembly. This gas undergoes a power
cycle, which is described below. The heat capacity is constant, cv = 1.5 R = (1.5) (8.314
J/mol⋅K).
1) A reversible, isothermal expansion from 10.1 bar to 8 bar.
2) A reversible, adiabatic expansion from 8 bar and 1000 K to 850 K.
3) A reversible, isothermal compression at 850 K.
4) A reversible, adiabatic compression from 850 K to 1000 K and 10.1 bar.
Perform the following analysis:
a) Calculate Q, W, and ∆U for each of the steps in the cycle.
b) Draw the cycle on a pv diagram.
c) Calculate the efficiency of the cycle and compare with that of the Carnot
cycle.
Solution -----------------------------------------------------------------------------------------a) Calculate Q, W, and ∆U for each of the steps in the cycle.
5
Koretsky M.D., Engineering and Chemical Thermodynamics, Wiley, 2004, pg. 89
5-12
Step 1:
A reversible, isothermal expansion at 1000 K and10.1 bar to 8 bar.
∆U = 0 for isothermal, ideal gas
W=−
Since V =
nRT
p
8 bar
10.1 bar
dV = −
pdV
nRT
dp
p2
W=
8 bar
10.1 bar
nRT
dp = − n R T ln(10.1/8)
p
W = − (1 mol)(8.314 J/mol⋅K)(1000 K) ln(10.1/8) = − 1,938 J
∆U = 0 = QH + W
V1 =
Step 2:
QH = − W = 1,938 J
nRT1
(1 mol)(8.314 J/mol ⋅ K)(1000 K)
=
= 0.00823 m3
5
p1
10.1× 10 Pa
A reversible, adiabatic expansion from 8 bar and 1000 K to 850 K.
Q = 0 for adiabatic process
∆U = n cv (T3 − T2) = (1 mol)(1.5*8.314 J/mol⋅K)(850 − 1000) K = − 1,871 J
W = ∆U = − 1,871 J
V2 =
Step 3:
nRT2
(1 mol)(8.314 J/mol ⋅ K)(1000 K)
=
= 0.01039 m3
5
p2
8.0 × 10 Pa
A reversible, isothermal compression at 850 K.
∆U = 0 for isothermal, ideal gas
W=−
p4
p3
pdV =
p4
p3
nRT
dp = n R T ln(p4/p3)
p
Since process 2-3 is adiabatic and reversible: p2V2k = p3V3k where k =
k=
1.5 R + R
5
= = 1.67
1.5 R
3
Using ideal gas law
5-13
cp
cv
=
cv + R
cv
p 2 V2
k
( nRT )
=
k
2
p
( nRT )
= 390 =
k
3
k −1
2
k −1
3
p
(Note: p2 = 8 bar, V2 = 0.01039 m3)
Solving for p3 we get
( nRT )
k
1.5
3
p3 =
= 6.7×105 Pa = 6.7 bar
390
Similarly for p4 we have p1V1k = p4V4k
p 1 V1
k
( nRT )
=
k
1
( nRT )
= 333.6 =
4
k −1
1
p
p
k −1
4
k
(Note: p2 = 10.1 bar, V1 = 0.00823 m3)
Solving for p4 we get
p4 =
( nRT )
k
1.5
3
333.6
= 7.2×105 Pa = 7.2 bar
W = n R T ln(p4/p3) = (1mol)(8.314 J/mol⋅K)(850 K) ln(7.2/6.7) = 1,656 J
∆U = 0 = QL + W
Step 4:
QL = − W = − 1,656 J
A reversible, adiabatic compression from 7.2 bar and 850 K to 10.1 bar and
1000 K.
Q = 0 for adiabatic process
∆U = n cv (T4 − T1) = (1 mol)(1.5*8.314 J/mol⋅K)(1000 − 850) K = 1,871 J
W = ∆U = 1,871 J
Process
State 1 to 2
State 2 to 3
State 3 to 4
State 4 to 1
Total
Table E5.5-1 Results of the calculation
W(J)
∆U(J)
0
− 1,938
− 1,871
− 1,871
0
1,656
1,871
1,871
0
− 282
5-14
Q(J)
1,938
0
− 1,656
0
282
State
(1)
(2)
(3)
(4)
Table E5.5-2 T , p, and V for the cycle
T(K)
p(bat)
1000
10.1
1000
8.0
850
6.7
850
7.2
b) Draw the cycle on a pv diagram.
The pv diagram can be plotted from the following Matlab program
% pv diagram
Th=1000;Tc=850;R=8.314;
ph=10.1e5;pl=8e5;
v1=R*Th/ph;
v2=R*Th/pl;
% Process 1 to 2 (isothermal expansion)
v12=linspace(v1,v2,50);
p12=R*Th./v12;
% Process 2 to 3 (adiabatic expansion)
k=5/3;con=(R*Th)^k/pl^(k-1);
T=linspace(Th,Tc,50);
p23=(((R*T).^k)/con).^1.5;
v23=R*T./p23;v3=v23(50);
% Process 3 to 4 (isothermal compression)
con2=(R*Th)^k/ph^(k-1);
p4=(((R*Tc)^k)/con2)^1.5;
v4=R*Tc/p4;
v34=linspace(v4,v3);
p34=R*Tc./v34;
% Process 4 to 1 (adiabatic compression)
p41=(((R*T).^k)/con2).^1.5;
v41=R*T./p41;
plot(v12,p12,v23,p23,v34,p34,v41,p41)
grid on
ylabel('p(Pa)');xlabel('v(m^3)')
5-15
V(m3)
0.00823
0.01039
0.0124
0.00981
Figure E5.5-1 pv diagram
c) Calculate the efficiency of the cycle and compare with that of the Carnot
cycle.
The thermal efficiency η of the power cycle is given by
η=
Wcycle
QH
=
282
= 0.1455
1938
The thermal efficiency η of the Carnot power cycle is given by
η=
Wcycle
QH
=1−
QC
T
850
=1− C =1−
= 0.150
QH
TH
1000
Therefore the given power cycle is a Carnot cycle. The small difference is due to round off
error.
5-16