5-9

Chapter 5 The maximum coefficient of performance β for the refrigeration cycle is given by

β = C

cycle

QW

= C

H C

QQ Q−

= C

H C

TT T−

The maximum coefficient of performance γfor the heat pump cycle is given by

γ = H

cycle

QW

= H

H C

QQ Q−

= H

H C

TT T−

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 ------------------------------------------------------------------------------------------

System

High T = 295 KHot reservoir

Low T = 268KCold reservoir

QH

QC

W = 3200 kJ/hcycle

Refrigeration cycle

Q = 8000 kJ/hc

The coefficient of performance β for the refrigeration cycle is given by

β = C

cycle

QW

= C

cycle

QW

�

� =

8,000 kJ/h3,200 kJ/h

= 2.5

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

5-10

β max = C

cycle

QW

= C

H C

QQ Q−

= C

H C

TT T−

= 268 K

295 K-268 K = 9.93

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 ------------------------------------------------------------------------------------------

System

High T = 530 RHot reservoir

Low T = 492 RCold reservoir

QH

QC

W = ?cycle

Heat Pump cycle

Q = 6x10 Btu/hH 5

The maximum coefficient of performance γ for the heat pump cycle is given by

γmax = H

cycle

QW

= H

H C

QQ Q−

= H

H C

TT T−

The minimum theoretical work input is then: Wmin =

max

HQγ

= H CH

H

T TQ

T−

Wmin = H CH

H

T TQ

T−

= (6×105 Btu/day) ( )530 - 492 R

530 R = 4.3×104 Btu/day

The minimum cost is

Cost = (4.3×104 Btu/day)1 kW h3413 Btu

⋅� �� �� �

0.08 $kW h

� �� �⋅� �

= 1.01 $/day

4 Moran, M. J. and Shapiro H. N., Fundamentals of Engineering Thermodynamics, Wiley, 2008, pg. 238

5-11

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.

2 1

2 3

3 4

1 4

Adiabaticcompression

Isothermalexpansion

Adiabaticexpansion

Isothermalcompression

QH

QC

Constant TH

Constant TC

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 = 23W + 34W − 41W − 12W

5-12

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 5.5-1 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, vc = 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

5-13

Step 1: A reversible, isothermal expansion at 1000 K and10.1 bar to 8 bar. ∆U = 0 for isothermal, ideal gas

W = − 8 bar

10.1barpdV�

Since V = nRT

p � dV = − 2

nRTp

dp � W = 8 bar

10.1bar

nRTdp

p� = − n R T ln(10.1/8)

W = − (1 mol)(8.314 J/mol⋅K)(1000 K) ln(10.1/8) = − 1,938 J ∆U = 0 = QH + W � QH = − W = 1,938 J

V1 = 1

1

nRTp

= 5

(1 mol)(8.314 J/mol K)(1000 K)10.1 10 Pa

⋅×

= 0.00823 m3

Step 2: A reversible, adiabatic expansion from 8 bar and 1000 K to 850 K. Q = 0 for adiabatic process ∆U = n vc (T3 − T2) = (1 mol)(1.5*8.314 J/mol⋅K)(850 − 1000) K = − 1,871 J W = ∆U = − 1,871 J

V2 = 2

2

nRTp

= 5

(1 mol)(8.314 J/mol K)(1000 K)8.0 10 Pa

⋅×

= 0.01039 m3

Step 3: A reversible, isothermal compression at 850 K. ∆U = 0 for isothermal, ideal gas

W = − 4

3

p

ppdV� =

4

3

p

p

nRTdp

p� = n R T ln(p4/p3)

Since process 2-3 is adiabatic and reversible: p2V2k = p3V3

k where k = p

v

c

c = v

v

c Rc+

k = 1.5

1.5R R

R+

= 53

= 1.67

Using ideal gas law

5-14

p2V2k =

( )2

12

k

k

nRT

p − = 390 = ( )3

13

k

k

nRT

p − (Note: p2 = 8 bar, V2 = 0.01039 m3)

Solving for p3 we get

p3 = ( )

1.5

3

390

knRT

� �� � �

= 6.7×105 Pa = 6.7 bar

Similarly for p4 we have p1V1

k = p4V4k

p1V1k =

( )1

11

k

k

nRT

p − = 333.6 = ( )4

14

k

k

nRT

p − (Note: p2 = 10.1 bar, V1 = 0.00823 m3)

Solving for p4 we get

p4 = ( )

1.5

3

333.6

knRT

� �� � �

= 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 � QL = − W = − 1,656 J Step 4: 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 vc (T4 − T1) = (1 mol)(1.5*8.314 J/mol⋅K)(1000 − 850) K = 1,871 J

W = ∆U = 1,871 J

Table E5.5-1 Results of the calculation

Process ∆U(J) W(J) Q(J) State 1 to 2 0 − 1,938 1,938 State 2 to 3 − 1,871 − 1,871 0 State 3 to 4 0 1,656 − 1,656 State 4 to 1 1,871 1,871 0 Total 0 − 282 282

5-15

Table E5.5-2 T , p, and V for the cycle State T(K) p(bat) V(m3) (1) 1000 10.1 0.00823 (2) 1000 8.0 0.01039 (3) 850 6.7 0.0124 (4) 850 7.2 0.00981

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)')

5-16

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 = cycle

H

W

Q =

2821938

= 0.1455

The thermal efficiency η of the Carnot power cycle is given by

η = cycle

H

W

Q = 1 − C

H

QQ

= 1 − C

H

TT

= 1 − 850

1000 = 0.150

Therefore the given power cycle is a Carnot cycle. The small difference is due to round off error.