Chapter 11 Refrigeration Cycles PDF

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Solutions Manual forThermodynamics: An Engineering ApproachSeventh EditionYunus A. Cengel, Michael A. BolesMcGraw-Hill, 2011Chapter 11REFRIGERATION CYCLESPROPRIETARY AND CONFIDENTIALThis Manual is the proprietary property of The McGraw-Hill Companies, Inc. (“McGraw-Hill”) andprotected by copyright a...

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11-1

Solutions Manual for

Thermodynamics: An Engineering Approach Seventh Edition Yunus A. Cengel, Michael A. Boles McGraw-Hill, 2011

Chapter 11 REFRIGERATION CYCLES

PROPRIETARY AND CONFIDENTIAL

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11-2

The Reversed Carnot Cycle

11-1C The reversed Carnot cycle serves as a standard against which actual refrigeration cycles can be compared. Also, the COP of the reversed Carnot cycle provides the upper limit for the COP of a refrigeration cycle operating between the specified temperature limits.

11-2C Because the compression process involves the compression of a liquid-vapor mixture which requires a compressor that will handle two phases, and the expansion process involves the expansion of high-moisture content refrigerant.

11-3 A steady-flow Carnot refrigeration cycle with refrigerant-134a as the working fluid is considered. The coefficient of performance, the amount of heat absorbed from the refrigerated space, and the net work input are to be determined. Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible. Analysis (a) Noting that TH = 40°C = 313 K and TL = Tsat @ 100 kPa = −26.37°C = 246.6 K, the COP of this Carnot refrigerator is determined from

COPR,C =

TH

1 1 = = 3.72 / TL − 1 (313 K ) / (246.6 K ) − 1

T

(b) From the refrigerant tables (Table A-11), 4

h3 = hg @ 40°C = 271.27 kJ/kg h 4 = h f @40°C = 108.26 kJ/kg

QH

3 40°C

Thus,

100 kPa q H = h3 − h4 = 271.27 − 108.26 = 163.0 kJ/kg

and qH qL

=

TH TL

⎯ ⎯→ qL =

TL TH

qH

1 QL

2

s

⎛ 246.6 K ⎞ = ⎜⎜ ⎟⎟(163.0 kJ/kg ) = 128.4 kJ/kg ⎝ 313 K ⎠

(c) The net work input is determined from wnet = q H − q L = 163.0 − 128.4 = 34.6 kJ/kg

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11-3

11-4E A steady-flow Carnot refrigeration cycle with refrigerant-134a as the working fluid is considered. The coefficient of performance, the quality at the beginning of the heat-absorption process, and the net work input are to be determined. Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible. Analysis (a) Noting that TH = Tsat @ 90 psia = 72.78°F = 532.8 R and TL = Tsat @ 30 psia = 15.37°F = 475.4 R.

COPR,C =

1 1 = = 8.28 T H / TL − 1 (532.8 R )/ (475.4 R ) − 1

T

(b) Process 4-1 is isentropic, and thus

(

s1 = s 4 = s f + x 4 s fg

) @ 90 psia = 0.07481 + (0.05)(0.14525)

= 0.08207 Btu/lbm ⋅ R ⎛ s1 − s f x1 = ⎜ ⎜ s fg ⎝

⎞ 0.08207 − 0.03793 ⎟ = = 0.2374 ⎟ 0.18589 ⎠ @ 30 psia

4

1

QH

QL

3

2

(c) Remembering that on a T-s diagram the area enclosed represents the net work, and s3 = sg @ 90 psia = 0.22006 Btu/lbm·R,

s

wnet,in = (T H − T L )( s3 − s 4 ) = (72.78− 15.37)( 0.22006− 0.08207) Btu/lbm⋅ R = 7.92 Btu/lbm

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11-4

Ideal and Actual Vapor-Compression Refrigeration Cycles

11-5C Yes; the throttling process is an internally irreversible process.

11-6C To make the ideal vapor-compression refrigeration cycle more closely approximate the actual cycle.

11-7C No. Assuming the water is maintained at 10°C in the evaporator, the evaporator pressure will be the saturation pressure corresponding to this pressure, which is 1.2 kPa. It is not practical to design refrigeration or air-conditioning devices that involve such extremely low pressures.

11-8C Allowing a temperature difference of 10°C for effective heat transfer, the condensation temperature of the refrigerant should be 25°C. The saturation pressure corresponding to 25°C is 0.67 MPa. Therefore, the recommended pressure would be 0.7 MPa.

11-9C The area enclosed by the cyclic curve on a T-s diagram represents the net work input for the reversed Carnot cycle, but not so for the ideal vapor-compression refrigeration cycle. This is because the latter cycle involves an irreversible process for which the process path is not known.

11-10C The cycle that involves saturated liquid at 30°C will have a higher COP because, judging from the T-s diagram, it will require a smaller work input for the same refrigeration capacity.

11-11C The minimum temperature that the refrigerant can be cooled to before throttling is the temperature of the sink (the cooling medium) since heat is transferred from the refrigerant to the cooling medium.

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11-5

11-12E A refrigerator operating on the ideal vapor-compression refrigeration cycle with refrigerant-134a as the working fluid is considered. The increase in the COP if the throttling process were replaced by an isentropic expansion is to be determined. Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible. Analysis In an ideal vapor-compression refrigeration cycle, the compression process is isentropic, the refrigerant enters the compressor as a saturated vapor at the evaporator pressure, and leaves the condenser as saturated liquid at the condenser pressure. From the refrigerant tables (Tables A-11E, A-12E, and A-13E), T1 = 20° F ⎫ h1 = hg @ 20°F = 105.98 Btu/lbm sat. vapor ⎬⎭ s 1 = s g @ 20°F = 0.22341 Btu/lbm ⋅ R P2 = 300 psia ⎫ ⎬ h2 = 125.68 Btu/lbm s 2 = s1 ⎭ P3 = 300 psia ⎫ h3 = hf ⎬ s =s sat. liquid f ⎭ 3

@ 300 psia @ 300 psia

T · QH

2 · Win

3 300 psia

= 66.339 Btu/lbm = 0.12715 Btu/lbm ⋅ R

20°F

h4 ≅ h3 = 66.339 Btu/lbm ( throttling ) T4 = 20°F ⎫ h4s = 59.80 Btu/lbm (isentropic expansion) s 4 = s 3 ⎬⎭ x 4 s = 0.4723

4s

4

· QL

1

s

The COP of the refrigerator for the throttling case is COPR =

qL h -h 105.98 − 66.339 = 2.012 = 1 4= w in h 2-h 1 125.68 −105.98

The COP of the refrigerator for the isentropic expansion case is COPR =

105.98 − 59.80 qL h -h = 1 4s = = 2.344 h 2-h 1 125.68 − 105.98 w in

The increase in the COP by isentropic expansion is 16.5%.

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11-6

11-13 An ideal vapor-compression refrigeration cycle with refrigerant-134a as the working fluid is considered. The COP and the power requirement are to be determined. Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible. Analysis In an ideal vapor-compression refrigeration cycle, the compression process is isentropic, the refrigerant enters the compressor as a saturated vapor at the evaporator pressure, and leaves the condenser as saturated liquid at the condenser pressure. From the refrigerant tables (Tables A-11, A-12, and A-13), T1 = 4 °C ⎫ h1 = hg sat. vapor ⎬⎭ s 1 = s g

= 252.77 kJ/kg @ 4° C = 0.92927 kJ/kg ⋅ K @ 4° C

T

P2 = 1 MPa ⎫ ⎬ h2 = 275.29 kJ/kg s 2 = s1 ⎭ P3 = 1 MPa ⎫ ⎬ h = hf sat. liquid ⎭ 3

@ 1 MPa

· QH

· Win

= 107.32 kJ/kg

h 4 ≅ h 3 = 107.32 kJ/kg ( throttling )

The mass flow rate of the refrigerant is

Q& L = m&( h1 − h 4 ) ⎯⎯→ m& =

2

3 1 MPa

& 400 kJ/s Q L = = 2.750 kg/s h1 − h 4 (252.77 −107.32) kJ/kg

4°C 4s

4

· QL

1

s

The power requirement is

&Win = m& ( h2− h1 )= (2.750 kg/s)(275.29− 252.77) kJ/kg = 61.93 kW The COP of the refrigerator is determined from its definition, 400 kW Q& COPR = L = = 6.46 & W in 61.93 kW

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11-7

11-14 An ideal vapor-compression refrigeration cycle with refrigerant-134a as the working fluid is considered. The rate of heat removal from the refrigerated space, the power input to the compressor, the rate of heat rejection to the environment, and the COP are to be determined. Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible. Analysis (a) In an ideal vapor-compression refrigeration cycle, the compression process is isentropic, the refrigerant enters the compressor as a saturated vapor at the evaporator pressure, and leaves the condenser as saturated liquid at the condenser pressure. From the refrigerant tables (Tables A-12 and A-13), P1 = 120 kPa sat. vapor

⎫ h1 = hg @ 120 kPa = 236.97 kJ/kg ⎬s ⎭ 1 = s g @ 120 kPa = 0.94779 kJ/kg ⋅ K

P2 = 0.7 MPa ⎫ ⎬ h 2 = 273.50 kJ/kg (T 2 = 34.95°C) s 2 = s1 ⎭ P3 = 0.7 MPa ⎫ ⎬ h3 = hf sat. liquid ⎭

@ 0 .7 MPa

T · QH

2

3 0.7 MPa

· Win

= 88.82 kJ/kg

h4 ≅ h3 = 88.82 kJ/kg (throttling )

Then the rate of heat removal from the refrigerated space and the power input to the compressor are determined from

0.12 4s

4

· QL

1 s

Q& L = m& ( h1 − h4 ) = ( 0.05 kg/s)(236.97 − 88.82) kJ/kg = 7.41 kW

and W& in = m& (h 2 − h1) = ( 0.05 kg/s)( 273.50 − 236.97) kJ/kg = 1.83 kW

(b) The rate of heat rejection to the environment is determined from

Q& H = Q& L + W& in = 7.41 + 1.83 = 9.23 kW (c) The COP of the refrigerator is determined from its definition, COPR =

Q& L 7.41 kW = = 4.06 W& in 1.83 kW

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11-8

11-15 An ideal vapor-compression refrigeration cycle with refrigerant-134a as the working fluid is considered. The rate of heat removal from the refrigerated space, the power input to the compressor, the rate of heat rejection to the environment, and the COP are to be determined. Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible. Analysis (a) In an ideal vapor-compression refrigeration cycle, the compression process is isentropic, the refrigerant enters the compressor as a saturated vapor at the evaporator pressure, and leaves the condenser as saturated liquid at the condenser pressure. From the refrigerant tables (Tables A-12 and A-13), P1 = 120 kPa sat. vapor

⎫ h1 = hg @ 120 kPa = 236.97 kJ/kg ⎬s ⎭ 1 = s g @ 120 kPa = 0.94779 kJ/kg ⋅ K

P2 = 0.9 MPa ⎫ ⎬h 2 = 278.93 kJ/kg (T 2 = 44.45°C) s 2 = s1 ⎭ P3 = 0.9 MPa ⎫ ⎬ h 3 = hf sat. liquid ⎭

@ 0.9 MPa

T · QH

2

3 0.9 MPa

· Win

= 101.61 kJ/kg

h 4 ≅ h 3 = 101.61 kJ/kg ( throttling)

Then the rate of heat removal from the refrigerated space and the power input to the compressor are determined from

0.12 MPa 4s

4

· QL

1 s

Q& L = m& ( h1 − h4 ) = ( 0.05 kg/s)( 236.97 −101.61) kJ/kg = 6.77 kW

and & (h 2 − h1) = ( 0.05 kg/s)( 278.93 − 236.97) kJ/kg = 2.10 kW W& in = m

(b) The rate of heat rejection to the environment is determined from

Q& H = Q& L + W& in = 6.77 + 2.10 = 8.87 kW (c) The COP of the refrigerator is determined from its definition, COPR =

Q& L 6.77 kW = = 3.23 W& in 2.10 kW

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11-9

11-16 An ideal vapor-compression refrigeration cycle with refrigerant-134a as the working fluid is considered. The throttling valve in the cycle is replaced by an isentropic turbine. The percentage increase in the COP and in the rate of heat removal from the refrigerated space due to this replacement are to be determined. Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible.

T

Analysis If the throttling valve in the previous problem is replaced by an isentropic turbine, we would have

s4s = s3 = sf @ 0.7 MPa = 0.33230 kJ/kg·K

· QH

2 · Win

3 0.7 MPa

and the enthalpy at the turbine exit would be ⎛ s3 −s x4 s = ⎜ ⎜ s fg ⎝

(

f

⎞ 0.33230 − 0.09275 ⎟ = = 0.2802 ⎟ 0.85503 ⎠ @ 120 kPa

h4 s = h f + x4 s h fg

)@ 120 kPa

= 22.49 + (0.2802)( 214.48) = 82.58 kJ/kg

0.12 MPa 4s

4

· QL

1 s

Then,

& = m&( h − h ) = ( 0.05 kg/s)( 236.97− 82.58) kJ/kg = 7.72 kW Q L 1 4s and COPR =

7.72 kW Q&L = = 4.23 W&in 1.83 kW

Then the percentage increase in Q& and COP becomes

∆Q& L 7.72 − 7.41 = 4.2% Increase in Q& L = & = 7.41 QL Increase in COPR =

∆ COPR 4.23− 4.06 = = 4.2% COP R 4.06

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11-10

11-17 A refrigerator with refrigerant-134a as the working fluid is considered. The rate of heat removal from the refrigerated space, the power input to the compressor, the isentropic efficiency of the compressor, and the COP of the refrigerator are to be determined. Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible. Analysis (a) From the refrigerant tables (Tables A-12 and A-13), P1 = 0.20 MPa ⎫ h1 = 248.80 kJ/kg ⎬ s = 0.95407 kJ/kg ⋅ K T1 = − 5° C ⎭ 1

T

P2 = 1.2 MPa ⎫ h 300.61 kJ/kg ⎬ 2= T2 = 70°C ⎭ P2 s = 1.2 MPa s 2 s = s1

⎫ ⎬h 2 s = 287.21 kJ/kg ⎭

P3 = 1.15 MPa ⎫ h h ⎬ 3= f T3 = 44°C ⎭

@ 44°C

= 114.28 kJ/kg

h4 ≅ h3 = 114.28 kJ/kg (throttling )

1.15 MPa

2s

· QH

2 1.2 MPa 70°C · Win

3 0.21 MPa 4

· QL

1

0.20 MPa -5°C s

Then the rate of heat removal from the refrigerated space and the power input to the compressor are determined from & (h1 − h4 ) = (0.07 kg/s )(248.80 − 114.28 ) kJ/kg = 9.42 kW Q& L = m

and & (h 2 − h1 ) = (0.07 kg/s )(300.61 − 248.80 ) kJ/kg = 3.63 kW W& in = m

(b) The isentropic efficiency of the compressor is determined from

ηC =

h2 s − h1 287.21 − 248.80 = = 0.741 = 74.1% 300.61 − 248.80 h2 − h1

(c) The COP of the refrigerator is determined from its definition,

COPR =

Q&L 9.42 kW = = 2.60 W&in 3.63 kW

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11-11

11-18E An ideal vapor-compression refrigeration cycle with refrigerant-134a as the working fluid is considered. The mass flow rate of the refrigerant and the power requirement are to be determined. Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible. Analysis In an ideal vapor-compression refrigeration cycle, the compression process is isentropic, the refrigerant enters the compressor as a saturated vapor at the evaporator pressure, and leaves the condenser as saturated liquid at the condenser pressure. From the refrigerant tables (Tables A-11E, A-12E, and A-13E), T1 = 5 °F ⎫ h1 = h g @ 50°F = 103.82 Btu/lbm sat. vapor ⎬⎭ s1 = s g @ 5 °F = 0.22485 Btu/lbm ⋅ R

T

P2 = 180 psia ⎫ ⎬ h 2 = 121.99 Btu/lbm s2 = s1 ⎭

· QH 3 180 psia

P3 = 180 psia ⎫ ⎬ h3 = h f @ 180 psia = 51.50 Btu/lbm sat. liquid ⎭ h4 ≅ h3 =51.50 Btu/lbm ( throttling )

2 · Win

5°F 4s

4

· QL

1

The mass flow rate of the refrigerant is & ( h1 − h4 ) ⎯ & = ⎯→ m Q& L = m

Q& L 45,000 Btu/h = = 860.1 lbm/h h1 − h4 (103.82 − 51.50) Btu/lbm

s

The power requirement is

1 kW ⎛ W&in = m& (h2 − h1 ) = (860.1 lbm/h)(121.99 − 103.82) Btu/lbm⎜ ⎝ 3412.14 Btu/h

⎞ ⎟ = 4.582 kW ⎠

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11-12

11-19E

Problem 11-18E is to be repeated if ammonia is used as the refrigerant.

Analysis The problem is solved using EES, and the solution is given below. "Given" x[1]=1 T[1]=5 [F] x[3]=0 P[3]=180 [psia] Q_dot_L=45000 [Btu/h] "Analysis" Fluid$='ammonia' "compressor" h[1]=enthalpy(Fluid$, T=T[1], x=x[1]) s[1]=entropy(Fluid$, T=T[1], x=x[1]) s[2]=s[1] P[2]=P[3] h[2]=enthalpy(Fluid$, P=P[2], s=s[2]) "expansion valve" h[3]=enthalpy(Fluid$, P=P[3], x=x[3]) h[4]=h[3] "cycle" m_dot_R=Q_dot_L/(h[1]-h[4]) W_dot_in=m_dot_R*(h[2]-h[1])*Convert(Btu/h, kW) Solution for ammonia COP_R=4.515 Fluid$='ammonia' m_dot_R=95.8 [lbm/h] Q_dot_L=45000 [Btu/h] W_dot_in=2.921 [kW] Solution for R-134a COP_R=2.878 Fluid$='R134a' m_dot_R=860.1 [lbm/h] Q_dot_L=45000 [Btu/h] W_dot_in=4.582 [kW]

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