Chapter 11 REFRIGERATION CYCLES PDF

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11-1 Chapter 11 REFRIGERATION CYCLES The Reversed Carnot Cycle 11-1C 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. 11-2...

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11-1

Chapter 11 REFRIGERATION CYCLES The Reversed Carnot Cycle

11-1C 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.

11-2 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 = 30°C = 303 K and TL = Tsat @ 160 kPa = -15.60°C = 257.4 K, the COP of this Carnot refrigerator is determined from COPR,C =

1 1 = = 5.64 TH / TL − 1 (303 K ) / (257.4 K ) − 1

T

(b) From the refrigerant tables (Table A-11), h3 = h g @30°C = 266.66 kJ/kg h4 = h f @30°C = 93.58 kJ/kg

Thus,

4

QH

3 30°C

160 kPa q H = h3 − h4 = 266.66 − 93.58 = 173.08 kJ/kg

1 QL

2

and ⎛ 257.4 K ⎞ q H TH T ⎟⎟(173.08 kJ/kg ) = 147.03 kJ/kg = ⎯ ⎯→ q L = L q H = ⎜⎜ q L TL TH ⎝ 303 K ⎠

(c) The net work input is determined from wnet = q H − q L = 173.08 − 147.03 = 26.05 kJ/kg

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s

11-2

11-3E 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 TH / 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

(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,

4

1

QH

QL

3

2

s

w net,in = (T H − T L )(s 3 − s 4 ) = (72.78 − 15.37)(0.22006 − 0.08207 ) Btu/lbm ⋅ R = 7.92 Btu/lbm

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11-3

Ideal and Actual Vapor-Compression Cycles

11-4C Yes; the throttling process is an internally irreversible process.

11-5C To make the ideal vapor-compression refrigeration cycle more closely approximate the actual cycle.

11-6C 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.

11-7C 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.

11-8C 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.

11-9C 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.

11-10C 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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11-4

11-11 A commercial refrigerator with refrigerant-134a as the working fluid is considered. The quality of the refrigerant at the evaporator inlet, the refrigeration load, the COP of the refrigerator, and the theoretical maximum refrigeration load for the same power input to the compressor are to be determined. Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible. Analysis (a) From refrigerant-134a tables (Tables A-11 through A-13) P1 = 60 kPa ⎫ ⎬ h1 = 230.03 kJ/kg T1 = −34°C ⎭ P2 = 1200 kPa ⎫ ⎬ h2 = 295.16 kJ/kg T2 = 65°C ⎭ P3 = 1200 kPa ⎫ ⎬ h3 = 111.23 kJ/kg T3 = 42°C ⎭ h4 = h3 = 111.23 kJ/kg

Water 18°C

26°C QH

1.2 MPa 65°C

42°C

P4 = 60 kPa

⎫ ⎬ x 4 = 0.4795 h4 = 111.23 kJ/kg ⎭

Using saturated liquid enthalpy at the given temperature, for water we have (Table A-4)

Condenser 3

2

Expansion valve

Win Compressor

4

60 kPa -34°C

1 Evaporator

hw1 = h f @ 18°C = 75.47 kJ/kg

QL

hw 2 = h f @ 26°C = 108.94 kJ/kg

(b) The mass flow rate of the refrigerant may be determined from an energy balance on the compressor m& R (h2 − h3 ) = m& w (hw 2 − hw1 ) m& R (295.16 − 111.23)kJ/kg = (0.25 kg/s)(108.94 − 75.47)kJ/kg ⎯ ⎯→ m& R = 0.0455 kg/s

The waste heat transferred from the refrigerant, the compressor power input, and the refrigeration load are Q& H = m& R (h2 − h3 ) = (0.0455 kg/s)(295.16 − 111.23)kJ/kg = 8.367 kW W& in = m& R (h2 − h1 ) − Q& in = (0.0455 kg/s)(295.16 − 230.03)kJ/kg − 0.45 kW = 2.513 kW Q& L = Q& H − W& in = 8.367 − 2.513 = 5.85 kW

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

T

Q& 5.85 COP = L = = 2.33 & Win 2.513

2 · QH

2 · Win

3

(d) The reversible COP of the refrigerator for the same temperature limits is COPmax =

1 1 = = 5.063 T H / T L − 1 (18 + 273) /(−30 + 273) − 1

4

· QL

1 s

Then, the maximum refrigeration load becomes Q& L,max = COPmax W& in = (5.063)(2.513 kW) = 12.72 kW

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11-5

11-12 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 = h g @ 4°C = 252.77 kJ/kg ⎬ sat. vapor ⎭ s1 = s g @ 4°C = 0.92927 kJ/kg ⋅ K P2 = 1 MPa ⎫ ⎬ h2 = 275.29 kJ/kg s 2 = s1 ⎭ P3 = 1 MPa ⎫ ⎬ h = hf sat. liquid ⎭ 3

@ 1 MPa

T · QH

2

3 1 MPa

· Win

= 107.32 kJ/kg

h4 ≅ h3 = 107.32 kJ/kg ( throttling)

4°C 4s

4

· QL

1

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

Q& L 400 kJ/s = = 2.750 kg/s h1 − h4 (252.77 − 107.32) kJ/kg

s

The power requirement is W& in = 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, COPR =

Q& L 400 kW = = 6.46 & Win 61.93 kW

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11-6

11-13 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-11, A-12, and A-13), P1 = 400 kPa ⎫ h1 = h g @ 400 kPa = 255.55 kJ/kg ⎬ s =s sat. vapor g @ 400 kPa = 0.92691 kJ/kg ⋅ K ⎭ 1 P2 = 800 kPa ⎫ ⎬ h2 = 269.90 kJ/kg s 2 = s1 ⎭ P3 = 800 kPa ⎫ ⎬ h3 = h f sat. liquid ⎭

@ 800 kPa

T · QH

2

3 0.8 MPa

· Win

= 95.47 kJ/kg

0.4 MPa

h4 ≅ h3 = 95.47 kJ/kg ( throttling )

4s

4

· QL

1

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

Q& L 10 kJ/s = = 0.06247 kg/s h1 − h4 (255.55 − 95.47) kJ/kg

s

The power requirement is W& in = m& (h2 − h1 ) = (0.06247 kg/s)(269.90 − 255.55) kJ/kg = 0.8964 kW

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11-7

11-14E 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 A13E), T1 = −10°F ⎫ h1 = h g @ −10° F = 101.61 Btu/lbm ⎬ sat. vapor ⎭ s1 = s g @ −10°F = 0.22660 Btu/lbm ⋅ R P2 = 100 psia ⎫ ⎬ h2 = 117.57 Btu/lbm s 2 = s1 ⎭ P3 = 100 psia ⎫ ⎬ h3 = h f sat. liquid ⎭

@ 100 psia

T · QH

= 37.869 Btu/lbm

h4 ≅ h3 = 37.869 Btu/lbm ( throttling )

2

3 100 psia

· Win

-10°F 4s

4

· QL

1

The mass flow rate of the refrigerant is

s

Q& L 24,000 Btu/h Q& L = m& (h1 − h4 ) ⎯ ⎯→ m& = = = 376.5 lbm/h h1 − h4 (101.61 − 37.869) Btu/lbm

The power requirement is

1 kW ⎛ ⎞ W& in = m& (h2 − h1 ) = (376.5 lbm/h)(117.57 − 101.61) Btu/lbm⎜ ⎟ = 1.761 kW 3412.14 Btu/h ⎝ ⎠

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11-8

11-15E Problem 11-14E 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]=-10 [F] x[3]=0 P[3]=100 [psia] Q_dot_L=24000 [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$, P=P[3], x=x[3]) h[4]=h[3] "cycle" m_dot_R=Q_dot_L/(h[1]-h[4]) W_dot_in=m_dot_R*(h[2]-h[1])*Convert(Btu/h, kW) Solution for ammonia Fluid$='ammonia' h[2]=701.99 [Btu/lb_m] m_dot_R=47.69 [lbm/h] Q_dot_L=24000 [Btu/h] T[1]=-10 [F] x[3]=0

COP_R=5.847 h[3]=112.67 [Btu/lb_m] P[2]=100 [psia] s[1]=1.42220 [Btu/lb_m-R] W_dot_in=1.203 [kW]

h[1]=615.92 [Btu/lb_m] h[4]=112.67 [Btu/lb_m] P[3]=100 [psia] s[2]=1.42220 [Btu/lb_m-R] x[1]=1

Solution for R-134a Fluid$='R134a' h[2]=117.58 [Btu/lb_m] m_dot_R=376.5 [lbm/h] Q_dot_L=24000 [Btu/h] T[1]=-10 [F] x[3]=0

COP_R=3.993 h[3]=37.87 [Btu/lb_m] P[2]=100 [psia] s[1]=0.22662 [Btu/lb_m-R] W_dot_in=1.761 [kW]

h[1]=101.62 [Btu/lb_m] h[4]=37.87 [Btu/lb_m] P[3]=100 [psia] s[2]=0.22662 [Btu/lb_m-R] x[1]=1

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11-9

11-16 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 = h g @ 120 kPa = 236.97 kJ/kg ⎬s = s g @ 120 kPa = 0.94779 kJ/kg ⋅ K ⎭ 1

P2 = 0.7 MPa s 2 = s1

⎫ ⎬ h2 = 273.50 kJ/kg (T2 = 34.95°C ) ⎭

P3 = 0.7 MPa sat. liquid

⎫ ⎬ h3 = h f ⎭

@ 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

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

and W& in = m& (h2 − 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 & 1.83 kW Win

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s

11-10

11-17 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 = h g @ 120 kPa = 236.97 kJ/kg ⎬s = s g @ 120 kPa = 0.94779 kJ/kg ⋅ K ⎭ 1

P2 = 0.9 MPa s 2 = s1

⎫ ⎬ h2 = 278.93 kJ/kg (T2 = 44.45°C ) ⎭

P3 = 0.9 MPa sat. liquid

⎫ ⎬ h3 = h f ⎭

@ 0.9 MPa

T · QH

2

3 0.9 MPa

· Win

= 101.61 kJ/kg

h4 ≅ h3 = 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 W& in = m& (h2 − h1 ) = (0.05 kg/s )(278.93 − 236.97 ) kJ/kg = 2.10 kW

(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 & Win 2.10 kW

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11-11

11-18 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, and the enthalpy at the turbine exit would be ⎛ s3 − s f x4s = ⎜ ⎜ s fg ⎝

(

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

h4 s = h f + x 4 s h fg

· QH

2

3 0.7 MPa

· Win

0.12 MPa 4s

4

· QL

1 s

)@ 120 kPa = 22.49 + (0.2802)(214.48) = 82.58 kJ/kg

Then, Q& L = m& (h1 − h4 s ) = (0.05 kg/s )(236.97 − 82.58) kJ/kg = 7.72 kW

and COPR =

Q& L 7.72 kW = = 4.23 & Win 1.83 kW

Then the percentage increase in Q& and COP becomes ΔQ& L 7.72 − 7.41 Increase in Q& L = = = 4.2% 7.41 Q& L Increase in COPR =

ΔCOPR 4.23 − 4.06 = = 4.2% COPR 4.06

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11-12

11-19 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 = 0.14 MPa ⎫ h1 = 246.36 kJ/kg ⎬ s = 0.97236 kJ/kg ⋅ K T1 = −10°C ⎭ 1

T

P2 = 0.7 MPa ⎫ ⎬ h2 = 288.53 kJ/kg T2 = 50°C ⎭

0.65 MPa

P2 s = 0.7 MPa ⎫ ⎬ h2 s = 281.16 kJ/kg s 2 s = s1 ⎭ P3 = 0.65 ...