THERMODYNAMICS-SI ED. EBOOK >I<
THERMODYNAMICS-SI ED. EBOOK >I<
9th Edition
ISBN: 9781307573022
Author: CENGEL
Publisher: MCG/CREATE
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Chapter 11.10, Problem 21P

A refrigerator uses refrigerant-134a as the working fluid and operates on the vapor-compression refrigeration cycle. The evaporator and condenser pressures are 200 kPa and 1400 kPa, respectively. The isentropic efficiency of the compressor is 88 percent. The refrigerant enters the compressor at a rate of 0.025 kg/s superheated by 10.1°C and leaves the condenser subcooled by 4.4°C. Determine (a) the rate of cooling provided by the evaporator, the power input, and the COP. Determine (b) the same parameters if the cycle operated on the ideal vapor-compression refrigeration cycle between the same pressure limits.

(a)

Expert Solution
Check Mark
To determine

The rate of cooling provided by the evaporator, the power input and the COP.

Answer to Problem 21P

The rate of cooling provided by the evaporator is 3.317kWand4.535kW respectively, the power input is 1.218kW and the COP is 2.723.

Explanation of Solution

Show the T-s diagram for the vapor-compression refrigeration cycle as in Figure (1).

THERMODYNAMICS-SI ED. EBOOK >I<, Chapter 11.10, Problem 21P , additional homework tip  1

Express specific enthalpy at state 2 by using the formula of Carnot efficiency.

ηC=h2sh1h2h1 (I)

Here, Carnot efficiency is ηC, specific enthalpy at state 1, 2 and 2s is h1,h2andh2s respectively.

Express rate of heat lost by the evaporator.

Q˙L=m˙(h1h4) (II)

Here, mass flow rate of refrigerant is m˙, specific enthalpy at state 4 is h4.

Express rate of heat supplied to the evaporator.

Q˙H=m˙(h2h3) (III)

Here, specific enthalpy at state 3 is h3.

Express power input.

W˙in=m˙(h2h1) (IV)

Express coefficient of performance.

COP=Q˙LW˙in (V)

Express initial temperature.

T1=Tsat@200kPa+TIR (VI)

Here, saturated temperature at initial pressure of 200kPa is Tsat@200kPa and initial temperature of refrigerant is TIR.

Express temperature at state 3.

T3=Tsat@1400kPaTFR (VII)

Here, saturated temperature at pressure at state 3 of 1400kPa is Tsat@1400kPa and final temperature of refrigerant is TFR.

Conclusion:

Refer Table A-12, “saturated refrigerant 134a-pressure table”, and write the saturated temperature at initial pressure of 200kPa.

Tsat@200kPa=10.1°C

Substitute 10.1°C for Tsat@200kPa and 10.1°C for TIR in Equation (VI).

T1=10.1°C+(10.1°C)=0°C

Refer Table A-12, “saturated refrigerant 134a-pressure table”, and write the saturated temperature at pressure at state 3 of 1400kPa.

Tsat@1400kPa=52.4°C

Substitute 52.4°C for Tsat@1400kPa and 4.4°C for TFR in Equation (VII).

T3=52.4°C4.4°C=48°C

Refer Table A-13, “superheated refrigerant-134a”, and write the properties corresponding to initial pressure (P1) of 200kPa and initial temperature of 0°C.

h1=253.07kJ/kgs1=0.9699kJ/kgK

Here, initial specific entropy is s1.

From Figure (1), the initial specific entropy is equal to specific entropy at state 2.

s1=s2=0.9699kJ/kgK

Perform unit conversion of pressure at state 2 from kPatoMPa.

P2=1400kPa[MPa1000kPa]=1.4MPa

Refer Table A-13, “superheated refrigerant 134a”, and write the specific enthalpy at state 2s corresponding to pressure at state 2 of 1.4MPa and specific entropy at state 2 (s2=s1) of 0.9699kJ/kgK using interpolation method.

Write the formula of interpolation method of two variables.

y2=(x2x1)(y3y1)(x3x1)+y1 (VIII)

Here, the variables denote by x and y is specific entropy at state 2 and specific enthalpy at state 2 respectively.

Show the specific enthalpy at state 2s corresponding to specific entropy as in Table (1).

Specific entropy at state 2

s2(kJ/kgK)

Specific enthalpy at state 2s

h2s(kJ/kg)

0.9389 (x1)285.47 (y1)
0.9699 (x2)(y2=?)
0.9733 (x3)297.10 (y3)

Substitute 0.9389kJ/kgK,0.9699kJ/kgKand0.9733kJ/kgK for x1,x2andx3 respectively, 285.47kJ/kg for y1 and 297.10kJ/kg for y3 in Equation (VIII).

y2=[(0.96990.9389)kJ/kgK][(297.10285.47)kJ/kg](0.97330.9389)kJ/kgK+285.47kJ/kg=295.95kJ/kg=h2s

From Figure (1), write the specific enthalpy at state 3 is equal to state 4 due to throttling process.

h3h4

Here, specific enthalpy at state 3 is h3.

Refer Table A-11, “saturated refrigerant-134a-pressure table”, and write the specific enthalpy at state 3 corresponding to temperature at state 3 of 48°C

h3=h4=hf=120.41kJ/kg

Substitute 295.95kJ/kg for h2s, 0.88 for ηC, and 253.07kJ/kg for h1 in Equation (I).

0.88=(295.95253.07)kJ/kgh2253.07kJ/kgh2253.07kJ/kg=48.7272kJ/kgh2=301.80kJ/kg

Substitute 0.025kg/s for m˙, 253.07kJ/kgand120.41kJ/kg for h1andh4 respectively in Equation (II).

Q˙L=0.025kg/s(253.07kJ/kg120.41kJ/kg)=3.317kJ/s[kWkJ/s]=3.317kW

Substitute 0.025kg/s for m˙, 301.80kJ/kgand120.41kJ/kg for h2andh3 respectively in Equation (III).

Q˙H=0.025kg/s(301.80kJ/kg120.41kJ/kg)=4.535kJ/s[kWkJ/s]=4.535kW

Hence, the rate of cooling provided by the evaporator is 3.317kWand4.535kW respectively.

Substitute 0.025kg/s for m˙, 301.80kJ/kgand253.07kJ/kg for h2andh1 respectively in Equation (IV).

W˙in=0.025kg/s(301.80kJ/kg253.07kJ/kg)=1.218kJ/s[kWkJ/s]=1.218kW

Hence, the power input is 1.218kW.

Substitute 1.218kW for W˙in and 3.317kW for Q˙L in Equation (V).

COP=3.317kW1.218kW=2.723

Hence, the COP is 2.723.

(b)

Expert Solution
Check Mark
To determine

The rate of cooling provided by the evaporator, the power input and the COP.

Answer to Problem 21P

The rate of cooling provided by the evaporator is 2.931kWand3.947kW respectively, the power input is 1.016kW and the COP is 2.885.

Explanation of Solution

Show the T-s diagram for the ideal vapor-compression refrigeration cycle as in Figure (2).

THERMODYNAMICS-SI ED. EBOOK >I<, Chapter 11.10, Problem 21P , additional homework tip  2

Express rate of heat lost by the evaporator.

Q˙L=m˙(h1h4) (IX)

Here, mass flow rate of refrigerant is m˙, specific enthalpy at state 4 is h4.

Express rate of heat supplied to the evaporator.

Q˙H=m˙(h2h3) (X)

Here, specific enthalpy at state 3 is h3.

Express power input.

W˙in=m˙(h2h1) (XI)

Express coefficient of performance.

COP=Q˙LW˙in (XII)

Conclusion:

From Figure (2), write the specific enthalpy at state 3 is equal to state 4 due to throttling process.

h3h4

Here, specific enthalpy at state 3 is h3.

Refer Table A-12, “saturated refrigerant-134a-pressure table”, and write the properties corresponding to pressure at state 1 (P1) of 200kPa.

h1=hg=244.50kJ/kgs1=sg=0.9378kJ/kgK

Here, specific entropy at state 1 is s1, specific enthalpy and entropy at saturated vapor is hgandsg respectively.

Refer Table A-13, “superheated refrigerant 134a”, and write the specific enthalpy at state 2 corresponding to pressure at state 2 of 1.4MPa and specific entropy at state 2 (s2=s1) of 0.9378kJ/kgK using interpolation method.

Show the specific enthalpy at state 2 corresponding to specific entropy as in Table (2).

Specific entropy at state 2

s2(kJ/kgK)

Specific enthalpy at state 2

h2(kJ/kg)

0.9107 (x1)276.17 (y1)
0.9378 (x2)(y2=?)
0.9389 (x3)285.47 (y3)

Substitute 0.9107kJ/kgK,0.9378kJ/kgKand0.9389kJ/kgK for x1,x2andx3 respectively, 276.17kJ/kg for y1 and 285.47kJ/kg for y3 in Equation (VIII).

y2=[(0.93780.9107)kJ/kgK][(285.47276.17)kJ/kg](0.93890.9107)kJ/kgK+276.17kJ/kg=285.13kJ/kg=h2

Refer Table A-12, “saturated refrigerant-134a-pressure table”, and write the specific enthalpy at state 3 corresponding to pressure at state 3 (P3) of 1400kPa.

h3=h4=hf=127.25kJ/kg

Here, specific enthalpy at saturated liquid is hf.

Substitute 0.025kg/s for m˙, 244.50kJ/kgand127.25kJ/kg for h1andh4 respectively in Equation (IX).

Q˙L=0.025kg/s(244.50kJ/kg127.25kJ/kg)=2.931kJ/s[kWkJ/s]=2.931kW

Substitute 0.025kg/s for m˙, 285.13kJ/kgand127.25kJ/kg for h2andh3 respectively in Equation (X).

Q˙H=0.025kg/s(285.13kJ/kg127.25kJ/kg)=3.947kJ/s[kWkJ/s]=3.947kW

Hence, the rate of cooling provided by the evaporator is 2.931kWand3.947kW respectively.

Substitute 0.025kg/s for m˙, 285.13kJ/kgand244.50kJ/kg for h2andh1 respectively in Equation (XI).

W˙in=0.025kg/s(285.13kJ/kg244.50kJ/kg)=1.016kJ/s[kWkJ/s]=1.016kW

Hence, the power input is 1.016kW.

Substitute 1.016kW for W˙in and 2.931kW for Q˙L in Equation (XII).

COP=2.931kW1.016kW=2.885

Hence, the COP is 2.885.

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Chapter 11 Solutions

THERMODYNAMICS-SI ED. EBOOK >I<

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