The mass flow rate of the refrigerant through the upper cycle.

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Answer 1

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Chapter 11.10, Problem 59P

(a)

To determine

The mass flow rate of the refrigerant through the upper cycle.

(a)

Expert Solution

Answer to Problem 59P

The mass flow rate of the refrigerant through the upper cycle is 0.384 kg/s.

Explanation of Solution

Express the specific enthalpy at state 2 using Carnot efficiency.

ηC=h2s−h1h2−h1 (I)

Here, specific enthalpy at state 1, 2 and 2s is h1,h2 and h2s respectively and Carnot efficiency is ηC.

Express specific enthalpy at state 3.

h3=hf@500 kPa (II)

Here, specific enthalpy at saturated liquid and pressure of 500 kPa is hf@500 kPa.

Write the specific enthalpy at state 3 is equal to state 4 due to throttling process.

h3≅h4 (III)

Here, specific enthalpy at state 3 and 4 is h3 and h4 respectively.

Express specific enthalpy at state 5.

h5=hg@400 kPa (IV)

Here, specific enthalpy at saturated vapor and pressure of 400 kPa is hg@400 kPa.

Express specific entropy at state 5.

s5=sg@400 kPa (V)

Here, specific entropy at saturated vapor and pressure of 400 kPa is sg@400 kPa.

Express the specific enthalpy at state 6 using Carnot efficiency.

ηC=h6s−h5h6−h5 (VI)

Here, specific enthalpy at state 5, 6 and 6s is h5,h6 and h6s respectively.

Express specific enthalpy at state 7.

h7=hf@1400 kPa (VII)

Here, specific enthalpy at saturated liquid and pressure of 1400 kPa is hf@1400 kPa.

Write the specific enthalpy at state 7 is equal to state 8 due to throttling process.

h7≅h8 (VIII)

Here, specific enthalpy at state 7 and 8 is h7 and h8 respectively.

Express the mass flow rate of the refrigerant through the upper cycle.

m˙A(h5−h8)=m˙B(h2−h3) (IX)

Here, mass flow rate of the refrigerant through the lower cycle is m˙B.

Conclusion:

Refer Table A-12, “saturated refrigerant-134a-pressure table”, and write the properties corresponding to pressure of 160 kPa.

h1=hg=241.14 kJ/kgs1=sg=0.420 kJ/kg⋅K

Here, specific entropy and enthalpy at state 1 is s1 and h1 respectively, specific enthalpy and entropy at saturated vapor is hg and sg respectively.

Refer Table A-13, “superheated refrigerant 134a”, and write the specific enthalpy at state 2s corresponding to pressure at state 2 of 0.50 MPa and specific entropy at state 2 (s2=s1) of 0.9420 kJ/kg⋅K using interpolation method.

Write the formula of interpolation method of two variables.

y2=(x2−x1)(y3−y1)(x3−x1)+y1 (X)

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

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

Specific entropy at state 2 s2(kJ/kg⋅K)	Specific enthalpy at state 2s h2s(kJ/kg)
0.9384 (x1)	263.48 (y1)
0.9420 (x2)	(y2=?)
0.9704 (x3)	273.03 (y3)

Substitute 0.9384 kJ/kg⋅K, 0.9420 kJ/kg⋅K and 0.9704 kJ/kg⋅K for x1, x2 and x3 respectively, 263.48 kJ/kg for y1 and 273.03 kJ/kg for y3 in Equation (X).

y2=[(0.9420−0.9384) kJ/kg⋅K][(273.03−263.48) kJ/kg](0.9704−0.9384) kJ/kg⋅K+263.48 kJ/kg=264.55 kJ/kg=h2s

Thus, the specific enthalpy at state 2s is,

h2s=264.55 kJ/kg

Substitute 264.55 kJ/kg for h2s, 241.14 kJ/kg for h1 and 0.80 for ηC in Equation (I).

0.80=264.55 kJ/kg−241.14 kJ/kgh2−241.14 kJ/kg0.80(h2−241.14 kJ/kg)=23.41 kJ/kg0.80h2−192.912 kJ/kg=23.41 kJ/kgh2=270.41 kJ/kg

Perform unit conversion of pressure at state 3 and 5 from MPa to kPa.

P3=0.5 MPa=0.5 MPa[1000 kPa MPa]=500 kPa

P5=0.4 MPa=0.4 MPa[1000 kPa MPa]=400 kPa

Refer Table A-12, “saturated refrigerant-134a-pressure table”, and write the property corresponding to pressure at state 3 (P3) of 500 kPa.

h3=hf@500 kPa=73.32 kJ/kg

Substitute 73.32 kJ/kg for hf@500 kPa in Equation (II).

h3=73.32 kJ/kg

Substitute 73.32 kJ/kg for h4 in Equation (III).

h4=73.32 kJ/kg

Refer Table A-12, “saturated refrigerant-134a-pressure table”, and write the properties corresponding to pressure of 400 kPa.

h5=hg@400 kPa=255.61 kJ/kgs5=sg@400 kPa=0.9271 kJ/kg⋅K

Substitute 255.61 kJ/kg for hg@400 kPa in Equation (IV).

h5=255.61 kJ/kg

Substitute 0.9271 kJ/kg⋅K for sg@400 kPa in Equation (V).

s5=0.9271 kJ/kg⋅K

Perform unit conversion of pressure at state 6 from kPa to MPa.

P6=1.4 MPa=1.4 MPa[1000 kPa MPa]=1400 kPa

Refer Table A-13, “superheated refrigerant 134a”, and write the specific enthalpy at state 6s corresponding to pressure at state 6 of 1.4 MPa and specific entropy at state 6 (s6=s5) of 0.9271 kJ/kg⋅K using interpolation method.

Show the specific enthalpy at state 6s corresponding to specific entropy as in Table (2).

Specific entropy at state 2 s6(kJ/kg⋅K)	Specific enthalpy at state 6s h6s(kJ/kg)
0.9107 (x1)	276.17 (y1)
0.9271 (x2)	(y2=?)
0.9389 (x3)	285.47 (y3)

Use excels and tabulates the values from Table (2) in Equation (X) to get,

h6s=281.56 kJ/kg

Substitute 281.56 kJ/kg for h6s, 255.61 kJ/kg for h5 and 0.80 for ηC in Equation (VI).

0.80=281.56 kJ/kg−255.61 kJ/kgh6−255.61 kJ/kg0.80(h6−255.61 kJ/kg)=281.56 kJ/kg−255.61 kJ/kg0.80h6−204.488 kJ/kg=25.95 kJ/kgh6=288.04 kJ/kg

Refer Table A-12, “saturated refrigerant-134a-pressure table”, and write the property corresponding to pressure at state 7 (P7) of 1400 kPa.

h7=hf@1400 kPa=127.25 kJ/kg

Substitute 127.25 kJ/kg for hf@1400 kPa in Equation (VII).

h7=127.25 kJ/kg

Substitute 127.25 kJ/kg for h7 in Equation (VIII).

h8=127.25 kJ/kg

Substitute 255.61 kJ/kg for h5, 127.25 kJ/kg for h8, 0.25 kg/s for m˙B, 270.41 kJ/kg for h2 and 73.32 kJ/kg for h3 in Equation (IX).

m˙A(255.61−127.25)=(0.25 kg/s)(270.41−73.32) kJ/kgm˙A(128.36 kJ/kg)=(0.25 kg/s)(197.09 kJ/kg)m˙A=0.3839 kg/s ≃0.384 kg/s

Hence, the mass flow rate of the refrigerant through the upper cycle is 0.384 kg/s.

(b)

To determine

The rate of heat removal from the refrigerated space.

(b)

Expert Solution

Answer to Problem 59P

The rate of heat removal from the refrigerated space is 42 kW.

Explanation of Solution

Express the rate of heat removal from the refrigerated space.

Q˙L=m˙B(h1−h4) (XI)

Conclusion:

Substitute 0.25 kg/s for m˙B, 241.14 kJ/kg for h1 and 73.32 kJ/kg for h4 in Equation (XI).

Q˙L=(0.25 kg/s)(241.14−73.32) kJ/kg=41.96 kJ/s[kWkJ/s]=41.96 kW≃42 kW

Hence, the rate of heat removal from the refrigerated space is 42 kW.

(c)

To determine

The COP of the refrigerator.

(c)

Expert Solution

Answer to Problem 59P

The COP of the refrigerator is 2.12.

Explanation of Solution

Express the power input.

W˙in=m˙A(h6−h5)+m˙B(h2−h1) (XII)

Express the COP of the refrigerator.

COP=Q˙LW˙in (XIII)

Conclusion:

Substitute 0.3839 kg/s and 0.25 kg/s for m˙A and m˙B respectively, 288.04 kJ/kg and 255.61 kJ/kg for h6 and h5 respectively, and 270.41 kJ/kg and 241.14 kJ/kg for h2 and h1 respectively in Equation (XII).

W˙in=(0.3839 kg/s)(288.04−255.61) kJ/kg+(0.25 kg/s)(270.41−241.14) kJ/kg=19.77 kJ/s[kWkJ/s]=19.77 kW

Substitute 19.77 kW for W˙in and 41.96 kW for Q˙L in Equation (XIII).

COP=41.96 kW19.77 kW=2.12

Hence, the COP of the refrigerator is 2.12.

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Answer 2

Question

100%

Chapter 11.10, Problem 59P

(a)

To determine

The mass flow rate of the refrigerant through the upper cycle.

(a)

Expert Solution

Answer to Problem 59P

The mass flow rate of the refrigerant through the upper cycle is 0.384 kg/s.

Explanation of Solution

Express the specific enthalpy at state 2 using Carnot efficiency.

ηC=h2s−h1h2−h1 (I)

Here, specific enthalpy at state 1, 2 and 2s is h1,h2 and h2s respectively and Carnot efficiency is ηC.

Express specific enthalpy at state 3.

h3=hf@500 kPa (II)

Here, specific enthalpy at saturated liquid and pressure of 500 kPa is hf@500 kPa.

Write the specific enthalpy at state 3 is equal to state 4 due to throttling process.

h3≅h4 (III)

Here, specific enthalpy at state 3 and 4 is h3 and h4 respectively.

Express specific enthalpy at state 5.

h5=hg@400 kPa (IV)

Here, specific enthalpy at saturated vapor and pressure of 400 kPa is hg@400 kPa.

Express specific entropy at state 5.

s5=sg@400 kPa (V)

Here, specific entropy at saturated vapor and pressure of 400 kPa is sg@400 kPa.

Express the specific enthalpy at state 6 using Carnot efficiency.

ηC=h6s−h5h6−h5 (VI)

Here, specific enthalpy at state 5, 6 and 6s is h5,h6 and h6s respectively.

Express specific enthalpy at state 7.

h7=hf@1400 kPa (VII)

Here, specific enthalpy at saturated liquid and pressure of 1400 kPa is hf@1400 kPa.

Write the specific enthalpy at state 7 is equal to state 8 due to throttling process.

h7≅h8 (VIII)

Here, specific enthalpy at state 7 and 8 is h7 and h8 respectively.

Express the mass flow rate of the refrigerant through the upper cycle.

m˙A(h5−h8)=m˙B(h2−h3) (IX)

Here, mass flow rate of the refrigerant through the lower cycle is m˙B.

Conclusion:

Refer Table A-12, “saturated refrigerant-134a-pressure table”, and write the properties corresponding to pressure of 160 kPa.

h1=hg=241.14 kJ/kgs1=sg=0.420 kJ/kg⋅K

Here, specific entropy and enthalpy at state 1 is s1 and h1 respectively, specific enthalpy and entropy at saturated vapor is hg and sg respectively.

Refer Table A-13, “superheated refrigerant 134a”, and write the specific enthalpy at state 2s corresponding to pressure at state 2 of 0.50 MPa and specific entropy at state 2 (s2=s1) of 0.9420 kJ/kg⋅K using interpolation method.

Write the formula of interpolation method of two variables.

y2=(x2−x1)(y3−y1)(x3−x1)+y1 (X)

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

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

Specific entropy at state 2 s2(kJ/kg⋅K)	Specific enthalpy at state 2s h2s(kJ/kg)
0.9384 (x1)	263.48 (y1)
0.9420 (x2)	(y2=?)
0.9704 (x3)	273.03 (y3)

Substitute 0.9384 kJ/kg⋅K, 0.9420 kJ/kg⋅K and 0.9704 kJ/kg⋅K for x1, x2 and x3 respectively, 263.48 kJ/kg for y1 and 273.03 kJ/kg for y3 in Equation (X).

y2=[(0.9420−0.9384) kJ/kg⋅K][(273.03−263.48) kJ/kg](0.9704−0.9384) kJ/kg⋅K+263.48 kJ/kg=264.55 kJ/kg=h2s

Thus, the specific enthalpy at state 2s is,

h2s=264.55 kJ/kg

Substitute 264.55 kJ/kg for h2s, 241.14 kJ/kg for h1 and 0.80 for ηC in Equation (I).

0.80=264.55 kJ/kg−241.14 kJ/kgh2−241.14 kJ/kg0.80(h2−241.14 kJ/kg)=23.41 kJ/kg0.80h2−192.912 kJ/kg=23.41 kJ/kgh2=270.41 kJ/kg

Perform unit conversion of pressure at state 3 and 5 from MPa to kPa.

P3=0.5 MPa=0.5 MPa[1000 kPa MPa]=500 kPa

P5=0.4 MPa=0.4 MPa[1000 kPa MPa]=400 kPa

Refer Table A-12, “saturated refrigerant-134a-pressure table”, and write the property corresponding to pressure at state 3 (P3) of 500 kPa.

h3=hf@500 kPa=73.32 kJ/kg

Substitute 73.32 kJ/kg for hf@500 kPa in Equation (II).

h3=73.32 kJ/kg

Substitute 73.32 kJ/kg for h4 in Equation (III).

h4=73.32 kJ/kg

Refer Table A-12, “saturated refrigerant-134a-pressure table”, and write the properties corresponding to pressure of 400 kPa.

h5=hg@400 kPa=255.61 kJ/kgs5=sg@400 kPa=0.9271 kJ/kg⋅K

Substitute 255.61 kJ/kg for hg@400 kPa in Equation (IV).

h5=255.61 kJ/kg

Substitute 0.9271 kJ/kg⋅K for sg@400 kPa in Equation (V).

s5=0.9271 kJ/kg⋅K

Perform unit conversion of pressure at state 6 from kPa to MPa.

P6=1.4 MPa=1.4 MPa[1000 kPa MPa]=1400 kPa

Refer Table A-13, “superheated refrigerant 134a”, and write the specific enthalpy at state 6s corresponding to pressure at state 6 of 1.4 MPa and specific entropy at state 6 (s6=s5) of 0.9271 kJ/kg⋅K using interpolation method.

Show the specific enthalpy at state 6s corresponding to specific entropy as in Table (2).

Specific entropy at state 2 s6(kJ/kg⋅K)	Specific enthalpy at state 6s h6s(kJ/kg)
0.9107 (x1)	276.17 (y1)
0.9271 (x2)	(y2=?)
0.9389 (x3)	285.47 (y3)

Use excels and tabulates the values from Table (2) in Equation (X) to get,

h6s=281.56 kJ/kg

Substitute 281.56 kJ/kg for h6s, 255.61 kJ/kg for h5 and 0.80 for ηC in Equation (VI).

0.80=281.56 kJ/kg−255.61 kJ/kgh6−255.61 kJ/kg0.80(h6−255.61 kJ/kg)=281.56 kJ/kg−255.61 kJ/kg0.80h6−204.488 kJ/kg=25.95 kJ/kgh6=288.04 kJ/kg

Refer Table A-12, “saturated refrigerant-134a-pressure table”, and write the property corresponding to pressure at state 7 (P7) of 1400 kPa.

h7=hf@1400 kPa=127.25 kJ/kg

Substitute 127.25 kJ/kg for hf@1400 kPa in Equation (VII).

h7=127.25 kJ/kg

Substitute 127.25 kJ/kg for h7 in Equation (VIII).

h8=127.25 kJ/kg

Substitute 255.61 kJ/kg for h5, 127.25 kJ/kg for h8, 0.25 kg/s for m˙B, 270.41 kJ/kg for h2 and 73.32 kJ/kg for h3 in Equation (IX).

m˙A(255.61−127.25)=(0.25 kg/s)(270.41−73.32) kJ/kgm˙A(128.36 kJ/kg)=(0.25 kg/s)(197.09 kJ/kg)m˙A=0.3839 kg/s ≃0.384 kg/s

Hence, the mass flow rate of the refrigerant through the upper cycle is 0.384 kg/s.

(b)

To determine

The rate of heat removal from the refrigerated space.

(b)

Expert Solution

Answer to Problem 59P

The rate of heat removal from the refrigerated space is 42 kW.

Explanation of Solution

Express the rate of heat removal from the refrigerated space.

Q˙L=m˙B(h1−h4) (XI)

Conclusion:

Substitute 0.25 kg/s for m˙B, 241.14 kJ/kg for h1 and 73.32 kJ/kg for h4 in Equation (XI).

Q˙L=(0.25 kg/s)(241.14−73.32) kJ/kg=41.96 kJ/s[kWkJ/s]=41.96 kW≃42 kW

Hence, the rate of heat removal from the refrigerated space is 42 kW.

(c)

To determine

The COP of the refrigerator.

(c)

Expert Solution

Answer to Problem 59P

The COP of the refrigerator is 2.12.

Explanation of Solution

Express the power input.

W˙in=m˙A(h6−h5)+m˙B(h2−h1) (XII)

Express the COP of the refrigerator.

COP=Q˙LW˙in (XIII)

Conclusion:

Substitute 0.3839 kg/s and 0.25 kg/s for m˙A and m˙B respectively, 288.04 kJ/kg and 255.61 kJ/kg for h6 and h5 respectively, and 270.41 kJ/kg and 241.14 kJ/kg for h2 and h1 respectively in Equation (XII).

W˙in=(0.3839 kg/s)(288.04−255.61) kJ/kg+(0.25 kg/s)(270.41−241.14) kJ/kg=19.77 kJ/s[kWkJ/s]=19.77 kW

Substitute 19.77 kW for W˙in and 41.96 kW for Q˙L in Equation (XIII).

COP=41.96 kW19.77 kW=2.12

Hence, the COP of the refrigerator is 2.12.

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Answer 3

Question

100%

Chapter 11.10, Problem 59P

(a)

To determine

The mass flow rate of the refrigerant through the upper cycle.

(a)

Expert Solution

Answer to Problem 59P

The mass flow rate of the refrigerant through the upper cycle is 0.384 kg/s.

Explanation of Solution

Express the specific enthalpy at state 2 using Carnot efficiency.

ηC=h2s−h1h2−h1 (I)

Here, specific enthalpy at state 1, 2 and 2s is h1,h2 and h2s respectively and Carnot efficiency is ηC.

Express specific enthalpy at state 3.

h3=hf@500 kPa (II)

Here, specific enthalpy at saturated liquid and pressure of 500 kPa is hf@500 kPa.

Write the specific enthalpy at state 3 is equal to state 4 due to throttling process.

h3≅h4 (III)

Here, specific enthalpy at state 3 and 4 is h3 and h4 respectively.

Express specific enthalpy at state 5.

h5=hg@400 kPa (IV)

Here, specific enthalpy at saturated vapor and pressure of 400 kPa is hg@400 kPa.

Express specific entropy at state 5.

s5=sg@400 kPa (V)

Here, specific entropy at saturated vapor and pressure of 400 kPa is sg@400 kPa.

Express the specific enthalpy at state 6 using Carnot efficiency.

ηC=h6s−h5h6−h5 (VI)

Here, specific enthalpy at state 5, 6 and 6s is h5,h6 and h6s respectively.

Express specific enthalpy at state 7.

h7=hf@1400 kPa (VII)

Here, specific enthalpy at saturated liquid and pressure of 1400 kPa is hf@1400 kPa.

Write the specific enthalpy at state 7 is equal to state 8 due to throttling process.

h7≅h8 (VIII)

Here, specific enthalpy at state 7 and 8 is h7 and h8 respectively.

Express the mass flow rate of the refrigerant through the upper cycle.

m˙A(h5−h8)=m˙B(h2−h3) (IX)

Here, mass flow rate of the refrigerant through the lower cycle is m˙B.

Conclusion:

Refer Table A-12, “saturated refrigerant-134a-pressure table”, and write the properties corresponding to pressure of 160 kPa.

h1=hg=241.14 kJ/kgs1=sg=0.420 kJ/kg⋅K

Here, specific entropy and enthalpy at state 1 is s1 and h1 respectively, specific enthalpy and entropy at saturated vapor is hg and sg respectively.

Refer Table A-13, “superheated refrigerant 134a”, and write the specific enthalpy at state 2s corresponding to pressure at state 2 of 0.50 MPa and specific entropy at state 2 (s2=s1) of 0.9420 kJ/kg⋅K using interpolation method.

Write the formula of interpolation method of two variables.

y2=(x2−x1)(y3−y1)(x3−x1)+y1 (X)

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

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

Specific entropy at state 2 s2(kJ/kg⋅K)	Specific enthalpy at state 2s h2s(kJ/kg)
0.9384 (x1)	263.48 (y1)
0.9420 (x2)	(y2=?)
0.9704 (x3)	273.03 (y3)

Substitute 0.9384 kJ/kg⋅K, 0.9420 kJ/kg⋅K and 0.9704 kJ/kg⋅K for x1, x2 and x3 respectively, 263.48 kJ/kg for y1 and 273.03 kJ/kg for y3 in Equation (X).

y2=[(0.9420−0.9384) kJ/kg⋅K][(273.03−263.48) kJ/kg](0.9704−0.9384) kJ/kg⋅K+263.48 kJ/kg=264.55 kJ/kg=h2s

Thus, the specific enthalpy at state 2s is,

h2s=264.55 kJ/kg

Substitute 264.55 kJ/kg for h2s, 241.14 kJ/kg for h1 and 0.80 for ηC in Equation (I).

0.80=264.55 kJ/kg−241.14 kJ/kgh2−241.14 kJ/kg0.80(h2−241.14 kJ/kg)=23.41 kJ/kg0.80h2−192.912 kJ/kg=23.41 kJ/kgh2=270.41 kJ/kg

Perform unit conversion of pressure at state 3 and 5 from MPa to kPa.

P3=0.5 MPa=0.5 MPa[1000 kPa MPa]=500 kPa

P5=0.4 MPa=0.4 MPa[1000 kPa MPa]=400 kPa

Refer Table A-12, “saturated refrigerant-134a-pressure table”, and write the property corresponding to pressure at state 3 (P3) of 500 kPa.

h3=hf@500 kPa=73.32 kJ/kg

Substitute 73.32 kJ/kg for hf@500 kPa in Equation (II).

h3=73.32 kJ/kg

Substitute 73.32 kJ/kg for h4 in Equation (III).

h4=73.32 kJ/kg

Refer Table A-12, “saturated refrigerant-134a-pressure table”, and write the properties corresponding to pressure of 400 kPa.

h5=hg@400 kPa=255.61 kJ/kgs5=sg@400 kPa=0.9271 kJ/kg⋅K

Substitute 255.61 kJ/kg for hg@400 kPa in Equation (IV).

h5=255.61 kJ/kg

Substitute 0.9271 kJ/kg⋅K for sg@400 kPa in Equation (V).

s5=0.9271 kJ/kg⋅K

Perform unit conversion of pressure at state 6 from kPa to MPa.

P6=1.4 MPa=1.4 MPa[1000 kPa MPa]=1400 kPa

Refer Table A-13, “superheated refrigerant 134a”, and write the specific enthalpy at state 6s corresponding to pressure at state 6 of 1.4 MPa and specific entropy at state 6 (s6=s5) of 0.9271 kJ/kg⋅K using interpolation method.

Show the specific enthalpy at state 6s corresponding to specific entropy as in Table (2).

Specific entropy at state 2 s6(kJ/kg⋅K)	Specific enthalpy at state 6s h6s(kJ/kg)
0.9107 (x1)	276.17 (y1)
0.9271 (x2)	(y2=?)
0.9389 (x3)	285.47 (y3)

Use excels and tabulates the values from Table (2) in Equation (X) to get,

h6s=281.56 kJ/kg

Substitute 281.56 kJ/kg for h6s, 255.61 kJ/kg for h5 and 0.80 for ηC in Equation (VI).

0.80=281.56 kJ/kg−255.61 kJ/kgh6−255.61 kJ/kg0.80(h6−255.61 kJ/kg)=281.56 kJ/kg−255.61 kJ/kg0.80h6−204.488 kJ/kg=25.95 kJ/kgh6=288.04 kJ/kg

Refer Table A-12, “saturated refrigerant-134a-pressure table”, and write the property corresponding to pressure at state 7 (P7) of 1400 kPa.

h7=hf@1400 kPa=127.25 kJ/kg

Substitute 127.25 kJ/kg for hf@1400 kPa in Equation (VII).

h7=127.25 kJ/kg

Substitute 127.25 kJ/kg for h7 in Equation (VIII).

h8=127.25 kJ/kg

Substitute 255.61 kJ/kg for h5, 127.25 kJ/kg for h8, 0.25 kg/s for m˙B, 270.41 kJ/kg for h2 and 73.32 kJ/kg for h3 in Equation (IX).

m˙A(255.61−127.25)=(0.25 kg/s)(270.41−73.32) kJ/kgm˙A(128.36 kJ/kg)=(0.25 kg/s)(197.09 kJ/kg)m˙A=0.3839 kg/s ≃0.384 kg/s

Hence, the mass flow rate of the refrigerant through the upper cycle is 0.384 kg/s.

(b)

To determine

The rate of heat removal from the refrigerated space.

(b)

Expert Solution

Answer to Problem 59P

The rate of heat removal from the refrigerated space is 42 kW.

Explanation of Solution

Express the rate of heat removal from the refrigerated space.

Q˙L=m˙B(h1−h4) (XI)

Conclusion:

Substitute 0.25 kg/s for m˙B, 241.14 kJ/kg for h1 and 73.32 kJ/kg for h4 in Equation (XI).

Q˙L=(0.25 kg/s)(241.14−73.32) kJ/kg=41.96 kJ/s[kWkJ/s]=41.96 kW≃42 kW

Hence, the rate of heat removal from the refrigerated space is 42 kW.

(c)

To determine

The COP of the refrigerator.

(c)

Expert Solution

Answer to Problem 59P

The COP of the refrigerator is 2.12.

Explanation of Solution

Express the power input.

W˙in=m˙A(h6−h5)+m˙B(h2−h1) (XII)

Express the COP of the refrigerator.

COP=Q˙LW˙in (XIII)

Conclusion:

Substitute 0.3839 kg/s and 0.25 kg/s for m˙A and m˙B respectively, 288.04 kJ/kg and 255.61 kJ/kg for h6 and h5 respectively, and 270.41 kJ/kg and 241.14 kJ/kg for h2 and h1 respectively in Equation (XII).

W˙in=(0.3839 kg/s)(288.04−255.61) kJ/kg+(0.25 kg/s)(270.41−241.14) kJ/kg=19.77 kJ/s[kWkJ/s]=19.77 kW

Substitute 19.77 kW for W˙in and 41.96 kW for Q˙L in Equation (XIII).

COP=41.96 kW19.77 kW=2.12

Hence, the COP of the refrigerator is 2.12.

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Answer 4

Question

100%

Chapter 11.10, Problem 59P

(a)

To determine

The mass flow rate of the refrigerant through the upper cycle.

(a)

Expert Solution

Answer to Problem 59P

The mass flow rate of the refrigerant through the upper cycle is 0.384 kg/s.

Explanation of Solution

Express the specific enthalpy at state 2 using Carnot efficiency.

ηC=h2s−h1h2−h1 (I)

Here, specific enthalpy at state 1, 2 and 2s is h1,h2 and h2s respectively and Carnot efficiency is ηC.

Express specific enthalpy at state 3.

h3=hf@500 kPa (II)

Here, specific enthalpy at saturated liquid and pressure of 500 kPa is hf@500 kPa.

Write the specific enthalpy at state 3 is equal to state 4 due to throttling process.

h3≅h4 (III)

Here, specific enthalpy at state 3 and 4 is h3 and h4 respectively.

Express specific enthalpy at state 5.

h5=hg@400 kPa (IV)

Here, specific enthalpy at saturated vapor and pressure of 400 kPa is hg@400 kPa.

Express specific entropy at state 5.

s5=sg@400 kPa (V)

Here, specific entropy at saturated vapor and pressure of 400 kPa is sg@400 kPa.

Express the specific enthalpy at state 6 using Carnot efficiency.

ηC=h6s−h5h6−h5 (VI)

Here, specific enthalpy at state 5, 6 and 6s is h5,h6 and h6s respectively.

Express specific enthalpy at state 7.

h7=hf@1400 kPa (VII)

Here, specific enthalpy at saturated liquid and pressure of 1400 kPa is hf@1400 kPa.

Write the specific enthalpy at state 7 is equal to state 8 due to throttling process.

h7≅h8 (VIII)

Here, specific enthalpy at state 7 and 8 is h7 and h8 respectively.

Express the mass flow rate of the refrigerant through the upper cycle.

m˙A(h5−h8)=m˙B(h2−h3) (IX)

Here, mass flow rate of the refrigerant through the lower cycle is m˙B.

Conclusion:

Refer Table A-12, “saturated refrigerant-134a-pressure table”, and write the properties corresponding to pressure of 160 kPa.

h1=hg=241.14 kJ/kgs1=sg=0.420 kJ/kg⋅K

Here, specific entropy and enthalpy at state 1 is s1 and h1 respectively, specific enthalpy and entropy at saturated vapor is hg and sg respectively.

Refer Table A-13, “superheated refrigerant 134a”, and write the specific enthalpy at state 2s corresponding to pressure at state 2 of 0.50 MPa and specific entropy at state 2 (s2=s1) of 0.9420 kJ/kg⋅K using interpolation method.

Write the formula of interpolation method of two variables.

y2=(x2−x1)(y3−y1)(x3−x1)+y1 (X)

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

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

Specific entropy at state 2 s2(kJ/kg⋅K)	Specific enthalpy at state 2s h2s(kJ/kg)
0.9384 (x1)	263.48 (y1)
0.9420 (x2)	(y2=?)
0.9704 (x3)	273.03 (y3)

Substitute 0.9384 kJ/kg⋅K, 0.9420 kJ/kg⋅K and 0.9704 kJ/kg⋅K for x1, x2 and x3 respectively, 263.48 kJ/kg for y1 and 273.03 kJ/kg for y3 in Equation (X).

y2=[(0.9420−0.9384) kJ/kg⋅K][(273.03−263.48) kJ/kg](0.9704−0.9384) kJ/kg⋅K+263.48 kJ/kg=264.55 kJ/kg=h2s

Thus, the specific enthalpy at state 2s is,

h2s=264.55 kJ/kg

Substitute 264.55 kJ/kg for h2s, 241.14 kJ/kg for h1 and 0.80 for ηC in Equation (I).

0.80=264.55 kJ/kg−241.14 kJ/kgh2−241.14 kJ/kg0.80(h2−241.14 kJ/kg)=23.41 kJ/kg0.80h2−192.912 kJ/kg=23.41 kJ/kgh2=270.41 kJ/kg

Perform unit conversion of pressure at state 3 and 5 from MPa to kPa.

P3=0.5 MPa=0.5 MPa[1000 kPa MPa]=500 kPa

P5=0.4 MPa=0.4 MPa[1000 kPa MPa]=400 kPa

Refer Table A-12, “saturated refrigerant-134a-pressure table”, and write the property corresponding to pressure at state 3 (P3) of 500 kPa.

h3=hf@500 kPa=73.32 kJ/kg

Substitute 73.32 kJ/kg for hf@500 kPa in Equation (II).

h3=73.32 kJ/kg

Substitute 73.32 kJ/kg for h4 in Equation (III).

h4=73.32 kJ/kg

Refer Table A-12, “saturated refrigerant-134a-pressure table”, and write the properties corresponding to pressure of 400 kPa.

h5=hg@400 kPa=255.61 kJ/kgs5=sg@400 kPa=0.9271 kJ/kg⋅K

Substitute 255.61 kJ/kg for hg@400 kPa in Equation (IV).

h5=255.61 kJ/kg

Substitute 0.9271 kJ/kg⋅K for sg@400 kPa in Equation (V).

s5=0.9271 kJ/kg⋅K

Perform unit conversion of pressure at state 6 from kPa to MPa.

P6=1.4 MPa=1.4 MPa[1000 kPa MPa]=1400 kPa

Refer Table A-13, “superheated refrigerant 134a”, and write the specific enthalpy at state 6s corresponding to pressure at state 6 of 1.4 MPa and specific entropy at state 6 (s6=s5) of 0.9271 kJ/kg⋅K using interpolation method.

Show the specific enthalpy at state 6s corresponding to specific entropy as in Table (2).

Specific entropy at state 2 s6(kJ/kg⋅K)	Specific enthalpy at state 6s h6s(kJ/kg)
0.9107 (x1)	276.17 (y1)
0.9271 (x2)	(y2=?)
0.9389 (x3)	285.47 (y3)

Use excels and tabulates the values from Table (2) in Equation (X) to get,

h6s=281.56 kJ/kg

Substitute 281.56 kJ/kg for h6s, 255.61 kJ/kg for h5 and 0.80 for ηC in Equation (VI).

0.80=281.56 kJ/kg−255.61 kJ/kgh6−255.61 kJ/kg0.80(h6−255.61 kJ/kg)=281.56 kJ/kg−255.61 kJ/kg0.80h6−204.488 kJ/kg=25.95 kJ/kgh6=288.04 kJ/kg

Refer Table A-12, “saturated refrigerant-134a-pressure table”, and write the property corresponding to pressure at state 7 (P7) of 1400 kPa.

h7=hf@1400 kPa=127.25 kJ/kg

Substitute 127.25 kJ/kg for hf@1400 kPa in Equation (VII).

h7=127.25 kJ/kg

Substitute 127.25 kJ/kg for h7 in Equation (VIII).

h8=127.25 kJ/kg

Substitute 255.61 kJ/kg for h5, 127.25 kJ/kg for h8, 0.25 kg/s for m˙B, 270.41 kJ/kg for h2 and 73.32 kJ/kg for h3 in Equation (IX).

m˙A(255.61−127.25)=(0.25 kg/s)(270.41−73.32) kJ/kgm˙A(128.36 kJ/kg)=(0.25 kg/s)(197.09 kJ/kg)m˙A=0.3839 kg/s ≃0.384 kg/s

Hence, the mass flow rate of the refrigerant through the upper cycle is 0.384 kg/s.

(b)

To determine

The rate of heat removal from the refrigerated space.

(b)

Expert Solution

Answer to Problem 59P

The rate of heat removal from the refrigerated space is 42 kW.

Explanation of Solution

Express the rate of heat removal from the refrigerated space.

Q˙L=m˙B(h1−h4) (XI)

Conclusion:

Substitute 0.25 kg/s for m˙B, 241.14 kJ/kg for h1 and 73.32 kJ/kg for h4 in Equation (XI).

Q˙L=(0.25 kg/s)(241.14−73.32) kJ/kg=41.96 kJ/s[kWkJ/s]=41.96 kW≃42 kW

Hence, the rate of heat removal from the refrigerated space is 42 kW.

(c)

To determine

The COP of the refrigerator.

(c)

Expert Solution

Answer to Problem 59P

The COP of the refrigerator is 2.12.

Explanation of Solution

Express the power input.

W˙in=m˙A(h6−h5)+m˙B(h2−h1) (XII)

Express the COP of the refrigerator.

COP=Q˙LW˙in (XIII)

Conclusion:

Substitute 0.3839 kg/s and 0.25 kg/s for m˙A and m˙B respectively, 288.04 kJ/kg and 255.61 kJ/kg for h6 and h5 respectively, and 270.41 kJ/kg and 241.14 kJ/kg for h2 and h1 respectively in Equation (XII).

W˙in=(0.3839 kg/s)(288.04−255.61) kJ/kg+(0.25 kg/s)(270.41−241.14) kJ/kg=19.77 kJ/s[kWkJ/s]=19.77 kW

Substitute 19.77 kW for W˙in and 41.96 kW for Q˙L in Equation (XIII).

COP=41.96 kW19.77 kW=2.12

Hence, the COP of the refrigerator is 2.12.

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Answer 5

Question

100%

Answer 6

Question

100%

Answer 7

Chapter 11.10, Problem 59P

(a)

To determine

The mass flow rate of the refrigerant through the upper cycle.

(a)

Expert Solution

Answer to Problem 59P

The mass flow rate of the refrigerant through the upper cycle is 0.384 kg/s.

Explanation of Solution

Express the specific enthalpy at state 2 using Carnot efficiency.

ηC=h2s−h1h2−h1 (I)

Here, specific enthalpy at state 1, 2 and 2s is h1,h2 and h2s respectively and Carnot efficiency is ηC.

Express specific enthalpy at state 3.

h3=hf@500 kPa (II)

Here, specific enthalpy at saturated liquid and pressure of 500 kPa is hf@500 kPa.

Write the specific enthalpy at state 3 is equal to state 4 due to throttling process.

h3≅h4 (III)

Here, specific enthalpy at state 3 and 4 is h3 and h4 respectively.

Express specific enthalpy at state 5.

h5=hg@400 kPa (IV)

Here, specific enthalpy at saturated vapor and pressure of 400 kPa is hg@400 kPa.

Express specific entropy at state 5.

s5=sg@400 kPa (V)

Here, specific entropy at saturated vapor and pressure of 400 kPa is sg@400 kPa.

Express the specific enthalpy at state 6 using Carnot efficiency.

ηC=h6s−h5h6−h5 (VI)

Here, specific enthalpy at state 5, 6 and 6s is h5,h6 and h6s respectively.

Express specific enthalpy at state 7.

h7=hf@1400 kPa (VII)

Here, specific enthalpy at saturated liquid and pressure of 1400 kPa is hf@1400 kPa.

Write the specific enthalpy at state 7 is equal to state 8 due to throttling process.

h7≅h8 (VIII)

Here, specific enthalpy at state 7 and 8 is h7 and h8 respectively.

Express the mass flow rate of the refrigerant through the upper cycle.

m˙A(h5−h8)=m˙B(h2−h3) (IX)

Here, mass flow rate of the refrigerant through the lower cycle is m˙B.

Conclusion:

Refer Table A-12, “saturated refrigerant-134a-pressure table”, and write the properties corresponding to pressure of 160 kPa.

h1=hg=241.14 kJ/kgs1=sg=0.420 kJ/kg⋅K

Here, specific entropy and enthalpy at state 1 is s1 and h1 respectively, specific enthalpy and entropy at saturated vapor is hg and sg respectively.

Refer Table A-13, “superheated refrigerant 134a”, and write the specific enthalpy at state 2s corresponding to pressure at state 2 of 0.50 MPa and specific entropy at state 2 (s2=s1) of 0.9420 kJ/kg⋅K using interpolation method.

Write the formula of interpolation method of two variables.

y2=(x2−x1)(y3−y1)(x3−x1)+y1 (X)

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

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

Specific entropy at state 2 s2(kJ/kg⋅K)	Specific enthalpy at state 2s h2s(kJ/kg)
0.9384 (x1)	263.48 (y1)
0.9420 (x2)	(y2=?)
0.9704 (x3)	273.03 (y3)

Substitute 0.9384 kJ/kg⋅K, 0.9420 kJ/kg⋅K and 0.9704 kJ/kg⋅K for x1, x2 and x3 respectively, 263.48 kJ/kg for y1 and 273.03 kJ/kg for y3 in Equation (X).

y2=[(0.9420−0.9384) kJ/kg⋅K][(273.03−263.48) kJ/kg](0.9704−0.9384) kJ/kg⋅K+263.48 kJ/kg=264.55 kJ/kg=h2s

Thus, the specific enthalpy at state 2s is,

h2s=264.55 kJ/kg

Substitute 264.55 kJ/kg for h2s, 241.14 kJ/kg for h1 and 0.80 for ηC in Equation (I).

0.80=264.55 kJ/kg−241.14 kJ/kgh2−241.14 kJ/kg0.80(h2−241.14 kJ/kg)=23.41 kJ/kg0.80h2−192.912 kJ/kg=23.41 kJ/kgh2=270.41 kJ/kg

Perform unit conversion of pressure at state 3 and 5 from MPa to kPa.

P3=0.5 MPa=0.5 MPa[1000 kPa MPa]=500 kPa

P5=0.4 MPa=0.4 MPa[1000 kPa MPa]=400 kPa

Refer Table A-12, “saturated refrigerant-134a-pressure table”, and write the property corresponding to pressure at state 3 (P3) of 500 kPa.

h3=hf@500 kPa=73.32 kJ/kg

Substitute 73.32 kJ/kg for hf@500 kPa in Equation (II).

h3=73.32 kJ/kg

Substitute 73.32 kJ/kg for h4 in Equation (III).

h4=73.32 kJ/kg

Refer Table A-12, “saturated refrigerant-134a-pressure table”, and write the properties corresponding to pressure of 400 kPa.

h5=hg@400 kPa=255.61 kJ/kgs5=sg@400 kPa=0.9271 kJ/kg⋅K

Substitute 255.61 kJ/kg for hg@400 kPa in Equation (IV).

h5=255.61 kJ/kg

Substitute 0.9271 kJ/kg⋅K for sg@400 kPa in Equation (V).

s5=0.9271 kJ/kg⋅K

Perform unit conversion of pressure at state 6 from kPa to MPa.

P6=1.4 MPa=1.4 MPa[1000 kPa MPa]=1400 kPa

Refer Table A-13, “superheated refrigerant 134a”, and write the specific enthalpy at state 6s corresponding to pressure at state 6 of 1.4 MPa and specific entropy at state 6 (s6=s5) of 0.9271 kJ/kg⋅K using interpolation method.

Show the specific enthalpy at state 6s corresponding to specific entropy as in Table (2).

Specific entropy at state 2 s6(kJ/kg⋅K)	Specific enthalpy at state 6s h6s(kJ/kg)
0.9107 (x1)	276.17 (y1)
0.9271 (x2)	(y2=?)
0.9389 (x3)	285.47 (y3)

Use excels and tabulates the values from Table (2) in Equation (X) to get,

h6s=281.56 kJ/kg

Substitute 281.56 kJ/kg for h6s, 255.61 kJ/kg for h5 and 0.80 for ηC in Equation (VI).

0.80=281.56 kJ/kg−255.61 kJ/kgh6−255.61 kJ/kg0.80(h6−255.61 kJ/kg)=281.56 kJ/kg−255.61 kJ/kg0.80h6−204.488 kJ/kg=25.95 kJ/kgh6=288.04 kJ/kg

Refer Table A-12, “saturated refrigerant-134a-pressure table”, and write the property corresponding to pressure at state 7 (P7) of 1400 kPa.

h7=hf@1400 kPa=127.25 kJ/kg

Substitute 127.25 kJ/kg for hf@1400 kPa in Equation (VII).

h7=127.25 kJ/kg

Substitute 127.25 kJ/kg for h7 in Equation (VIII).

h8=127.25 kJ/kg

Substitute 255.61 kJ/kg for h5, 127.25 kJ/kg for h8, 0.25 kg/s for m˙B, 270.41 kJ/kg for h2 and 73.32 kJ/kg for h3 in Equation (IX).

m˙A(255.61−127.25)=(0.25 kg/s)(270.41−73.32) kJ/kgm˙A(128.36 kJ/kg)=(0.25 kg/s)(197.09 kJ/kg)m˙A=0.3839 kg/s ≃0.384 kg/s

Hence, the mass flow rate of the refrigerant through the upper cycle is 0.384 kg/s.

(b)

To determine

The rate of heat removal from the refrigerated space.

(b)

Expert Solution

Answer to Problem 59P

The rate of heat removal from the refrigerated space is 42 kW.

Explanation of Solution

Express the rate of heat removal from the refrigerated space.

Q˙L=m˙B(h1−h4) (XI)

Conclusion:

Substitute 0.25 kg/s for m˙B, 241.14 kJ/kg for h1 and 73.32 kJ/kg for h4 in Equation (XI).

Q˙L=(0.25 kg/s)(241.14−73.32) kJ/kg=41.96 kJ/s[kWkJ/s]=41.96 kW≃42 kW

Hence, the rate of heat removal from the refrigerated space is 42 kW.

(c)

To determine

The COP of the refrigerator.

(c)

Expert Solution

Answer to Problem 59P

The COP of the refrigerator is 2.12.

Explanation of Solution

Express the power input.

W˙in=m˙A(h6−h5)+m˙B(h2−h1) (XII)

Express the COP of the refrigerator.

COP=Q˙LW˙in (XIII)

Conclusion:

Substitute 0.3839 kg/s and 0.25 kg/s for m˙A and m˙B respectively, 288.04 kJ/kg and 255.61 kJ/kg for h6 and h5 respectively, and 270.41 kJ/kg and 241.14 kJ/kg for h2 and h1 respectively in Equation (XII).

W˙in=(0.3839 kg/s)(288.04−255.61) kJ/kg+(0.25 kg/s)(270.41−241.14) kJ/kg=19.77 kJ/s[kWkJ/s]=19.77 kW

Substitute 19.77 kW for W˙in and 41.96 kW for Q˙L in Equation (XIII).

COP=41.96 kW19.77 kW=2.12

Hence, the COP of the refrigerator is 2.12.

Answer 8

Chapter 11.10, Problem 59P

(a)

To determine

The mass flow rate of the refrigerant through the upper cycle.

(a)

Expert Solution

Answer to Problem 59P

The mass flow rate of the refrigerant through the upper cycle is 0.384 kg/s.

Explanation of Solution

Express the specific enthalpy at state 2 using Carnot efficiency.

ηC=h2s−h1h2−h1 (I)

Here, specific enthalpy at state 1, 2 and 2s is h1,h2 and h2s respectively and Carnot efficiency is ηC.

Express specific enthalpy at state 3.

h3=hf@500 kPa (II)

Here, specific enthalpy at saturated liquid and pressure of 500 kPa is hf@500 kPa.

Write the specific enthalpy at state 3 is equal to state 4 due to throttling process.

h3≅h4 (III)

Here, specific enthalpy at state 3 and 4 is h3 and h4 respectively.

Express specific enthalpy at state 5.

h5=hg@400 kPa (IV)

Here, specific enthalpy at saturated vapor and pressure of 400 kPa is hg@400 kPa.

Express specific entropy at state 5.

s5=sg@400 kPa (V)

Here, specific entropy at saturated vapor and pressure of 400 kPa is sg@400 kPa.

Express the specific enthalpy at state 6 using Carnot efficiency.

ηC=h6s−h5h6−h5 (VI)

Here, specific enthalpy at state 5, 6 and 6s is h5,h6 and h6s respectively.

Express specific enthalpy at state 7.

h7=hf@1400 kPa (VII)

Here, specific enthalpy at saturated liquid and pressure of 1400 kPa is hf@1400 kPa.

Write the specific enthalpy at state 7 is equal to state 8 due to throttling process.

h7≅h8 (VIII)

Here, specific enthalpy at state 7 and 8 is h7 and h8 respectively.

Express the mass flow rate of the refrigerant through the upper cycle.

m˙A(h5−h8)=m˙B(h2−h3) (IX)

Here, mass flow rate of the refrigerant through the lower cycle is m˙B.

Conclusion:

Refer Table A-12, “saturated refrigerant-134a-pressure table”, and write the properties corresponding to pressure of 160 kPa.

h1=hg=241.14 kJ/kgs1=sg=0.420 kJ/kg⋅K

Here, specific entropy and enthalpy at state 1 is s1 and h1 respectively, specific enthalpy and entropy at saturated vapor is hg and sg respectively.

Refer Table A-13, “superheated refrigerant 134a”, and write the specific enthalpy at state 2s corresponding to pressure at state 2 of 0.50 MPa and specific entropy at state 2 (s2=s1) of 0.9420 kJ/kg⋅K using interpolation method.

Write the formula of interpolation method of two variables.

y2=(x2−x1)(y3−y1)(x3−x1)+y1 (X)

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

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

Specific entropy at state 2 s2(kJ/kg⋅K)	Specific enthalpy at state 2s h2s(kJ/kg)
0.9384 (x1)	263.48 (y1)
0.9420 (x2)	(y2=?)
0.9704 (x3)	273.03 (y3)

Substitute 0.9384 kJ/kg⋅K, 0.9420 kJ/kg⋅K and 0.9704 kJ/kg⋅K for x1, x2 and x3 respectively, 263.48 kJ/kg for y1 and 273.03 kJ/kg for y3 in Equation (X).

y2=[(0.9420−0.9384) kJ/kg⋅K][(273.03−263.48) kJ/kg](0.9704−0.9384) kJ/kg⋅K+263.48 kJ/kg=264.55 kJ/kg=h2s

Thus, the specific enthalpy at state 2s is,

h2s=264.55 kJ/kg

Substitute 264.55 kJ/kg for h2s, 241.14 kJ/kg for h1 and 0.80 for ηC in Equation (I).

0.80=264.55 kJ/kg−241.14 kJ/kgh2−241.14 kJ/kg0.80(h2−241.14 kJ/kg)=23.41 kJ/kg0.80h2−192.912 kJ/kg=23.41 kJ/kgh2=270.41 kJ/kg

Perform unit conversion of pressure at state 3 and 5 from MPa to kPa.

P3=0.5 MPa=0.5 MPa[1000 kPa MPa]=500 kPa

P5=0.4 MPa=0.4 MPa[1000 kPa MPa]=400 kPa

Refer Table A-12, “saturated refrigerant-134a-pressure table”, and write the property corresponding to pressure at state 3 (P3) of 500 kPa.

h3=hf@500 kPa=73.32 kJ/kg

Substitute 73.32 kJ/kg for hf@500 kPa in Equation (II).

h3=73.32 kJ/kg

Substitute 73.32 kJ/kg for h4 in Equation (III).

h4=73.32 kJ/kg

Refer Table A-12, “saturated refrigerant-134a-pressure table”, and write the properties corresponding to pressure of 400 kPa.

h5=hg@400 kPa=255.61 kJ/kgs5=sg@400 kPa=0.9271 kJ/kg⋅K

Substitute 255.61 kJ/kg for hg@400 kPa in Equation (IV).

h5=255.61 kJ/kg

Substitute 0.9271 kJ/kg⋅K for sg@400 kPa in Equation (V).

s5=0.9271 kJ/kg⋅K

Perform unit conversion of pressure at state 6 from kPa to MPa.

P6=1.4 MPa=1.4 MPa[1000 kPa MPa]=1400 kPa

Refer Table A-13, “superheated refrigerant 134a”, and write the specific enthalpy at state 6s corresponding to pressure at state 6 of 1.4 MPa and specific entropy at state 6 (s6=s5) of 0.9271 kJ/kg⋅K using interpolation method.

Show the specific enthalpy at state 6s corresponding to specific entropy as in Table (2).

Specific entropy at state 2 s6(kJ/kg⋅K)	Specific enthalpy at state 6s h6s(kJ/kg)
0.9107 (x1)	276.17 (y1)
0.9271 (x2)	(y2=?)
0.9389 (x3)	285.47 (y3)

Use excels and tabulates the values from Table (2) in Equation (X) to get,

h6s=281.56 kJ/kg

Substitute 281.56 kJ/kg for h6s, 255.61 kJ/kg for h5 and 0.80 for ηC in Equation (VI).

0.80=281.56 kJ/kg−255.61 kJ/kgh6−255.61 kJ/kg0.80(h6−255.61 kJ/kg)=281.56 kJ/kg−255.61 kJ/kg0.80h6−204.488 kJ/kg=25.95 kJ/kgh6=288.04 kJ/kg

Refer Table A-12, “saturated refrigerant-134a-pressure table”, and write the property corresponding to pressure at state 7 (P7) of 1400 kPa.

h7=hf@1400 kPa=127.25 kJ/kg

Substitute 127.25 kJ/kg for hf@1400 kPa in Equation (VII).

h7=127.25 kJ/kg

Substitute 127.25 kJ/kg for h7 in Equation (VIII).

h8=127.25 kJ/kg

Substitute 255.61 kJ/kg for h5, 127.25 kJ/kg for h8, 0.25 kg/s for m˙B, 270.41 kJ/kg for h2 and 73.32 kJ/kg for h3 in Equation (IX).

m˙A(255.61−127.25)=(0.25 kg/s)(270.41−73.32) kJ/kgm˙A(128.36 kJ/kg)=(0.25 kg/s)(197.09 kJ/kg)m˙A=0.3839 kg/s ≃0.384 kg/s

Hence, the mass flow rate of the refrigerant through the upper cycle is 0.384 kg/s.

Answer 9

Chapter 11.10, Problem 59P

(a)

To determine

The mass flow rate of the refrigerant through the upper cycle.

Answer 10

(a)

Expert Solution

Answer to Problem 59P

The mass flow rate of the refrigerant through the upper cycle is 0.384 kg/s.

Answer 11

(a)

Expert Solution

Answer 12

(a)

Expert Solution

Answer 13

Expert Solution

Answer 14

Explanation of Solution

Express the specific enthalpy at state 2 using Carnot efficiency.

ηC=h2s−h1h2−h1 (I)

Here, specific enthalpy at state 1, 2 and 2s is h1,h2 and h2s respectively and Carnot efficiency is ηC.

Express specific enthalpy at state 3.

h3=hf@500 kPa (II)

Here, specific enthalpy at saturated liquid and pressure of 500 kPa is hf@500 kPa.

Write the specific enthalpy at state 3 is equal to state 4 due to throttling process.

h3≅h4 (III)

Here, specific enthalpy at state 3 and 4 is h3 and h4 respectively.

Express specific enthalpy at state 5.

h5=hg@400 kPa (IV)

Here, specific enthalpy at saturated vapor and pressure of 400 kPa is hg@400 kPa.

Express specific entropy at state 5.

s5=sg@400 kPa (V)

Here, specific entropy at saturated vapor and pressure of 400 kPa is sg@400 kPa.

Express the specific enthalpy at state 6 using Carnot efficiency.

ηC=h6s−h5h6−h5 (VI)

Here, specific enthalpy at state 5, 6 and 6s is h5,h6 and h6s respectively.

Express specific enthalpy at state 7.

h7=hf@1400 kPa (VII)

Here, specific enthalpy at saturated liquid and pressure of 1400 kPa is hf@1400 kPa.

Write the specific enthalpy at state 7 is equal to state 8 due to throttling process.

h7≅h8 (VIII)

Here, specific enthalpy at state 7 and 8 is h7 and h8 respectively.

Express the mass flow rate of the refrigerant through the upper cycle.

m˙A(h5−h8)=m˙B(h2−h3) (IX)

Here, mass flow rate of the refrigerant through the lower cycle is m˙B.

Conclusion:

Refer Table A-12, “saturated refrigerant-134a-pressure table”, and write the properties corresponding to pressure of 160 kPa.

h1=hg=241.14 kJ/kgs1=sg=0.420 kJ/kg⋅K

Here, specific entropy and enthalpy at state 1 is s1 and h1 respectively, specific enthalpy and entropy at saturated vapor is hg and sg respectively.

Refer Table A-13, “superheated refrigerant 134a”, and write the specific enthalpy at state 2s corresponding to pressure at state 2 of 0.50 MPa and specific entropy at state 2 (s2=s1) of 0.9420 kJ/kg⋅K using interpolation method.

Write the formula of interpolation method of two variables.

y2=(x2−x1)(y3−y1)(x3−x1)+y1 (X)

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

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

Specific entropy at state 2 s2(kJ/kg⋅K)	Specific enthalpy at state 2s h2s(kJ/kg)
0.9384 (x1)	263.48 (y1)
0.9420 (x2)	(y2=?)
0.9704 (x3)	273.03 (y3)

Substitute 0.9384 kJ/kg⋅K, 0.9420 kJ/kg⋅K and 0.9704 kJ/kg⋅K for x1, x2 and x3 respectively, 263.48 kJ/kg for y1 and 273.03 kJ/kg for y3 in Equation (X).

y2=[(0.9420−0.9384) kJ/kg⋅K][(273.03−263.48) kJ/kg](0.9704−0.9384) kJ/kg⋅K+263.48 kJ/kg=264.55 kJ/kg=h2s

Thus, the specific enthalpy at state 2s is,

h2s=264.55 kJ/kg

Substitute 264.55 kJ/kg for h2s, 241.14 kJ/kg for h1 and 0.80 for ηC in Equation (I).

0.80=264.55 kJ/kg−241.14 kJ/kgh2−241.14 kJ/kg0.80(h2−241.14 kJ/kg)=23.41 kJ/kg0.80h2−192.912 kJ/kg=23.41 kJ/kgh2=270.41 kJ/kg

Perform unit conversion of pressure at state 3 and 5 from MPa to kPa.

P3=0.5 MPa=0.5 MPa[1000 kPa MPa]=500 kPa

P5=0.4 MPa=0.4 MPa[1000 kPa MPa]=400 kPa

Refer Table A-12, “saturated refrigerant-134a-pressure table”, and write the property corresponding to pressure at state 3 (P3) of 500 kPa.

h3=hf@500 kPa=73.32 kJ/kg

Substitute 73.32 kJ/kg for hf@500 kPa in Equation (II).

h3=73.32 kJ/kg

Substitute 73.32 kJ/kg for h4 in Equation (III).

h4=73.32 kJ/kg

Refer Table A-12, “saturated refrigerant-134a-pressure table”, and write the properties corresponding to pressure of 400 kPa.

h5=hg@400 kPa=255.61 kJ/kgs5=sg@400 kPa=0.9271 kJ/kg⋅K

Substitute 255.61 kJ/kg for hg@400 kPa in Equation (IV).

h5=255.61 kJ/kg

Substitute 0.9271 kJ/kg⋅K for sg@400 kPa in Equation (V).

s5=0.9271 kJ/kg⋅K

Perform unit conversion of pressure at state 6 from kPa to MPa.

P6=1.4 MPa=1.4 MPa[1000 kPa MPa]=1400 kPa

Refer Table A-13, “superheated refrigerant 134a”, and write the specific enthalpy at state 6s corresponding to pressure at state 6 of 1.4 MPa and specific entropy at state 6 (s6=s5) of 0.9271 kJ/kg⋅K using interpolation method.

Show the specific enthalpy at state 6s corresponding to specific entropy as in Table (2).

Specific entropy at state 2 s6(kJ/kg⋅K)	Specific enthalpy at state 6s h6s(kJ/kg)
0.9107 (x1)	276.17 (y1)
0.9271 (x2)	(y2=?)
0.9389 (x3)	285.47 (y3)

Use excels and tabulates the values from Table (2) in Equation (X) to get,

h6s=281.56 kJ/kg

Substitute 281.56 kJ/kg for h6s, 255.61 kJ/kg for h5 and 0.80 for ηC in Equation (VI).

0.80=281.56 kJ/kg−255.61 kJ/kgh6−255.61 kJ/kg0.80(h6−255.61 kJ/kg)=281.56 kJ/kg−255.61 kJ/kg0.80h6−204.488 kJ/kg=25.95 kJ/kgh6=288.04 kJ/kg

Refer Table A-12, “saturated refrigerant-134a-pressure table”, and write the property corresponding to pressure at state 7 (P7) of 1400 kPa.

h7=hf@1400 kPa=127.25 kJ/kg

Substitute 127.25 kJ/kg for hf@1400 kPa in Equation (VII).

h7=127.25 kJ/kg

Substitute 127.25 kJ/kg for h7 in Equation (VIII).

h8=127.25 kJ/kg

Substitute 255.61 kJ/kg for h5, 127.25 kJ/kg for h8, 0.25 kg/s for m˙B, 270.41 kJ/kg for h2 and 73.32 kJ/kg for h3 in Equation (IX).

m˙A(255.61−127.25)=(0.25 kg/s)(270.41−73.32) kJ/kgm˙A(128.36 kJ/kg)=(0.25 kg/s)(197.09 kJ/kg)m˙A=0.3839 kg/s ≃0.384 kg/s

Hence, the mass flow rate of the refrigerant through the upper cycle is 0.384 kg/s.

Answer 15

(b)

To determine

The rate of heat removal from the refrigerated space.

(b)

Expert Solution

Answer to Problem 59P

The rate of heat removal from the refrigerated space is 42 kW.

Explanation of Solution

Express the rate of heat removal from the refrigerated space.

Q˙L=m˙B(h1−h4) (XI)

Conclusion:

Substitute 0.25 kg/s for m˙B, 241.14 kJ/kg for h1 and 73.32 kJ/kg for h4 in Equation (XI).

Q˙L=(0.25 kg/s)(241.14−73.32) kJ/kg=41.96 kJ/s[kWkJ/s]=41.96 kW≃42 kW

Hence, the rate of heat removal from the refrigerated space is 42 kW.

Answer 16

(b)

To determine

The rate of heat removal from the refrigerated space.

Answer 17

(b)

Expert Solution

Answer to Problem 59P

The rate of heat removal from the refrigerated space is 42 kW.

Answer 18

(b)

Expert Solution

Answer 19

(b)

Expert Solution

Answer 20

Expert Solution

Answer 21

Explanation of Solution

Express the rate of heat removal from the refrigerated space.

Q˙L=m˙B(h1−h4) (XI)

Conclusion:

Substitute 0.25 kg/s for m˙B, 241.14 kJ/kg for h1 and 73.32 kJ/kg for h4 in Equation (XI).

Q˙L=(0.25 kg/s)(241.14−73.32) kJ/kg=41.96 kJ/s[kWkJ/s]=41.96 kW≃42 kW

Hence, the rate of heat removal from the refrigerated space is 42 kW.

Answer 22

(c)

To determine

The COP of the refrigerator.

(c)

Expert Solution

Answer to Problem 59P

The COP of the refrigerator is 2.12.

Explanation of Solution

Express the power input.

W˙in=m˙A(h6−h5)+m˙B(h2−h1) (XII)

Express the COP of the refrigerator.

COP=Q˙LW˙in (XIII)

Conclusion:

Substitute 0.3839 kg/s and 0.25 kg/s for m˙A and m˙B respectively, 288.04 kJ/kg and 255.61 kJ/kg for h6 and h5 respectively, and 270.41 kJ/kg and 241.14 kJ/kg for h2 and h1 respectively in Equation (XII).

W˙in=(0.3839 kg/s)(288.04−255.61) kJ/kg+(0.25 kg/s)(270.41−241.14) kJ/kg=19.77 kJ/s[kWkJ/s]=19.77 kW

Substitute 19.77 kW for W˙in and 41.96 kW for Q˙L in Equation (XIII).

COP=41.96 kW19.77 kW=2.12

Hence, the COP of the refrigerator is 2.12.

Answer 23

(c)

To determine

The COP of the refrigerator.

Answer 24

(c)

Expert Solution

Answer to Problem 59P

The COP of the refrigerator is 2.12.

Answer 25

(c)

Expert Solution

Answer 26

(c)

Expert Solution

Answer 27

Expert Solution

Answer 28

Explanation of Solution

Express the power input.

W˙in=m˙A(h6−h5)+m˙B(h2−h1) (XII)

Express the COP of the refrigerator.

COP=Q˙LW˙in (XIII)

Conclusion:

Substitute 0.3839 kg/s and 0.25 kg/s for m˙A and m˙B respectively, 288.04 kJ/kg and 255.61 kJ/kg for h6 and h5 respectively, and 270.41 kJ/kg and 241.14 kJ/kg for h2 and h1 respectively in Equation (XII).