THERMODYNAMICS(SI UNITS,INTL.ED)EBOOK>I
THERMODYNAMICS(SI UNITS,INTL.ED)EBOOK>I
8th Edition
ISBN: 9781307434316
Author: CENGEL
Publisher: INTER MCG
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Chapter 11.10, Problem 59P

(a)

To determine

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

(a)

Expert Solution
Check Mark

Answer to Problem 59P

The mass flow rate of the refrigerant through the upper compression cycle is 0.2026kg/s.

Explanation of Solution

Sketch the schematic diagram for the two stage cascade refrigeration system as in Figure (1).

THERMODYNAMICS(SI UNITS,INTL.ED)EBOOK>I, Chapter 11.10, Problem 59P

Write the relation between the specific enthalpies at the inlet and exit of throttling process.

hin=hout (I)

Here, specific enthalpy at the inlet of throttling is hin, and the specific enthalpy at the exit of throttling is hout.

Write the expression for the isentropic efficiency of the compressor (ηC).

ηC=h2sh1h2h1 (II)

Here, specific enthalpy at the isentropic exit of compressor is h2s, and the specific enthalpy at the inlet of compressor is h1.

Write the formula to calculate the dryness fraction at the exit of expansion valve (x6) in high pressure cycle.

x6=h6hfhfg (III)

Here, specific enthalpy of refrigerant at expansion valve exit is h6, specific enthalpy of saturated liquid is hf, and the specific enthalpy of the saturated liquid vapor mixture is hfg.

Write the expression for the mass flow rate of refrigerant (m˙) from dryness fraction.

m˙=m˙71x6 (IV)

Here, mass flow rate of refrigerant at the inlet of low pressure compressor is m˙7 and the dryness fraction at exit of upper pressure compression is x6.

Conclusion:

From the Table A-12 of “Saturated refrigerant R-134a: Pressure”, obtain the properties of refrigerant at low pressure compressor inlet pressure (P1) of 200kPa as follows:

h1=hg=244.50kJ/kgs1=sg=0.9379kJ/kgK

Here, specific enthalpy of the saturated vapor is hg and the specific entropy of the saturated vapor is sg.

The specific entropy at the end of isentropic compression (s2) is calculated as

s2=s1=0.9379kJ/kgK

Refer to Table A-13, “Superheated R-134a”, and obtain the values of R-134a at pressure of 450kPa and specific entropy of 0.9379kJ/kgK by using interpolation method as,

h2s=261.13kJ/kg

Substitute 0.80 for ηC, 261.13kJ/kg for h2s, and 244.50kJ/kg for h1 in Equation (II).

0.80=261.13kJ/kg244.50kJ/kgh2244.50kJ/kgh2=265.28kJ/kg

From the Table A-12 of “Saturated refrigerant R-134a: Pressure”, obtain the properties of refrigerant at throttling inlet pressure (P3) of 450kPa as follows:

h3=hg=257.58kJ/kg

From the Table A-12 of “Saturated refrigerant R-134a: Pressure”, obtain the properties of refrigerant at second stage compressor inlet pressure (P5) of 1200kPa as follows:

h5=hf=117.79kJ/kg

Substitute 117.79kJ/kg for h5 in Equation (I).

h6=117.79kJ/kg

From the Table A-12 of “Saturated refrigerant R-134a: Pressure”, obtain the properties of refrigerant at second stage expansion inlet pressure (P7) of 450kPa as follows:

h7=hf=68.80kJ/kg

Substitute 68.80kJ/kg for h7 in Equation (II).

h8=68.80kJ/kg

From the Table A-12 of “Saturated refrigerant R-134a: Pressure”, obtain the properties of refrigerant at pressure (P6) of 450kPa as follows:

hg=257.58kJ/kg

hf=68.80kJ/kg

Substitute 117.79kJ/kg for h6, 68.81kJ/kg for hf, and 188.71kJ/kg for hfg in Equation (III).

x6=117.7968.81188.71=0.2595

Substitute 0.15kg/s for m˙7, and 0.2595 for x6 in Equation (IV).

m˙=0.15kg/s10.2595=0.2026kg/s

The mass flow rate of the refrigerant through the upper compression cycle is 0.2026kg/s.

(b)

To determine

The rate at which heat removed from the refrigerated space.

(b)

Expert Solution
Check Mark

Answer to Problem 59P

The rate at which heat removed from the refrigerated space is 26.35kW.

Explanation of Solution

Write the mass balance equation for the mass transfer in flash chamber.

m˙3=m˙m˙7 (V)

Write the energy balance equation for the flash chamber.

m˙h9=m˙7h2+m˙3h3 (VI)

Here, specific enthalpy at state 9 is h9.

Write the formula to calculate the rate of heat transfer from the refrigerated space (Q˙L).

Q˙L=m˙7(h1h8) (VII)

Conclusion:

Refer to Table A-13, “Superheated R-134a”, and obtain the values of R-134a at pressure of 450kPa and specific enthalpy of 263.28kJ/kg by using interpolation method as,

s9=0.9453kJ/kgK

The specific entropy at the end of isentropic compression (s4) is calculated as

s4=s9=0.9453kJ/kgK

Refer to Table A-13, “Superheated R-134a”, and obtain the values of R-134a at pressure of 1200kPa and specific entropy of 0.9453kJ/kgK by using interpolation method as,

h4s=284.32kJ/kg

Substitute 284.32kJ/kgK for h4s, 263.28kJ/kgK for h9, and 0.8 for ηC in Equation (II).

0.80=284.32kJ/kgK263.28kJ/kgKh4263.28kJ/kgKh4=289.57kJ/kg

Substitute 0.15kg/s for m˙7, 244.50kJ/kg for h1, and 68.80kJ/kg for h8 in Equation (VII).

Q˙L=(0.15kg/s)(244.5068.80)kJ/kg=26.35kW

Thus, the rate at which heat removed from the refrigerated space is 26.35kW.

(c)

To determine

The power input required to the two stage cascade refrigeration system.

The coefficient of refrigeration for the two-stage cascade refrigeration system.

(c)

Expert Solution
Check Mark

Answer to Problem 59P

The power input required to the two stage cascade refrigeration system is 8.443kW.

The coefficient of refrigeration for the two-stage cascade refrigeration system is 3.12.

Explanation of Solution

Write the formula to calculate the total required work input (W˙in) to the compression.

W˙in=W˙comp I,in+W˙comp II,in

W˙in=m˙7(h2h1)+m˙(h4h9) (VIII)

Here, required work input to the first stage compression is W˙comp I,in, and required work input to the second stage compression is W˙comp II,in.

Write the formula to calculate the COP of the two-stage cascade refrigeration system.

COPR=Q˙LW˙in (IX)

Conclusion:

Substitute 0.15kg/s for m˙7, 265.28kJ/kg for h2, 244.50kJ/kg for h1, 0.2026kg/s for m˙, 289.57kJ/kg for h4, and 263.28kJ/kg for h9 in Equation (VIII).

W˙in=((0.15kg/s)(265.28244.50)kJ/kg+(0.2026kg/s)(289.57263.28)kJ/kg)=8.443kW

Thus, the power input required to two stage cascade refrigeration system is 8.443kW.

Substitute 26.35 kW for Q˙L, and 8.443 kW for W˙in in Equation (IX).

COP=26.35kW8.443kW=3.12

Thus, the coefficient of refrigeration for the two-stage cascade refrigeration system is 3.12.

(d)

To determine

The rate at which heat removed from the refrigerated space.

The coefficient of refrigeration for the two-stage cascade refrigeration system.

(d)

Expert Solution
Check Mark

Answer to Problem 59P

The rate at which heat removed from the refrigerated space is 25.67kW.

The coefficient of refrigeration for the two-stage cascade refrigeration system is 2.71.

Explanation of Solution

Write the formula to calculate the rate of heat transfer from the refrigerated space (Q˙L).

Q˙L=m˙(h1h4) (X)

Here, the specific enthalpy of refrigerant at the exit of expansion valve is h4.

Write the formula to calculate the required work input (W˙in) to the compression.

W˙in=m˙(h2h1) (XI)

Conclusion:

From the Table A-12 of “Saturated refrigerant R-134a: Pressure”, obtain the properties of refrigerant at low pressure compressor inlet pressure (P1) of 200kPa as follows:

h1=hg=244.50kJ/kgs1=sg=0.9379kJ/kgK

Here, specific enthalpy of the saturated vapor is hg and the specific entropy of the saturated vapor is sg.

The specific entropy at the end of isentropic compression (s2) is calculated as

s2=s1=0.9379kJ/kgK

Refer to Table A-13, “Superheated R-134a”, and obtain the values of R-134a at pressure of 1200kPa and specific entropy of 0.9379kJ/kgK by using interpolation method as,

h2s=281.88kJ/kg

Substitute 0.80 for ηC, 281.88kJ/kg for h2s, and 244.50kJ/kg for h1 in Equation (II).

0.80=281.88kJ/kg244.50kJ/kgh2244.50kJ/kgh2=291.23kJ/kg

From the Table A-12 of “Saturated refrigerant R-134a: Pressure”, obtain the properties of refrigerant at expansion valve inlet pressure (P3) of 1200kPa as follows:

h3=hf=117.79kJ/kg

Substitute 117.79kJ/kg for h3 in Equation (I).

h4=117.79kJ/kg

Substitute 0.2026kg/s for m˙, 244.50kJ/kg for h1, and 117.79kJ/kg for h4 in Equation (X).

Q˙L=(0.2026kg/s)(244.50117.79)kJ/kg=25.67kW

Thus, the rate at which heat removed from the refrigerated space is 25.67kW.

Substitute 0.2026kg/s for m˙, 291.23kJ/kg for h2, and 244.50kJ/kg for h1 in Equation (XI).

W˙in=(0.2026kg/s)(291.23244.50)kJ/kg=9.467kW

Substitute 25.67 kW for Q˙L, and 9.467 kW for W˙in in Equation (IX).

COP=25.67kW9.467kW=2.71

Thus, the coefficient of refrigeration for the two-stage cascade refrigeration system is 2.71.

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Chapter 11 Solutions

THERMODYNAMICS(SI UNITS,INTL.ED)EBOOK>I

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