EBK THERMODYNAMICS: AN ENGINEERING APPR
EBK THERMODYNAMICS: AN ENGINEERING APPR
9th Edition
ISBN: 8220106796979
Author: CENGEL
Publisher: YUZU
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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
Check Mark

Answer to Problem 59P

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

Explanation of Solution

Express the specific enthalpy at state 2 using Carnot efficiency.

ηC=h2sh1h2h1 (I)

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

Express specific enthalpy at state 3.

h3=hf@500kPa (II)

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

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

h3h4 (III)

Here, specific enthalpy at state 3 and 4 is h3andh4 respectively.

Express specific enthalpy at state 5.

h5=hg@400kPa (IV)

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

Express specific entropy at state 5.

s5=sg@400kPa (V)

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

Express the specific enthalpy at state 6 using Carnot efficiency.

ηC=h6sh5h6h5 (VI)

Here, specific enthalpy at state 5, 6 and 6s is h5,h6andh6s respectively.

Express specific enthalpy at state 7.

h7=hf@1400kPa (VII)

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

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

h7h8 (VIII)

Here, specific enthalpy at state 7 and 8 is h7andh8 respectively.

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

m˙A(h5h8)=m˙B(h2h3) (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 160kPa.

h1=hg=241.14kJ/kgs1=sg=0.420kJ/kgK

Here, specific entropy and enthalpy at state 1 is s1andh1 respectively, specific enthalpy and entropy at saturated vapor is hgandsg respectively.

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

Write the formula of interpolation method of two variables.

y2=(x2x1)(y3y1)(x3x1)+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/kgK)

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 0.9384kJ/kgK,0.9420kJ/kgKand0.9704kJ/kgK for x1,x2andx3 respectively, 263.48kJ/kg for y1 and 273.03kJ/kg for y3 in Equation (X).

y2=[(0.94200.9384)kJ/kgK][(273.03263.48)kJ/kg](0.97040.9384)kJ/kgK+263.48kJ/kg=264.55kJ/kg=h2s

Thus, the specific enthalpy at state 2s is,

h2s=264.55kJ/kg

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

0.80=264.55kJ/kg241.14kJ/kgh2241.14kJ/kg0.80(h2241.14kJ/kg)=23.41kJ/kg0.80h2192.912kJ/kg=23.41kJ/kgh2=270.41kJ/kg

Perform unit conversion of pressure at state 3 and 5 from MPatokPa.

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

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

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

h3=hf@500kPa=73.32kJ/kg

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

h3=73.32kJ/kg

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

h4=73.32kJ/kg

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

h5=hg@400kPa=255.61kJ/kgs5=sg@400kPa=0.9271kJ/kgK

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

h5=255.61kJ/kg

Substitute 0.9271kJ/kgK for sg@400kPa in Equation (V).

s5=0.9271kJ/kgK

Perform unit conversion of pressure at state 6 from kPatoMPa.

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

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

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=281.56kJ/kg

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

0.80=281.56kJ/kg255.61kJ/kgh6255.61kJ/kg0.80(h6255.61kJ/kg)=281.56kJ/kg255.61kJ/kg0.80h6204.488kJ/kg=25.95kJ/kgh6=288.04kJ/kg

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

h7=hf@1400kPa=127.25kJ/kg

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

h7=127.25kJ/kg

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

h8=127.25kJ/kg

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

m˙A(255.61127.25)=(0.25kg/s)(270.4173.32)kJ/kgm˙A(128.36kJ/kg)=(0.25kg/s)(197.09kJ/kg)m˙A=0.3839kg/s0.384kg/s

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

(b)

To determine

The rate of heat removal from the refrigerated space.

(b)

Expert Solution
Check Mark

Answer to Problem 59P

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

Explanation of Solution

Express the rate of heat removal from the refrigerated space.

Q˙L=m˙B(h1h4) (XI)

Conclusion:

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

Q˙L=(0.25kg/s)(241.1473.32)kJ/kg=41.96kJ/s[kWkJ/s]=41.96kW42kW

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

(c)

To determine

The COP of the refrigerator.

(c)

Expert Solution
Check Mark

Answer to Problem 59P

The COP of the refrigerator is 2.12.

Explanation of Solution

Express the power input.

W˙in=m˙A(h6h5)+m˙B(h2h1) (XII)

Express the COP of the refrigerator.

COP=Q˙LW˙in (XIII)

Conclusion:

Substitute 0.3839kg/sand0.25kg/s for m˙Aandm˙B respectively, 288.04kJ/kgand255.61kJ/kg for h6andh5 respectively, and 270.41kJ/kgand241.14kJ/kg for h2andh1 respectively in Equation (XII).

W˙in=(0.3839kg/s)(288.04255.61)kJ/kg+(0.25kg/s)(270.41241.14)kJ/kg=19.77kJ/s[kWkJ/s]=19.77kW

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

COP=41.96kW19.77kW=2.12

Hence, the COP of the refrigerator is 2.12.

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Chapter 11 Solutions

EBK THERMODYNAMICS: AN ENGINEERING APPR

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