THERMODYNAMICS-SI ED. EBOOK >I<
THERMODYNAMICS-SI ED. EBOOK >I<
9th Edition
ISBN: 9781307573022
Author: CENGEL
Publisher: MCG/CREATE
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Chapter 6.11, Problem 58P

Refrigerant-134a enters the evaporator coils placed at the back of the freezer section of a household refrigerator at 100 kPa with a quality of 20 percent and leaves at 100 kPa and −26°C. If the compressor consumes 600 W of power and the COP of the refrigerator is 1.2, determine (a) the mass flow rate of the refrigerant and (b) the rate of heat rejected to the kitchen air.

FIGURE P6–58

Chapter 6.11, Problem 58P, Refrigerant-134a enters the evaporator coils placed at the back of the freezer section of a

(a)

Expert Solution
Check Mark
To determine

The mass flow rate of the refrigerant.

Answer to Problem 58P

The mass flow rate of the refrigerant is 0.00414kg/s_.

Explanation of Solution

Write the expression for the energy balance equation.

EinEout=ΔEsystem (I)

Here, the total energy entering the system is Ein, the total energy leaving the system is Eout, and the change in the total energy of the system is ΔEsystem.

Simplify Equation (II) and write energy balance relation of refrigrent-134a.

W˙in+m˙h1=Q˙R+m˙h2 (II)

Here, the rate of work to be done into the system is W˙in, the mass flow rate of refrigerant is m˙, the rate of heat rejected in the refrigerant is Q˙R, the initial specific enthalpy of the refrigerant is h1 and the final specific enthalpy of the refrigerant is h2.

Substitute 0 for W˙in in Equation (II), write the expression for the energy balance of mass flow rate of the refrigerant.

Q˙L=m˙(h1h2)m˙=Q˙Lh1h2 (III)

Here, the rate of heat transfer with low-temperature body is Q˙L.

Write the expression for the rate of coefficient performance of a refrigerant.

COP=Q˙LW˙inQ˙L=(COP)W˙in (IV)

Conclusion:

Determine the initial specific enthalpy of refrigerant.

h1=hf+xhfg (V)

Here, the specific enthalpy of saturated liquid is hf , the dryness fraction is x, and the specific enthalpy change upon vaporization is hfg.

Refer to Table A-13, “Saturated refrigerant-134a-Pressure table”, obtain the value of specific enthalpy of saturated liquid and specific enthalpy change upon vaporization at 100 kPa pressure.

hf=17.27kJ/kghfg=217.19kJ/kg

Substitute 17.27kJ/kg for hf, 0.2 for x, and 217.19kJ/kg for hfg in Equation (V).

h1=(17.27kJ/kg)+(0.2)(217.19kJ/kg)=(17.27kJ/kg)+(43.438kJ/kg)=60.708kJ/kg60.71kJ/kg

Refer to Table A-13, “Saturated refrigerant-134a”, obtain the below properties at the final pressure and saturated temperature of 100 kPa and (-26 C) using interpolation method of two variables.

Write the formula of interpolation method of two variables.

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

Here, the variables denote by x and y are saturated temperature and specific enthalpy.

Show the temperature at 31.31 C and 40 C as in Table (1).

Temperature,  CSpecific enthalpy, kJ/kg
-26.65 C234.46
-26 C?
-22.32 C236.99

Calculate final pressure and saturated temperature of 100 kPa and (-26 C) for liquid phase using interpolation method.

Substitute 26.65°C for x1, 26°C for x2, 22.32°C for x3, 234.46kJ/kg for y1, and 236.99kJ/kg for y3 in Equation (VI).

y2=((26°C)(26.65°C))(236.99kJ/kg234.46kJ/kg)((22.32°C)(26.65°C))+234.46kJ/kg=234.84kJ/kg

From above calculation the final enthalpy of refrigerant is 234.84kJ/kg.

Substitute 1.2 for COP and 0.600kW for W˙in in Equation (IV).

Q˙L=(1.2)(0.600kW)=0.72kW

Substitute 0.72 kW for Q˙L, 234.84kJ/kg for h2, and 60.71kJ/kg for h1 in Equation (III).

m˙=0.72kW(234.8460.71)kJ/kg=0.72kW(174.13)kJ/kg=0.004135kg/s0.00414kg/s

Thus, the mass flow rate of the refrigerant is 0.00414kg/s_.

(b)

Expert Solution
Check Mark
To determine

The rate of heat rejected from refrigerant.

Answer to Problem 58P

The rate of heat rejected from refrigerant is 1.32kW_.

Explanation of Solution

Write the expression for the rate of conversation of energy principle for refrigerant 134a.

W˙in=Q˙HQ˙LQ˙H=Q˙L+W˙in (V)

Conclusion:

Substitute 0.72kW for Q˙L and 600W for W˙in in Equation (V).

Q˙L=0.72kW600W=0.72kW600W×(103kW1W)=1.32kW

Thus, the rate of heat rejected from refrigerant is 1.32kW_.

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Chapter 6 Solutions

THERMODYNAMICS-SI ED. EBOOK >I<

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