Fundamentals of Thermal-Fluid Sciences
Fundamentals of Thermal-Fluid Sciences
5th Edition
ISBN: 9780078027680
Author: Yunus A. Cengel Dr., Robert H. Turner, John M. Cimbala
Publisher: McGraw-Hill Education
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Chapter 7, Problem 97P

(a)

To determine

The mass flow rate of the refrigerant.

(a)

Expert Solution
Check Mark

Explanation of Solution

Given:

The refrigerated temperature is 35°C.

The cooling water inlet temperature (Tw1) is 18°C.

The cooling water outlet temperature (Tw2) is 26°C.

The mass flow rate of the water (m˙w) is 0.25kg/s.

The inlet and outlet pressure of the condenser is 1.2 MPa.

The power input consumed by compressor (W˙in) is 3.3kW.

Calculation:

First find out the state properties of the system as shown below:

From the Table A-13, “Superheated refrigerant R-134a” obtain the value of enthalpy of the refrigerant at the inlet of the condenser at the 1.2 MPa of pressure and 50°C of temperature as,

  h1=278.28kJ/kg.

At state 2, the refrigerant is subcooled by 5°C. Thus the exit temperature of the refrigerant from the condenser is expressed as follows:

  T2=Tsat@1.2MPa+ΔTsubcool        (I)

Here, the temperature leave from the condenser is ΔTsubcool.

From the Table A-13, “Superheated refrigerant R-134a” obtain the value of temperature of the refrigerant at the inlet of the condenser at the 1.2 MPa of pressure as,

  Tsat@1.2MPa=46.29°C

Calculate the exit temperature (T2) of the condenser using the equation (I).

  T2=Tsat@1.2MPa+ΔTsubcool

  T2=46.23°C+(5°C)=41.29°C

Refer to Table A-11, “Saturated refrigerant R-134a”, obtain the below exit enthalpy of the condenser at compressed liquid state on the basis of exit temperature of 41.29°C using interpolation method of two variables.

S. No

Temperature, °C

(x)

enthalpy of vaporization kJ/kg

(y)

140°C108.28kJ/kg
241.29°Cy2=?
342°C111.28kJ/kg

Calculate exit enthalpy of the condenser using interpolation method.

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

Substitute 40°C for x1, 41.29°C for x2, 42°C for x3, 108.28kJ/kg for y1, and 111.28kJ/kg for y3 in Equation (IV).

  y2=(41.29°C40°C)(111.28kJ/kg108.28kJ/kg)(42°C40°C)+108.28kJ/kg=110.19kJ/kg

From above calculation the exit enthalpy of the condenser at compressed liquid state on the basis of exit temperature of 41.29°C is 110.19kJ/kg.

Repeat the above Equation (II) to obtain the value of enthalpy of saturated liquid that entering the inlet of the condenser at the 18°C of temperature from the Table A-4, “Saturated water-Temperature “as,

  hw,1=75.54kJ/kg.

Repeat the above Equation (IV) to obtain the value of enthalpy of saturated liquid which is leaving the condenser at the 26°C of temperature from the Table A-4, “Saturated water-Temperature “as,

  hw,2=109.01kJ/kg.

Calculate the rate of heat transferred to the water.

  Q˙H=m˙w(h2h1)w

  Q˙H=(0.25kg/s)(109.01kJ/kg75.54kJ/kg)=(0.25kg/s)(33.47kJ/kg)=8.367kJ/s=8.367kJ/s×(1kW1kJ/s)

        =8.367kW

Calculate the mass flow rate of a refrigerant.

  m˙R=Q˙Hh1h2

  m˙R=8.367kW(278.28110.2)kJ/kg=8.367kW×(1kg/s1kW)168.08kJ/kg=0.04977kg/s0.0498kg/s

Thus, the mass flow rate of the refrigerant is 0.0498kg/s_.

(b)

To determine

The refrigeration load of the refrigerator.

(b)

Expert Solution
Check Mark

Explanation of Solution

Calculate the refrigeration load of the refrigerator.

  Q˙L=Q˙HW˙in

  Q˙L=(8.367kW)(3.3kW)=5.067kW5.07kW

Thus, the refrigeration load of the refrigerator is 5.07kW_.

(c)

To determine

The COP of a reversible refrigerator operating between the same temperature limits.

(c)

Expert Solution
Check Mark

Explanation of Solution

Determine the coefficient of performance of the refrigerator.

  COPR=Q˙LW˙in

  COPR=5.07kW3.3kW=1.5361.54

Thus, the COP of a reversible refrigerator operating between the same temperature limits is 1.54_.

(d)

To determine

The minimum power input to the compressor.

(d)

Expert Solution
Check Mark

Explanation of Solution

Calculate the maximum coefficient of performance of the reversible refrigerator operating between the same temperature limits.

  COPmax=1THTL1

  COPmax=1(18°C)(35°C)1=1(18°C+273)(35°C+273)1=1(291K238K)1=4.49

Here, the temperature of higher temperature body is TH and the temperature of lower temperature body is TL.

Calculate the minimum power input to the condenser for the same refrigerator load.

  W˙in,min=Q˙LCOPmax

  W˙in,min=5.07kW4.49=1.129kW1.13kW

Thus, the minimum power input to the compressor is 1.13kW_.

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Chapter 7 Solutions

Fundamentals of Thermal-Fluid Sciences

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