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 53P

(a)

To determine

The COP of the heat pump.

(a)

Expert Solution
Check Mark

Explanation of Solution

Given:

The initial pressure of the heat pump (P1) is 800 kPa.

The initial temperature of the heat pump (T1) is 35°C.

The mass flow rate of heat pump is (m˙) is 0.018kg/s.

The final pressure of the heat pump (P2) is 800 kPa.

The quality of the refrigerant at the exit is 0 (saturated liquid).

The rate of required input of the heat pump (W˙net,in) is 1.2 kW.

Calculation:

Convert the unit of pressure from kPa to MPa.

  P=800kPa=800kPa×103MPakPa=0.80MPa

Refer to Table A-13, “Superheated refrigerant-134a”, obtain the below properties at the superheated pressure and temperature of 800 kPa (0.80 MPa) and 35 C using interpolation method of two variables.

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

Temperature,  CSpecific enthalpy, kJ/kg
Saturated liquid, vf
31.31 C267.34
35 C?
40 C276.46

Calculate superheated pressure and temperature of 800 kPa (0.80 MPa) and 35 C for liquid phase using interpolation method.

  y2=(x2x1)(y3y1)(x3x1)+y1

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

Substitute x1= 31.31 °C, x2=35 °C, x3=40 °C, y1=267.34kJ/kg, and y3=  276.46 kJ/kg in the above Equation.

  y2=(35°C31.31°C)(276.46kJ/kg267.34kJ/kg)(40°C31.31°C)+267.34kJ/kg=271.21kJ/kg

From above calculation the initial enthalpy (h1) of condenser is 271.21kJ/kg.

Refer to Table A-12, “Saturated pressure table”, at final state of 800 kPa and 0.

  hf=h2=95.48kJ/kg

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˙H+m˙h2        (II)

Here, The heat rejected in the condenser is Q˙H,

The initial specific enthalpy of the condenser is h1 and the final specific enthalpy of the condenser is h2.

Substitute W˙in=0 in Equation (II) and write the expression for the energy balance on the heat rejected in the condenser.

  Q˙H=m˙(h1h2)        (III)

  Q˙H=(0.018kg/s)(271.21kJ/kg95.48kJ/kg)=(0.018kg/s)(175.73kJ/kg)=3.163kJ/s×(kW1kJ/s)=3.163kW

Write the expression for the rate of coefficient performance of a heat pump.

  COPHP=Q˙HW˙net,in

  COP=3.163kW1.2kW=2.63592.64

Thus, the COP of the heat pump is 2.64_.

(b)

To determine

The rate of heat absorbed from the outside air.

(b)

Expert Solution
Check Mark

Explanation of Solution

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

  W˙in=Q˙HQ˙LQ˙L=Q˙HW˙in

  Q˙L=(3.1631.2)kW=1.96kW

Thus, the rate of heat absorbed from the outside air is 1.96kW_.

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Chapter 7 Solutions

Fundamentals of Thermal-Fluid Sciences

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