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 123RQ
To determine

The maximum money saved by using the lake water instead of outside air as the heat source.

Expert Solution & Answer
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Explanation of Solution

Given:

The lower temperature of the reversible heat pump (TL)Air is 0°C.

The lower temperature of the reversible heat pump (TL)Lake is 25°C.

The higher temperature of the reversible heat pump (TH) is 25°C.

The rate of heat reject by the heat pump (Q˙H) is 140,000kJ/h.

The cost of electricity is $0.105/kWh.

The number of operating hours is 100 hours.

Calculation:

Initially calculate for outside air:

Calculate coefficient of performance for the reversible heat pump.

  COPHP,rev=11((TL)air/TH)        (I)

  COPHP,rev=11(0°C/25°C)=11(0+273K/25+273K)=11.92

Calculate the minimum power input required to operate the heat pump.

  W˙in,min=Q˙HCOPHP,rev        (II)

  W˙in,min=(140,000kJ/h)(11.92)=(140,000kJ/h)×(1kW3600kJ/h)(11.92)=38.88889kW11.92=3.2624kW

Calculate the cost of the energy in a heat pump (Costair).

  Cost=[(W˙in,min)×(number of hours operates by heat pump)×(cost of electricity)]        (III)

  Costair=(3.2624kW)×(100h)×($0.105/kWh)=$34.255$34.26

Similarly,

For lake water:

Substitute (TL)Lake=10°C and TH=25°C in Equation (I).

  COPHP,rev=11(10°C/25°C)=11(10+273K/25+273K)=19.86619.87

Substitute Q˙H=140,000kJ/h and COPHP,rev=19.87 in Equation (II).

  W˙in,min=(140,000kJ/h)(19.87)=(140,000kJ/h)×(1kW3600kJ/h)(19.87)=38.88889kW19.87=1.957kW

Substitute W˙in,min=1.957kW, 100 h for number of hours operates by heat pump, and $0.105/kWh for cost of electricity in Equation (III).

  Costlake=(1.9571kW)×(100h)×($0.105/kWh)=$20.549$20.55

Calculate the maximum money saved by using the lake water (Costlake) instead of outside air as the heat source.

  Moneysaved=CostairCostlake        (IV)

  Moneysaved=($34.26)($20.55)=$13.71$13.7

Thus, the maximum money saved by using the lake water instead of outside air as the heat source is $13.7_.

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Chapter 7 Solutions

Fundamentals of Thermal-Fluid Sciences

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