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 3, Problem 86RQ
To determine

The potential revenue generated by the pump-turbine system.

Expert Solution & Answer
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Explanation of Solution

Given:

Volumetric flow rate of the water (V˙) is 2m3/s.

Elevation difference of the water (Δz) is 40m.

Efficiency of the turbine-generator (ηturbine-gen) is 0.75.

Efficiency of the pump-motor (ηpump-motor) is 0.75.

Operating hours (Δt) is 10hr/year.

Unit cost of electric power produced at night is $0.05/kWh.

Unit cost of electric power produced at day is $0.12/kWh.

Density of the water (ρ) is 1000kg/m3.

Acceleration due to gravity (g) is 9.81m/s2.

Calculation:

Calculate the minimum power required to pump water from the lower to higher reservoir (W˙min,pump).

  W˙min,pump=W˙max,turbine=ΔE˙mech=m˙Δemech=m˙Δpe=m˙gΔz

  =ρV˙gΔz=(1000kg/m3)(2m3/s)(9.81m/s2)(40m)=(1000kg/m3)(2m3/s)(9.81m/s2)(40m)(1N1kgm/s2)(1kW1000Nm/s)=784.8kW

Calculate the actual pump electric power (W˙pump,elect).

  W˙pump,elect=W˙idealηpump-motor=784.8kW0.75=1046kW

Calculate the actual turbine electric power (W˙turbine).

  W˙turbine=ηturbine-genW˙ideal=(0.75)(784.8kW)=588.6kW

Calculate the power consumption cost of the pump.

  Cost=W˙pump,electΔt×Unitprice=(1046kW)(365×10hr/year)($0.05/kWh)=$190,968/year

Calculate the revenue generated by the turbine.

  Revenue=W˙turbineΔt×Unitprice=(588.6kW)(365×10hr/year)($0.05/kWh)=$257,807/year

Calculate the net income per year.

  Netincome=RevenueCost=$257,807/year$190,968/year=$66,839/year

Thus, the potential revenue generated by the pump-turbine system is $66,839/year_.

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Chapter 3 Solutions

Fundamentals of Thermal-Fluid Sciences

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