# HW16s - Spring 2006 Thermodynamics Homework 16 Solution 1...

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Spring 2006 1 ME 201 Thermodynamics Homework 16 Solution 1. Consider a power plant that is producing 1 MW of electric power as it operates with a high temperature of 1800 K and a low temperature of 290 K. If we can sell the electric power for \$0.04 per kW hr, and the heat transfer from the high temperature reservoir costs \$0.0075 per kW hr, what is the maximum income per year the plant can generate? Solution: We begin with our interaction diagram for a heat engine. Our appropriate equations are: η th net H = W Q Q = W + Q H net L η Carnot L H = 1- T T Our income is given by Income = (Electric Energy Generated)(0.04) - (Heat Energy Consumed)(0.0075) High Temperature Reservoir at T H Low Temperature Reservoir at T L Heat Engine Q H Q L W net

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ME 201 Thermodynamics Spring 2006 2 where Electric Energy Generated = W net x # of hours in a year = (1000 kW)(24 x 365) = 8,760,000 kW hr Heat Energy Consumed = Q # of hours in a year H × We can determine Q H from our thermal efficiency. The thermal efficiency is
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## This note was uploaded on 07/25/2008 for the course ME 201 taught by Professor Somerton during the Spring '06 term at Michigan State University.

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HW16s - Spring 2006 Thermodynamics Homework 16 Solution 1...

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