Consider a $1.0MW$ power plant (this is the useful output in the form of electric energy) that operates between 30^oC and 450^oC at 65% of the Carnot efficiency. This is enough electric energy for about 750 homes. One way to use energy more efficiently would be to use the 30^oC“waste” energy to heat the homes rather than releasing that heat energy into the environment. This is called cogeneration, and it is used in some parts of Europe but rarely in the United States. The average home uses 70GJof energy per year for heating. For estimating purposes, assume that all the power plant’s exhaust energy can be transported to homes without loss and that home heating takes place at a steady rate for half a year each year. How many homes could be heated by the power plant?

Step by step solution

01

Introduction

Carnot Efficiency

Carnot efficiency refers to a heat engine's greatest efficiency while operating between two temperatures.

02

Find Number of homes 

The amount of energy required in each home will be calculated by dividing the annual energy requirement by half of a year. That is,

$P_h=\frac{Q_h}{t}$

$=\frac{70\mathrm{GJ}}{3600·24·182.5}$

$=4.5\mathrm{kW}$

This is heat requirement per house.

The power plant's heat efficiency

$\eta =\frac{P_e}{P_H}\Rightarrow P_H=\frac{P_e}{\eta }$

P_eis electrical output and P_H is heat input of plant.

$\eta =0.65{\eta }_c=0.65\times \left(1-\frac{303}{723}\right)=0.65\times 0.58$

03

Limited heat

The power plant's heat output power is,

$P_H=\frac{1\mathrm{MW}}{0.65·0.58}=\underset{¯}{2.648\mathrm{MW}}$

Waste energy P_w= 2.648 MW - 1 MW=1.648 MW.

As a result, the number of homes that can be heated is calculated as follows.

$n=\frac{P_W}{P_h}=\frac{1.648MW}{4.5kW}=366$

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