91影视

Efficiency of a power plant is the ratio of output power to input power . To calculate waste , output power will be subtracted from input power .

efficiency =$\frac{outputpower}{inputpower}$

wasted power = input power - output power .

The $900MW$of electric power generated represents $32\%$of the energy taken from the fuel.

input power = $\frac{900MW}{.32}=2812.5MW.$

Waste = input power - output power = 2812.5 MW - 900 MW

= 1912.5 MW.

This is distributed equally to each home , with each home taking a share of 20 kW .

Then their maximal number n would be such that

$n=\frac{P_c}{P_h}$

Therefore, substituting numerically the expression for the cooling power and the power used by each home, we have

$n=\frac{1912.5MW}{20kW}=\frac{1912.5脳{10}^6}{20脳{10}^3}=95625$

With power used by each home ,we can have $95625$homes