Question

A coal-burning power plant produces 500 megawatts of electricity while burning 225,000 kg of coal per hour (almost 250 tons each hour!). If the coal has an average energy content of 20 million Joules per kg, determine the efficiency of the power plant.

Answer 1

40%

Step-by-step explanation:

The efficiency of a plant is determined by

`eff=(E_(produced))/(E_(in))`

Our produced energy is

`500 \ MW`

While the input energy can be calculated

`E_(in)=number\ of \ coal \ in \ * energy\ per\ coal\ unit\\ =(225000)/(3600)coal/ s\ *20\ MJ/coal\\ = 1250\ MW`

So, the efficiency of plant is

`eff=(500)/(1250)}\\ 0.4`

or 40% efficient