Homework Solution

TP 3-5

Given:

    A coal plant burns 2 tons of coal to generate 6000 kWh of electricity.

Find:

    efficiency of the power plant

Science:

    e = useful energy out
               energy in

Solve:

The expression for the efficiency of an automobile engine in words:

    e = electrical energy generated by the power plant
          chemical energy of the coal burned by the plant

    E_elect = (6000 kWh) = (6000x10³ Wh) = (6000x10³ Jh/s) = (6000x10³ Wh)(3600 s/h) = (2.16x10¹⁰ J)

    E_chem = (2 tons)(25x10⁶ Btu/ton)(1055 J/Btu) = 5.275x10¹⁰ J

    e = (2.16x10¹⁰ J)/(.275x10¹⁰ J) = 0.409

    efficiency is 41%