Answer to Question #249012 in Molecular Physics | Thermodynamics for Prince

Question

.A coal power plant consumes 100000 kg of coal per hour and produces 500 MW of power. If the heat combustion of coal is 30 MJ/kg, what is the efficiency of the power plant?

Answer 1

Efficiency in terms of power

$e=\frac{P_0}{P_i}$

The power input by the coal is

$P_i =AH \\ P_i = (10^5 \;kg/hr)(30\;Mj/kg) \\ = 30 \times 10^5 \;mJ/hr \\ P_i = [30 \times 10^5 \;mJ/hr][\frac{1 \;hr}{3600 \;s}] \\ = 833.3 \;Mj/s \\ = 833.3 \;MW$

The efficiency of power plant is

$e =\frac{P_0}{P_i} \\ = \frac{500 \;MW}{833.3 \;MW} \\ = 0.6$