Answer to Question #156901 in Physics for Arif Saleem

Question

a steam power plant with a power output of 250 MW consumes coal at a rate of 60 tons/h. If the heating value of the coal is 30,000 kj/kg, determine the overall efficiency of this plant

Answer 1

By definition, the efficiency is the ratio of the production $P_{prod}$ and consumption $P_{cons}$ powers:

$\eta = \dfrac{P_{prod}}{P_{cons}}$

The production power is $P_{prod} = 250MW$. The consumption power is:

$P_{cons} = qm$

where $q = 30000\dfrac{kJ}{kg} = 30\dfrac{MJ}{kg}$ is the heating value of the coil, and $m = 60\dfrac{tons}{h} = 60\cdot \dfrac{1000kg}{3600s} = \dfrac{150}{9}\dfrac{kg}{s}$ is the rate of coil consumption. Thus, obtain;

$\eta = \dfrac{P_{prod}}{qm}\\ \eta = \dfrac{250MW}{30\dfrac{MJ}{kg}\cdot \dfrac{150}{9}\dfrac{kg}{s}} = 0.5$

Answer. 0.5.