Question

A 500 MW steam power plant consumes coal at a rate of 65 tons/h. If the heating value of the coal is 30,000 kJ/kg, what is the overall efficiency of this plant? 

Answer 1

$Q_{in} =M_{coal} ×q_{hv}$

Where:

$Q_{in} =$ Amount of heat input

$M_{coal} =$ Mass of coal consumed per day

$q_{hv} =$ heating value

$Q_{in} = 65000kg/hr ×24hr×30MJ/Kg=4.68×10^7 MJ$

Further $Q^.= \frac{Q_{in}}{24 hr}$

Where $Q^.=$ Coal consumption rate

$Q^.= \frac{4.68×10^7MJ}{24×3600}=541.67MJ/s$

Efficiency$n=\frac{ W_{output}}{Q^.}$

$n=\frac{500MW}{541.67MJ/s}=0.923$

$n=$ 92.3%