Question

A steam power plant with a power output of 101.393 MW consumes coal at a rate of 55,309.154 kg/h. If the heating value of 23,888.055 kJ/kg, determine the overall efficiency of this plant. in percent (%) with three decimals.

Answer 1

Answer:

27.626%

Explanation:

First we find the amount of energy the plant consumes

Q=(55309.154)(23888.055)=1321228113KJ/h  

then we convert kj / h to MW

Q=1321228113/(3600*1000)=367.008Mw

then we calculate the efficiency

η=101.393/367.008=0.27626=27.626%