Question

A steam power plant with a power output of 230 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

Answer:

$\eta =46\%$

Explanation:

Hello!

In this case, we compute the heat output from coal, given its heating value and the mass flow:

$Q_H=60\frac{tons}{h}*\frac{1000kg}{1ton}*\frac{1h}{3600s}*\frac{30,000kJ}{kg}\\\\Q_H=500,000\frac{kJ}{s}*\frac{1MJ}{1000J} =500MW$

Next, since the work done by the power plant is 230 MW, we compute the efficiency as shown below:

$\eta =\frac{230MW}{500MW}*100\% \\\\\eta =46\%$

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