A coal-burning steam power plant produces a net power of 300 MW with an overall

Question

A coal-burning steam power plant produces a net power of 300 MW with an overall thermal efficiency of 32 percent The actual gravimeter air-fuel ration in the furnance ic calculated to be 12 kg air/kg fuel. The heating value of coal is 28,000 kJ/kg. Determine (a) the amount of coal consumed during a 24-hour period and (b) the rate of air flowing through the furance.

Explanation / Answer

power plant produces a net power of 300 MW overall thermal efficiency = 32 percent = 32/100 = .32 => actual power produced = 300 /.32 = 937.5 MW. The heating value of coal = 28,000 kJ/kg . mass of the coal burnt per second= 937.5MW/28000KJ kg = 33.48214 kg/s hence mass of coal burnt in 24 hours is 33.48214 * 24 * (60*60) =2892857.142857 kg air-fuel ration in the furnace is given to be 12 kg air/kg fuel => rate of air flow = 12 * 33.48214 kg/s = 401.78565 kg/s

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