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A steam power plant with a power output of 200 MW consumes coal at a rate of 700 tons/h. If the heating value of the coal is 40,000 kJ/kg, determine the overall efficiency of this plant.

Respuesta :

Answer:

η = 2.57%

Explanation:

Given:

Output power of the steam power plant, P(out) = 200 MW

Consumption of coal, m = 700 tons/h = (700 × 10³)/3600 kg/s= 194.44 kg/s

Heating capacity of the coal, Cv = 40000 kJ/kg

now,

Input power,  P(in) = mCv

or

P(in) =  194.44 kg/s × 40000 = 7777.77 MW

thus,

efficiency, η = P(out)/P(in) = (200 MW) / (7777.77 MW) = 0.02571

or

η = 2.57%