Question

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.

Answer 1

η = 2.57%

Step-by-step 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%