Question

A coal power plant consumes 100,000 kg of coal per hour and produces 500 MW of power. If the heat of combustion of coal is 30 MJ/kg, what is the efficiency of the power plant?

a) 15%
b) 20%
c) 25%
d) 30%

Answer 1

The efficiency of the coal power plant is 16.67%.

Step-by-step explanation:

The efficiency of a coal power plant can be calculated using the equation:

Efficiency = (Useful energy output / Energy input) x 100%

In this case, the useful energy output is the power produced by the power plant. Given that the power output is 500 MW, we can convert it to joules using the conversion factor 1 MW = 1 x 10^6 J/s. Therefore, the power output is 500 x 10^6 J/s.

The energy input is the heat of combustion of the coal used by the power plant. The heat of combustion is given as 30 MJ/kg. Therefore, the energy input can be calculated by multiplying the heat of combustion by the mass of coal consumed per hour: 30 MJ/kg x 100,000 kg/hour = 3 x 10 9 J/s

Now we can calculate the efficiency:

Efficiency = 16.67%