Question

A steam power plant with a power output of \(150 \mathrm{MW}\) consumes coal at a rate of 60 tons/h. If the heating value of the coal is \(30,000 \mathrm{kJ} / \mathrm{kg},\) determine the overall efficiency of this plant.

Short answer

Answer: The overall efficiency of the steam power plant is 30%.

Step-by-step solution

Convert the coal consumption rate to power input

First, we need to convert the rate of coal consumption (60 tons/h) to a power input (in MW). To do this, we will use the heating value of the coal (30,000 kJ/kg). We will also need to convert the units of coal consumption to kg/h and then determine the power input in kW. 1 ton = 1000 kg 60 tons/h = 60 * 1000 kg/h = 60,000 kg/h Power input (kJ/h) = (60,000 kg/h) * (30,000 kJ/kg) = 1,800,000,000 kJ/h Now, we need to convert the power input to Megawatts (MW). 1 MW = 1,000,000 W = 1,000 kW 1,800,000,000 kJ/h * (1 kW/1000 W) * (3600s/1h) = 500,000 kW = 500 MW

Calculate the efficiency

Now that we have the power input (500 MW) and the power output (150 MW), we can determine the efficiency of the steam power plant using the following formula: Efficiency = (Power output / Power input) * 100 Efficiency = (150 MW / 500 MW) * 100 = 30% The overall efficiency of the steam power plant is 30%.