Question

a 100-mw coal-burning electricity generating station operates at 33% efficiency burning coal containing 5% ash and 2% sulfur. assume a coal heat content of 30 kj/g. if 95% of the ash and 50% of the sulfur are trapped before emission from the stack by a filter, calculate (a) the emission rate to the atmosphere of ash and sulfur, in g/s and (b) the removal rate to the filter of ash and sulfur, in g/s .

Answer 1

The emission rate of ash and sulfur to the atmosphere is 166.67 g/s and 66.67 g/s respectively. The removal rate to the filter is 158.33 g/s for ash and 33.33 g/s for sulfur.

Step-by-step explanation:

To calculate the emission rate to the atmosphere of ash and sulfur, we need to start by finding the amount of coal being burned per second. We can do this by converting the power output of the station from MW to W and dividing it by the heat content of the coal. So, the coal consumption rate would be:

Coal consumption rate = (100 MW x 10^6 W/MW) / (30 kJ/g x 10^3 J/kJ) = 3333.33 g/s

Next, we need to find the amount of ash and sulfur present in the coal consumed per second. The ash content is 5% and the sulfur content is 2%. So, the ash emission rate and sulfur emission rate would be:

Ash emission rate = 3333.33 g/s x 0.05 = 166.67 g/s

Sulfur emission rate = 3333.33 g/s x 0.02 = 66.67 g/s

To calculate the removal rate to the filter, we need to multiply the emission rates by the respective removal percentages:

Ash removal rate = 166.67 g/s x 0.95 = 158.33 g/s

Sulfur removal rate = 66.67 g/s x 0.50 = 33.33 g/s