A coal-fired 500 MW Power Plant has an efficiency of 35%, burns coal with an energy content of 10,000 BTU/lb, with 60% carbon content, 1.5% sulfur content. The coal burned produces 10% ash, and of the ash, 70% is fly ash. A) What are the carbon emissions (in lb C/hr)?

B) What are the sulfur emissions if a 70% efficient scrubbe

C) What are the particulate emissions released if a 99% efficient scrubber is used (in lb Fly Ash/hr) is used (in lb S/hr)?

carbon emission  = 292,468.8 lb/hr

sulfur emission = 2,193.526 lb/hr

emission release  = 3,412.13 lb/hr

Explanation:

given data

energy = 500 MW

efficiency of 35%

solution

As we know for generate 500 MW energy we need to generate 1706071225.06 Btu/hr.

for efficiency of 35% we need total energy is

total energy = 1706071225.06 ÷ 0.35

total energy = 4874489214 Btu/hr.

and

For 4,874,489,214 Btu/hr = 487,448 lb/hr

So, for coal has 60% of carbon

so carbon emission = 487,448 × 0.60

carbon emission  = 292,468.8 lb/hr

and

Total sulfur emission = 487,448 × 0.015

Total sulfur emission = 7,311.72 lb/hr

70% of this sulfur is removed before emission so the sulfur emission is

sulfur emission = 7,311.73 × 0.30

sulfur emission = 2,193.526 lb/hr

and

The burned coal will produce 487,448 × 0.70 = 341,213.6 lb/hr fly ash.

so 99% of this amount will be removed before emission, so emission is

emission release = 341,213.6 × 0.01

emission release  = 3,412.13 lb/hr

7.03-1.70=5.33 and 6.80+5.09=11.89 and 5.5-3.01=2.49