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

the answer is option c.

explanation:

recycled water alludes to the water that being utilized again after it already  used for farming purposes.   this kind of water frequently contains pharmaceutical gathering due to the compound compost that is utilized for the plants is consumed into the water.

show options, then i will gladly answer you question.

have a good day.