A 440-MW coal-fired power plant is burning coal that contains 185 ppb of Mercury

Question

A 440-MW coal-fired power plant is burning coal that contains 185 ppb of Mercury. The coal firing rate is 300,000 lb/hr, and the flue gas flow rate is 1.0 million SCFM. The mercury standard that must be met is 0.015 lb/GWh of gross energy output. Calculate the overall mercury capture percentage that is required to meet this standard. If the wet scrubber can remove 40% of the mercury that comes into it, what removal efficiency is needed for the activated carbon injection/baghouse system. Answer: 82%

Explanation / Answer

Power plant = 440 MW

Concentration of Hg in Coal = 185 ppb

Coal firing rate = 300,000 lb/hr

Mercury standard to be maintained = 0.015 lb/GWh

Wet scrubber removes 40% of mercury.

About 0.015 lb of Hg is allowed for 1 GWh(103 MWh).Then for running 440 MW plat for an hour, (440 MWh),

0.015 Hg = 103 MWh

x = 440 MWh

x = 0.015*440/1000 = 0.0066 lb

Mercury should be maintained at 0.0066 lb/hr for 440 MW plant

Amount of Hg = 185 ppb = 185 lb per 109 lb

for 300000 lb/hr of coal , Hg = 185*300000/109 = 0.0555 lb/hr

Percentage of Hg to be removed = (Hg in coal - Hg to be maintained /Hg in coal)*100 = (0.0555 - 0.0066/0.0555)*100 = 88.10 %

Flue gas entering the scrubber.

300000 lb/hr of coal = 1 million SCFM

0.0555 lb of Hg is present in 1 million SCFM flue gas enters the scrubber.

In scrubber 40% of Hg is removed. Thus Hg coming out of scrubber = (1-0.4)*0.0555 = 0.0333 lb

Amount of Hg entering activated carbon injection / baghouse system = 0.0333 lb in 1 million SCFM flue gas

Percentage of Hg to be removed = (Hg entering - Hg to be maintained / Hg entering) * 100 = (0.0333 - 0.0066/ 0.0333 )*100 = 80.2 %

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