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 1

coal contains 185 ppb of mercury .

The coal firing rate is 300000 lb/hr

then total mercury content is (185/10⁹)*3*10⁵=0.0555lb/hr

mercury standard that must be met is 0.015 lb/GWh

but it contains = 0.0555/0.44=0.126136lb/GWh

as scrubber removes only 40%

after scrubbing mercury contain remains=0.6*0.126136 =0.07568 lb/GWh

what removal efficiency is needed of activated carbon to standardise mercury =(0.07568-0.015)/0.07568=0.8017

approx 80% efficiency of activated carbon is needed.