Question

An electric power generation burns bituminous coal in a dry furnace at a rate of 200 tons/h. The plnat operates continuoulsy 8760 h/yr. The coal has higher heating value of 12,500 Btu/lbm, 3% sulfur and 8% ash, and 0.001% mercury (by weight). An electrostatic precipitator (ESP) removes sufficient flyash to satisfy the new source performance standards.
(a) If all the mercury condenses on the minuscule flyash particles that are nor captured by ESP, estimate the mercury emissions (tons/yr).
(b) If thermal efficiency of the power plant is 30%, what is the capacity of the power plant?
(c) Determine SO2 emission using (1) mass-balance balance approach; and (2) emission factor
(d) If 95% of SO2 is captured by a wets crubber using lime (CaO) converting SO2 into gypsum, how much (1) lime will be needed and (2) amount of gypsum produced

Answer 1

1.) Answer of (a) mercury emission tons year =? Total amount of Cool used per year = 200 Tons/x Totelhous/ - 200 x 8760 = 1752000 Tons/year. mercury = 0.001% of coal = 0.00) X 1752000 Tons year. 100 X 172 - 17.52 Tons/year 2.) Answer of (b) Capacity of pour plant = ? Heating Value (Hu) of wal = 12500/8 tup Lbm = 29075 kg/kg Power (k/m) = Hux Goal) hi =29075 kJ/kg x 200 xpoolne = 29075X200x1000 = 5815000000 kJ/h. Power (kwatt or kJ/s) = 5815000000 = 1615.27 mw. 360o Topacity of Power with 30%) Thermal officiany. = 30x1615.27 = 484.58 mw.

3.) Answer of (c ) from given data 3% of Good Sulphur is burt & Considering 1kg of coal. amount of sulphur (5) = 31.x 10oog - 3000 - 30gm According to mars balance opposach amount of O₂ = 'Sx164/32) constant S0₂= 30gmx be 32 SO2 = 60 gm for per kg coal, emission factor = CEF) eos weight of polutant divided by unit weight or volume of the activity emitting the pollutant Total weight of So in Total 1 kg of coal EF = = emission factor 60gm = Too Ogn 0.06 -

64gm 4) Answer of (d) element molecular weight ingen Cao 56m SO2 Cason T 136 gm O P 32gm » Chemical reaiton CaO + 5O₂ + %0. = Casou (Gypsom) 56sm Gugm 16gm = 136gm Por gm of 50₂ 56 google + 64/64 son = 136/64 Soo for 1 kg of Coal = 502= 60g with 95% absorbtion S02= 60x95/100 SOLF 57gm. generated required for 1 kg of coal = 57.8 136 Casou = 121, 125 gm / kg of coal = 121.125 Toms x Total coal year. 1000x1000 = 121.125 x 1752000 10% 1000 = 212.2 Ton/year. Cao require for tag coal = 57% 56/4 = 49.87 gm. /kg ofwal = (49.87 Tons X Total coal year. 1000X1000 49.87 x 1752000 100 x 1000 = 87.37 Ton/year.