A power plant rated at 500 MW e can burn a bituminouscoal from either Kentucky o

Question

A power plant rated at 500 MWe can burn a bituminouscoal from either Kentucky or Wyoming. The power plant thermalefficient is 0.38 based on the higher heating value (HHV) of thefuel. Kentucky coal has a HHV = 28079 kJ/kg and a sulfur content of0.5 percent by mass. For the Wyoming coal, HHV = 30125 kJ/kg and asulfur content of 3.5 percent by mass. For continuous fullload operation, determine the annual SO2 production(kg/year) for these two coals. If you are concerned about theenvironment, which coal will you choose?

Explanation / Answer

500 MJ/s /0.38 = 1315.78947 MJ/s of heat needed For Kentucky 1315.78947 MJ/s *(1000 kJ/1MJ) *(1 kg/28079 kJ)*0.005 = 0.234 kgS/s 1 year *(0.234 kg S/s) *(3600 s/ hr) *(24 hr/day)*(365 day/yr) *(1kmol S/32.1 kg S) *(1 kmol SO2/ 1 kmolS)*(64.1 kg SO2/1 kmolSO2) = 1.48 x 107 kg    (in one year) For Wyoming 1315.78947 MJ/s *(1000 kJ/1MJ) *(1 kg/30125 kJ)*0.035 = 1.52871806kg S/s 1 year *(1.5287 kg S/s) *(3600 s/ hr) *(24 hr/day)*(365 day/yr) *(1kmol S/32.1 kg S) *(1 kmol SO2/ 1 kmolS)*(64.1 kg SO2/1 kmolSO2) = 96267982.3 kg = 9.63 x 107 kg (in one year) Pick the Kentucky

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