Question

A 600 MW coal-fired power plant has an overall thermal efficiency of 38%. It is burning coal that has a heating value of 12,000 Btu/lb, an ash content of 5%, a sulfur content of 3.0%, and a CO2 emission factor of 220lb/million Btu. Calculate the heat emitted to the environment (Btu/sec), the coal feed rate (tons/day), the degree (%) of sulfur dioxide control needed to meet an emission standard of 0.15 lb SO2/million Btu of heat input, and the CO2 emission rate (metric tons/day).
I know that the heat emitted in Btu/seconds = 927,865 Btu/second and that the feed rate is 5805 tons/day however I am having trouble with the SO2 content and the COs emission rate.

Answer 1

Answer:

See step by step explanations for answer.

Explanation:

600 megawatts =

568 690.272 btu / second

thermal eficiency=work done/Heat supllied

0.38=568690.272/Heat supplied

Heat supplied=1496553.35btu /s

heat emmitted to the atmosphere=heat supplied -work done=(1496553.35-568690.272)=927863.1 btu/s

feed rate=(1496553.35)/12000=124.71 lb/s =10775184.1056 lb/day=5 387.472 ton / day

sulphur content released=(0.03*124.71)/(1.496553)=2.5 lb SO2/million Btu of heat input

so

the degree (%) of sulfur dioxide control needed to meet an emission standard=(2.5/0.15)*100=1666.67 %

the CO2 emission rate=220*(1.496553) =329.241 lb/s =12 903.0802 metric ton / day