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12 March, 17:29

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.

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Answers (1)
  1. 12 March, 18:15
    0
    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
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