A coal fired steam-electric plant burns 90 kg of coal per second. The coal has a

Question

A coal fired steam-electric plant burns 90 kg of coal per second. The coal has a heating value of 14000 Btu/lb (1 BTU = 1054J), an ash content of 7% and a sulfure content of 1.1%. Thirty percent of the ash remains in the boiler, and the remainder becomes entrained in the exhaust gases. All of the sulfur is converted to SO2. The plant is equipped with an electrostatic precipitator that has a particulate removal effeciency of 99% and an SO2 scrubber that has an SO2 removal efficiency of 80%. the Thermal efficiency of the power plant is 33%.

Question: what is the emission rate of particulate matter (in kg/s) to the atmosphere?

I've been using this equation from class:   S = Mass(coal)*Fs*Fox*r(1-)

Fs= fraction of coal that is sulfur

Fox = fraction of sulfur that is oxidized

r= ratio of molecular weight (SO2)/S

=efficiency of removing SO2

I'm not sure I have done it right. Thanks.

Explanation / Answer

Rate of burning of coal = 90 kg/sec

Sulfur content = 1.1 %

fraction of coal that is sulfur, Fs = 1.1 / 100 * 90

= 0.99 Kg/sec

As all of the sulfur is converted to SO2.

So, fraction of sulfur that is oxidized, Fox = 1

Efficiency of removing SO2 = 80 %

ratio of molecular weight (SO2)/S, r = 64 / 32 = 2

emission rate of particulate matter,

S = Mass(coal) * Fs * Fox * r (1 - )

= 90 * 0.99 * 1 * 2 * (1 - 0.8)

= 35.64 kg / sec

ash content = 7 %

= 7 / 100 * 90

= 6.3 Kg/sec

particulate removal effeciency = 99%

amount of ash content removed = 0.99 * 6.3

= 6.237 kg/sec

So, emission rate of particulate matter = 35.64 kg / sec + 6.237 kg/sec

= 41.877 kg/sec

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