Question

Penn State Civil and Environmental Engineering Department just built a super-duper coal-fired power plant in Altoona, PA. It is 1,100 MW (i.e., large) with 40% efficiency. We used local coal that is 14,000 Btu/ pound. The sulfur content is 2%. We added a scrubber that is 65% efficient. A) How much sulfur will we release per hour (pounds/hour)? B) If all of the sulfur is oxidized to SO2 and the EPA insists on a 80% efficiency scrubber how much SO2 would be released (report value in pounds SO2/ kWh generated)?

Answer 1

Answer:

Part -A:

Sulfur released is $1.338 \times 10^{4}pounds/hour$.

Part-B:

$SO_{2}$ released of per EPA norms is $4.865 /times 10^{-3}P/kw$.

Explanation:

From the given,

The output = 1100 Mw

Efficiency = η = 0.4

Substitute the values in the following

Input = Output / η

$= \frac{1100}{0.4}= 2750 Mw$

Let's converts between joules to BTU.

1 BTU = 1056 J

Part-A;

$Energy\,\, required = Power \times time$

$2750 \times 10^{6} \,\, J/s \times 3600s = 9.9 \times 10^{12}J$

$Energy \,\, required = \frac{9.9 \times 10^{22}}{1056}= 9.375 \times 10^{9} BTU$

$Mass\,of\, coal\,required = \frac{9.375 \times 10^{9}}{14,000}=6.69 \times 10^{5}pounds$

But it has 2%  sulfur

The mass of sulfur released

$6.69 \times 10^{5}pounds \,coal \times 0.02 = 1.338 \times 10^{4}pound$

Therefore, released sulfur is $1.338 \times 10^{4}pounds/hour$.

Part -B;

One pound of sulfur produce two pounds of sulfur dioxide

Initial amount of produced sulfur =

$2 \times 1.338 \times 10^{4}pound = 2.676\times 10^{4}pound/hour$

Assuming we added a 80% efficiency then,

Released sulfur dioxide = $(1-0.8) \times SO_{2} \,produced =0.2 \times 2.676 \times 10^{4}Pounds/hr= 5.352 \times 10^{3}Pounds/hr$

Energy produced in an hour = $400 mw \times 1 hr = 1.1 \times 10^{6}kwh$

$SO_{2}\,released \, as \, per EPA = \frac{5.352 \times 10^{3} \,pounds}{1.1 \times 10^{6}kwh}= 4.865 \times 10^{-3}p\kwh$

Therefore, $SO_{2}$ released of per EPA norms is $4.865 /times 10^{-3}P/kw$.