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

Part -A:

Sulfur released is
`1.338 * 10^(4)pounds/hour`.

Part-B:

`SO_(2)` released of per EPA norms is
`4.865 /times 10^(-3)P/kw`.

Step-by-step explanation:

From the given,

The output = 1100 Mw

Efficiency = η = 0.4

Substitute the values in the following

Input = Output / η

`= (1100)/(0.4)= 2750 Mw`

Let's converts between joules to BTU.

1 BTU = 1056 J

Part-A;

`Energy\,\, required = Power * time`

`2750 * 10^(6) \,\, J/s * 3600s = 9.9 * 10^(12)J`

`Energy \,\, required = (9.9 * 10^(22))/(1056)= 9.375 * 10^(9) BTU`

`Mass\,of\, coal\,required = (9.375 * 10^(9))/(14,000)=6.69 * 10^(5)pounds`

But it has 2% sulfur

The mass of sulfur released

`6.69 * 10^(5)pounds \,coal * 0.02 = 1.338 * 10^(4)pound`

Therefore, released sulfur is
`1.338 * 10^(4)pounds/hour`.

Part -B;

One pound of sulfur produce two pounds of sulfur dioxide

Initial amount of produced sulfur =

`2 * 1.338 * 10^(4)pound = 2.676* 10^(4)pound/hour`

Assuming we added a 80% efficiency then,

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

Energy produced in an hour =
`400 mw * 1 hr = 1.1 * 10^(6)kwh`

`SO_(2)\,released \, as \, per EPA = (5.352 * 10^(3) \,pounds)/(1.1 * 10^(6)kwh)= 4.865 * 10^(-3)p\kwh`

Therefore,
`SO_(2)` released of per EPA norms is
`4.865 /times 10^(-3)P/kw`.