Answer:
<u>Part -A:</u>
Sulfur released is
.
<u>Part-B:</u>
released of per EPA norms is
.
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](https://tex.z-dn.net/?f=%3D%20%5Cfrac%7B1100%7D%7B0.4%7D%3D%202750%20Mw)
Let's converts between joules to BTU.
1 BTU = 1056 J
<u>Part-A;</u>
![Energy\,\, required = Power \times time](https://tex.z-dn.net/?f=Energy%5C%2C%5C%2C%20required%20%3D%20Power%20%5Ctimes%20time)
![2750 \times 10^{6} \,\, J/s \times 3600s = 9.9 \times 10^{12}J](https://tex.z-dn.net/?f=2750%20%5Ctimes%2010%5E%7B6%7D%20%5C%2C%5C%2C%20J%2Fs%20%5Ctimes%203600s%20%3D%209.9%20%5Ctimes%2010%5E%7B12%7DJ)
![Energy \,\, required = \frac{9.9 \times 10^{22}}{1056}= 9.375 \times 10^{9} BTU](https://tex.z-dn.net/?f=Energy%20%5C%2C%5C%2C%20required%20%3D%20%5Cfrac%7B9.9%20%5Ctimes%2010%5E%7B22%7D%7D%7B1056%7D%3D%209.375%20%5Ctimes%2010%5E%7B9%7D%20BTU)
![Mass\,of\, coal\,required = \frac{9.375 \times 10^{9}}{14,000}=6.69 \times 10^{5}pounds](https://tex.z-dn.net/?f=Mass%5C%2Cof%5C%2C%20coal%5C%2Crequired%20%3D%20%5Cfrac%7B9.375%20%5Ctimes%2010%5E%7B9%7D%7D%7B14%2C000%7D%3D6.69%20%5Ctimes%2010%5E%7B5%7Dpounds)
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](https://tex.z-dn.net/?f=6.69%20%5Ctimes%2010%5E%7B5%7Dpounds%20%5C%2Ccoal%20%5Ctimes%200.02%20%3D%201.338%20%5Ctimes%2010%5E%7B4%7Dpound)
Therefore, released sulfur is
.
<u>Part -B;</u>
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](https://tex.z-dn.net/?f=2%20%5Ctimes%201.338%20%5Ctimes%2010%5E%7B4%7Dpound%20%3D%202.676%5Ctimes%2010%5E%7B4%7Dpound%2Fhour)
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](https://tex.z-dn.net/?f=%281-0.8%29%20%5Ctimes%20SO_%7B2%7D%20%5C%2Cproduced%20%3D0.2%20%5Ctimes%202.676%20%5Ctimes%2010%5E%7B4%7DPounds%2Fhr%3D%205.352%20%5Ctimes%2010%5E%7B3%7DPounds%2Fhr)
Energy produced in an hour = ![400 mw \times 1 hr = 1.1 \times 10^{6}kwh](https://tex.z-dn.net/?f=400%20mw%20%5Ctimes%201%20hr%20%3D%201.1%20%5Ctimes%2010%5E%7B6%7Dkwh)
![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](https://tex.z-dn.net/?f=SO_%7B2%7D%5C%2Creleased%20%5C%2C%20as%20%5C%2C%20per%20EPA%20%3D%20%5Cfrac%7B5.352%20%5Ctimes%2010%5E%7B3%7D%20%5C%2Cpounds%7D%7B1.1%20%5Ctimes%2010%5E%7B6%7Dkwh%7D%3D%204.865%20%5Ctimes%2010%5E%7B-3%7Dp%5Ckwh)
Therefore,
released of per EPA norms is
.