Answer :
Answer:
Part -A:
Sulfur released is [tex]1.338 \times 10^{4}pounds/hour[/tex].
Part-B:
[tex]SO_{2}[/tex] released of per EPA norms is [tex]4.865 /times 10^{-3}P/kw[/tex].
Explanation:
From the given,
The output = 1100 Mw
Efficiency = η = 0.4
Substitute the values in the following
Input = Output / η
[tex]= \frac{1100}{0.4}= 2750 Mw[/tex]
Let's converts between joules to BTU.
1 BTU = 1056 J
Part-A;
[tex]Energy\,\, required = Power \times time[/tex]
[tex]2750 \times 10^{6} \,\, J/s \times 3600s = 9.9 \times 10^{12}J[/tex]
[tex]Energy \,\, required = \frac{9.9 \times 10^{22}}{1056}= 9.375 \times 10^{9} BTU[/tex]
[tex]Mass\,of\, coal\,required = \frac{9.375 \times 10^{9}}{14,000}=6.69 \times 10^{5}pounds[/tex]
But it has 2% sulfur
The mass of sulfur released
[tex]6.69 \times 10^{5}pounds \,coal \times 0.02 = 1.338 \times 10^{4}pound[/tex]
Therefore, released sulfur is [tex]1.338 \times 10^{4}pounds/hour[/tex].
Part -B;
One pound of sulfur produce two pounds of sulfur dioxide
Initial amount of produced sulfur =
[tex]2 \times 1.338 \times 10^{4}pound = 2.676\times 10^{4}pound/hour[/tex]
Assuming we added a 80% efficiency then,
Released sulfur dioxide = [tex](1-0.8) \times SO_{2} \,produced =0.2 \times 2.676 \times 10^{4}Pounds/hr= 5.352 \times 10^{3}Pounds/hr[/tex]
Energy produced in an hour = [tex]400 mw \times 1 hr = 1.1 \times 10^{6}kwh[/tex]
[tex]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[/tex]
Therefore, [tex]SO_{2}[/tex] released of per EPA norms is [tex]4.865 /times 10^{-3}P/kw[/tex].