Consider a 28% efficient 500 MW power plant burning 10,000 Btu/lb coal containin

Question

Consider a 28% efficient 500 MW power plant burning 10,000 Btu/lb coal containing 60% carbon and 2% sulfur by mass (a) What is the rate of sulfur production at the power plant in lbs S/hr? (b) Suppose you install a caustic scrubber with SO_2 removal efficiency of 80%, how many lbs/hr of SO_2 are released? (c) Given that the SO_2 emissions are limited to 0.6 lbs SO_2 per million Btu of heat input, does the plant meet the EPA standards? If not, what would you suggest as the new efficiency for the caustic scrubber? (d) How much cooling water would be required (in m^3/s) to carry away the waste heat at the power plant if only 10% of waste heat escaped with stack gases, and we allowed the water temperature of the cooling water to increase in temperature by 12 degree C? (e) Compare the carbon emissions from the coal-fired power plant with those estimated as 0.0206 g C/kJ for petroleum in Problem 2.7, and as 0.0134 g C/kJ for methane in Example 2.3. How would you rank the three fossil fuels from cleanest to dirtiest based on carbon emissions?

Explanation / Answer

a)

Energy in 1 hour = 500 MW = 500*10^6 J/s * 3600 s /h = 1.8*10^12 J/h

change to BTU

1 BTU = 1055 J

1.8*10^12 J/1055 J BTU = 1706161137.44 BTU/h

if 28% efficient

Total BTU reuqired --> 1706161137.44/0.28 = 6093432633.71 BTU/h at 28%

Total Coal = 6093432633.71 BTU/ 10000 BTU/lb coal = 609343.26 lb coal required

of which

Mass of S = 2%/100% * 609343.26 = 12186.8652 lb of S per hour

b)

if eff 80% how much SO2 is released

12186.8652 lb of S per h

lbmol = 12186.8652 /32 = 380.839 lbmol of S

1:1 ratio of S:SO2 so

380.839 lbmol of S = 380.839 lbmol of SO2

now.. 20% escapes

380.839 *0.20 = 76.1678 lbmol of SO2

MW of S02 = 64.066 lbmol/lb

mass = 64.066 *76.1678 = 4879.7 lb of SO2

c)

if 0.6 lb per SO2 / 10^6 BTU --> 0.6*10^-6

Our rate:

mass / Energy in = 4879.7 / 6093432633.71 = 8.00812*10^-7

this is correct... since

0.6*10^-6 > 8.00812*10^-7 so we are not exceeding

d)

assume

(1-0.1)*6093432633.71 = 5484089370.34 BTU are lost due to heat

change to J --> 5484089370.34 *1055 = 5.78571*10^12 J/h

change to s

5.78571*10^12 J/h * 1h /3600s = 1607141666.67J/s

now..

Q = m*C*(Tf-Ti)

1607141666.67= m * 4.184*(12)

m = 1607141666.67/(4.184*12) = 32009673.0933 grams = 32009.673 kg/s

1 kg = 1 m3 s

V =  32009.673 m3/s

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