In: Finance
rom Rubin, Problem 13.3: A 120. MW coal
plant wants to add a SOX emissions control system to reduce its emissions. The emission
control system has a capital cost of $815 per kW and reduces SOX emissions by 98% to
0.0918 kg/MWh. Assume the life of the emission control system is 20 years and the power
plant produces 650 GWh/y.
a)What is the annualized cost of this system in dollars per year if the discount rate (interest
rate) is 7.0 percent/year? What if the discount rate is 4.0 percent/year?
b)What is the cost per 1.0399 x 10^4kWh of electricity usage (the average annual U.S.
residential electricity consumption in 2017 according to the EIA) at each of the discount
rates specified in part (a)?
c) What is the cost per metric ton of SOXremoved?
(1 metric ton = 1,000 kg = 10^6g)
Solution:
We know that one GWH = 1000000 KWH
Power plant produces 650 GWH/y i.e. it produces 650 GW per year hourly.
Converting the same to KWH we get = 650 X 1000000 = 650,000,000
Cost of the emission control system is $815 per KW
Total Annualized cost will be 815 X 650,000,000 = 529,750,000,000 $ --- 1)
The same cost is an yearly cost and should be incurred after the plant is through with the yearly production so to find the cost today we will be discounting the same.
a) Annualized cost at a discount rate of 7% if incurred will be : Total cost / (1+7%)^1
i.e. 529,750,000,000 / (1.07) = 495,093,457,900 $
Cost at 4% = 529,750,000,000 / 1.04 = 509,375,000,000 $
b) Cost of 650,000,000 kWh = 529,750,000,000
Cost of 1 KWH = 815 KWh
Cost per 10399kwh = 815 X 10399 = 8475185
@ 7% = 8475185 / (1.07) = 7920,733.645 $
@ 4% = 8475185/1.04 = 8149216.346
C) Since 98% of the SOX emmisions are removed it amounts to be :
x - 98% of x = .00918
x = 4.59 kg/Mwh and the removed proportion = 98% of x = .98 * 4.59 = 4.49 Kg/Mwh
Cost of removing SOX from one MW = 815/1000 = .815
4.49 Kg removed in .815 $, 1000 kg will be removed in $= (.815/4.49)*1000 = 181.51$
So the cost of removing per metric ton of SOX emissions is = 181.5$