Please to problem 4 HOME INSERT DESIGN Read Print Web Mode Layout Layout PAGE 1
ID: 1715687 • Letter: P
Question
Please to problem 4
HOME INSERT DESIGN Read Print Web Mode Layout Layout PAGE 1 OF 1 WORDS PICTURETOOLS test1 [Read-Onlyl [Compatibility Model Word PAGE LAYOUT REFERENCES MAILINGS REVIEW VIEW FORMAT One Page EE Multiple Pages Zoom 100% Switch Macros New Arrange Split page width Navigation Pane Window All Reset Window Po e If the energy content in coal is 24,000kJ/kg, what would be the carbon emission rate and CO2 emission rate. 4. (25) Consider two power plants with Fixed Rate Charge 30.1 and he following parameters Capital cost SkW Variable cost $/kWh Capacity Factor 0.3 Combustion turbine 600 0.05 2000 0.02 0.9 Coal steam. a) What should be the price of energy for both if operated separately? b) Draw on one plot the screening cost graph for each plant c What is the optimal mix of two plants e. for what capacity factor (and hour r operation) is coal steam and combustion turbine plants more efficient? lyou can do it algebraically or graphically) Extra credit (5p) For the load duration curve on the right, how many MW of power of each plant should utility have in the optimal mix? you can do it algebraically or graphically) Microsoft accountExplanation / Answer
If Installed capaity is of 1 kW each and time considered one year of operation. one year has 365 days. Based on capacity factor of each plant we can obtain average load on the system seperately.
Capacity factor = Average demand / Rated Capacity.
for Plant 1 :
Installed Capacity = 1 kW, Capital Cost = 600 $/kW
Combustion Turbine
Capacity factor is equal to 0.3 , there fore Average load = 0.3*1 = 300 W.
No.of units generated per year = 300*24(hours/day)*365(days/year) = 2628 kWh/year.
Fixed rate charge per year is 0.1*600 = 60 $/kW.
Running Cost = 0.05*2628 = 131.4 $/kWh.
Total Cost of Plant = Fixed Charge + Running Cost = (60 + 131.4 ) $/year = 191.4 $/year
Cost of one unit of energy generation = Total cost / No.of Units of Energy Utilized = 191.4/2628 = 0.0728 $/kWh.
for Plant 2 :
Installed Capacity = 1 kW, Capital Cost = 2000 $/kW
Coal Steam
Capacity factor is equal to 0.9 , there fore Average load = 0.9*1 = 900 W.
No.of units generated per year = 900*24(hours/day)*365(days/year) = 7884 kWh/year.
Fixed rate charge per year is 0.1*2000 = 200 $/kW.
Running Cost = 0.02*7884 = 157.68 $/kWh.
Total Cost of Plant = Fixed Charge + Running Cost = (200 + 157.68 ) $/year = 357.68 $/year
Cost of one unit of energy generation = Total cost / No.of Units of Energy Utilized = 357.68/7884 = 0.0453 $/kWh.