Determine the construction costs of coal.
Construction costcoal=[(the amount of elelctricity generated by a plant)×(construction cost for coal power plant)] (I)
Determine the construction costs of IGCC.
Construction costIGCC=[(the amount of elelctricity generated by a plant)×(construction cost for IGCC power plant)] (II)
Determine the construction costs difference of coal and IGCC.
Construction cost diference=[(Construction costcoal)−(Construction costIGCC)] (III)
Determine the amount of electricity produced by either plant in 5 years.
We=W˙Δt (IV)
Here, the power generation capacity of power plant is W˙ and the rate of time is Δt.
Determine the amount of fuel needed to generate a specified amount of power.
η=WeQinη=We(mfuel)×HVmfuel=Weη×HV (V)
Here, the efficiency of plant is η, the mass of the fuel is mfuel, and the heating value of coal is HV.
Determine the mass difference between plants.
Δmcoal=mcoal,coal plant−mcoal,IGCC plant (VI)
Here, the mass of the coal in coal plant is mcoal,coal plant and the mass of coal in the IGCC plant is mcoal,IGCC plant.
Determine the unit cost of coal per ton used in the power plant.
unit costcoal=ΔcostΔmcoal . (VII)
Here, the construction cost difference is Δcost.
Conclusion:
Substitute 150,000,000 kW for the amount of electricity generated by a plant and $1300/kW for construction cost for coal power plant in Equation (I).
Construction costcoal=(150,000,000 kW)×($1300/kW)=1.95×1011
Substitute 150,000,000 kW for the amount of electricity generated by a plant and $1500/kW for construction cost for coal power plant in Equation (II).
Construction costcoal=(150,000,000 kW)×($1500/kW)=2.25×1011
Substitute 2.25×1011 for construction cost of coal and 1.95×1011 for construction cost of IGCC in Equation (III).
Construction cost diference=(2.25×1011)−(1.95×1011)=3×1010
Substitute 150,000,000 kW for the amount of electricity generated by a plant and 5 year for Δt in Equation (IV).
We=(150,000,000 kW)×(5 year)=(150,000,000 kW)×(5 year)×(356 days1 year)(24 hours1 day)=6.570×1012 kWh
For coal plant,
Substitute 6.570×1012 kWh for We, 0.4 for ηcoal, and 28×106 kJ/ton for HV in Equation (V)
mcoal=(6.570×1012 kWh)(0.4)×(28×106 kJ/ton)=(6.570×1012 kWh)(0.4)×(28×106 kJ/ton)×(3600 kJ1 kWh)=2.1117×109 tons≅2.112×109 tons
For IGCC plant,
Substitute 6.570×1012 kWh for We, 0.48 for ηcoal, and 28×106 kJ/ton for HV in Equation (V)
mIGCC=(6.570×1012 kWh)(0.48)×(28×106 kJ/ton)=(6.570×1012 kWh)(0.48)×(28×106 kJ/ton)×(3600 kJ1 kWh)=1.7598×109 tons≅1.760×109 tons
Substitute 2.112×109 tons for mcoal and 1.760×109 tons for mIGCC in Equation (VI).
Δmcoal=(2.112×109 tons)−(1.760×109 tons)=0.352×109 tons
Substitute 3×1010 for Δcost and 0.352×109 tons for Δmcoal in Equation (VII).
unit costcoal=(3×1010)(0.352×109 tons)=$85.2/ton
Thus, the price of coal that will enable the IGCC plant to recover their cost difference from fuel saving in 5 year is $85.2/ton_.