Thermodynamics: An Engineering Approach
Thermodynamics: An Engineering Approach
9th Edition
ISBN: 9781259822674
Author: Yunus A. Cengel Dr., Michael A. Boles
Publisher: McGraw-Hill Education
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Question
Chapter 6.11, Problem 27P
To determine

The price of coal that will enable the IGCC plant to recover their cost difference from fuel saving in 5 year.

Expert Solution & Answer
Check Mark

Answer to Problem 27P

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_.

Explanation of Solution

Determine the construction costs of coal.

Constructioncostcoal=[(the amount of elelctricity generatedby a plant)×(constructioncostforcoalpowerplant)] (I)

Determine the construction costs of IGCC.

ConstructioncostIGCC=[(the amount of elelctricity generatedby a plant)×(constructioncostforIGCCpowerplant)] (II)

Determine the construction costs difference of coal and IGCC.

Constructioncost diference=[(Constructioncostcoal)(ConstructioncostIGCC)] (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,coalplantmcoal,IGCCplant (VI)

Here, the mass of the coal in coal plant is mcoal,coalplant and the mass of coal in the IGCC plant is mcoal,IGCCplant.

Determine the unit cost of coal per ton used in the power plant.

unitcostcoal=ΔcostΔmcoal . (VII)

Here, the construction cost difference is Δcost.

Conclusion:

Substitute 150,000,000kW for the amount of electricity generated by a plant and $1300/kW for construction cost for coal power plant in Equation (I).

Constructioncostcoal=(150,000,000kW)×($1300/kW)=1.95×1011

Substitute 150,000,000kW for the amount of electricity generated by a plant and $1500/kW for construction cost for coal power plant in Equation (II).

Constructioncostcoal=(150,000,000kW)×($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).

Constructioncost diference=(2.25×1011)(1.95×1011)=3×1010

Substitute 150,000,000kW for the amount of electricity generated by a plant and 5 year for Δt in Equation (IV).

We=(150,000,000kW)×(5year)=(150,000,000kW)×(5year)×(356days1year)(24hours1day)=6.570×1012kWh

For coal plant,

Substitute 6.570×1012kWh for We, 0.4 for ηcoal, and 28×106kJ/ton for HV in Equation (V)

mcoal=(6.570×1012kWh)(0.4)×(28×106kJ/ton)=(6.570×1012kWh)(0.4)×(28×106kJ/ton)×(3600kJ1kWh)=2.1117×109tons2.112×109tons

For IGCC plant,

Substitute 6.570×1012kWh for We, 0.48 for ηcoal, and 28×106kJ/ton for HV in Equation (V)

mIGCC=(6.570×1012kWh)(0.48)×(28×106kJ/ton)=(6.570×1012kWh)(0.48)×(28×106kJ/ton)×(3600kJ1kWh)=1.7598×109tons1.760×109tons

Substitute 2.112×109tons for mcoal and 1.760×109tons for mIGCC in Equation (VI).

Δmcoal=(2.112×109tons)(1.760×109tons)=0.352×109tons

Substitute 3×1010 for Δcost and 0.352×109tons for Δmcoal in Equation (VII).

unitcostcoal=(3×1010)(0.352×109tons)=$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_.

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Chapter 6 Solutions

Thermodynamics: An Engineering Approach

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