EBK THERMODYNAMICS: AN ENGINEERING APPR
EBK THERMODYNAMICS: AN ENGINEERING APPR
8th Edition
ISBN: 9780100257054
Author: CENGEL
Publisher: YUZU
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Chapter 15.7, Problem 49P
To determine

The higher heating value of the coal.

The lower heating value of the coal.

Expert Solution & Answer
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Answer to Problem 49P

The higher heating value of the coal is 33,650 kJ/kg coal.

The lower heating value of the coal is 31,370 kJ/kg coal.

Explanation of Solution

Write the expression to calculate the number of moles of constituent (N) in a mixture.

N=mM (I)

Here, mass of the gas constituent is m and the molecular mass of the constitute is M.

Write the expression to calculate the total number of moles of gases (Nm) in products.

Nm=NA+NB+..............+NZ (II)

Here, number of moles of compound A is NA, number of moles of compound B is NB, and the number of moles of compound Z is NZ.

Write the expression to calculate the mole fraction of a constituent (y) in a mixture.

y=NNm (III)

Write the expression to calculate the mass of the mixture of gases (mtotal).

mtotal=NAMA+NBMB+.......+NZMZ (IV)

Here, number of moles of the compound A in the product is NA, the molecular weight of the compound A is MA, number of moles of compound B is NB, molecular weight of compound B is MB

Write the expression to calculate the apparent molecular weight of the product gas (Mm).

Mm=mmNm (V)

Here, mass of the product gas is mm.

Write the expression to calculate the heat transfer (q) for the combustion.

q=hC=HPHR

q=NPh¯f,PoNRh¯f,Ro (VI)

Here, enthalpy of combustion is hC, enthalpy of the products is HP, enthalpy of reactants is HR, number of moles of product gas is NP, enthalpy of formation of product gas is h¯f,Po, number of moles of reactants is NR, and the enthalpy of formation of reactants is h¯f,Ro.

Write the expression to calculate the heating value of coal.

HV=hCMm (VII)

Conclusion:

From the Table A-1 of “Molar mass, gas constants, and critical-point properties”, select the molar masses of carbon, hydrogen, oxygen, Sulphur, and air as,

MC=12kg/kmolMH2=2kg/kmolMO2=32kg/kmolMS=32kg/kmolMair=29kg/kmol

Consider for 100kg of coal. Substitute 67.40kg for mC and 12kg/kmol for MC in Equation (I).

NC=67.40kg12kg/kmol=5.617kmol

Substitute 5.31kg for mH2 and 2kg/kmol for MH2 in Equation (I).

NH2=5.31kg2kg/kmol=2.655kmol

Substitute 15.11kg for mO2 and 32kg/kmol for MO2 in Equation (I).

NO2=15.11kg32kg/kmol=0.4722kmol

Substitute 1.44kg for mN2 and 28kg/kmol for MN2 in Equation (I).

NN2=1.44kg28kg/kmol=0.05143kmol

Substitute 2.36kg for mS and 32kg/kmol for MS in Equation (I).

NS=2.36kg32kg/kmol=0.07375kmol

Substitute 5.617kmol for NC, 2.655kmol for NH2, 0.4722kmol for NO2, 0.05143kmol for NN2, and 0.07375kmol for NS in Equation (II).

(Nm)reactants=(5.617kmol+2.655kmol+0.4722kmol+0.05143kmol+0.07375kmol)=8.869kmol

Substitute 5.617kmol for NC and 8.869kmol for Nm in Equation (III).

yC=5.617kmol8.869kmol=0.6333

Substitute 2.655kmol for NH2 and 8.869kmol for Nm in Equation (III).

yH2=2.655kmol8.869kmol=0.2994

Substitute 0.4722kmol for NO2 and 8.869kmol for Nm in Equation (III).

yO2=0.4722kmol8.869kmol=0.05323

Substitute 0.05143kmol for NN2 and 8.869kmol for Nm in Equation (III).

yN2=0.05143kmol8.869kmol=0.00580

Substitute 0.07375kmol for NS and 8.869kmol for Nm in Equation (III).

yS=0.07375kmol8.869kmol=0.00832

Write the chemical reaction equation for complete combustion.

{0.6333C+0.2994H2+0.05323O2+0.00580N2+0.00832S+ath(O2+3.76N2)}(xCO2+yH2O+zSO2+kN2) (VIII)

Balance for the Carbon from Combustion reaction Equation (VIII).

x=0.6333

Balance for the Hydrogen from Combustion reaction Equation (VIII).

y=0.2994

Balance for the Sulphur from Combustion reaction Equation (VIII).

z=0.00832

Balance for the Oxygen from Combustion reaction Equation (VIII).

0.05323+ath=x+0.5y+z

Substitute 0.6333 for x, 0.2994 for y, and 0.00832 for z.

0.05323+ath=0.6333+0.5(0.2994)+0.00832ath=0.7381

Balance for the Nitrogen from Combustion reaction Equation (VIII).

k=0.00580+3.76ath

Substitute 0.7381 for ath.

k=0.00580+3.76(0.7381)k=2.781

Rewrite the complete balanced chemical reaction for combustion as follows:

{0.6333C+0.2994H2+0.05323O2+0.00580N2+0.00832S+0.7381(O2+3.76N2)}(0.6333CO2+0.2994H2O+0.00832SO2+2.781N2) (IX)

Substitute 0.6333kmol for NC, 0.2994kmol for NH2, 0.00832kmol for NS, 0.05323kmol for NO2, and 0.00580kmol for NN2 in Equation (II).

(Nm)reactants=(0.6333kmol+0.2994kmol+0.00832kmol+0.05323kmol+0.000580kmol)1.0kmol

Substitute 0.6333kmol for NC, 0.2994kmol for NH2, 0.00832kmol for NS, 0.05323kmol for NO2, 0.00580kmol for NN2, 12kg/kmol for MC, 2kg/kmol for MH2, 32kg/kmol for MO2, 28kg/kmol for MN2, and 32kg/kmol for MS in Equation (IV).

mtotal={(0.6333kmol)12kg/kmol+(0.2994kmol)2kg/kmol+(0.05323kmol)32kg/kmol+(0.00580kmol)28kg/kmol+(0.00832kmol)32kg/kmol}=10.3304kg

Substitute 10.3304kg for mm and 1kmol for Nm in Equation (V).

Mm=10.3304kg1kmol=10.33kg/kmol

From the Table A-26 of “Enthalpy of formation, Gibbs function of formation, and the absolute entropy at 25°C, 1 atm”, obtain the enthalpy of formations as,

(h¯fo)N2=0(h¯fo)CO2=393,520 kJ/kmol(h¯fo)H2O,l=285,830 kJ/kmol(h¯fo)H2O,g=241,820 kJ/kmol(h¯fo)SO2=297,100 kJ/kmol

Here, the enthalpy of formation of N2 is (h¯fo)N2, the enthalpy of formation of CO2 is (h¯fo)CO2, the enthalpy of formation for H2O in vapor is (h¯fo)H2O,g, the enthalpy of formation for H2O in liquid is (h¯fo)H2O,l, and the enthalpy of formation of SO2 is (h¯fo)SO2.

Rewrite the Equation (VI) for the higher heating value.

hC=NCO2(h¯fo)CO2+NH2O(h¯fo)H2O,l+NSO2(h¯fo)SO2+NN2(h¯fo)N2 (X)

Substitute 0.6333kmol for NCO2, 393,520 kJ/kmol for (h¯fo)CO2, 285,830 kJ/kmol for (h¯fo)H2O,l, 0 for (h¯fo)N2, 0.2994kmol for NH2O, 297,100 kJ/kmol for (h¯fo)SO2, and 0.00832kmol for NSO2 in Equation (X).

hC={0.6333kmol(393,520 kJ/kmol)+0.2994kmol(285,830 kJ/kmol)+0.00832kmol(297,100 kJ/kmol)+NN2(0)}hC=337,270kJ/kmol coal

Substitute 10.33kg/kmol for Mm and 337,270kJ/kmol coal for hC in Equation (VII).

HHV=(337,270kJ/kmol coal)10.33kg/kmol33,650 kJ/kg coal

Thus, the higher heating value of the coal is 33,650 kJ/kg coal.

Rewrite the Equation (VI) for the higher heating value.

hC=NCO2(h¯fo)CO2+NH2O(h¯fo)H2O,g+NSO2(h¯fo)SO2+NN2(h¯fo)N2 (XI)

Substitute 0.6333kmol for NCO2, 393,520 kJ/kmol for (h¯fo)CO2, 241,820 kJ/kmol for (h¯fo)H2O,g, 0 for (h¯fo)N2, 0.2994kmol for NH2O, 297,100 kJ/kmol for (h¯fo)SO2, and 0.00832kmol for NSO2 in Equation (XI).

hC={0.6333kmol(393,520 kJ/kmol)+0.2994kmol(241,820 kJ/kmol)+0.00832kmol(297,100 kJ/kmol)+NN2(0)}hC=324,090kJ/kmol coal

Substitute 10.33kg/kmol for Mm and 324,090kJ/kmol coal for hC in Equation (VII).

LHV=(324,090kJ/kmol coal)10.33kg/kmol31,370 kJ/kg coal

Thus, the lower heating value of the coal is 31,370 kJ/kg coal.

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Chapter 15 Solutions

EBK THERMODYNAMICS: AN ENGINEERING APPR

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