THERMODYNAMICS (LL)-W/ACCESS >CUSTOM<
THERMODYNAMICS (LL)-W/ACCESS >CUSTOM<
9th Edition
ISBN: 9781266657610
Author: CENGEL
Publisher: MCG CUSTOM
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Chapter 15.7, Problem 38P
To determine

The mass fraction and the apparent molecular weight of the products and the mass of air required per unit mass of fuel burned.

Expert Solution & Answer
Check Mark

Answer to Problem 38P

The mass fraction of carbon dioxide is 0.2441, carbon monoxide is 0.0082, water is 0.0595, sulphur dioxide is 0.0032, nitrogen is 0.6849, the apparent molecular weight of the products is 29.71kg/kmol and the mass of air required per unit mass of fuel burned is 8.27kgair/kgfuel.

Explanation of Solution

Express the total mass of the coal when the ash is substituted.

mtotal=100mash (I)

Here, mass of ash is mash.

Express the mass fraction of carbon.

mfC=mCmtotal (II)

Here, mass of carbon is mC.

Express the mass fraction of hydrogen.

mfH2=mH2mtotal (III)

Here, mass of hydrogen is mH2.

Express the mass fraction of oxygen.

mfO2=mO2mtotal (IV)

Here, mass of oxygen is mO2.

Express the mass fraction of nitrogen.

mfN2=mN2mtotal (V)

Here, mass of nitrogen is mN2.

Express the mass fraction of sulphur.

mfS=mSmtotal (VI)

Here, mass of sulphur is mS.

Express the number of moles of carbon.

NC=mfCMC (VII)

Here, molar mass of carbon is MC.

Express the number of moles of hydrogen.

NH2=mfH2MH2 (VIII)

Here, molar mass of hydrogen is MH2.

Express the number of moles of oxygen.

NO2=mfO2MO2 (IX)

Here, molar mass of oxygen is MO2.

Express the number of moles of nitrogen.

NN2=mfN2MN2 (X)

Here, molar mass of nitrogen is MN2.

Express the number of moles of sulphur.

NS=mfSMS (XI)

Here, molar mass of sulphur is MS.

Express the total number of moles.

Nm=NC+NH2+NO2+NN2+NS (XII)

Express the mole fraction of carbon.

yC=NCNm (XIII)

Express the mole fraction of hydrogen.

yH2=NH2Nm (XIV)

Express the mole fraction of oxygen.

yO2=NO2Nm (XV)

Express the mole fraction of nitrogen.

yN2=NN2Nm (XVI)

Express the mole fraction of sulphur.

yS=NSNm (XVII)

Express the total molar mass of the products.

mtotal=NCO2MCO2+NCOMCO+NH2OMH2O+NSO2MSO2+NN2MN2 (XVIII)

Here, number of moles of carbon dioxide, carbon monoxide, water, sulphur dioxide, and nitrogen is NCO2,NCO,NH2O,NSO2andNN2 respectively, molar masses of carbon dioxide, carbon monoxide, water, sulphur dioxide, and nitrogen is MCO2,MCO,MH2O,MSO2andMN2 respectively.

Express the mole fraction of carbon dioxide.

mfCO2=NCO2MCO2mtotal (XIX)

Here, molar mass of carbon dioxide is MCO2.

Express the mole fraction of carbon monoxide.

mfCO=NCOMCOmtotal (XX)

Here, molar mass of carbon monoxide is MCO.

Express the mole fraction of water.

mfH2O=NH2OMH2Omtotal (XXI)

Here, molar mass of water is MH2O.

Express the mole fraction of sulphur dioxide.

mfSO2=NSO2MSO2mtotal (XXII)

Here, molar mass of sulphur dioxide is MSO2.

Express the mole fraction of nitrogen.

mfN2=NN2MN2mtotal (XXIII)

Here, molar mass of nitrogen is MN2.

Express the total number of moles of product.

Nm,P=NCO2+NCO+NH2O+NSO2+NN2 (XXIV)

Express the apparent molecular weight of the product gas.

Mm=mtotalNm,P (XXV)

Express the air-fuel mass ratio.

FA=0.6403NairMairNCMC+NH2MH2+NO2MO2+NN2MN2+NSMS (XXVI)

Conclusion:

Refer Table A-1, “molar mass, gas constant, and the critical point properties”, and write the molar masses.

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

Mair=29kg/kmolMN2=28kg/kmol

Here, molar mass of air is Mair.

Substitute 5 for mash in Equation (I).

mtotal=1005=95kg

Substitute 61.40kg for mC and 95kg for mtotal in Equation (II).

mfC=61.40kg95kg=0.6463=64.63kg

Substitute 5.79kg for mH2 and 95kg for mtotal in Equation (III).

mfH2=5.79kg95kg=0.06095=6.095kg

Substitute 25.31kg for mO2 and 95kg for mtotal in Equation (IV).

mfO2=25.31kg95kg=0.2664=26.64kg

Substitute 1.09kg for mN2 and 95kg for mtotal in Equation (V).

mfN2=1.09kg95kg=0.01147=1.147kg

Substitute 1.41kg for mS and 95kg for mtotal in Equation (VI).

mfS=1.41kg95kg=0.01484=1.484kg

Substitute 64.63kg for mC and 12kg/kmol for MC in Equation (VII).

NC=64.63kg12kg/kmol=5.386kmol

Substitute 6.095kg for mH2 and 2kg/kmol for MH2 in Equation (VIII).

NH2=6.095kg2kg/kmol=3.048kmol

Substitute 26.64kg for mO2 and 32kg/kmol for MO2 in Equation (IX).

NO2=26.64kg32kg/kmol=0.8325kmol

Substitute 1.147kg for mN2 and 28kg/kmol for MN2 in Equation (X).

NN2=1.147kg28kg/kmol=0.04096kmol

Substitute 1.484kg for mS and 32kg/kmol for MS in Equation (XI).

NS=1.484kg32kg/kmol=0.04638kmol

Substitute 5.386kmol for NC, 3.048kmol for NH2, 0.8325kmol for NO2, 0.04096kmol for NN2 and 0.04638kmol for NS in Equation (XII).

Nm=5.386kmol+3.048kmol+0.8325kmol+0.04096kmol+0.04638kmol=9.354kmol

Substitute 5.386kmol for NC and 9.354kmol for Nm in Equation (XIII).

yC=5.386kmol9.354kmol=0.5758

Substitute 3.048kmol for NH2 and 9.354kmol for Nm in Equation (XIV).

yH2=3.048kmol9.354kmol=0.3258

Substitute 0.8325kmol for NO2 and 9.354kmol for Nm in Equation (XV).

yO2=0.8325kmol9.354kmol=0.0890

Substitute 0.04096kmol for NN2 and 9.354kmol for Nm in Equation (XVI).

yN2=0.04096kmol9.354kmol=0.00438

Substitute 0.04638kmol for NS and 9.354kmol for Nm in Equation (XVII).

yS=0.04638kmol9.354kmol=0.00496

Express the combustion equation.

[0.5758C+0.3258H2+0.0890O2+0.00438N2+0.00496S+ath(O2+3.76N2)x(0.95CO2+0.05CO)+yH2O+zSO2+kN2] (XXVII)

Perform the species balancing:

Carbon balance:

x=0.5758

Hydrogen balance:

y=0.3258

Sulphur balance:

z=0.00496

Oxygen balance:

0.0890+ath=0.95x+(0.5×0.05x)+0.5y+z0.0890+ath=0.95(0.5758)+(0.5×0.05×0.5758)+0.5(0.3258)+0.004960.0890+ath=0.72925ath=0.6403

Nitrogen balance:

k=0.00438+3.76ath=0.00438+3.76(0.6403)=2.412

Substitute 0.5758 for x, 0.3258 for y, 0.00496 for z, 0.6403 for ath and 2.412 for k in Equation (XXVII).

[0.5758C+0.3258H2+0.0890O2+0.00438N2+0.00496S+0.6403(O2+3.76N2)0.5470CO2+0.0288CO+0.3258H2O+0.00496SO2+2.412N2] (XXVIII)

Refer Equation (XXVIII), and write the number of moles of products.

NCO2=0.5470kmolNCO=0.0288kmolNH2O=0.3258kmol

NSO2=0.00496kmolNN2=2.412kmol

Refer Table A-1, “molar mass, gas constant, and the critical point properties”, and write the molar masses.

MCO2=44kg/kmolMCO=28kg/kmolMH2O=18kg/kmol

MSO2=64kg/kmolMN2=28kg/kmol

Substitute 0.5470kmol for NCO2, 0.0288kmol for NCO, 0.3258kmol for NH2O, 0.00496kmol for NSO2, 2.412kmol for NN2, 44kg/kmol for MCO2, 28kg/kmol for MCO, 18kg/kmol for MH2O, 64kg/kmol for MSO2 and 28kg/kmol for MN2 in Equation (XVIII).

mtotal=[(0.5470kmol)(44kg/kmol)+(0.0288kmol)(28kg/kmol)+(0.3258kmol)(18kg/kmol)+(0.00496kmol)(64kg/kmol)+(2.412kmol)(28kg/kmol)]=98.6kg

Substitute 0.5470kmol for NCO2, 44kg/kmol for MCO2 and 98.6kg for mtotal in Equation (XIX).

mfCO2=(0.5470kmol)(44kg/kmol)98.6kg=0.2441

Hence, the mass fraction of carbon dioxide is 0.2441.

Substitute 0.0288kmol for NCO, 28kg/kmol for MCO and 98.6kg for mtotal in Equation (XX).

mfCO=(0.0288kmol)(28kg/kmol)98.6kg=0.0082

Hence, the mass fraction of carbon monoxide is 0.0082.

Substitute 0.3258kmol for NH2O, 18kg/kmol for MH2O and 98.6kg for mtotal in Equation (XXI).

mfH2O=(0.3258kmol)(18kg/kmol)98.6kg=0.0595

Hence, the mass fraction of water is 0.0595.

Substitute 0.00496kmol for NSO2, 64kg/kmol for MSO2 and 98.6kg for mtotal in Equation (XXII).

mfSO2=(0.00496kmol)(64kg/kmol)98.6kg=0.0032

Hence, the mass fraction of sulphur dioxide is 0.0032.

Substitute 2.412kmol for NN2, 28kg/kmol for MN2 and 98.6kg for mtotal in Equation (XXIII).

mfN2=(2.412kmol)(28kg/kmol)98.6kg=0.6849

Hence, the mass fraction of nitrogen is 0.6849.

Substitute 0.5470kmol for NCO2, 0.0288kmol for NCO, 0.3258kmol for NH2O, 0.00496kmol for NSO2 and 2.412kmol for NN2 in Equation (XXIV).

Nm,P=0.5470kmol+0.0288kmol+0.3258kmol+0.00496kmol+2.412kmol=3.319kmol

Substitute 98.6kg for mtotal and 3.319kmol for Nm,P in Equation (XXV).

Mm=98.6kg3.319kmol=29.71kg/kmol

Hence, the apparent molecular weight of the products is 29.71kg/kmol.

Since each kmol of O2 in air is accompanied by 3.76kmol of N2, thus

Nair=1kmolO2+3.76kmolN2=4.76kmolair

Refer Equation (XXVIII), and write the number of moles of reactants.

NC=0.5758kmolNH2=0.3258kmolNO2=0.0890kmol

NN2=0.00438kmolNS=0.00496kmol

Substitute 4.76kmolair for Nair, 29kg/kmol for Mair, 0.5758kmol for NC, 0.3258kmol for NH2, 0.0890kmol for NO2, 0.00438kmol for NN2, 0.00496kmol for NS, 12kg/kmol for MC, 2kg/kmol for MH2, 32kg/kmol for MO2, 32kg/kmol for MS and 28kg/kmol for MN2 in Equation (XXVI).

FA=0.6403(4.76kmolair)(29kg/kmol)[(0.5758kmol)(12kg/kmol)+(0.3258kmol)(2kg/kmol)+(0.0890kmol)(32kg/kmol)+(0.00438kmol)(28kg/kmol)+(0.00496kmol)(32kg/kmol)]=88.39kg10.69kg=8.27kgair/kgfuel

Hence, the mass of air required per unit mass of fuel burned is 8.27kgair/kgfuel.

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Chapter 15 Solutions

THERMODYNAMICS (LL)-W/ACCESS >CUSTOM<

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