The heat transfer of fuel in the boiler.

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Answer 1

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Chapter 15.7, Problem 58P

To determine

The heat transfer of fuel in the boiler.

Expert Solution & Answer

Answer to Problem 58P

The heat transfer of fuel in the boiler is $23,022 kJ/kg fuel$ .

Explanation of Solution

Express the number of moles of carbon.

$NC=mfCMC$ (I)

Here, molar mass of carbon is $MC$ and mole fraction of carbon is $mfC$ .

Express the number of moles of hydrogen.

$NH2=mfH2MH2$ (II)

Here, molar mass of hydrogen is $MH2$ and mole fraction of hydrogen is $mfH2$ .

Express the number of moles of oxygen.

$NO2=mfO2MO2$ (III)

Here, molar mass of oxygen is $MO2$ and mole fraction of oxygen is $mfO2$ .

Express the number of moles of nitrogen.

$NN2=mfN2MN2$ (IV)

Here, molar mass of nitrogen is $MN2$ and mole fraction of nitrogen is $mfN2$ .

Express the number of moles of sulfur.

$NS=mfSMS$ (V)

Here, molar mass of sulphur is $MS$ and mole fraction of sulfur is $mfS$ .

Express the total number of moles.

$Nm=NC+NH2+NO2+NN2+NS$ (VI)

Here, number of moles of carbon, hydrogen, oxygen, nitrogen and sulfur is $NC$ , $NH2$ , $NO2$ , $NN2$ and $NS$ respectively.

Express the mole fraction of carbon.

$yC=NCNm$ (VII)

Express the mole fraction of hydrogen.

$yH2=NH2Nm$ (VIII)

Express the mole fraction of oxygen.

$yO2=NO2Nm$ (IX)

Express the mole fraction of nitrogen.

$yN2=NN2Nm$ (X)

Express the mole fraction of sulfur.

$yS=NSNm$ (XI)

Express the heat transfer from the combustion chamber.

$−Q¯out=∑NP(h¯fo+h¯−h¯o)P−∑NR(h¯fo+h¯−h¯o)R=∑NP(h¯fo+h¯400 K−h¯298 Ko)P−∑NR(h¯fo+h¯400 K−h¯298 Ko)R$ (XII)

Here, number of moles of products is $NP$ , number of moles of reactants is $NR$ , enthalpy of formation is $h¯f$ .

Express the enthalpy change of sulfur dioxide between the standard temperature and the product temperature.

$Δh¯SO2=cpΔT=cp(T2−Tstand)$ (XIII)

Here, specific heat constant pressure is $cp$ net temperature is $ΔT$ , temperature at state 2 is $T2$ and temperature at standard condition is $Tstand$ .

Express the heat transfer of fuel in the boiler.

$Qout=Q¯outNCMC+NH2MH2+NO2MO2+NN2MN2+NSMS$ (XIV)

Here, number of moles of carbon, hydrogen, oxygen, nitrogen and sulfur is $MC, MH2$ , $MO2$ , $MN2 and MS$ respectively.

Conclusion:

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

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

$MN2=28 kg/kmol$

Substitute $39.25 kg$ for $mC$ and $12 kg/kmol$ for $MC$ in Equation (I).

$NC=39.25 kg12 kg/kmol=3.271 kmol$

Substitute $6.93 kg$ for $mH2$ and $2 kg/kmol$ for $MH2$ in Equation (II).

$NH2=6.93 kg2 kg/kmol=3.465 kmol$

Substitute $41.11 kg$ for $mO2$ and $32 kg/kmol$ for $MO2$ in Equation (III).

$NO2=41.11 kg32 kg/kmol=1.285 kmol$

Substitute $0.72 kg$ for $mN2$ and $28 kg/kmol$ for $MN2$ in Equation (IV).

$NN2=0.72 kg28 kg/kmol=0.0257 kmol$

Substitute $0.79 kg$ for $mS$ and $32 kg/kmol$ for $MS$ in Equation (V).

$NS=0.79 kg32 kg/kmol=0.0247 kmol$

Substitute $3.271 kmol$ for $NC$ , $3.465 kmol$ for $NH2$ , $1.285 kmol$ for $NO2$ , $0.0257 kmol$ for $NN2$ and $0.00247 kmol$ for $NS$ in Equation (VI).

$Nm=3.271 kmol+3.465 kmol+1.285 kmol+0.0257 kmol+0.0247 kmol=8.071 kmol$

Substitute $3.271 kmol$ for $NC$ and $8.071 kmol$ for $Nm$ in Equation (VII).

$yC=3.271 kmol8.071 kmol=0.4052$

Substitute $3.465 kmol$ for $NH2$ and $8.071 kmol$ for $Nm$ in Equation (VIII).

$yH2=3.465 kmol8.071 kmol=0.4293$

Substitute $1.285 kmol$ for $NO2$ and $8.071 kmol$ for $Nm$ in Equation (IX).

$yO2=1.285 kmol8.071 kmol=0.1592$

Substitute $0.0257 kmol$ for $NN2$ and $8.071 kmol$ for $Nm$ in Equation (X).

$yN2=0.0257 kmol8.071 kmol=0.00319$

Substitute $0.0247 kmol$ for $NS$ and $8.071 kmol$ for $Nm$ in Equation (XI).

$yS=0.0247 kmol8.071 kmol=0.00306$

Express the combustion equation.

$[{0.4052C+0.4293H2+0.1592O2+0.00319N2+0.00306S+1.4ath(O2+3.76N2)}→{0.4052CO2+0.4293H2O+0.4athO2+0.00306SO2+1.4ath×3.76N2}]$ (XV)

Here, stoichiometric coefficient of air is $ath$ , oxygen is $O2$ , nitrogen is $N2$ , carbon dioxide is $CO2$ , water is $H2O$ and hydrogen is $H2$ .

Express the stoichiometric coefficient of air by $O2$ balancing.

$0.1592+1.4ath=0.4052+0.5×0.4293+0.4ath+0.00306ath=0.4637$

Substitute $0.4637$ for $ath$ in Equation (XV).

$[{0.4052C+0.4293H2+0.1592O2+0.00319N2+0.00306S+1.4×0.4637(O2+3.76N2)}→{0.4052CO2+0.4293H2O+0.4×0.4637O2+0.00306SO2+1.4×0.4637×3.76N2}][{0.4052C+0.4293H2+0.1592O2+0.00319N2+0.00306S+0.6492(O2+3.76N2)}→{0.4052CO2+0.4293H2O+0.1855O2+0.00306SO2+2.441N2}]$ (XVI)

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

$MC=0.4052 kmolMH2=0.4293 kmolMO2=0.15920 kmol$

$MN2=0.00319 kmolMS=0.00306 kmol$

Perform unit conversion of temperature at state 2 from degree Celsius to Kelvin.

$T2=127°C=(127+273) K=400 K$

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

$NP,CO2=0.4052 kmolNP,H2O=0.4293 kmolNP,O2=0.1855 kmolNP,SO2=0.00306 kmol$

$NP,N2=2.441 kmol$

Here, number of moles of product carbon dioxide, water, oxygen, sulfur dioxide and nitrogen is $NP,CO2, NP,H2O,NP,O2,NP,SO2 and NP,N2$ respectively.

Take the temperature at standard condition as,

$Tstand=25°C$

Substitute $41.7 kJ/kmol⋅K$ for $cp$ and $127°C$ for $T2$ and $25°C$ for $Tstand$ in Equation (XIII).

$Δh¯SO2=(41.7 kJ/kmol⋅K)(127°C−25°C)=(41.7 kJ/kmol⋅K)[(127−25) K]=4253 kJ/kmol$

Refer Appendix Table A-18, A-19, A-20, A-23 and A-26 and write the property table for products and reactants as in Table (1).

Substance	$hfo¯$ $(kJ/kmol)$	$h¯298 K$ $(kJ/kmol)$	$h¯400 K$ $(kJ/kmol)$
$O2$	0	8682	11,711
$N2$	0	8669	11,640
$H2O(g)$	$−241,820$	9904	13,356
$CO2$	$−393,520$	9364	13,372
$SO2$	$−297,100$

Substitute the values form Table (I) into Equation (XII) to get,

$−Q¯out={[(0.4052)(−393,520+13,372−9364)]+[(04293)(−241,820+13,356−9904)]+[(0.1855)(0+11,711−8682)]+[(2.441)(0+11,640−8669)]+[(0.00306)(−297,100+4253−0)]}=−253,244 kJ/kmol fuelQ¯out=253,244 kJ/kmol fuel$

Substitute $253,244 kJ/kmol fuel$ for $Q¯out$ , $0.4052 kmol$ for $MC$ , $0.4293 kmol$ for $MH2$ , $0.15920 kmol$ for $MO2$ , $0.00319 kmol$ for $MN2$ , $0.00306 kmol$ for $MS$ , $12 kg/kmol$ for $MC$ , $2 kg/kmol$ for $MH2$ , $32 kg/kmol$ for $MO2$ , $28 kg/kmol$ for $MN2$ and $32 kg/kmol$ for $MS$ in Equation (XIV).

$Qout=253,244 kJ/kmol fuel[(0.4052×12)+(0.4293×2)+(0.1592×32)+(0.00319×28)+(0.00306×32)] kg/kmol=253,244 kJ/kmol fuel11 kg/kmol=23,022 kJ/kg fuel$

Hence, the heat transfer of fuel in the boiler is $23,022 kJ/kg fuel$ .

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Answer 2

Question

100%

Chapter 15.7, Problem 58P

To determine

The heat transfer of fuel in the boiler.

Expert Solution & Answer

Answer to Problem 58P

The heat transfer of fuel in the boiler is $23,022 kJ/kg fuel$ .

Explanation of Solution

Express the number of moles of carbon.

$NC=mfCMC$ (I)

Here, molar mass of carbon is $MC$ and mole fraction of carbon is $mfC$ .

Express the number of moles of hydrogen.

$NH2=mfH2MH2$ (II)

Here, molar mass of hydrogen is $MH2$ and mole fraction of hydrogen is $mfH2$ .

Express the number of moles of oxygen.

$NO2=mfO2MO2$ (III)

Here, molar mass of oxygen is $MO2$ and mole fraction of oxygen is $mfO2$ .

Express the number of moles of nitrogen.

$NN2=mfN2MN2$ (IV)

Here, molar mass of nitrogen is $MN2$ and mole fraction of nitrogen is $mfN2$ .

Express the number of moles of sulfur.

$NS=mfSMS$ (V)

Here, molar mass of sulphur is $MS$ and mole fraction of sulfur is $mfS$ .

Express the total number of moles.

$Nm=NC+NH2+NO2+NN2+NS$ (VI)

Here, number of moles of carbon, hydrogen, oxygen, nitrogen and sulfur is $NC$ , $NH2$ , $NO2$ , $NN2$ and $NS$ respectively.

Express the mole fraction of carbon.

$yC=NCNm$ (VII)

Express the mole fraction of hydrogen.

$yH2=NH2Nm$ (VIII)

Express the mole fraction of oxygen.

$yO2=NO2Nm$ (IX)

Express the mole fraction of nitrogen.

$yN2=NN2Nm$ (X)

Express the mole fraction of sulfur.

$yS=NSNm$ (XI)

Express the heat transfer from the combustion chamber.

$−Q¯out=∑NP(h¯fo+h¯−h¯o)P−∑NR(h¯fo+h¯−h¯o)R=∑NP(h¯fo+h¯400 K−h¯298 Ko)P−∑NR(h¯fo+h¯400 K−h¯298 Ko)R$ (XII)

Here, number of moles of products is $NP$ , number of moles of reactants is $NR$ , enthalpy of formation is $h¯f$ .

Express the enthalpy change of sulfur dioxide between the standard temperature and the product temperature.

$Δh¯SO2=cpΔT=cp(T2−Tstand)$ (XIII)

Here, specific heat constant pressure is $cp$ net temperature is $ΔT$ , temperature at state 2 is $T2$ and temperature at standard condition is $Tstand$ .

Express the heat transfer of fuel in the boiler.

$Qout=Q¯outNCMC+NH2MH2+NO2MO2+NN2MN2+NSMS$ (XIV)

Here, number of moles of carbon, hydrogen, oxygen, nitrogen and sulfur is $MC, MH2$ , $MO2$ , $MN2 and MS$ respectively.

Conclusion:

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

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

$MN2=28 kg/kmol$

Substitute $39.25 kg$ for $mC$ and $12 kg/kmol$ for $MC$ in Equation (I).

$NC=39.25 kg12 kg/kmol=3.271 kmol$

Substitute $6.93 kg$ for $mH2$ and $2 kg/kmol$ for $MH2$ in Equation (II).

$NH2=6.93 kg2 kg/kmol=3.465 kmol$

Substitute $41.11 kg$ for $mO2$ and $32 kg/kmol$ for $MO2$ in Equation (III).

$NO2=41.11 kg32 kg/kmol=1.285 kmol$

Substitute $0.72 kg$ for $mN2$ and $28 kg/kmol$ for $MN2$ in Equation (IV).

$NN2=0.72 kg28 kg/kmol=0.0257 kmol$

Substitute $0.79 kg$ for $mS$ and $32 kg/kmol$ for $MS$ in Equation (V).

$NS=0.79 kg32 kg/kmol=0.0247 kmol$

Substitute $3.271 kmol$ for $NC$ , $3.465 kmol$ for $NH2$ , $1.285 kmol$ for $NO2$ , $0.0257 kmol$ for $NN2$ and $0.00247 kmol$ for $NS$ in Equation (VI).

$Nm=3.271 kmol+3.465 kmol+1.285 kmol+0.0257 kmol+0.0247 kmol=8.071 kmol$

Substitute $3.271 kmol$ for $NC$ and $8.071 kmol$ for $Nm$ in Equation (VII).

$yC=3.271 kmol8.071 kmol=0.4052$

Substitute $3.465 kmol$ for $NH2$ and $8.071 kmol$ for $Nm$ in Equation (VIII).

$yH2=3.465 kmol8.071 kmol=0.4293$

Substitute $1.285 kmol$ for $NO2$ and $8.071 kmol$ for $Nm$ in Equation (IX).

$yO2=1.285 kmol8.071 kmol=0.1592$

Substitute $0.0257 kmol$ for $NN2$ and $8.071 kmol$ for $Nm$ in Equation (X).

$yN2=0.0257 kmol8.071 kmol=0.00319$

Substitute $0.0247 kmol$ for $NS$ and $8.071 kmol$ for $Nm$ in Equation (XI).

$yS=0.0247 kmol8.071 kmol=0.00306$

Express the combustion equation.

$[{0.4052C+0.4293H2+0.1592O2+0.00319N2+0.00306S+1.4ath(O2+3.76N2)}→{0.4052CO2+0.4293H2O+0.4athO2+0.00306SO2+1.4ath×3.76N2}]$ (XV)

Here, stoichiometric coefficient of air is $ath$ , oxygen is $O2$ , nitrogen is $N2$ , carbon dioxide is $CO2$ , water is $H2O$ and hydrogen is $H2$ .

Express the stoichiometric coefficient of air by $O2$ balancing.

$0.1592+1.4ath=0.4052+0.5×0.4293+0.4ath+0.00306ath=0.4637$

Substitute $0.4637$ for $ath$ in Equation (XV).

$[{0.4052C+0.4293H2+0.1592O2+0.00319N2+0.00306S+1.4×0.4637(O2+3.76N2)}→{0.4052CO2+0.4293H2O+0.4×0.4637O2+0.00306SO2+1.4×0.4637×3.76N2}][{0.4052C+0.4293H2+0.1592O2+0.00319N2+0.00306S+0.6492(O2+3.76N2)}→{0.4052CO2+0.4293H2O+0.1855O2+0.00306SO2+2.441N2}]$ (XVI)

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

$MC=0.4052 kmolMH2=0.4293 kmolMO2=0.15920 kmol$

$MN2=0.00319 kmolMS=0.00306 kmol$

Perform unit conversion of temperature at state 2 from degree Celsius to Kelvin.

$T2=127°C=(127+273) K=400 K$

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

$NP,CO2=0.4052 kmolNP,H2O=0.4293 kmolNP,O2=0.1855 kmolNP,SO2=0.00306 kmol$

$NP,N2=2.441 kmol$

Here, number of moles of product carbon dioxide, water, oxygen, sulfur dioxide and nitrogen is $NP,CO2, NP,H2O,NP,O2,NP,SO2 and NP,N2$ respectively.

Take the temperature at standard condition as,

$Tstand=25°C$

Substitute $41.7 kJ/kmol⋅K$ for $cp$ and $127°C$ for $T2$ and $25°C$ for $Tstand$ in Equation (XIII).

$Δh¯SO2=(41.7 kJ/kmol⋅K)(127°C−25°C)=(41.7 kJ/kmol⋅K)[(127−25) K]=4253 kJ/kmol$

Refer Appendix Table A-18, A-19, A-20, A-23 and A-26 and write the property table for products and reactants as in Table (1).

Substance	$hfo¯$ $(kJ/kmol)$	$h¯298 K$ $(kJ/kmol)$	$h¯400 K$ $(kJ/kmol)$
$O2$	0	8682	11,711
$N2$	0	8669	11,640
$H2O(g)$	$−241,820$	9904	13,356
$CO2$	$−393,520$	9364	13,372
$SO2$	$−297,100$

Substitute the values form Table (I) into Equation (XII) to get,

$−Q¯out={[(0.4052)(−393,520+13,372−9364)]+[(04293)(−241,820+13,356−9904)]+[(0.1855)(0+11,711−8682)]+[(2.441)(0+11,640−8669)]+[(0.00306)(−297,100+4253−0)]}=−253,244 kJ/kmol fuelQ¯out=253,244 kJ/kmol fuel$

Substitute $253,244 kJ/kmol fuel$ for $Q¯out$ , $0.4052 kmol$ for $MC$ , $0.4293 kmol$ for $MH2$ , $0.15920 kmol$ for $MO2$ , $0.00319 kmol$ for $MN2$ , $0.00306 kmol$ for $MS$ , $12 kg/kmol$ for $MC$ , $2 kg/kmol$ for $MH2$ , $32 kg/kmol$ for $MO2$ , $28 kg/kmol$ for $MN2$ and $32 kg/kmol$ for $MS$ in Equation (XIV).

$Qout=253,244 kJ/kmol fuel[(0.4052×12)+(0.4293×2)+(0.1592×32)+(0.00319×28)+(0.00306×32)] kg/kmol=253,244 kJ/kmol fuel11 kg/kmol=23,022 kJ/kg fuel$

Hence, the heat transfer of fuel in the boiler is $23,022 kJ/kg fuel$ .

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Answer 3

Question

100%

Chapter 15.7, Problem 58P

To determine

The heat transfer of fuel in the boiler.

Expert Solution & Answer

Answer to Problem 58P

The heat transfer of fuel in the boiler is $23,022 kJ/kg fuel$ .

Explanation of Solution

Express the number of moles of carbon.

$NC=mfCMC$ (I)

Here, molar mass of carbon is $MC$ and mole fraction of carbon is $mfC$ .

Express the number of moles of hydrogen.

$NH2=mfH2MH2$ (II)

Here, molar mass of hydrogen is $MH2$ and mole fraction of hydrogen is $mfH2$ .

Express the number of moles of oxygen.

$NO2=mfO2MO2$ (III)

Here, molar mass of oxygen is $MO2$ and mole fraction of oxygen is $mfO2$ .

Express the number of moles of nitrogen.

$NN2=mfN2MN2$ (IV)

Here, molar mass of nitrogen is $MN2$ and mole fraction of nitrogen is $mfN2$ .

Express the number of moles of sulfur.

$NS=mfSMS$ (V)

Here, molar mass of sulphur is $MS$ and mole fraction of sulfur is $mfS$ .

Express the total number of moles.

$Nm=NC+NH2+NO2+NN2+NS$ (VI)

Here, number of moles of carbon, hydrogen, oxygen, nitrogen and sulfur is $NC$ , $NH2$ , $NO2$ , $NN2$ and $NS$ respectively.

Express the mole fraction of carbon.

$yC=NCNm$ (VII)

Express the mole fraction of hydrogen.

$yH2=NH2Nm$ (VIII)

Express the mole fraction of oxygen.

$yO2=NO2Nm$ (IX)

Express the mole fraction of nitrogen.

$yN2=NN2Nm$ (X)

Express the mole fraction of sulfur.

$yS=NSNm$ (XI)

Express the heat transfer from the combustion chamber.

$−Q¯out=∑NP(h¯fo+h¯−h¯o)P−∑NR(h¯fo+h¯−h¯o)R=∑NP(h¯fo+h¯400 K−h¯298 Ko)P−∑NR(h¯fo+h¯400 K−h¯298 Ko)R$ (XII)

Here, number of moles of products is $NP$ , number of moles of reactants is $NR$ , enthalpy of formation is $h¯f$ .

Express the enthalpy change of sulfur dioxide between the standard temperature and the product temperature.

$Δh¯SO2=cpΔT=cp(T2−Tstand)$ (XIII)

Here, specific heat constant pressure is $cp$ net temperature is $ΔT$ , temperature at state 2 is $T2$ and temperature at standard condition is $Tstand$ .

Express the heat transfer of fuel in the boiler.

$Qout=Q¯outNCMC+NH2MH2+NO2MO2+NN2MN2+NSMS$ (XIV)

Here, number of moles of carbon, hydrogen, oxygen, nitrogen and sulfur is $MC, MH2$ , $MO2$ , $MN2 and MS$ respectively.

Conclusion:

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

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

$MN2=28 kg/kmol$

Substitute $39.25 kg$ for $mC$ and $12 kg/kmol$ for $MC$ in Equation (I).

$NC=39.25 kg12 kg/kmol=3.271 kmol$

Substitute $6.93 kg$ for $mH2$ and $2 kg/kmol$ for $MH2$ in Equation (II).

$NH2=6.93 kg2 kg/kmol=3.465 kmol$

Substitute $41.11 kg$ for $mO2$ and $32 kg/kmol$ for $MO2$ in Equation (III).

$NO2=41.11 kg32 kg/kmol=1.285 kmol$

Substitute $0.72 kg$ for $mN2$ and $28 kg/kmol$ for $MN2$ in Equation (IV).

$NN2=0.72 kg28 kg/kmol=0.0257 kmol$

Substitute $0.79 kg$ for $mS$ and $32 kg/kmol$ for $MS$ in Equation (V).

$NS=0.79 kg32 kg/kmol=0.0247 kmol$

Substitute $3.271 kmol$ for $NC$ , $3.465 kmol$ for $NH2$ , $1.285 kmol$ for $NO2$ , $0.0257 kmol$ for $NN2$ and $0.00247 kmol$ for $NS$ in Equation (VI).

$Nm=3.271 kmol+3.465 kmol+1.285 kmol+0.0257 kmol+0.0247 kmol=8.071 kmol$

Substitute $3.271 kmol$ for $NC$ and $8.071 kmol$ for $Nm$ in Equation (VII).

$yC=3.271 kmol8.071 kmol=0.4052$

Substitute $3.465 kmol$ for $NH2$ and $8.071 kmol$ for $Nm$ in Equation (VIII).

$yH2=3.465 kmol8.071 kmol=0.4293$

Substitute $1.285 kmol$ for $NO2$ and $8.071 kmol$ for $Nm$ in Equation (IX).

$yO2=1.285 kmol8.071 kmol=0.1592$

Substitute $0.0257 kmol$ for $NN2$ and $8.071 kmol$ for $Nm$ in Equation (X).

$yN2=0.0257 kmol8.071 kmol=0.00319$

Substitute $0.0247 kmol$ for $NS$ and $8.071 kmol$ for $Nm$ in Equation (XI).

$yS=0.0247 kmol8.071 kmol=0.00306$

Express the combustion equation.

$[{0.4052C+0.4293H2+0.1592O2+0.00319N2+0.00306S+1.4ath(O2+3.76N2)}→{0.4052CO2+0.4293H2O+0.4athO2+0.00306SO2+1.4ath×3.76N2}]$ (XV)

Here, stoichiometric coefficient of air is $ath$ , oxygen is $O2$ , nitrogen is $N2$ , carbon dioxide is $CO2$ , water is $H2O$ and hydrogen is $H2$ .

Express the stoichiometric coefficient of air by $O2$ balancing.

$0.1592+1.4ath=0.4052+0.5×0.4293+0.4ath+0.00306ath=0.4637$

Substitute $0.4637$ for $ath$ in Equation (XV).

$[{0.4052C+0.4293H2+0.1592O2+0.00319N2+0.00306S+1.4×0.4637(O2+3.76N2)}→{0.4052CO2+0.4293H2O+0.4×0.4637O2+0.00306SO2+1.4×0.4637×3.76N2}][{0.4052C+0.4293H2+0.1592O2+0.00319N2+0.00306S+0.6492(O2+3.76N2)}→{0.4052CO2+0.4293H2O+0.1855O2+0.00306SO2+2.441N2}]$ (XVI)

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

$MC=0.4052 kmolMH2=0.4293 kmolMO2=0.15920 kmol$

$MN2=0.00319 kmolMS=0.00306 kmol$

Perform unit conversion of temperature at state 2 from degree Celsius to Kelvin.

$T2=127°C=(127+273) K=400 K$

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

$NP,CO2=0.4052 kmolNP,H2O=0.4293 kmolNP,O2=0.1855 kmolNP,SO2=0.00306 kmol$

$NP,N2=2.441 kmol$

Here, number of moles of product carbon dioxide, water, oxygen, sulfur dioxide and nitrogen is $NP,CO2, NP,H2O,NP,O2,NP,SO2 and NP,N2$ respectively.

Take the temperature at standard condition as,

$Tstand=25°C$

Substitute $41.7 kJ/kmol⋅K$ for $cp$ and $127°C$ for $T2$ and $25°C$ for $Tstand$ in Equation (XIII).

$Δh¯SO2=(41.7 kJ/kmol⋅K)(127°C−25°C)=(41.7 kJ/kmol⋅K)[(127−25) K]=4253 kJ/kmol$

Refer Appendix Table A-18, A-19, A-20, A-23 and A-26 and write the property table for products and reactants as in Table (1).

Substance	$hfo¯$ $(kJ/kmol)$	$h¯298 K$ $(kJ/kmol)$	$h¯400 K$ $(kJ/kmol)$
$O2$	0	8682	11,711
$N2$	0	8669	11,640
$H2O(g)$	$−241,820$	9904	13,356
$CO2$	$−393,520$	9364	13,372
$SO2$	$−297,100$

Substitute the values form Table (I) into Equation (XII) to get,

$−Q¯out={[(0.4052)(−393,520+13,372−9364)]+[(04293)(−241,820+13,356−9904)]+[(0.1855)(0+11,711−8682)]+[(2.441)(0+11,640−8669)]+[(0.00306)(−297,100+4253−0)]}=−253,244 kJ/kmol fuelQ¯out=253,244 kJ/kmol fuel$

Substitute $253,244 kJ/kmol fuel$ for $Q¯out$ , $0.4052 kmol$ for $MC$ , $0.4293 kmol$ for $MH2$ , $0.15920 kmol$ for $MO2$ , $0.00319 kmol$ for $MN2$ , $0.00306 kmol$ for $MS$ , $12 kg/kmol$ for $MC$ , $2 kg/kmol$ for $MH2$ , $32 kg/kmol$ for $MO2$ , $28 kg/kmol$ for $MN2$ and $32 kg/kmol$ for $MS$ in Equation (XIV).

$Qout=253,244 kJ/kmol fuel[(0.4052×12)+(0.4293×2)+(0.1592×32)+(0.00319×28)+(0.00306×32)] kg/kmol=253,244 kJ/kmol fuel11 kg/kmol=23,022 kJ/kg fuel$

Hence, the heat transfer of fuel in the boiler is $23,022 kJ/kg fuel$ .

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Answer 4

Question

100%

Chapter 15.7, Problem 58P

To determine

The heat transfer of fuel in the boiler.

Expert Solution & Answer

Answer to Problem 58P

The heat transfer of fuel in the boiler is $23,022 kJ/kg fuel$ .

Explanation of Solution

Express the number of moles of carbon.

$NC=mfCMC$ (I)

Here, molar mass of carbon is $MC$ and mole fraction of carbon is $mfC$ .

Express the number of moles of hydrogen.

$NH2=mfH2MH2$ (II)

Here, molar mass of hydrogen is $MH2$ and mole fraction of hydrogen is $mfH2$ .

Express the number of moles of oxygen.

$NO2=mfO2MO2$ (III)

Here, molar mass of oxygen is $MO2$ and mole fraction of oxygen is $mfO2$ .

Express the number of moles of nitrogen.

$NN2=mfN2MN2$ (IV)

Here, molar mass of nitrogen is $MN2$ and mole fraction of nitrogen is $mfN2$ .

Express the number of moles of sulfur.

$NS=mfSMS$ (V)

Here, molar mass of sulphur is $MS$ and mole fraction of sulfur is $mfS$ .

Express the total number of moles.

$Nm=NC+NH2+NO2+NN2+NS$ (VI)

Here, number of moles of carbon, hydrogen, oxygen, nitrogen and sulfur is $NC$ , $NH2$ , $NO2$ , $NN2$ and $NS$ respectively.

Express the mole fraction of carbon.

$yC=NCNm$ (VII)

Express the mole fraction of hydrogen.

$yH2=NH2Nm$ (VIII)

Express the mole fraction of oxygen.

$yO2=NO2Nm$ (IX)

Express the mole fraction of nitrogen.

$yN2=NN2Nm$ (X)

Express the mole fraction of sulfur.

$yS=NSNm$ (XI)

Express the heat transfer from the combustion chamber.

$−Q¯out=∑NP(h¯fo+h¯−h¯o)P−∑NR(h¯fo+h¯−h¯o)R=∑NP(h¯fo+h¯400 K−h¯298 Ko)P−∑NR(h¯fo+h¯400 K−h¯298 Ko)R$ (XII)

Here, number of moles of products is $NP$ , number of moles of reactants is $NR$ , enthalpy of formation is $h¯f$ .

Express the enthalpy change of sulfur dioxide between the standard temperature and the product temperature.

$Δh¯SO2=cpΔT=cp(T2−Tstand)$ (XIII)

Here, specific heat constant pressure is $cp$ net temperature is $ΔT$ , temperature at state 2 is $T2$ and temperature at standard condition is $Tstand$ .

Express the heat transfer of fuel in the boiler.

$Qout=Q¯outNCMC+NH2MH2+NO2MO2+NN2MN2+NSMS$ (XIV)

Here, number of moles of carbon, hydrogen, oxygen, nitrogen and sulfur is $MC, MH2$ , $MO2$ , $MN2 and MS$ respectively.

Conclusion:

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

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

$MN2=28 kg/kmol$

Substitute $39.25 kg$ for $mC$ and $12 kg/kmol$ for $MC$ in Equation (I).

$NC=39.25 kg12 kg/kmol=3.271 kmol$

Substitute $6.93 kg$ for $mH2$ and $2 kg/kmol$ for $MH2$ in Equation (II).

$NH2=6.93 kg2 kg/kmol=3.465 kmol$

Substitute $41.11 kg$ for $mO2$ and $32 kg/kmol$ for $MO2$ in Equation (III).

$NO2=41.11 kg32 kg/kmol=1.285 kmol$

Substitute $0.72 kg$ for $mN2$ and $28 kg/kmol$ for $MN2$ in Equation (IV).

$NN2=0.72 kg28 kg/kmol=0.0257 kmol$

Substitute $0.79 kg$ for $mS$ and $32 kg/kmol$ for $MS$ in Equation (V).

$NS=0.79 kg32 kg/kmol=0.0247 kmol$

Substitute $3.271 kmol$ for $NC$ , $3.465 kmol$ for $NH2$ , $1.285 kmol$ for $NO2$ , $0.0257 kmol$ for $NN2$ and $0.00247 kmol$ for $NS$ in Equation (VI).

$Nm=3.271 kmol+3.465 kmol+1.285 kmol+0.0257 kmol+0.0247 kmol=8.071 kmol$

Substitute $3.271 kmol$ for $NC$ and $8.071 kmol$ for $Nm$ in Equation (VII).

$yC=3.271 kmol8.071 kmol=0.4052$

Substitute $3.465 kmol$ for $NH2$ and $8.071 kmol$ for $Nm$ in Equation (VIII).

$yH2=3.465 kmol8.071 kmol=0.4293$

Substitute $1.285 kmol$ for $NO2$ and $8.071 kmol$ for $Nm$ in Equation (IX).

$yO2=1.285 kmol8.071 kmol=0.1592$

Substitute $0.0257 kmol$ for $NN2$ and $8.071 kmol$ for $Nm$ in Equation (X).

$yN2=0.0257 kmol8.071 kmol=0.00319$

Substitute $0.0247 kmol$ for $NS$ and $8.071 kmol$ for $Nm$ in Equation (XI).

$yS=0.0247 kmol8.071 kmol=0.00306$

Express the combustion equation.

$[{0.4052C+0.4293H2+0.1592O2+0.00319N2+0.00306S+1.4ath(O2+3.76N2)}→{0.4052CO2+0.4293H2O+0.4athO2+0.00306SO2+1.4ath×3.76N2}]$ (XV)

Here, stoichiometric coefficient of air is $ath$ , oxygen is $O2$ , nitrogen is $N2$ , carbon dioxide is $CO2$ , water is $H2O$ and hydrogen is $H2$ .

Express the stoichiometric coefficient of air by $O2$ balancing.

$0.1592+1.4ath=0.4052+0.5×0.4293+0.4ath+0.00306ath=0.4637$

Substitute $0.4637$ for $ath$ in Equation (XV).

$[{0.4052C+0.4293H2+0.1592O2+0.00319N2+0.00306S+1.4×0.4637(O2+3.76N2)}→{0.4052CO2+0.4293H2O+0.4×0.4637O2+0.00306SO2+1.4×0.4637×3.76N2}][{0.4052C+0.4293H2+0.1592O2+0.00319N2+0.00306S+0.6492(O2+3.76N2)}→{0.4052CO2+0.4293H2O+0.1855O2+0.00306SO2+2.441N2}]$ (XVI)

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

$MC=0.4052 kmolMH2=0.4293 kmolMO2=0.15920 kmol$

$MN2=0.00319 kmolMS=0.00306 kmol$

Perform unit conversion of temperature at state 2 from degree Celsius to Kelvin.

$T2=127°C=(127+273) K=400 K$

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

$NP,CO2=0.4052 kmolNP,H2O=0.4293 kmolNP,O2=0.1855 kmolNP,SO2=0.00306 kmol$

$NP,N2=2.441 kmol$

Here, number of moles of product carbon dioxide, water, oxygen, sulfur dioxide and nitrogen is $NP,CO2, NP,H2O,NP,O2,NP,SO2 and NP,N2$ respectively.

Take the temperature at standard condition as,

$Tstand=25°C$

Substitute $41.7 kJ/kmol⋅K$ for $cp$ and $127°C$ for $T2$ and $25°C$ for $Tstand$ in Equation (XIII).

$Δh¯SO2=(41.7 kJ/kmol⋅K)(127°C−25°C)=(41.7 kJ/kmol⋅K)[(127−25) K]=4253 kJ/kmol$

Refer Appendix Table A-18, A-19, A-20, A-23 and A-26 and write the property table for products and reactants as in Table (1).

Substance	$hfo¯$ $(kJ/kmol)$	$h¯298 K$ $(kJ/kmol)$	$h¯400 K$ $(kJ/kmol)$
$O2$	0	8682	11,711
$N2$	0	8669	11,640
$H2O(g)$	$−241,820$	9904	13,356
$CO2$	$−393,520$	9364	13,372
$SO2$	$−297,100$

Substitute the values form Table (I) into Equation (XII) to get,

$−Q¯out={[(0.4052)(−393,520+13,372−9364)]+[(04293)(−241,820+13,356−9904)]+[(0.1855)(0+11,711−8682)]+[(2.441)(0+11,640−8669)]+[(0.00306)(−297,100+4253−0)]}=−253,244 kJ/kmol fuelQ¯out=253,244 kJ/kmol fuel$

Substitute $253,244 kJ/kmol fuel$ for $Q¯out$ , $0.4052 kmol$ for $MC$ , $0.4293 kmol$ for $MH2$ , $0.15920 kmol$ for $MO2$ , $0.00319 kmol$ for $MN2$ , $0.00306 kmol$ for $MS$ , $12 kg/kmol$ for $MC$ , $2 kg/kmol$ for $MH2$ , $32 kg/kmol$ for $MO2$ , $28 kg/kmol$ for $MN2$ and $32 kg/kmol$ for $MS$ in Equation (XIV).

$Qout=253,244 kJ/kmol fuel[(0.4052×12)+(0.4293×2)+(0.1592×32)+(0.00319×28)+(0.00306×32)] kg/kmol=253,244 kJ/kmol fuel11 kg/kmol=23,022 kJ/kg fuel$

Hence, the heat transfer of fuel in the boiler is $23,022 kJ/kg fuel$ .

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Answer 5

Question

100%

Answer 6

Question

100%

Answer 7

Chapter 15.7, Problem 58P

To determine

The heat transfer of fuel in the boiler.

Expert Solution & Answer

Answer to Problem 58P

The heat transfer of fuel in the boiler is $23,022 kJ/kg fuel$ .

Explanation of Solution

Express the number of moles of carbon.

$NC=mfCMC$ (I)

Here, molar mass of carbon is $MC$ and mole fraction of carbon is $mfC$ .

Express the number of moles of hydrogen.

$NH2=mfH2MH2$ (II)

Here, molar mass of hydrogen is $MH2$ and mole fraction of hydrogen is $mfH2$ .

Express the number of moles of oxygen.

$NO2=mfO2MO2$ (III)

Here, molar mass of oxygen is $MO2$ and mole fraction of oxygen is $mfO2$ .

Express the number of moles of nitrogen.

$NN2=mfN2MN2$ (IV)

Here, molar mass of nitrogen is $MN2$ and mole fraction of nitrogen is $mfN2$ .

Express the number of moles of sulfur.

$NS=mfSMS$ (V)

Here, molar mass of sulphur is $MS$ and mole fraction of sulfur is $mfS$ .

Express the total number of moles.

$Nm=NC+NH2+NO2+NN2+NS$ (VI)

Here, number of moles of carbon, hydrogen, oxygen, nitrogen and sulfur is $NC$ , $NH2$ , $NO2$ , $NN2$ and $NS$ respectively.

Express the mole fraction of carbon.

$yC=NCNm$ (VII)

Express the mole fraction of hydrogen.

$yH2=NH2Nm$ (VIII)

Express the mole fraction of oxygen.

$yO2=NO2Nm$ (IX)

Express the mole fraction of nitrogen.

$yN2=NN2Nm$ (X)

Express the mole fraction of sulfur.

$yS=NSNm$ (XI)

Express the heat transfer from the combustion chamber.

$−Q¯out=∑NP(h¯fo+h¯−h¯o)P−∑NR(h¯fo+h¯−h¯o)R=∑NP(h¯fo+h¯400 K−h¯298 Ko)P−∑NR(h¯fo+h¯400 K−h¯298 Ko)R$ (XII)

Here, number of moles of products is $NP$ , number of moles of reactants is $NR$ , enthalpy of formation is $h¯f$ .

Express the enthalpy change of sulfur dioxide between the standard temperature and the product temperature.

$Δh¯SO2=cpΔT=cp(T2−Tstand)$ (XIII)

Here, specific heat constant pressure is $cp$ net temperature is $ΔT$ , temperature at state 2 is $T2$ and temperature at standard condition is $Tstand$ .

Express the heat transfer of fuel in the boiler.

$Qout=Q¯outNCMC+NH2MH2+NO2MO2+NN2MN2+NSMS$ (XIV)

Here, number of moles of carbon, hydrogen, oxygen, nitrogen and sulfur is $MC, MH2$ , $MO2$ , $MN2 and MS$ respectively.

Conclusion:

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

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

$MN2=28 kg/kmol$

Substitute $39.25 kg$ for $mC$ and $12 kg/kmol$ for $MC$ in Equation (I).

$NC=39.25 kg12 kg/kmol=3.271 kmol$

Substitute $6.93 kg$ for $mH2$ and $2 kg/kmol$ for $MH2$ in Equation (II).

$NH2=6.93 kg2 kg/kmol=3.465 kmol$

Substitute $41.11 kg$ for $mO2$ and $32 kg/kmol$ for $MO2$ in Equation (III).

$NO2=41.11 kg32 kg/kmol=1.285 kmol$

Substitute $0.72 kg$ for $mN2$ and $28 kg/kmol$ for $MN2$ in Equation (IV).

$NN2=0.72 kg28 kg/kmol=0.0257 kmol$

Substitute $0.79 kg$ for $mS$ and $32 kg/kmol$ for $MS$ in Equation (V).

$NS=0.79 kg32 kg/kmol=0.0247 kmol$

Substitute $3.271 kmol$ for $NC$ , $3.465 kmol$ for $NH2$ , $1.285 kmol$ for $NO2$ , $0.0257 kmol$ for $NN2$ and $0.00247 kmol$ for $NS$ in Equation (VI).

$Nm=3.271 kmol+3.465 kmol+1.285 kmol+0.0257 kmol+0.0247 kmol=8.071 kmol$

Substitute $3.271 kmol$ for $NC$ and $8.071 kmol$ for $Nm$ in Equation (VII).

$yC=3.271 kmol8.071 kmol=0.4052$

Substitute $3.465 kmol$ for $NH2$ and $8.071 kmol$ for $Nm$ in Equation (VIII).

$yH2=3.465 kmol8.071 kmol=0.4293$

Substitute $1.285 kmol$ for $NO2$ and $8.071 kmol$ for $Nm$ in Equation (IX).

$yO2=1.285 kmol8.071 kmol=0.1592$

Substitute $0.0257 kmol$ for $NN2$ and $8.071 kmol$ for $Nm$ in Equation (X).

$yN2=0.0257 kmol8.071 kmol=0.00319$

Substitute $0.0247 kmol$ for $NS$ and $8.071 kmol$ for $Nm$ in Equation (XI).

$yS=0.0247 kmol8.071 kmol=0.00306$

Express the combustion equation.

$[{0.4052C+0.4293H2+0.1592O2+0.00319N2+0.00306S+1.4ath(O2+3.76N2)}→{0.4052CO2+0.4293H2O+0.4athO2+0.00306SO2+1.4ath×3.76N2}]$ (XV)

Here, stoichiometric coefficient of air is $ath$ , oxygen is $O2$ , nitrogen is $N2$ , carbon dioxide is $CO2$ , water is $H2O$ and hydrogen is $H2$ .

Express the stoichiometric coefficient of air by $O2$ balancing.

$0.1592+1.4ath=0.4052+0.5×0.4293+0.4ath+0.00306ath=0.4637$

Substitute $0.4637$ for $ath$ in Equation (XV).

$[{0.4052C+0.4293H2+0.1592O2+0.00319N2+0.00306S+1.4×0.4637(O2+3.76N2)}→{0.4052CO2+0.4293H2O+0.4×0.4637O2+0.00306SO2+1.4×0.4637×3.76N2}][{0.4052C+0.4293H2+0.1592O2+0.00319N2+0.00306S+0.6492(O2+3.76N2)}→{0.4052CO2+0.4293H2O+0.1855O2+0.00306SO2+2.441N2}]$ (XVI)

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

$MC=0.4052 kmolMH2=0.4293 kmolMO2=0.15920 kmol$

$MN2=0.00319 kmolMS=0.00306 kmol$

Perform unit conversion of temperature at state 2 from degree Celsius to Kelvin.

$T2=127°C=(127+273) K=400 K$

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

$NP,CO2=0.4052 kmolNP,H2O=0.4293 kmolNP,O2=0.1855 kmolNP,SO2=0.00306 kmol$

$NP,N2=2.441 kmol$

Here, number of moles of product carbon dioxide, water, oxygen, sulfur dioxide and nitrogen is $NP,CO2, NP,H2O,NP,O2,NP,SO2 and NP,N2$ respectively.

Take the temperature at standard condition as,

$Tstand=25°C$

Substitute $41.7 kJ/kmol⋅K$ for $cp$ and $127°C$ for $T2$ and $25°C$ for $Tstand$ in Equation (XIII).

$Δh¯SO2=(41.7 kJ/kmol⋅K)(127°C−25°C)=(41.7 kJ/kmol⋅K)[(127−25) K]=4253 kJ/kmol$

Refer Appendix Table A-18, A-19, A-20, A-23 and A-26 and write the property table for products and reactants as in Table (1).

Substance	$hfo¯$ $(kJ/kmol)$	$h¯298 K$ $(kJ/kmol)$	$h¯400 K$ $(kJ/kmol)$
$O2$	0	8682	11,711
$N2$	0	8669	11,640
$H2O(g)$	$−241,820$	9904	13,356
$CO2$	$−393,520$	9364	13,372
$SO2$	$−297,100$

Substitute the values form Table (I) into Equation (XII) to get,

$−Q¯out={[(0.4052)(−393,520+13,372−9364)]+[(04293)(−241,820+13,356−9904)]+[(0.1855)(0+11,711−8682)]+[(2.441)(0+11,640−8669)]+[(0.00306)(−297,100+4253−0)]}=−253,244 kJ/kmol fuelQ¯out=253,244 kJ/kmol fuel$

Substitute $253,244 kJ/kmol fuel$ for $Q¯out$ , $0.4052 kmol$ for $MC$ , $0.4293 kmol$ for $MH2$ , $0.15920 kmol$ for $MO2$ , $0.00319 kmol$ for $MN2$ , $0.00306 kmol$ for $MS$ , $12 kg/kmol$ for $MC$ , $2 kg/kmol$ for $MH2$ , $32 kg/kmol$ for $MO2$ , $28 kg/kmol$ for $MN2$ and $32 kg/kmol$ for $MS$ in Equation (XIV).

$Qout=253,244 kJ/kmol fuel[(0.4052×12)+(0.4293×2)+(0.1592×32)+(0.00319×28)+(0.00306×32)] kg/kmol=253,244 kJ/kmol fuel11 kg/kmol=23,022 kJ/kg fuel$

Hence, the heat transfer of fuel in the boiler is $23,022 kJ/kg fuel$ .

Answer 8

Chapter 15.7, Problem 58P

To determine

The heat transfer of fuel in the boiler.

Expert Solution & Answer

Answer to Problem 58P

The heat transfer of fuel in the boiler is $23,022 kJ/kg fuel$ .

Explanation of Solution

Express the number of moles of carbon.

$NC=mfCMC$ (I)

Here, molar mass of carbon is $MC$ and mole fraction of carbon is $mfC$ .

Express the number of moles of hydrogen.

$NH2=mfH2MH2$ (II)

Here, molar mass of hydrogen is $MH2$ and mole fraction of hydrogen is $mfH2$ .

Express the number of moles of oxygen.

$NO2=mfO2MO2$ (III)

Here, molar mass of oxygen is $MO2$ and mole fraction of oxygen is $mfO2$ .

Express the number of moles of nitrogen.

$NN2=mfN2MN2$ (IV)

Here, molar mass of nitrogen is $MN2$ and mole fraction of nitrogen is $mfN2$ .

Express the number of moles of sulfur.

$NS=mfSMS$ (V)

Here, molar mass of sulphur is $MS$ and mole fraction of sulfur is $mfS$ .

Express the total number of moles.

$Nm=NC+NH2+NO2+NN2+NS$ (VI)

Here, number of moles of carbon, hydrogen, oxygen, nitrogen and sulfur is $NC$ , $NH2$ , $NO2$ , $NN2$ and $NS$ respectively.

Express the mole fraction of carbon.

$yC=NCNm$ (VII)

Express the mole fraction of hydrogen.

$yH2=NH2Nm$ (VIII)

Express the mole fraction of oxygen.

$yO2=NO2Nm$ (IX)

Express the mole fraction of nitrogen.

$yN2=NN2Nm$ (X)

Express the mole fraction of sulfur.

$yS=NSNm$ (XI)

Express the heat transfer from the combustion chamber.

$−Q¯out=∑NP(h¯fo+h¯−h¯o)P−∑NR(h¯fo+h¯−h¯o)R=∑NP(h¯fo+h¯400 K−h¯298 Ko)P−∑NR(h¯fo+h¯400 K−h¯298 Ko)R$ (XII)

Here, number of moles of products is $NP$ , number of moles of reactants is $NR$ , enthalpy of formation is $h¯f$ .

Express the enthalpy change of sulfur dioxide between the standard temperature and the product temperature.

$Δh¯SO2=cpΔT=cp(T2−Tstand)$ (XIII)

Here, specific heat constant pressure is $cp$ net temperature is $ΔT$ , temperature at state 2 is $T2$ and temperature at standard condition is $Tstand$ .

Express the heat transfer of fuel in the boiler.

$Qout=Q¯outNCMC+NH2MH2+NO2MO2+NN2MN2+NSMS$ (XIV)

Here, number of moles of carbon, hydrogen, oxygen, nitrogen and sulfur is $MC, MH2$ , $MO2$ , $MN2 and MS$ respectively.

Conclusion:

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

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

$MN2=28 kg/kmol$

Substitute $39.25 kg$ for $mC$ and $12 kg/kmol$ for $MC$ in Equation (I).

$NC=39.25 kg12 kg/kmol=3.271 kmol$

Substitute $6.93 kg$ for $mH2$ and $2 kg/kmol$ for $MH2$ in Equation (II).

$NH2=6.93 kg2 kg/kmol=3.465 kmol$

Substitute $41.11 kg$ for $mO2$ and $32 kg/kmol$ for $MO2$ in Equation (III).

$NO2=41.11 kg32 kg/kmol=1.285 kmol$

Substitute $0.72 kg$ for $mN2$ and $28 kg/kmol$ for $MN2$ in Equation (IV).

$NN2=0.72 kg28 kg/kmol=0.0257 kmol$

Substitute $0.79 kg$ for $mS$ and $32 kg/kmol$ for $MS$ in Equation (V).

$NS=0.79 kg32 kg/kmol=0.0247 kmol$

Substitute $3.271 kmol$ for $NC$ , $3.465 kmol$ for $NH2$ , $1.285 kmol$ for $NO2$ , $0.0257 kmol$ for $NN2$ and $0.00247 kmol$ for $NS$ in Equation (VI).

$Nm=3.271 kmol+3.465 kmol+1.285 kmol+0.0257 kmol+0.0247 kmol=8.071 kmol$

Substitute $3.271 kmol$ for $NC$ and $8.071 kmol$ for $Nm$ in Equation (VII).

$yC=3.271 kmol8.071 kmol=0.4052$

Substitute $3.465 kmol$ for $NH2$ and $8.071 kmol$ for $Nm$ in Equation (VIII).

$yH2=3.465 kmol8.071 kmol=0.4293$

Substitute $1.285 kmol$ for $NO2$ and $8.071 kmol$ for $Nm$ in Equation (IX).

$yO2=1.285 kmol8.071 kmol=0.1592$

Substitute $0.0257 kmol$ for $NN2$ and $8.071 kmol$ for $Nm$ in Equation (X).

$yN2=0.0257 kmol8.071 kmol=0.00319$

Substitute $0.0247 kmol$ for $NS$ and $8.071 kmol$ for $Nm$ in Equation (XI).

$yS=0.0247 kmol8.071 kmol=0.00306$

Express the combustion equation.

$[{0.4052C+0.4293H2+0.1592O2+0.00319N2+0.00306S+1.4ath(O2+3.76N2)}→{0.4052CO2+0.4293H2O+0.4athO2+0.00306SO2+1.4ath×3.76N2}]$ (XV)

Here, stoichiometric coefficient of air is $ath$ , oxygen is $O2$ , nitrogen is $N2$ , carbon dioxide is $CO2$ , water is $H2O$ and hydrogen is $H2$ .

Express the stoichiometric coefficient of air by $O2$ balancing.

$0.1592+1.4ath=0.4052+0.5×0.4293+0.4ath+0.00306ath=0.4637$

Substitute $0.4637$ for $ath$ in Equation (XV).

$[{0.4052C+0.4293H2+0.1592O2+0.00319N2+0.00306S+1.4×0.4637(O2+3.76N2)}→{0.4052CO2+0.4293H2O+0.4×0.4637O2+0.00306SO2+1.4×0.4637×3.76N2}][{0.4052C+0.4293H2+0.1592O2+0.00319N2+0.00306S+0.6492(O2+3.76N2)}→{0.4052CO2+0.4293H2O+0.1855O2+0.00306SO2+2.441N2}]$ (XVI)

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

$MC=0.4052 kmolMH2=0.4293 kmolMO2=0.15920 kmol$

$MN2=0.00319 kmolMS=0.00306 kmol$

Perform unit conversion of temperature at state 2 from degree Celsius to Kelvin.

$T2=127°C=(127+273) K=400 K$

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

$NP,CO2=0.4052 kmolNP,H2O=0.4293 kmolNP,O2=0.1855 kmolNP,SO2=0.00306 kmol$

$NP,N2=2.441 kmol$

Here, number of moles of product carbon dioxide, water, oxygen, sulfur dioxide and nitrogen is $NP,CO2, NP,H2O,NP,O2,NP,SO2 and NP,N2$ respectively.

Take the temperature at standard condition as,

$Tstand=25°C$

Substitute $41.7 kJ/kmol⋅K$ for $cp$ and $127°C$ for $T2$ and $25°C$ for $Tstand$ in Equation (XIII).

$Δh¯SO2=(41.7 kJ/kmol⋅K)(127°C−25°C)=(41.7 kJ/kmol⋅K)[(127−25) K]=4253 kJ/kmol$

Refer Appendix Table A-18, A-19, A-20, A-23 and A-26 and write the property table for products and reactants as in Table (1).

Substance	$hfo¯$ $(kJ/kmol)$	$h¯298 K$ $(kJ/kmol)$	$h¯400 K$ $(kJ/kmol)$
$O2$	0	8682	11,711
$N2$	0	8669	11,640
$H2O(g)$	$−241,820$	9904	13,356
$CO2$	$−393,520$	9364	13,372
$SO2$	$−297,100$

Substitute the values form Table (I) into Equation (XII) to get,

$−Q¯out={[(0.4052)(−393,520+13,372−9364)]+[(04293)(−241,820+13,356−9904)]+[(0.1855)(0+11,711−8682)]+[(2.441)(0+11,640−8669)]+[(0.00306)(−297,100+4253−0)]}=−253,244 kJ/kmol fuelQ¯out=253,244 kJ/kmol fuel$

Substitute $253,244 kJ/kmol fuel$ for $Q¯out$ , $0.4052 kmol$ for $MC$ , $0.4293 kmol$ for $MH2$ , $0.15920 kmol$ for $MO2$ , $0.00319 kmol$ for $MN2$ , $0.00306 kmol$ for $MS$ , $12 kg/kmol$ for $MC$ , $2 kg/kmol$ for $MH2$ , $32 kg/kmol$ for $MO2$ , $28 kg/kmol$ for $MN2$ and $32 kg/kmol$ for $MS$ in Equation (XIV).

$Qout=253,244 kJ/kmol fuel[(0.4052×12)+(0.4293×2)+(0.1592×32)+(0.00319×28)+(0.00306×32)] kg/kmol=253,244 kJ/kmol fuel11 kg/kmol=23,022 kJ/kg fuel$

Hence, the heat transfer of fuel in the boiler is $23,022 kJ/kg fuel$ .

Answer 9

Chapter 15.7, Problem 58P

To determine

The heat transfer of fuel in the boiler.

Answer 10

Expert Solution & Answer

Answer to Problem 58P

The heat transfer of fuel in the boiler is $23,022 kJ/kg fuel$ .

Answer 11

Expert Solution & Answer

Answer 12

Expert Solution & Answer

Answer 13

Explanation of Solution

Express the number of moles of carbon.

$NC=mfCMC$ (I)

Here, molar mass of carbon is $MC$ and mole fraction of carbon is $mfC$ .

Express the number of moles of hydrogen.

$NH2=mfH2MH2$ (II)

Here, molar mass of hydrogen is $MH2$ and mole fraction of hydrogen is $mfH2$ .

Express the number of moles of oxygen.

$NO2=mfO2MO2$ (III)

Here, molar mass of oxygen is $MO2$ and mole fraction of oxygen is $mfO2$ .

Express the number of moles of nitrogen.

$NN2=mfN2MN2$ (IV)

Here, molar mass of nitrogen is $MN2$ and mole fraction of nitrogen is $mfN2$ .

Express the number of moles of sulfur.

$NS=mfSMS$ (V)

Here, molar mass of sulphur is $MS$ and mole fraction of sulfur is $mfS$ .

Express the total number of moles.

$Nm=NC+NH2+NO2+NN2+NS$ (VI)

Here, number of moles of carbon, hydrogen, oxygen, nitrogen and sulfur is $NC$ , $NH2$ , $NO2$ , $NN2$ and $NS$ respectively.

Express the mole fraction of carbon.

$yC=NCNm$ (VII)

Express the mole fraction of hydrogen.

$yH2=NH2Nm$ (VIII)

Express the mole fraction of oxygen.

$yO2=NO2Nm$ (IX)

Express the mole fraction of nitrogen.

$yN2=NN2Nm$ (X)

Express the mole fraction of sulfur.

$yS=NSNm$ (XI)

Express the heat transfer from the combustion chamber.

$−Q¯out=∑NP(h¯fo+h¯−h¯o)P−∑NR(h¯fo+h¯−h¯o)R=∑NP(h¯fo+h¯400 K−h¯298 Ko)P−∑NR(h¯fo+h¯400 K−h¯298 Ko)R$ (XII)

Here, number of moles of products is $NP$ , number of moles of reactants is $NR$ , enthalpy of formation is $h¯f$ .

Express the enthalpy change of sulfur dioxide between the standard temperature and the product temperature.

$Δh¯SO2=cpΔT=cp(T2−Tstand)$ (XIII)

Here, specific heat constant pressure is $cp$ net temperature is $ΔT$ , temperature at state 2 is $T2$ and temperature at standard condition is $Tstand$ .

Express the heat transfer of fuel in the boiler.

$Qout=Q¯outNCMC+NH2MH2+NO2MO2+NN2MN2+NSMS$ (XIV)

Here, number of moles of carbon, hydrogen, oxygen, nitrogen and sulfur is $MC, MH2$ , $MO2$ , $MN2 and MS$ respectively.

Conclusion:

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

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

$MN2=28 kg/kmol$

Substitute $39.25 kg$ for $mC$ and $12 kg/kmol$ for $MC$ in Equation (I).

$NC=39.25 kg12 kg/kmol=3.271 kmol$

Substitute $6.93 kg$ for $mH2$ and $2 kg/kmol$ for $MH2$ in Equation (II).

$NH2=6.93 kg2 kg/kmol=3.465 kmol$

Substitute $41.11 kg$ for $mO2$ and $32 kg/kmol$ for $MO2$ in Equation (III).

$NO2=41.11 kg32 kg/kmol=1.285 kmol$

Substitute $0.72 kg$ for $mN2$ and $28 kg/kmol$ for $MN2$ in Equation (IV).

$NN2=0.72 kg28 kg/kmol=0.0257 kmol$

Substitute $0.79 kg$ for $mS$ and $32 kg/kmol$ for $MS$ in Equation (V).

$NS=0.79 kg32 kg/kmol=0.0247 kmol$

Substitute $3.271 kmol$ for $NC$ , $3.465 kmol$ for $NH2$ , $1.285 kmol$ for $NO2$ , $0.0257 kmol$ for $NN2$ and $0.00247 kmol$ for $NS$ in Equation (VI).

$Nm=3.271 kmol+3.465 kmol+1.285 kmol+0.0257 kmol+0.0247 kmol=8.071 kmol$

Substitute $3.271 kmol$ for $NC$ and $8.071 kmol$ for $Nm$ in Equation (VII).

$yC=3.271 kmol8.071 kmol=0.4052$

Substitute $3.465 kmol$ for $NH2$ and $8.071 kmol$ for $Nm$ in Equation (VIII).

$yH2=3.465 kmol8.071 kmol=0.4293$

Substitute $1.285 kmol$ for $NO2$ and $8.071 kmol$ for $Nm$ in Equation (IX).

$yO2=1.285 kmol8.071 kmol=0.1592$

Substitute $0.0257 kmol$ for $NN2$ and $8.071 kmol$ for $Nm$ in Equation (X).

$yN2=0.0257 kmol8.071 kmol=0.00319$

Substitute $0.0247 kmol$ for $NS$ and $8.071 kmol$ for $Nm$ in Equation (XI).

$yS=0.0247 kmol8.071 kmol=0.00306$

Express the combustion equation.

$[{0.4052C+0.4293H2+0.1592O2+0.00319N2+0.00306S+1.4ath(O2+3.76N2)}→{0.4052CO2+0.4293H2O+0.4athO2+0.00306SO2+1.4ath×3.76N2}]$ (XV)

Here, stoichiometric coefficient of air is $ath$ , oxygen is $O2$ , nitrogen is $N2$ , carbon dioxide is $CO2$ , water is $H2O$ and hydrogen is $H2$ .

Express the stoichiometric coefficient of air by $O2$ balancing.

$0.1592+1.4ath=0.4052+0.5×0.4293+0.4ath+0.00306ath=0.4637$

Substitute $0.4637$ for $ath$ in Equation (XV).

$[{0.4052C+0.4293H2+0.1592O2+0.00319N2+0.00306S+1.4×0.4637(O2+3.76N2)}→{0.4052CO2+0.4293H2O+0.4×0.4637O2+0.00306SO2+1.4×0.4637×3.76N2}][{0.4052C+0.4293H2+0.1592O2+0.00319N2+0.00306S+0.6492(O2+3.76N2)}→{0.4052CO2+0.4293H2O+0.1855O2+0.00306SO2+2.441N2}]$ (XVI)

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

$MC=0.4052 kmolMH2=0.4293 kmolMO2=0.15920 kmol$

$MN2=0.00319 kmolMS=0.00306 kmol$

Perform unit conversion of temperature at state 2 from degree Celsius to Kelvin.

$T2=127°C=(127+273) K=400 K$

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

$NP,CO2=0.4052 kmolNP,H2O=0.4293 kmolNP,O2=0.1855 kmolNP,SO2=0.00306 kmol$

$NP,N2=2.441 kmol$

Here, number of moles of product carbon dioxide, water, oxygen, sulfur dioxide and nitrogen is $NP,CO2, NP,H2O,NP,O2,NP,SO2 and NP,N2$ respectively.

Take the temperature at standard condition as,

$Tstand=25°C$

Substitute $41.7 kJ/kmol⋅K$ for $cp$ and $127°C$ for $T2$ and $25°C$ for $Tstand$ in Equation (XIII).

$Δh¯SO2=(41.7 kJ/kmol⋅K)(127°C−25°C)=(41.7 kJ/kmol⋅K)[(127−25) K]=4253 kJ/kmol$

Refer Appendix Table A-18, A-19, A-20, A-23 and A-26 and write the property table for products and reactants as in Table (1).

Substance	$hfo¯$ $(kJ/kmol)$	$h¯298 K$ $(kJ/kmol)$	$h¯400 K$ $(kJ/kmol)$
$O2$	0	8682	11,711
$N2$	0	8669	11,640
$H2O(g)$	$−241,820$	9904	13,356
$CO2$	$−393,520$	9364	13,372
$SO2$	$−297,100$

Substitute the values form Table (I) into Equation (XII) to get,

$−Q¯out={[(0.4052)(−393,520+13,372−9364)]+[(04293)(−241,820+13,356−9904)]+[(0.1855)(0+11,711−8682)]+[(2.441)(0+11,640−8669)]+[(0.00306)(−297,100+4253−0)]}=−253,244 kJ/kmol fuelQ¯out=253,244 kJ/kmol fuel$

Substitute $253,244 kJ/kmol fuel$ for $Q¯out$ , $0.4052 kmol$ for $MC$ , $0.4293 kmol$ for $MH2$ , $0.15920 kmol$ for $MO2$ , $0.00319 kmol$ for $MN2$ , $0.00306 kmol$ for $MS$ , $12 kg/kmol$ for $MC$ , $2 kg/kmol$ for $MH2$ , $32 kg/kmol$ for $MO2$ , $28 kg/kmol$ for $MN2$ and $32 kg/kmol$ for $MS$ in Equation (XIV).