THERMODYNAMICS(SI UNITS,INTL.ED)EBOOK>I
THERMODYNAMICS(SI UNITS,INTL.ED)EBOOK>I
8th Edition
ISBN: 9781307434316
Author: CENGEL
Publisher: INTER MCG
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Chapter 15.7, Problem 99RP

(a)

To determine

The theoretical air fuel ratio required for the combustion reaction.

(a)

Expert Solution
Check Mark

Answer to Problem 99RP

The theoretical air fuel ratio required for the combustion reaction is 14.83kgair/kg fuel.

Explanation of Solution

Write the chemical reaction for the combustion of liquid gas fuel mixture of 90% octane and 10% alcohol with 200% excess theoretical amount of air.

0.9C8H18(l)+0.1C2H5OH+2ath(O2+3.76N2)xCO2+yH2O+zO2+wN2 (I)

Write the expression to calculate the theoretical air-fuel ratio (AFth).

AFth=mairmfuelAFth=(NairMair)(NC8H18MC8H18)+(NC2H5OHMC2H5OH) (II)

Here, mass of the air is mair, mass of the fuel is mfuel, number of moles of air is Nair, molar mass of air is Mair, number of moles of octane is NC8H18, molar mass of octane is MC8H18, number of moles of alcohol is NC2H5OH, and molar mass of alcohol is MC2H5OH.

Conclusion:

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

x=8×0.9+2×0.1x=7.4

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

18×0.9+6×0.1=2yy=8.4

Balance for the excess Oxygen from Combustion reaction Equation (I).

ath=z

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

0.1(1)+2×2ath=2x+y+2z

Substitute 7.4 for x, 8.4 for y, and ath for z.

0.1(1)+2×2ath=2(7.4)+8.4+2(ath)ath=11.55

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

3.76×2ath=w

Substitute 11.55 for ath.

3.76×2(11.55)=ww=86.86

Rewrite the complete balanced chemical reaction for combustion as follows:

0.9C8H18(l)+0.1C2H5OH+2(11.55)(O2+3.76N2)7.4CO2+8.4H2O+11.55O2+86.86N2

0.9C8H18(l)+0.1C2H5OH+23.1(O2+3.76N2)7.4CO2+8.4H2O+11.55O2+86.86N2 (III)

From the table A-2 of “Molar mass, gas constants, and critical-point properties”, select the molar masses of carbon dioxide, water, air, oxygen, nitrogen, octane and alcohol as,

MCO2=44kg/kmolMair=29kg/kmolMH2O=18kg/kmolMO2=32kg/kmol

MN2=28kg/kmolMC8H18=114kg/kmolMC2H5OH=46kg/kmol

Substitute (11.55×4.76kmol) for Nair, 29kg/kmol for Mair, 0.9kmol for NC8H18, 114kg/kmol for MC8H18, 0.1kmol for NC2H5OH, and 46kg/kmol for MC2H5OH in Equation (II).

AFth=(11.55×4.76kmol)(29kg/kmol)(0.9kmol)(114kg/kmol)+(0.1kmol)(46kg/kmol)=14.83kgair/kg fuel

Thus, the theoretical air fuel ratio required for the combustion reaction is 14.83kgair/kg fuel.

(b)

To determine

The ratio of products of the combustion to the fuel.

(b)

Expert Solution
Check Mark

Answer to Problem 99RP

The ratio of products of the combustion to the fuel is 30.54kg product/kg fuel.

Explanation of Solution

Write the formula to calculate the molar mass of the product gases (Mprod) .

Mprod=NCO2MCO2+NH2OMH2O+NO2MO2+NN2MN2NCO2+NH2O+NO2+NN2 (IV)

Here, number of moles of carbon dioxide is NCO2, molar mass of carbon dioxide is MCO2, number of moles of water is NH2O, molar mass of hydrogen is MH2O, number of moles of oxygen is NO2, molar mass of oxygen is MO2, number of moles of nitrogen is NN2, and molar mass of hydrogen is MN2.

Write the formula to calculate the mass of the product gases per unit mass of fuel (mprod).

mprod=NCO2+NH2O+NO2+NN2(NC8H18MC8H18)+(NC2H5OHMC2H5OH) (V)

Conclusion:

Substitute 7.4kmol for NCO2, 44kg/kmol for MCO2, 8.4kmol for NH2O, 18kg/kmol for MH2O, 11.55kmol for NO2, 32kg/kmol for MO2, 86.86kmol for NN2, and 28kg/kmol for MN2 in Equation (IV).

Mprod={(7.4kmol)(44kg/kmol)+(8.4kmol)(18kg/kmol)+(11.55kmol)(32kg/kmol)+(86.86kmol)(28kg/kmol)}7.4kmol+8.4kmol+11.55kmol+86.86kmol=28.72kg/kmol

Substitute 7.4kmol for NCO2, 8.4kmol for NH2O, 11.55kmol for NO2, 86.86kmol for NN2, 0.9kmol for NC8H18, 114kg/kmol for MC8H18, 0.1kmol for NC2H5OH, and 46kg/kmol for MC2H5OH in Equation (V).

mprod=7.4kmol+8.4kmol+11.55kmol+86.86kmol(0.9kmol)(114kg/kmol)+(0.1kmol)(46kg/kmol)=30.54kg product/kg fuel

Thus, the ratio of products of the combustion to the fuel is 30.54kg product/kg fuel.

(c)

To determine

The mass flow rate of the air required for the combustion.

(c)

Expert Solution
Check Mark

Answer to Problem 99RP

The mass flow rate of the air required for the combustion is 148.3kg/s.

Explanation of Solution

Write the formula to calculate the actual air-fuel ratio (AFact).

AFact=2(AFth) (VI)

Write the formula to calculate the mass flow rate of air (m˙air).

m˙air=AFact(m˙fuel) (VII)

Here, mass flow rate of fuel is m˙fuel.

Conclusion:

Substitute 14.83kgair/kg fuel for AFth in Equation (VI).

AFact=2(14.83kgair/kg fuel)=29.65kgair/kg fuel

Substitute 29.65kgair/kg fuel for AFact, and 5kg/s for m˙fuel in Equation (VII).

m˙air=(29.65kgair/kg fuel)(5kg/s)=148.3kg/s

Thus, the mass flow rate of the air required for the combustion is 148.3kg/s.

(d)

To determine

The lower heating value of the fuel mixture.

(d)

Expert Solution
Check Mark

Answer to Problem 99RP

The lower heating value of the fuel mixture is 43,760kJ/kg.

Explanation of Solution

Write the expression to calculate the equation for steady-flow energy balance.

HR=q¯LHV+HP (VIII)

Here, enthalpy of product is HP, lower heating value is q¯LHV, and enthalpy of reactant is HR.

Write the expression to calculate the enthalpy of reactant (HR).

HR={0.9(h¯foh¯298 K)C8H18+0.1(h¯foh¯298 K)C2H5OH+23.1(h¯fo)O2+86.86(h¯fo)N2} (IX)

Here, enthalpy of vaporization is h¯fo and the enthalpy of formation is h¯298 K.

Write the expression to calculate the enthalpy of product (HP).

HP=7.4(h¯fo)CO2+8.4(h¯fo)H2O+11.55(h¯fo)O2+86.86(h¯fo)N2 (X)

Calculate the lower heating value on mass basis (qLHV).

qLHV=q¯LHV(NC8H18MC8H18)+(NC2H5OHMC2H5OH) (XI)

Write the formula to calculate the lower heating value on mass basis (qLHV).

qLHV=q¯LHV(NC8H18MC8H18)+(NC2H5OHMC2H5OH) (XII)

Conclusion:

From the Table A-18 through Table A-26, obtain the enthalpies of vaporization and enthalpies of formation for different substances as in Table 1.

Substance

Enthalpy of vaporization

h¯fo (kJ/kmol)

Enthalpy of formation at 298 K

h¯298K (kJ/kmol)

C8H18130,84741,465
C2H5OH235,31042,340
O208682
N208669
CO2393,5209364
H2O(g)241,8209904

Substitute 130,847kJ/kmol for (h¯fo)C8H18, 41,465kJ/kmol for (h¯298K)C8H18, 235,310kJ/kmol for (h¯fo)C2H5OH, 42,340kJ/kmol for (h¯298K)C2H5OH, 0 for (h¯fo)O2, and 0 for (h¯fo)N2 in Equation (IX).

HR={0.9(130,847kJ/kmol41,465kJ/kmol)+0.1(235,310kJ/kmol42,340kJ/kmol)+23.1(0)+86.86(0)}=252,697kJ/kmol

Substitute 393,520kJ/kmol for (h¯fo)CO2, 241,820kJ/kmol for (h¯fo)H2O, 0 for (h¯fo)O2, and 0 for (h¯fo)N2 in Equation (X).

HP={7.4(393,520kJ/kmol)+8.4(241,820kJ/kmol)+11.55(0)+86.86(0)}=4.943×106kJ/kmol

Substitute 252,697kJ/kmol for HR, and 4.943×106kJ/kmol for HP in Equation (XI).

252,697kJ/kmol=q¯LHV+(4.943×106kJ/kmol)

q¯LHV=4.943×106kJ/kmol252,697kJ/kmol=4.691×106kJ/kmol

Substitute 4.691×106kJ/kmol for q¯LHV, 0.9kmol for NC8H18, 114kg/kmol for MC8H18, 0.1kmol for NC2H5OH, and 46kg/kmol for MC2H5OH in Equation (XII).

qLHV=4.691×106kJ/kmol(0.9kmol)(114kg/kmol)+(0.1kmol)(46kg/kmol)=43,760kJ/kg

Thus, the lower heating value of the fuel mixture is 43,760kJ/kg.

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Chapter 15 Solutions

THERMODYNAMICS(SI UNITS,INTL.ED)EBOOK>I

Ch. 15.7 - What are the causes of incomplete combustion?Ch. 15.7 - Which is more likely to be found in the products...Ch. 15.7 - Methane (CH4) is burned with the stoichiometric...Ch. 15.7 - Prob. 14PCh. 15.7 - n-Butane fuel (C4H10) is burned with the...Ch. 15.7 - Prob. 16PCh. 15.7 - Prob. 17PCh. 15.7 - 15–18 n-Octane (C8H18) is burned with 50 percent...Ch. 15.7 - In a combustion chamber, ethane (C2H6) is burned...Ch. 15.7 - Prob. 20PCh. 15.7 - Prob. 21PCh. 15.7 - 15–22 One kilogram of butane (C4H10) is burned...Ch. 15.7 - 15–23E One lbm of butane (C4H10) is burned with 25...Ch. 15.7 - Prob. 24PCh. 15.7 - A fuel mixture of 60 percent by mass methane (CH4)...Ch. 15.7 - A certain natural gas has the following volumetric...Ch. 15.7 - Prob. 27PCh. 15.7 - A gaseous fuel with a volumetric analysis of 45...Ch. 15.7 - Prob. 30PCh. 15.7 - 15–31 Octane (C8H18) is burned with dry air. The...Ch. 15.7 - Prob. 32PCh. 15.7 - Prob. 33PCh. 15.7 - Prob. 34PCh. 15.7 - Prob. 35PCh. 15.7 - Prob. 36PCh. 15.7 - Prob. 37PCh. 15.7 - Prob. 38PCh. 15.7 - Prob. 39PCh. 15.7 - Prob. 40PCh. 15.7 - Prob. 41PCh. 15.7 - Prob. 42PCh. 15.7 - Prob. 44PCh. 15.7 - Repeat Prob. 1546 for liquid octane (C8H18).Ch. 15.7 - Ethane (C2H6) is burned at atmospheric pressure...Ch. 15.7 - Reconsider Prob. 1550. What minimum pressure of...Ch. 15.7 - Calculate the HHV and LHV of gaseous n-octane fuel...Ch. 15.7 - Prob. 49PCh. 15.7 - Prob. 50PCh. 15.7 - Consider a complete combustion process during...Ch. 15.7 - Prob. 53PCh. 15.7 - Prob. 54PCh. 15.7 - Propane fuel (C3H8) is burned with an airfuel...Ch. 15.7 - 15–56 Hydrogen (H2) is burned completely with the...Ch. 15.7 - Prob. 57PCh. 15.7 - Prob. 58PCh. 15.7 - Octane gas (C8H18) at 25C is burned steadily with...Ch. 15.7 - Prob. 61PCh. 15.7 - Liquid ethyl alcohol [C2H5OH(l)] at 25C is burned...Ch. 15.7 - Prob. 63PCh. 15.7 - Prob. 64PCh. 15.7 - A constant-volume tank contains a mixture of 120 g...Ch. 15.7 - Prob. 67PCh. 15.7 - Prob. 68PCh. 15.7 - Prob. 69PCh. 15.7 - A fuel is completely burned first with the...Ch. 15.7 - Prob. 71PCh. 15.7 - Acetylene gas (C2H2) at 25C is burned during a...Ch. 15.7 - Octane gas (C8H18) at 25C is burned steadily with...Ch. 15.7 - Express the increase of entropy principle for...Ch. 15.7 - Prob. 81PCh. 15.7 - What does the Gibbs function of formation gf of a...Ch. 15.7 - Liquid octane (C8H18) enters a steady-flow...Ch. 15.7 - Benzene gas (C6H6) at 1 atm and 77F is burned...Ch. 15.7 - Prob. 87PCh. 15.7 - Prob. 88PCh. 15.7 - A steady-flow combustion chamber is supplied with...Ch. 15.7 - Prob. 91RPCh. 15.7 - 15–92 A gaseous fuel with 80 percent CH4, 15...Ch. 15.7 - Prob. 93RPCh. 15.7 - Prob. 94RPCh. 15.7 - Prob. 95RPCh. 15.7 - Prob. 96RPCh. 15.7 - Prob. 97RPCh. 15.7 - Prob. 98RPCh. 15.7 - Prob. 99RPCh. 15.7 - Prob. 100RPCh. 15.7 - A 6-m3 rigid tank initially contains a mixture of...Ch. 15.7 - Prob. 102RPCh. 15.7 - Propane gas (C3H8) enters a steady-flow combustion...Ch. 15.7 - Determine the highest possible temperature that...Ch. 15.7 - Prob. 106RPCh. 15.7 - Prob. 107RPCh. 15.7 - A steam boiler heats liquid water at 200C to...Ch. 15.7 - Repeat Prob. 15112 using a coal from Utah that has...Ch. 15.7 - Liquid octane (C8H18) enters a steady-flow...Ch. 15.7 - Prob. 111RPCh. 15.7 - Prob. 112RPCh. 15.7 - Prob. 113RPCh. 15.7 - Consider the combustion of a mixture of an...Ch. 15.7 - A fuel is burned steadily in a combustion chamber....Ch. 15.7 - A fuel is burned with 70 percent theoretical air....Ch. 15.7 - Prob. 123FEPCh. 15.7 - One kmol of methane (CH4) is burned with an...Ch. 15.7 - An equimolar mixture of carbon dioxide and water...Ch. 15.7 - The higher heating value of a hydrocarbon fuel...Ch. 15.7 - Acetylene gas (C2H2) is burned completely during a...Ch. 15.7 - Prob. 129FEPCh. 15.7 - A fuel is burned during a steady-flow combustion...
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