Fuel, Air, and Combustion Thermodynamics · Chapter 3 Fuel, Air, and Combustion Thermodynamics 3.1) What is the molecular weight, enthalpy (kJ/kg), and entropy (kJ/kg K) of a gas

Chapter 3

Fuel, Air, and CombustionThermodynamics

3.1) What is the molecular weight, enthalpy (kJ/kg), and entropy (kJ/kg K) of a gas mixture at P = 1000kPa and T = 500 K, if the mixture contains the following species and mole fractions?

a) A table of the given and computed parameters is:

i yi Mi hof hi − ho

f soi yi

(soi − Ru ln(yi)

)[kg/kmol] [MJ/kmol] [MJ/kmol] [kg/kmol−K] [kJ/kmol−K]

CO2 1 0.10 44.01 -393.52 33.40 269.30 28.84H2O 2 0.15 18.01 -241.83 26.00 232.74 37.28N2 3 0.70 28.01 0 21.46 228.17 161.79CO 4 0.05 28.01 -110.53 21.69 234.54 12.97

The mixture molecular mass is:

M =∑

yiMi = (0.10)(44.01) + (0.15)(18.01) + (0.70)(28.01) + (0.05)(28.01)

M = 28.11 kg/kmol

The specific mixture enthalpy is:

hf =∑

yihi = (0.10)(−393.52 + 33.40) + (0.15)(−241.83 + 26.00)

+ (0.70)(0 + 21.46) + (0.05)(−110.53 + 21.69)

hf = −57.8 MJ/kmol = −5.78× 104 kJ/kmol

The mixture enthalpy is:

h =h

m=−5.78× 104

28.11h = −2056 kJ/kg

The specific mixture entropy is:

s = −Ru ln

(P

P0

)+ yi (soi −Ru ln(yi))

s = (−8.314) ln

(1000

100

)+ 28.84 + 37.28 + 161.79 + 12.97

s = 221.74 kJ/kmol−K

1

2 CHAPTER 3. FUEL, AIR, AND COMBUSTION THERMODYNAMICS

The mixture entropy is:

s =s

m=

103.23

28.11s = 7.888 kJ/kg−K

3

3.2) What is the enthalpy h (kJ/kg) and entropy s (kJ/kg-K) of a mixture of 30% H2 and 70% CO2 byvolume at a temperature of 3000 K ?


i yi Mi hof hi − ho

f soi yi

(soi − Ru ln(yi)

)[kg/kmol] [MJ/kmol] [MJ/kmol] [kg/kmol−K] [kJ/kmol−K]

CO2 1 0.70 44.01 -393.52 152.85 334.17 235.99H2 2 0.30 2.016 0 88.72 202.90 63.87


M =∑

yiMi = (0.70)(44.01) + (0.30)(2.016)

M = 31.41 kg/kmol

The specific mixture enthalpy is:

hf =∑

yihi = (0.70)(−393.52 + 152.85) + (0.30)(0 + 88.73)

= −14.8 MJ/kmol = −1.418× 105 kJ/kmol

The mixture enthalpy is:

h =h

m=−1.418× 105

31.41h = −4516 kJ/kg

The specific mixture entropy is:

s = −Ru ln

(P

P0

)+ yi (soi −Ru ln(yi))

s = (−8.314) ln

(2000

100

)+ 235.99 + 63.87

s = 296.86 kJ/kmol−K

The mixture entropy is:

s =s

m=

296.86

31.41s = 9.451 kJ/kg−K


3.3) Using the Gordon and McBride equations, Equations (3.38) and (3.39), calculate the enthalpy h andentropy so of CO2 and compare with the gas table values used in Example 3.1.

A Matlab program for calculating the enthalpy h and entropy so of CO2 is

t=1000; % temp in K

R=8.31451 % univ. gas const.

a1=2.4007797;

a2=8.73509757e-3;

a3=-6.60707878e-6;

a4=2.0021861e-9;

a5=6.3274039e-16;

a6=-4.8377527e4;

a7=9.6951457;

nondimh=a1+a2/2*t+a3/3*t^2+a4/4*t^3+a5/5*t^4+a6/t

h=nondimh*t*R

nondims=a1*log(t)+a2*t+a3/2*t^2+a4/3*t^3+a5/4*t^4+a7

s=nondims*R

The calculated h = −3.6011 × 105 kJ/kmol and the entropy so = 269.21 kJ/kmol-K. The gas tablevalues are h = −3.6012 × 105 kJ/kmol and the entropy so = 269.30 kJ/kmol-K. The enthalpy valuesagree to 4 figures, and the entropy values agree to 3 figures.

5

3.4) Using the program Fuel.m, at what temperature is the specific heat cp of methane CH4 = 3.0 kJ/kg-K?


3.5) Why does Equation 3.27 contain yi ?

a) From the Gibbs equations, we have

si(T, P ) = soi (T )−Riln(PiP0

)Where soi (T ) is the standard entropy.

yi =PiP

=PiP0· P0

P

ln(yi) = ln

(PiP0

)+ ln

(P0

P

)ln

(PiP0

)= ln(yi) + ln

(P

P0

)Substituting back in:

si(T, P ) = soi (T )−Ri(

ln(yi) + ln

(P

P0

))The mole fraction yi originates from the pressure ratio term.

7

3.6) A system whose composition is given below is in equilibrium at P = 101 kPa and T = 298 K. Whatare the enthalpy (kJ/kg), specific volume (m3/kg), and quality χ of the mixture?


i yi Mi hf hi xi vi

[kg/kmol] [kJ/kmol] [kJ/kg][m3/kg]

H2O 1 0.141 18.01 -241,826 -13,424 0.0888 1.362CO2 2 0.125 44.01 -393,522 -8942 0.1924 0.557N2 3 0.734 28.01 0 0 0.7189 0.875


M =∑

yiMi = (0.141)(18.01) + (0.125)(44.01) + (0.734)(28.01)

= 28.60 kg/kmol

Sample calculations for CO2 (i=2):

h2 =hfMi

=−393,522

44.01= −8942 kJ/kg

x2 = y2

(Mi

M

)= 0.125

44.01

28.60= 0.1924

v2 =Ru · TMi · P

=8.314 · 298

44.01 · 101= 0.557 m3

/kg

If all of the water is vapor, the water partial pressure would be

PH2O = yi · P = 0.141(101) = 14.2 kPa

Whoever, at T = 298K, the saturation pressure of the water is 3.17 kPa (Table 3.1), so some fractionof the water will be liquid. The quality of the gas mixture is given in the Appendix:

X =

(1

yH2O− 1

)(P

Psat− 1

)−1=

(1

0.141− 1

)(101

3.17− 1

)−1X = 0.197

The enthalpy of vaporization hfg is 44.02/18.01 = 2444 kJ/kg (Table 3.1)The term vfg is vg-vf = (1.362-0.001) = 1.361 m3/kg (Table 3.1)

Therefore using equations (3.55) and (3.57)

The enthalpy is given by:

h = x1h1 + x2h2 + x3h3 − (1−X)x1hfg

h = (0.0888)(−13,424) + (0.1924)(−8942) + 0− (1− 0.197)(0.0888)(2444)

h = −3087 kJ/kg

The specific volume is given by:

v = x1v1 + x2v2 + x3v3 − (1−X)x1vfg

v = (0.0888)(1.362) + (0.1924)(0.557) + (0.7189)(0.875)− (1− 0.197)(0.0888)(1.361)

v = 0.760 m3/kg


3.7) A four cylinder four stroke 2.8 L port injected spark ignition engine is running at 2000 rpm on a lean(φ =0.9) mixture of octane and standard air (101 kPa, 298 K) If the octane flow rate is 2.5 g/s, whatis the mass of fuel entering each cylinder per cycle and the volumetric efficiency ?

a) The mass of fuel entering each cylinder per cycle for a four stroke engine

mf = mf

(2

N

)(1

nc

)= 2.5

(2 · 60

2000

)(1

4

)mf = 3.75× 10−2 g

b) Since the engine is port injected

ev =ma +mf

ρivd=mf (AF + 1)

ρivd=mf (AFs

φ + 1)

ρivd

From Table 3.5, the stoichiometric air-fuel ratio for octane is AFs = 15.03. Assume R = 0.287

ρi =P

RT=

101

(0.287)(298)= 1.18 kg/m3 = 1180 g/m3

Solving for the volumetric efficiency:

ηv =(3.75× 10−2)

(15.030.9 + 1

)(1180)

(2.8× 10−3

4

)ηv = 0.80

9

3.8) An engine cylinder has a 90 mm bore and a 85 mm stroke, and contains air and residual gases at 350K and 1 bar. If the engine is to operate on diesel fuel and run lean with an overall equivalence ratioof φ = 0.7, what is the mass of diesel fuel that needs to be injected during the compression stroke ?(Assume R of the air-residual gas mixture is 0.29 kJ/kg K).

a)

m = ma +mr

We know that

f =mr

m

Substituting we have

ma = (1− f)m

Since

φ =FA

FAs

Substituting

mf = φ(FAs)(ma) = φ(FAs)(1− f)m

Using the ideal gas law

mf =PvdRT

=P(π4

)b2s

RT=

100(π4

)(0.09)2(0.085)

(0.29)(350)(1000)

mf = 0.53 g

From Table 3.5, AFs=14.30 or FAs=0.0699

The mass of injected diesel fuel is:

mf = (0.7)(0.0699)(1− 0.015)(0.53)

mf = 2.5× 10−2 g


3.9) Using the low temperature (T < 1000K) combustion equations, what are the composition, enthalpy,and entropy of the combustion products of methanol, CH3OH, at φ = 1.1, T = 1200 K, and P = 101kPa? Compare with the results from the program ecp.m.

11

3.10) What are the mole fractions of CO2, H2O, CO, N2and H2 produced when methane (CH4) is burnedin rich conditions at φ = 1.1, T = 1000 K, and P = 101 kPa ?


3.11) If a lean (φ = 0.8) mixture of methane CH4 is burned at a temperature of 1500 K and pressure of500 kPa, what are the mole fractions of the products, and the product enthalpy, entropy, and specificheat? Use the program ecp.m.

13

3.12) At what temperature does the saturation pressure Psat of an octane droplet equal 0.5 bar ? At thattemperature, what is the enthalpy of vaporization hfg ?

a) Antoine’s equation is:

log10 (Psat) =

[a− b

T + c

]Solving for T and using coefficients from Table 3.3

T =b

a− log10 (Psat)− c

Using coefficients from Table 3.3

T =1355.1

4.0487− log10 (0.5)− (−63.633) = 375 K (octane)

T =1739.6

4.1373− log10 (0.5)− (−105.62) = 498 K (tetradecane)

This shows the greater volatility of octane relative to tetradecane

b) The molar enthalpy of vaporization hfg is:

hfg = Ae−dT/Tc

(1− T

Tc

)βUsing coefficients from Table 3.4:

hfg = 58.46 exp

(−0.1834 · 375

568.8

)(1− 375

568.8

)0.3324

= 36.2 MJ/kmol (octane)

hfg = 95.66 exp

(−0.2965 · 498

694

)(1− 498

694

)0.2965

= 53.1 MJ/kmol (tetradecane)

The octane value compares well with the value of 36.4 MJ/kmol in Table 3.2. The tetradecanerequires about 47% more energy to vaporize.


3.13) Compare the enthalpy of vaporization hfg (MJ/kmol) of nitromethane, methanol, octane, and tetrade-cane at 400 K.

a) The molar enthalpy of vaporization hfg is

hfg = A exp

(−α · T

Tc

)(1− T

Tc

)βUsing coefficients from Table 3.4 and molecular mass from Table 3.5

For Nitromethane:

hfg = (53.33) exp

(−0.2732 · 400

588

)(1− 400

588

)0.2732

= 32.43 MJ/kmol =32.43

61.04

hfg = 0.53 MJ/kg

For Methanol:

hfg = (45.30) exp

(0.31 · 400

512.6

)(1− 400

512.6

)0.4241

= 30.34 MJ/kmol =30.34

32.04

hfg = 0.95 MJ/kg

For Octane:

hfg = (58.46) exp

(−0.1834 · 400

568.8

)(1− 400

568.8

)0.3324

= 34.31 MJ/kmol =34.31

114.22

hfg = 0.30 MJ/kg

For Tetradecane:

hfg = (95.66) exp

(−0.2965 · 400

694

)(1− 400

694

)0.2965

= 62.50 MJ/kmol =62.50

198.39

hfg = 0.31 MJ/kg

The “charge cooling” effect of the vaporization of methanol and nitromethane is 2-3 times thatof octane and tetradecane on a per kg basis. The charge cooling produces greater volumetricefficiencies by increasing the density of the cylinder gases.

15

3.14) a.) If a rich (φ = 1.1) mixture of diesel fuel is burned at a temperature of 2000 K and pressure of 750kPa, what are the mole fractions of the products, and the product enthalpy, entropy, specific volume,and specific heat? b.) Repeat the calculation for φ = 1.25. Discuss the effects of equivalence ratio.


3.15) Using the program ecp.m, plot the product equilibrium mole fractions as a function of equivalenceratio (0.5 < φ < 2) resulting from the combustion of methane at 5000 kPa and 2500 K.

17

3.16) Derive Equation (??) for the species mole fractions of a mixture of air and residual gas.

a) The mixture m is composed of the residual mass mr and the premixed fuel-air mfa

m = mf +mfa

The residual mole fraction yr is

yr =nr

nfa + nr=

1nfa

nr+ 1

Since the residual fraction f = mr/m

1

f=

m

mr=mr +mfa

mr= 1 +

mfa

mr

or

mfa

mr=

1

f− 1

The mole ratio is

nfanr

=mfa

Mfa· Mr

mr=mfa

mr· M

′′

M ′=

(1

f− 1

)M ′′

M ′

So

yr =

[1 +

M ′′

M ′

(1

f− 1

)]−1The species mole fractions yi are

yi =niN

=ninfa· nfaN

+ninr· nrN

=

(ninfa

)yfa +

(ninr

)yr

Since the sum of the residual fractions must equal 1

yfa = 1− yr

yi′ =

ninfa

yi′′ =

ninr

So the species mole fractions accounting for both the residual gas and the inlet fuel- air

yi = (1− yr) yi′ + (yr) yi′′


3.17) At what equivalence ratio for octane-air mixtures does the carbon to oxygen ratio of the system equalone? Why is this of interest?

a) The combustion equation is

C8H18 +asφ

(O2 + 3.76N2) −→ Products

From Table 3.5, as = 12.50 for octane.For every mole of octane, there are 8 carbon atoms and 2as/φ oxygen atoms. The carbon/oxygenratio is:

[C]

[D]=

8

2(asφ

)So this ratio is equal to one, i.e. CO formation,

φ =2as8

=(2)(12.50)

8= 3.125

Therefore for φ > 3.125, there will be solid carbon in the products, since the carbon atoms are inexcess of those used to form CO.

19

3.18) At what temperature is the concentration of H2 a minimum for the combustion of gasoline and air atφ = 1.2 and 4500 kPa? What is that minimum value of H2 ?


3.19) At what equivalence ratio is the concentration of OH a maximum for the combustion of diesel and airat T = 2500 K and 4500 kPa? What is that maximum value of OH ?

21

3.20) At what temperature does the mole fraction of NO reach 0.010 for the equilibrium products resultingfrom the combustion of gasoline and air at φ = 1.0 and 5000 kPa?


3.21) At what temperature does the mole fraction of CO reach 0.080 for the equilibrium products resultingfrom the combustion of methane and air at φ = 1.1 and 3000 kPa?

23

3.22) What is the equilibrium and the frozen specific heat cp of the combustion products of gasoline at apressure of 2000 kPa and temperature of 2000 K burned at a.) an equivalence ratio of 1.1, and b.) anequivalence ratio of 0.9 ?

Fuel, Air, and Combustion Thermodynamics · Chapter 3 Fuel, Air, and Combustion Thermodynamics 3.1) What is the molecular weight, enthalpy (kJ/kg), and entropy (kJ/kg K) of a gas

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