Chapter 9

Fuels

9.1) What is the chemical structure of (a) 3-methyl-3-ethylpentane, and (b) 2, 4-diethylpentane?

a) 3-methyl-3-ethylpentane
This pentane has methyl CH3 and an ethyl C2 H5 on the third carbon. Referencing Section 9-2,
a pentane parraffin has five single bonded carbon atoms. Its chemical formula is C8 H18 , so it is
an isomer of octane. Its structure is:

CH3

C C C C C

C2 H5
b) 2, 4-diethylpentane
This pentane has diethyl C2 H5 on the second and fourth carbon. From Section 9-2, a pentane
parraffin has five single bonded carbon atoms. Its chemical formula is C9 H20 , so it is an isomer
of nonane. Its structure is:

C2 H5 C2 H5

C C C C C

1
2 CHAPTER 9. FUELS

9.2) If a hydrocarbon fuel is represented by the general formula Cx H2 x, what is its stoichiometric mass
air-fuel ratio?

The stoichiometric combustion equation is

   
3x 3x
1Cx H2x + (O2 + 3.76N2 ) → xCO2 + xH2 O + 3.76N2
2 2

The stoichiometric mass air-fuel ratio is

mair
AFs =
mf uel

nair ∗ Mair
=
nf uel ∗ Mf uel

4.76 ∗ 3x/2 ∗ 28.85

=
x ∗ 12.01 + 2x ∗ 1.008

= 14.7
 3

9.3) A fuel has the following composition by mass: 10% pentane, 35% heptane, 30% octane, and 25%
dodecane. If its general formula is of the form Cx Hy , find x and y.

The mass fractions need to be converted to mole fractions. The number of moles of species i per kg of
mixture is
ni mi/m xi
= =
m Mi Mi
P
The total number of moles is n = ni and the total mole fraction yi = ni/n

The mass fraction to mole fraction is shown in the table:

Species F ormula C atoms xi Mi ni/m yi

Pentane C5 H12 5 0.10 72 1.388 × 10−3 0.1544
Heptane C7 H16 7 0.35 100 3.500 × 10−3 0.3893
Octane C8 H18 8 0.30 114 2.630 × 10−3 0.2927
Dodecane C12 H26 12 0.25 170 1.470 × 10−3 0.1636
Total 1 8.99 × 10−3 1

The number of carbon atoms in the fuel’s chemical formula is

X
nc = yi Ci
= 0.1544 ∗ 5 + 0.3893 ∗ 7 + 0.2927 ∗ 8 + 0.1636 ∗ 12
= 7.8

Since the components are paraffins with formula Cn H2n+2

The number of hydrogen atoms in the fuel is

nH = 2nc + 2 = 17.6

Therefore the chemical formula is C7.8 H17.6

4 CHAPTER 9. FUELS

9.4) If the mass composition of a hydrocarbon fuel mixture is 55% paraffins, 30% aromatics, and 15%
monoolefins, what is its specific heat? Assume T = 1000 K.

Assume that T = 1000 K, the mass composition is 0.55 paraffin, 0.30 aromatics, and 0.15 monoolefins
From Equation 9.3
X
cp = xi cpi

And from Equation 9.2 with t = f racT (K)1000

cpi = ai + bi t + ci t2

Therefore,

cp = 0.55(0.33 + 5.0 − 1.5) + 0.3(0.21 + 4.2 − 1.3) + 0.15(0.33 + 4.6 − 1.3)

= 2.1 + 0.93 + 0.54
= 3.58 kJ/kg−K
 5

9.5) Compute the enthalpy of formation of C8 H18 .

On a per mole basis, Equation 9.1 is

b
hof = q i + ahf,CO2 +

hf,H2 O − (1 − x)hf g,H2 O
2

for C8 H18 a = 8 and b = 18

From the Tables in Chapter 3, assuming the higher heat of combustion with x= 0.

q c,l = qc,l · M = 47.9 · 114 = 5.47 × 106 kJ/kmol

hf,CO2 = −393,522 kJ/kmol
hf,H2 O = −241,826 kJ/kmol
hf g,H2 O = 44.02 kJ/kmol
so
18
hf (v) = (5.4711 × 106 ) + 8(−393,522) + (−241,826 − 1 · 44020)
2
= −2.4969 × 105 kJ/kmol
= −249.69 kJ/mol for vapor (v)

and for liquid (l) where hf g = −41.51 kJ/mol

hf (l) = hf (v) − hf g
= −249.66 − (−41.51)
= −208.18 kJ/mol

The value agrees closely with the value of −208.18 kJ/mol given in Chapter 3.
6 CHAPTER 9. FUELS

9.6) A four-stroke engine operates on methane with an equivalence ratio of 0.9. The air and fuel enter the
engine at 298 K, and the exhaust is at 800 K. The heat rejected to the coolant is 350 MJ/kmolf uel .
(a) What is the enthalpy of the exhaust combustion products? (b) What is the specific work output
of the engine? (c) What is the first law efficiency of the engine?

From the first law,

q − w = he − hi

Since the air and fuel enter at 298 K, the inlet enthalpy is the heat of formation of methane

hf −74,900
hi = hf = = = −4681 kJ/kgfuel
M 16
a) The exhaust enthalpy is found from the RunExp.m program with T = 800 K and P = 100 kPa

he = −2144 kJ/kgmix
Converting to a per kg of fuel basis, the actual air-fuel ratio is
AFS 17.12
AF = = = 19.02
φ 0.9

with AF + 1 = 20.02 kgmix/kgfuel

he = −2144 kJ/kgmix ∗ 20.02 kgmix/kgfuel = −42,923 kJ/kgfuel

Converting the heat transfer to a per kg fuel basis,

q −350,000 kJ/kmolfuel
q= = = −21,875 kJ/kgfuel
M 16

b) The first law is therefore

w = q − he + hi = −21,875 − (−42,863) + (−4681) = −16,300 kJ/kgfuel

c) The first law efficiency is

w 16,300
η= = = 0.326
qc 50,010
 7

9.7) A fuel blend has a density of 700 kg/m3 and a midpoint boiling temperature of 90◦ C. What is its
cetane index?

The density is given as 700 kg/m3 and the midpoint boiling temperature T50 = 90 ◦ C = 194 ◦ F

The calculated cetane index is

CCI = −420.34 + 0.016G2 + 0.192G log T50 + 65.01(log T50 )2 − 0.0001809T50

2

Where G is the API gravity

141.5
G= − 131.5
SG@60F
ρ 700
The specific gravity SG = ρH2 O = 1000 = 0.70, therefore,
141.5
G= − 131.5 = 70.64
0.70
CCI = 23.99
8 CHAPTER 9. FUELS

9.8) What is the decrease in volumetric efficiency of a 5 L gasoline fueled engine when it is retrofitted to
operate with propane ? The decrease is due to the displacement of a portion of the inlet air by the
propane fuel. Assume standard temperature and pressure inlet air conditions.

If a gaseous fuel is mixed with air in the inlet manifold, it will displace some fraction of the incoming
air, reducing the volumetric efficiency. (This is an advantage of direct injection into the cylinder).

The volumetric efficiency is

ma + mf ma (1 + F A) ma (1 + φF As )
ev = = =
ρi vd ρi vd ρi vd

If we label the air mass of the unretrofitted engine as m′a , the displacement of a portion of the air by
the gas is

m′a = ma + mf = (1 + F A)ma = (1 + φF As )ma

therefore,
ma 1
=
ma
′ 1 + φF As

For the case of stoichiometric operation with φ = 1, and for propane, from Table 3.5.
1 1
F As = = = 0.0642
AFS 15.57
ma 1
= = 0.940
m′a 1 + 1 · 0.0642

There will be a 6% decrease in the volumetric efficiency.

 9

9.9) Repeat Problem 9.8 for hydrogen and methane.

Based on Problem 9.8, the volumetric efficiency is directly proportional to the mass of the inlet air.

ma (1 + φF As )
ev =
ρi vd

the ratio of the retrofitted engine air mass ma to the unretrofitted air mass m′a is
ma 1
=
ma
′ 1 + φF As

For the case of the stoichiometric operation with φ = 1,

For hydrogen,
1 1
F As = = = 0.0294
AFs 34.06
ma 1
= = 0.971, a 3% decrease
m′a 1 + 1 · 0.0294

For methane,
1 1
F As = = = 0.0584
AFs 17.12
ma 1
= = 0.940, a 5.5% decrease
ma
′ 1 + 1 · 0.0642
 10 CHAPTER 9. FUELS

9.10) A flexible fuel vehicle operates with a mixture of 35% isooctane and 65% methanol, by volume. If the
combustion is to be stoichiometric, what should the mass air-fuel ratio be?

The volume fractions yi are to be converted to mass fractions. Since yi = Vi/V , we can write the
mass fraction xi in terms of ρ and y.

mi ρi · Vi ρi · Vi/V ρi · y i
xi = = P = P = P
m ρi · Vi ρi · Vi/V ρi · y i

In tabular form,

F uel F ormula yi ρi ρ i yi xi AFs,i

Iso-octane C8 H18 0.35 702.7 245.9 0.32 15.03
Methanol CH3 OH 0.65 791.1 514.2 0.68 6.43
Total 1.0 760.1 1

The mass air-fuel ratio is the weighted average of the two fuel’s air-fuel ratios
X
AFS = AFs,i · xi
= 15.03 · 0.32 + 6.43 · 0.68
= 9.18
 11

9.11) If a dragster fuel tank contains a mixture of 70% octane and 30% nitromethane, by volume, what
should the mass air-fuel ratio be to run rich at φ = 1.24 ?

The volume fractions yi are to be converted to mass fractions. Since yi = Vi/V , we can write the
mass fraction xi in terms of ρ and y.

mi ρi · Vi ρi · Vi/V ρi · y i
xi = = P = P = P
m ρi · Vi ρi · Vi/V ρi · y i

In tabular form,

F uel F ormula yi ρi ρ i yi xi AFs,i

Octane C8 H18 0.70 702.7 491.9 0.59 15.03
Nitro CH3 NO2 0.30 1138 341.3 0.41 1.69
Total 1.0 833.2 1.00

The stoichiometric mass air-fuel ratio is the weighted average of the two fuel’s air-fuel ratios
X
AFS = AFs,i · xi
= 15.03 · 0.59 + 1.69 · 0.41
= 9.56

and
AFs 9.56
AF = = = 7.71
φ 1.24
 12 CHAPTER 9. FUELS

9.12) A vehicle is equipped with a flex fuel 8 cylinder 6 L spark ignition engine running on a mixture of
octane and ethanol at φ = 0.95. At an operating condition of 2500 rpm, the thermal efficiency is 0.30
and the volumetric efficiency is 0.80. If the fuel mixture is changed from 10% ethanol (E10) to 85%
ethanol (E85), a.) What is the change in the overall mass air-fuel ratio, and b.) What is the change
in the engine power ?

a) The volume fractions yi are to be converted to mass fractions. Since yi = Vi/V , we can write the
mass fraction xi in terms of ρ and y.
mi ρi · Vi ρi · Vi/V ρi · y i
xi = = P =P = P
m ρi · Vi ρi · /V
Vi ρi · y i

For Ethanol (E10) in tabular form,

F uel yi ρi ρ i yi xi AFs,i
Ethanol (E10) 0.10 789.3 78.9 0.11 8.94
Octane 0.90 702.7 632.4 0.89 15.03
Total 1.0 711.3 1.00

For Ethanol (E85) in tabular form,

F uel yi ρi ρ i yi xi AFs,i
Ethanol (E85) 0.85 789.3 670.9 0.86 8.94
Octane 0.15 702.7 105.4 0.14 15.03
Total 1.0 776.3 1.00

Knowing that
P
AFs AFs,i · xi
AF = =
φ φ

For E10 fuel,

8.94 · 0.11 + 15.03 · 0.89
AF = = 15.11
0.95

For E85 fuel,

8.94 · 0.86 + 15.03 · 0.14
AF = = 10.31
0.95
As more ethanol is used, the air-fuel ratio will decrease since there is more fuel bound oxygen.

b) The thermal efficiency η is

ẇb
η= −→ ẇb = ηqc ṁf
ṁf · qc

The heat of combustion qc is defined on a mass basis

X
qc = qc,i · xi
 13

For E10 fuel,

qc = 29.7 · 0.11 + 47.9 · 0.89 = 45.8 MJ/kg

For E85 fuel,

qc = 29.7 · 0.86 + 47.9 · 0.14 = 32.2 MJ/kg

The fuel flowrate ṁf is obtained from the volumetric efficiency,

 
ṁa ev ρi Vd N
ṁf = = ∗
AF AF 2

For E10 fuel,

 
(0.8)(1.17)(6 × 10−3 ) 2500
ṁf = ∗ = 7.74 × 10−2 kg/s
15.11 60 · 2
ẇb = (0.3)(45,800)(7.74 × 10−2 ) = 106 kW

For E85 fuel,

 
(0.8)(1.17)(6 × 10−3 ) 2500
ṁf = ∗ = 1.13 × 10−2 kg/s
10.31 60 · 2
ẇb = (0.3)(32,200)(1.3 × 10−2 ) = 109 kW

The engine power increases by 2.8 kW (2.6 %)

 14 CHAPTER 9. FUELS

9.13) Verify the CO2 concentration values resulting from the combustion of propane, methane, methanol,
ethanol, and gasoline given in Table 10-8.

gCO2
The CO2 concentration values are in units of MJf uel so molecular mass M and heating value conver-
sions are needed
gCO2 gCO2 kgf uel 1000 g
= ∗ ∗
MJf uel gf uel MJf uel kg

nCO2 MCO2 1 1000 g

= ∗ ∗ ∗
nf uel Mf uel qc kg

In tabular form:

F uel F ormula M qc nCO2 /nf uel gCO

2 /MJf uel AFs,i
g/mol MJ/kg

Propane C3 H8 44.09 46.36 3 64.59 64.5

Methane CH4 16.04 50.01 1 54.93 54.9
Methanol CH4 O 32.04 19.91 1 68.99 69.0
Ethanol C2 H6 O 46.07 26.82 2 71.23 71.2
Gasoline C8 H15 110.0 44.51 8 71.91 71.9

The lowest normalized CO2 emissions are from methane, and the largest are from gasoline.