Sonntag, Borgnakke and van Wylen
14.36
The hot exhaust gas from an internal combustion engine is analyzed and found to
have the following percent composition on a volumetric basis at the engine
exhaust manifold. 10% CO2, 2% CO, 13% H2O, 3% O2 and 72% N2. This gas is
fed to an exhaust gas reactor and mixed with a certain amount of air to eliminate
the carbon monoxide, as shown in Fig. P14.36. It has been determined that a mole
fraction of 10% oxygen in the mixture at state 3 will ensure that no CO remains.
What must the ratio of flows be entering the reactor?
Exhaust gas at state 1: CO2 10 %, H2O 13%,
CO 2%, O2 3%, N2 72%
1 Exh. gas
Exhaust gas at state 3:
2 Air
3 gas
Reactor
CO = 0 %, O2 = 10 %
out
Reaction equation for the carbon monoxide
Ÿ 0.02 CO + x O2 + 3.76x N2 o 0.02 CO2 + (x-0.01) O2 + 3.76x N2
At 3:
QCO = 0.10 + 0.02 = 0.12,
QH
QO = (x-0.01) + 0.03 = x + 0.02
QN = 0.72 + 3.76x
2O
2
2
= 0.13
2
or
nTOT = 0.12 + 0.13 + x + 0.02 + 0.72 + 3.76x = 0.99 + 4.76x
yO2 = 0.10 =
or
x + 0.02
0.99 + 4.76x
o
x = 0.151
air 2
kmol air
4.76x
=
= 0.718 kmol Exh. gas
Exh. Gas 1
1
Sonntag, Borgnakke and van Wylen
14.42
In a test of rocket propellant performance, liquid hydrazine (N2H4) at 100 kPa,
25qC, and oxygen gas at 100 kPa, 25qC, are fed to a combustion chamber in the
ratio of 0.5 kg O2/kg N2H4. The heat transfer from the chamber to the
surroundings is estimated to be 100 kJ/kg N2H4. Determine the temperature of the
products exiting the chamber. Assume that only H2O, H2, and N2 are present. The
enthalpy of formation of liquid hydrazine is 50 417 kJ/kmol.
1
o
Liq. N2H4: 100 kPa, 25 C
Gas O2: 100 kPa, 25oC
Products
2
.
.
.
.
mO2/mN2H4 = 0.5 = 32nO2/32nN2H4
Energy Eq.:
3
Comb.
Chamber
and
. .
Q/mN2H4 = -100 kJ/kg
QCV = HP - HR = -100u32.045 = -3205 kJ/kmol fuel
1
1 N2H4 + 2 O2 o H2O + H2 + N2
Combustion eq.:
1
HR = 1(50417) + 2(0) = 50417 kJ
HP = -241 826 + 'hH2O + 'hH2 + 'hN2
Energy Eq. now reads
o
HP = HR + QCV = HP + 'HP
o
'HP = 'hH2O + 'hH2 + 'hN2 = -HP + HR + QCV
= 241 826 + 50 417 - 3205 = 289 038 kJ/kmol fuel
Table A.9 : Guess T and read for water, hydrogen and nitrogen
2800 K: 'HP = 115 463 + 81 355 + 85 323 = 282 141 too low
3000 K: 'HP = 126 548 + 88 725 + 92 715 = 307 988 too high
Interpolate to get
TP = 2854 K
Sonntag, Borgnakke and van Wylen
Enthalpy of Combustion and Heating Value
14.55
Liquid pentane is burned with dry air and the products are measured on a dry
basis as: 10.1% CO2, 0.2% CO, 5.9% O2 remainder N2. Find the enthalpy of
formation for the fuel and the actual equivalence ratio.
QFuC5H12 + QO O2 + 3.76 QO N2 o
2
2
x H2O + 10.1 CO2 + 0.2 CO + 5.9 O2 + 83.8 N2
Balance of C: 5 QFu = 10.1 + 0.2 Ÿ QFu = 2.06
Balance of H: 12 QFu = 2 x Ÿ x = 6 QFu = 12.36
Balance of O: 2 QO = x + 20.2 + 0.2 + 2u5.9 Ÿ QO = 22.28
2
2
Balance of N: 2u3.76 QO = 83.8u2 Ÿ QO = 22.287 ŸOK
2
2
QO for 1 kmol fuel = 10.816
2
I = 1, C5H12 + 8O2 + 8u3.76N2 o 6H2O + 5CO2 + 30.08N2
°
°
°
-°
-°
-°
HRP = HP - HR = 6 hf H2O + 5 hf CO2 - hf fuel
°
14.3: HRP = 44 983 u 7
Ÿ
-°
hf fuel = -172 998 kJ/kmol
I = AFs / AF = QO2 stoich/QO2 AC = 8/10.816 = 0.74
Sonntag, Borgnakke and van Wylen
14.81
Acetylene gas at 25qC, 100 kPa is fed to the head of a cutting torch. Calculate the
adiabatic flame temperature if the acetylene is burned with
a.
100% theoretical air at 25qC.
b.
100% theoretical oxygen at 25qC.
a)
C2H2 + 2.5 O2 + 2.5u3.76 N2 o 2 CO2 + 1 H2O + 9.4 N2
-o
HR = hf C2H2 = +226 731 kJ/kmol
from table A.10
-*
-*
-*
HP = 2(-393 522 + 'hCO2) + 1(-241 826 + 'hH2O) + 9.4 'hN2
-*
-*
-*
2 'hCO2 + 1 'hH2O + 9.4 'hN2 = 1 255 601 kJ
QCV = HP - HR = 0 Ÿ
b)
Trial and Error A.9:
LHS2800 = 1 198 369,
Linear interpolation:
TPROD = 2909 K
LHS3000 = 1 303 775
C2H2 + 2.5 O2 o 2 CO2 + H2O
HR = +226 731 kJ ;
-*
-*
HP = 2(-393 522 + 'hCO2) + 1(-241 826 + 'hH2O)
-*
-*
Ÿ 2 'hCO2 + 1 'hH2O = 1 255 601 kJ/kmol fuel
At 6000 K (limit of A.9)
2u343 782 + 302 295 = 989 859
At 5600 K
2u317 870 + 278 161 = 913 901
Slope 75 958/400 K change
Extrapolate to cover the difference above 989 859 kJ/kmol fuel
TPROD | 6000 + 400(265 742/75 958) | 7400 K
Sonntag, Borgnakke and van Wylen
15.46
Water from the combustion of hydrogen and pure oxygen is at 3800 K and 50
kPa. Assume we only have H2O, O2 and H2 as gases find the equilibrium
composition.
With only the given components we have the reaction
2 H2O œ 2H2 + O2
which at 3800 K has an equilibrium constant from A.11 as ln K = -1.906
Assume we start with 2 kmol water and let it dissociate x to the left then
Species
H2O
H2
O2
Initial
2
0
0
Change
-2x
2x
x
Final
2 2x
2x
x
Tot: 2 + x
Then we have
K = exp(-1.906) =
y2H yO
2
y2H O
2
2 § P ·2+1-2
¨ 0¸
©P ¹
2
§ 2x · x
©2 + x¹ 2 + x 50
=
§2 - 2x·2 100
©2 + x¹
which reduces to
1
4x3 1 1
0.148674 =
(1- x)2 2 + x 4 2
or
x3 = 0.297348 (1 – x)2 (2 + x)
Trial and error to solve for x = 0.54 then the concentrations are
2 - 2x
yH2O = 2 + x = 0.362;
x
yO2 = 2 + x = 0.213;
2x
yH2 = 2 + x = 0.425
Sonntag, Borgnakke and van Wylen
15.57
One approach to using hydrocarbon fuels in a fuel cell is to “reform” the
hydrocarbon to obtain hydrogen, which is then fed to the fuel cell. As a part of the
analysis of such a procedure, consider the reaction CH4 + H2O œ CO + 3H2.
One kilomole each of methane and water are fed to a catalytic reformer. A
mixture of CH4, H2O, H2, and CO exits in chemical equilibrium at 800 K, 100
kPa; determine the equilibrium composition of this mixture using an equilibrium
constant of K = 0.0237.
The reaction equation is:
CH4
+
initial
1
change
-x
equil.
(1-x)
nTOTAL = 2 + 2x
3
yH2yCO
K=y
y
(PP0)
2
H2O
1
-x
(1-x)
=
4u0.0237
= 0.003 51
27u1
x
x
(1-x
) (1+x
)
or
x2
= 0.003 51 = 0.059 25
1-x2
nCH4 = 0.7635
½° ­ y
y
= 0.7095
¾®y
= 0.2365
°¯y
= 2.473 ¿
nH2O = 0.7635
nH2
nCO
nTOT
0
+3x
3x
(3x)3x
100 2
2(100)
(1-x)(1-x)(2+2x)
or
2
3 H2
=
CH4 H2O
2
œ
Solving,
CH4
= 0.3087
H2O
= 0.3087
H2
= 0.2870
CO
= 0.0956
+
CO
0
+x
x
x = 0.2365