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APPLIED THERMODYNAMICS (ME-306) Reacting Mixtures and Combustion
Presentation by
Prof. Sreedhara Sheshadri
Department of Mechanical Engineering
Indian Institute of Technology Bombay
Powai, Mumbai, India- 400 076 1
TUTORIAL PROBLEM 1 Reacting Mixtures and Combustion
A vessel contains a mixture of 60% O2 and 40% CO on a
mass basis. Determine the percent excess or percent
deficiency of oxygen, as appropriate.
2
SOLUTION PROBLEM 1 (Slide 1 of 2)
Known
A vessel contains a mixture of 60% O2 and 40% CO on a mass basis
Find
The percent excess or percent deficiency of oxygen
Analysis
For complete combustion of CO with minimum amount of O2, the reaction is
written as
Therefore theoretical air fuel ratio is,
2 2
1
2CO O CO
2
2
0.5o
theo co
xA
F x
3
SOLUTION PROBLEM 1 (Slide 2 of 2)
Actual air fuel ratio
Percentage of excess
2 2 2 2 2
2 2 2 2 2
/ / 0.6 / 321.31
/ / 0.4 / 28.01
o o mix o o o
act co co mix co co co
x y MW MW y MWA
F x y MW MW y MW
2
1.31-0.5100 162
0.5O
4
TUTORIAL PROBLEM 2 Reacting Mixtures and Combustion
Propane (C3H8) is burned with air. For each case, obtain the
balanced reaction equation for complete combustion (a) with
the theoretical amount of air. (b) with 20% excess air. (c) with
20% excess air, but only 90% of the propane being
consumed in the reaction.
5
SOLUTION PROBLEM 2 (Slide 1 of 4)
Known
Propane (C3H8) is burned with air
Find
Balanced reaction equation for complete combustion
• With the theoretical amount of air
• With 20% excess air
• With 20% excess air, but only 90% of the propane being consumed in
the reaction
Assumptions
Each mole of oxygen in combustion air is accompanied by 3.76 moles of
nitrogen, which is inert
6
SOLUTION PROBLEM 2 (Slide 2 of 4)
Analysis
(a) For complete combustion of C8H18 with theoretical amount of air, the
products contain carbon dioxide, water and nitrogen only
Applying the conservation of mass principle to carbon, hydrogen,
oxygen and nitrogen respectively, gives
Solving these equations, a = 5, b = 3, c = 4, d = 18.8
The balanced equation is
3 8 2 2 2 2 2( 3.76 )C H a O N bCO cH O dN
C: 3
H: 2 8 4
O: 2 2
N: 3.76
b
c c
b c a
d a
3 8 2 2 2 2 25( 3.76 ) 3 4 18.8C H O N CO H O N 7
SOLUTION PROBLEM 2 (Slide 3 of 4)
(b) For complete combustion of C8H18 with 20% excess of air, the products
contain carbon dioxide, water, oxygen and nitrogen
Applying the conservation of mass principle to carbon, hydrogen,
oxygen and nitrogen respectively, gives
Solving these equations, b = 3, c = 4, d = 22.56, e = 1
The balanced equation is
3 8 2 2 2 2 2 2(5)(1.2)( 3.76 )C H O N bCO cH O dN eO
C: 3
H: 2 8 4
O: 2 2 2 5 1.2
N: 3.76 5 1.2
b
c c
b c e
d
3 8 2 2 2 2 2 26( 3.76 ) 3 4 22.56C H O N CO H O N O 8
SOLUTION PROBLEM 2 (Slide 4 of 4)
(c) With 20% excess of air and only 90% of consumption of fuel, the
products contain carbon dioxide, water, oxygen, nitrogen and traces of
fuel.
Applying the conservation of mass principle to carbon, hydrogen,
oxygen and nitrogen respectively, gives
Solving these equations, b = 2.7, c = 3.6, d = 22.56, e = 1.5
The balanced equation is
3 8 2 2 3 8 2 2 2 2(5)(1.2)( 3.76 ) 0.1C H O N C H bCO cH O dN eO
C: 0.3 3 2.7
H: 2 0.8 8 3.6
O: 2 2 2 5 1.2
N: 3.76 5 1.2
b b
c c
b c e
d
3 8 2 2 3 8 2 2 2 26( 3.76 ) 0.1 2.7 3.6 22.56 1.5C H O N C H CO H O N O 9
TUTORIAL PROBLEM 3 Reacting Mixtures and Combustion
A natural gas with the molar analysis 78% CH4, 13% C2H6,
6% C3H8, 1.7% C4H10, 1.3% N2 burns completely with 40%
excess air in a reactor operating at steady state. If the molar
flow rate of the fuel is 0.5 kmol/h, determine the molar flow
rate of the air, in kmol/h.
10
SOLUTION PROBLEM 3 (Slide 1 of 2)
Known
Natural gas burns completely with 40% excess air
The molar flow rate of the fuel is 0.5 kmol/h
Find
Molar flow rate of air in kmol/h
Assumptions
Each mole of oxygen in combustion air is accompanied by 3.76 moles of
nitrogen, which is inert
11
SOLUTION PROBLEM 3 (Slide 2 of 2)
Analysis
(a) For complete combustion of natural gas with stoichiometric A/F
ratio
4 2 6 3 8 4 10 2 2 2
2 2 2
0.5[0.78 0.13 0.06 0.017 0.013 ] [ 3.76 ]
(0.065 3.76 )
CH C H C H C H N a O N
bCO cH O a N
Cbalance: 0.644
Obalance: 2 2
Hbalance: 2 2.275 1.1375
1.21275
As natural gas burns with 40% excess air,
molar flow rate of air 1.4 1.21275 1.69785kmol/h
b
a b c
c c
a
12
TUTORIAL PROBLEM 4 Reacting Mixtures and Combustion
Hexane (C6H14) burns with air to give products with the dry
molar analysis of CO2, 11.5%; CO, 2.4%; O2, 2.0%; H2,
1.6%; N2, 82.5%. Determine the air-fuel ratio on a molar
basis.
13
SOLUTION PROBLEM 4 (Slide 1 of 2)
Known
C6H14 burns with air to give products with the dry molar composition of
CO2, 11.5%; CO, 2.4%; O2, 2.0%; H2, 1.6%; N2, 82.5%
Find
Air-fuel ratio on a molar basis
Assumptions
Each mole of oxygen in combustion air is accompanied by 3.76 moles of
nitrogen, which is inert
14
SOLUTION PROBLEM 4 (Slide 2 of 2)
Analysis
The analysis is on basis of 100 moles of dry products
Applying the conservation of mass principle to carbon, hydrogen and
oxygen respectively, gives
Check for closure;
Then,
6 14 2 2 2 2 2 2 2( 3.76 ) 11.5 2.4 2 1.6 82.5aC H b O N CO CO O H N cH O
C: 6 11.5 2.4 2.317
H: 14 2 1.6 2 14.619
O: 2 2 11.5 2.4 2 2 , 22.1
a a
a c c
b C b
2 22.1 3.76 82.5 ( )N acceptable
22.1 4.76 2.317 45.44 ( ) / ( )AF kmol fuel kmol air 15
TUTORIAL PROBLEM 5 Reacting Mixtures and Combustion
Liquid ethanol (C2H5OH) at 288 K, 1 atm enters a
combustion chamber operating at steady state and burns
with air entering at 500 K, 1 atm. The fuel flow rate is 25 kg/s
and the equivalence ratio is 1.2. Heat transfer from the
combustion chamber to the surroundings is at a rate of
3.75×105 kJ/s. Products of combustion, consisting of CO2,
CO, H2O(g), and N2, exit. Ignoring kinetic and potential
energy effects, determine (a) the exit temperature, in K. (b)
the air-fuel ratio on a mass basis.
16
SOLUTION PROBLEM 5 (Slide 1 of 3)
Known
Liquid ethanol at 288 K, 1 atm enters a combustion chamber and burns
with air entering at 500 K, 1 atm
The fuel flow rate is 25 kg/s and the equivalence ratio is 1.2
Heat transfer rate from the combustion chamber to the surroundings is
3.75×105 kJ/s.
Find
The exit temperature
The air-fuel ratio on a mass basis
Assumptions
Each mole of oxygen in combustion air is accompanied by 3.76 moles of
nitrogen, which is inert
Air and product gases behave as ideal gas mixtures
17
SOLUTION PROBLEM 5 (Slide 2 of 3)
Analysis
For the complete combustion of C2H5OH with theoretical air
The actual combustion reaction is
2 5 2 2 2 2 2( 3.76 ) 2 3 (3.76)C H OH b O N CO H O b N
O balance: 1 2 4 3 3
Thus, 3 4.76 14.28 (on a mole basis)
Now, Equivalance ratio =( ) ( )
14.28 1.2 11.9 (on a mole basis)
4.76 2.5
Theo
Theo Act
Act
Act Act
b b
AF
AF AF
AF
b AF
2 5 2 2 2 2 22.5( 3.76 ) 3 9.4C H OH O N cCO dCO H O N
n
C balance: 2
O balance: 1 2.5 2 2 3
Sol gives : 1, 1
c d
c d
c d
2 5 2 2 2 2 22.5( 3.76 ) 3 9.4C H OH O N CO CO H O N 18
SOLUTION PROBLEM 5 (Slide 3 of 3)
Analysis
An energy rate balance at steady state reduces to
Air fuel ratio on mass basis is
11.9 28.97 46.06 7.483 kg air/kg fuel
Act act air fuelmolebasis
Act
AF AF M M
AF
2 2 2 2 2
0
0 2.5 9.4 3 9.4
, 25 46.06 0.5427 /
,
Above equation can be solved iteratively to T using table data
99
cvfuel O N CO H O NCO
f
f f fuel
f
exit
exit
Qh h h h h h h
n
Where n m M kmol s
and h h h
T
4K
19
TUTORIAL PROBLEM 6 Reacting Mixtures and Combustion
Propane gas (C3H8) at 25C, 1 atm enters an insulated
reactor operating at steady state and burns completely with
air entering at 25C, 1 atm. Determine the adiabatic flame
temperatures when 100% and 200% of theoretical air is
supplied. What happens to adiabatic flame temperature
when excess is supplied and why?
20
SOLUTION PROBLEM 6 (Slide 1 of 4)
Known
Propane enters an insulated reactor at 25C and burns completely with
air entering at 25C
Find
Adiabatic flame temperature when 100% and 200% air is supplied
Assumptions
Each mole of oxygen in combustion air is accompanied by 3.76 moles of
nitrogen, which is inert
Air and product gases behave as ideal gas mixtures
The insulated reactor is at steady state
The kinetic and potential energy effects are negligible
21
SOLUTION PROBLEM 6 (Slide 2 of 4)
Analysis
For complete combustion of propane with theoretical amount of air is
given by
For N times the theoretical amount of air
An energy rate balance gives
3 8 2 2 2 2 25[ 3.76 ] 3 4 (5)(3.76)C H O N CO H O N
2222
2283
8.18)1(543
]76.3[5
nNOnOHCO
NOnHC
3 8 2 2 20 ( ) 3[ ] 5( 1)[ ] 18.8 [ ]O O
f C H f H O O Nh h h n h n h
22
SOLUTION PROBLEM 6 (Slide 3 of 4)
We can use
From the ideal gas tables and thermo-chemical properties table
2 2 2 3 8
2 2
2 2
3 ( ) 4 ( ) 5( 1) ( ) ( )
3[ (298 )] 4[ (298 )]
5( 1)[ (298 )] 18.8 [ (298 )]
O
CO H O O f C H
O O
f CO f H O
O N
h T h T n h T h
h h K h h K
n h K n h K
3 8
2
2
2
2
1
1
1
1
1
( ) 103,850 .
[ (298 )] [ 393,520 9364] .
[ (298 )] [ 241,820 9904] .
[ (298 )] 8682 .
[ (298 )] 8669 .
O
f C H
O
f CO
O
f H O
O
N
h kJ kmol
h h K kJ kmol
h h K kJ kmol
h K kJ kmol
h K kJ kmol
)298()(][ KhThh
23
SOLUTION PROBLEM 6 (Slide 4 of 4)
For 100% theoretical air, n=1
For T=2400 K, RHS-LHS = 48093 kJ.kmol-1
For T=2450 K, RHS-LHS = -60315 kJ.kmol-1
By linear interpolation, T=2422 K
For 200% theoretical air, n=2
For T=1500 K, RHS-LHS = 19427 kJ.kmol-1
For T=1520 K, RHS-LHS = -17754 kJ.kmol-1
By linear interpolation, T=1510 K
1.675,274,2)(8.18)(4)(3222
kmolkJThThTh NOHCO
2 2 2 2
13 ( ) 4 ( ) 5 ( ) 37.6 ( ) 2,481,062 .CO H O O Nh T h T h T h T kJ kmol
24
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