APPLIED THERMODYNAMICS (ME-306) R ea ea c t in i n g M ix ix t u r e s a n d C o m b u s t i o n
Presentation by Prof. Sreedhara Sheshadri
Department of Mechanical Engineering Indian Institute Instit ute of Technology echnol ogy Bombay Powai, Mumbai, India- 400 076
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TUTORIAL PROBLEM 1 R ea ea c t in i n g M ix ix t u r e s a n d C o m b u s t i o n
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.
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SOLUTION PROBLEM 1 (Slid e 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 O 2, the reaction is written as
1 CO O2 CO2 2
Therefore theoretical air fuel ratio is,
xo A F x theo co
2
2
0.5
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SOLUTION PROBLEM 1 (Slid e 2 of 2) Actual air fuel ratio
xo A F x act co
2
2
yo MWmix / MWo yo / MW o 0.6/32 1.31 yco / MW co 0.4 / 28.01 yco MWmix / MWco 2
2
2
Percentage of excess O2
2
2
1.31-0.5 0.5
2
2
2
100 162
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TUTORIAL PROBLEM 2 R ea c t in g M ix t u r e s a n d C o m b u s t i o n
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.
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SOLUTION PROBLEM 2 (Slid e 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 (Slid e 2 of 4)
Analysis (a) For complete combustion of C8H18 with theoretical amount of air, the products contain carbon dioxide, water and nitrogen only
C3 H8 a(O2 3.76 N2 ) bCO2 cH2O dN2 Applying the conservation of mass principle to carbon, hydrogen, oxygen and nitrogen respectively, gives
C:
b 3
H:
2c 8 c 4
O:
2b c 2a
N:
d 3.76a
Solving these equations, a = 5, b = 3, c = 4, d = 18.8 The balanced equation is 7
SOLUTION PROBLEM 2 (Slid e 3 of 4)
(b) For complete combustion of C 8H18 with 20% excess of air, the products contain carbon dioxide, water, oxygen and nitrogen
C3 H8 (5)(1.2)(O2 3.76 N2 ) bCO2 cH2O dN2 eO2 Applying the conservation of mass principle to carbon, hydrogen, oxygen and nitrogen respectively, gives
C:
b3
H:
2c 8 c 4
O:
2b c 2e 2 51.2
N:
d 3.76 5 1.2
Solving these equations, b = 3, c = 4, d = 22.56, e = 1 The balanced equation is
C3 H8 6(O2 3.76 N2 ) 3CO2 4 H2O 22.56 N2 O2
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SOLUTION PROBLEM 2 (Slid e 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.
C3 H8 (5)(1.2)(O2 3.76N2 ) 0.1C3 H8 bCO2 cH2 O dN2 eO2 Applying the conservation of mass principle to carbon, hydrogen, oxygen and nitrogen respectively, gives
C:
b 0.3 3 b 2.7
H:
2c 0.8 8 c 3.6
O:
2b c 2e 2 5 1.2
N:
d 3.76 5 1.2
Solving these equations, b = 2.7, c = 3.6, d = 22.56, e = 1.5 The balanced equation is
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TUTORIAL PROBLEM 3 R ea c t in g M ix t u r e s a n d C o m b u s t i o n
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.
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SOLUTION PROBLEM 3 (Slid e 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 Analysis
(Slid e 2 of 2)
(a) For complete combustion of natural gas with stoichiometric A/F ratio
0.5[0.78CH4 0.13C2 H6 0.06C3 H8 0.017 C4 H10 0.013 N2] a[ O2 3.76 N2]
bCO2 cH2O (0.065 3.76a) N2 Cbalance: b 0.644 O balance: 2a 2b c H balance: 2c 2.275 c 1.1375 a 1.21275 As natural gas burns with 40% excess air, molar flow rate of air
1.4 1.21275 1.69785kmol/h
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TUTORIAL PROBLEM 4 R ea c t in g M ix t u r e s a n d C o m b u s t i o n
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.
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SOLUTION PROBLEM 4 (Slid e 1 of 2)
Known C6H14 burns with air to give products with the dry molar composition of CO2, 11.5%; CO, 2.4%; O 2, 2.0%; H 2, 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
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SOLUTION PROBLEM 4 (Slide 2 of 2)
Analysis
The analysis is on basis of 100 moles of dry products
aC6 H14 b(O2 3.76 N2 ) 11.5CO2 2.4CO 2O2 1.6 H2 82.5 N2 cH2 O Applying the conservation of mass principle to carbon, hydrogen and oxygen respectively, gives
C:
6a 11.5 2.4 a 2.317
H:
14a 2 1.6 2c c 14.619
O:
2b 2 11.5 2.4 2 2 C, b 22.1
Check for closure; N2
Then, AF
22.1 3.76 82.5 (acceptable)
22.1 4.76 2.317 45.44 kmol ( fuel ) / kmol(air ) 15
TUTORIAL PROBLEM 5 R ea c t in g M ix t u r e s a n d C o m b u s t i o n
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.
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SOLUTION PROBLEM 5 Known
(Slid e 1 of 3)
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 (Slid e 2 of 3)
Analysis
For the complete combustion of C 2H5OH with theoretical air
C2 H5OH b(O2 3.76 N2 ) 2CO2 3H2 O b(3.76) N2 O balance:
1 2b 4 3 b 3
Thus, AF Theo 3 4.76 14.28 (on a mole basis) Now, Equivalance ratio = ( AFTheo ) ( AF Act ) AF Act 14.28 1.2 11.9 (on a mole basis) b Act AF Act 4.76 2.5
The actual combustion reaction is
C2 H5OH 2.5(O2 3.76 N2 ) cCO2 dCO 3H2 O 9.4 N2 C balance: 2 c d O balance:
1 2.5 2 2c d 3
Soln gives : c 1, d 1 18
SOLUTION PROBLEM 5 (Slid e 3 of 3)
Analysis
An energy rate balance at steady state reduces to
0
Qcv h fuel 2.5hO2 9.4hN2 hCO2 hCO 3hH2O 9.4hN 2 n f
Where, n f m f M fuel 25 46.06 0.5427 kmol / s and , h h f 0 h Above equation can be solved iteratively to Texit using table data T exit 994 K
Air fuel ratio on mass basis is
AF Act AFact molebasis M air M fuel AF Act 11.9 28.97 46.06 7.483 kg air/kg fuel 19
TUTORIAL PROBLEM 6 R ea c t in g M ix t u r e s a n d C o m b u s t i o n
Propane gas (C3H8) at 25 C, 1 atm enters an insulated reactor operating at steady state and burns completely with air entering at 25 C, 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?
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SOLUTION PROBLEM 6 (Slid e 1 of 4)
Known Propane enters an insulated reactor at 25 C and burns completely with air entering at 25 C
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
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SOLUTION PROBLEM 6 (Slid e 2 of 4)
Analysis
For complete combustion of propane with theoretical amount of air is given by
C3 H8 5[O2 3.76N2 ] 3CO2 4H2O (5)(3.76) N2
For N times the theoretical amount of air
C 3 H 8 5n[O2 3.76 N 2 ] 3CO2 4 H 2O 5(n 1)O2 18.8nN 2
An energy rate balance gives
0 (h fO )C3H8 3[hfO h ]H2O 5(n 1)[h ]O2 18.8n[h ]N 2 22
SOLUTION PROBLEM 6 (Slid e 3 of 4)
We can use [h ] h (T ) h (298K )
3hCO2 (T ) 4hH2O (T ) 5(n 1)hO2 (T ) (hfO )C3H8
3[h fO h (298K )]CO2
4[hfO h (298K )]H 2O
5(n 1)[h (298K)]O2 18.8n[h (298K )]N 2
From the ideal gas tables and thermo-chemical properties table
(h fO )C3H 8 103,850kJ .kmol 1 [h fO h (298K )]CO2 [393,520 9364]kJ .kmol 1 [h fO h (298K )]H2O [241,820 9904]kJ .kmol 1 [h (298K )]O2 8682kJ .kmol 1 [h (298K )]
8669kJ kmol 1
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SOLUTION PROBLEM 6 (Slid e 4 of 4)
For 100% theoretical air, n=1
3hCO2 (T ) 4h H 2O (T ) 18.8h N 2 (T ) 2,274,675kJ .kmol 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
3hCO2 (T ) 4hH2O (T ) 5hO2 (T ) 37.6hN2 (T ) 2, 481,062kJ .kmol 1 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 24
