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