Department of Mechanical Engineering ME 322 – Mechanical Engineering Thermodynamics Lecture 36 Combustion Reactions.

Department of Mechanical Engineering

ME 322 – Mechanical Engineering Thermodynamics

Lecture 36

Combustion Reactions

Combustion Processes

2

Why do mechanical engineers have to know about combustion? Consider a combustion chamber in a gas turbine cycle,

inQ

3 2in aQ m h h

2 3

Air from compressor

Air to turbine

Our current model What really happens

2 3

Air from compressor

Combustion products to the turbine

Fuel input

How is this analyzed??

Combustion

• Fuels– Stored chemical energy

• Combustion Reaction– Transforms the chemical energy stored in

the fuel to thermal energy (heat)

• Goals of this section of the course– Understand combustion chemistry– Use combustion chemistry to determine

the heat released during a combustion process

• Heat of reaction

3

Combustion

The combustion of a fuel requires oxygen,

fuel + oxidant products

FuelIn the most general sense, a fossil fuel makeup is,

C H S O N

4

The Greek letters signify the atomic composition of the fuel.

For example ... 8 18C H octane

2 5C H OH ethanol (ethyl alcohol)

Combustion

OxidantThe oxidant must contain oxygen. The most abundant ‘free’ source is atmospheric air. By molar percent, atmospheric air is considered to be ...

Ni-tro-gen79%

Oxygen21%

For every mole of oxygen involved in a combustion reaction, there are 79/21 = 3.76 moles of nitrogen.

5


Combustion

Products (for fuels with no sulfur content)

Complete Combustion: CO2, H2O, and N2

Incomplete Combustion: CO2, H2O, N2, CO, NOx

Combustion with Excess Oxygen: CO2, H2O, N2, and O2

NOTE: Fuels containing sulfur have the potential of introducing sulfuric acid into the product stream.

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Combustion Terminology

• Theoretical or Stoichiometric Air– The amount of air required for complete

combustion of the fuel• Determined by balancing the combustion

reaction

• Excess or Percent Theoretical Air– The amount of air actually used in the

combustion process relative to the stoichiometric value

• Can cause incomplete combustion or excess oxygen

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Combustion Terminology

8

In many combustion processes, one of the parameters we are interested in is how much air (or oxygen) is required per unit quantity (moles or mass) of fuel.

Air-Fuel and Fuel-Air Ratios

1/ /

/

1/ / /

/

mol molmol

airmass mol mass

fuel mass

A F F AA F

MA F A F F A

M A F

moles of air moles of fuel

moles of fuel moles of air

mass of fuel

mass of air

Equivalence Ratio

/

/actual

stoichiometric

F A

F A

Equivalence Ratio and Products

• Stoichiometric (F=1)– CO2, H2O, N2

• Lean ( F < 1 with T < 1800 R)– CO2, H2O, N2, O2

• Rich ( > F 1 with T < 1800 R)– CO2, H2O, N2, O2, CO, H2

• Rich ( > F 1 with T > 1800 R)– CO2, H2O, N2, O2, CO, H2, H, O,

OH, N, C(s), NO2, CH4

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ME 322

Advanced courses

ME 422 & 433

Stoichiometric (Complete) Combustion

0 1 2 3 4 2 2 2 2 2 2C H S O N + (O + 3.76 N ) CO + H O + SO + N

1

2

3

0 1 2 3

0 4

2

2 2 2

2(3.76) 2

C:

H:

S:

O:

N:

5 equations5 unknowns(n0 through n4)

Atomic Balances

Stoichiometric Combustion of a General Fuel in Air

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Lean Combustion

0 1 2 3 4 5 2 2 2 2 2 2 2C H S O N + (O + 3.76 N ) CO + H O + SO + O + N

0 0

1

2

3

0 1 2 3 4

0 5

2

2 2 2 2

2(3.76) 2

( )

C:

H:

S:

O:

N:

PTA PTA = Percent Theoretical Air expressed as a decimal

6 equations6 unknowns(x0 through x5)

Requires a stoichiometricbalance first (to get n0)

Lean Combustion of a General Fuel in Excess Air

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Example – Octane Combustion

Given: Gasoline (modeled as octane - C8H18) burns completely in 150% theoretical air (or 50% excess air).

Find:(a) the A/F ratios (mass and molar)(b) the equivalence ratio(c) the dew point of the products of combustion at assuming

that the products are at 1 atm

12


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In order to calculate the air-fuel ratios and the equivalence ratio, we need to know how much air is used in the combustion reaction. This is determined by balancing the combustion reaction.

In order to determine the dew point of the products, we need to know the molar composition of the products. This is also determined by balancing the combustion reaction.

Everything depends on the correct balance of the combustion reaction!


1. Balance the stoichiometric reaction to get n0

0 1 2 4 8 18 2 2 2 2 2C H + (O + 3.76 N ) CO + H O + N

0 1 2 4 5 8 18 2 2 2 2 2 2C H + 1.5 (O + 3.76 N ) CO + H O O + N

2. Balance the reaction with 150% theoretical air

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Solution strategy ...

3. Calculate the required (A/F) ratios, the equivalence ratio, and the dew point temperature of the products


0 1 2 4 8 18 2 2 2 2 2C H + (O + 3.76 N ) CO + H O + N

Stoichiometric Reaction

8 18 2 2 2 2 2C H + 12.5(O + 3.76 N ) 8CO + 9H O + 47N

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0 4 42(3.76) 2 3.76 12.5 47 N:0 1 2 3 02 2 2 8 9 / 2 12.5 O:

30 S:2 218 2 9 H:

18 C:


Combustion in 150% theoretical air

0 1 2 4 5 8 18 2 2 2 2 2 2C H + 1.5 (O + 3.76 N ) CO + H O O + N

0 0 0 12.5 1.5 18.75 ( )PTA

8 18 2 2 2 2 2 2C H + 18.75(O + 3.76 N ) 8CO + 9H O + 6.25O + 70.5N

16

1

2 2

3

0 1 2 3 4 4

0 5 5

8

18 2 9

0

2 2 2 2 18.75 8 9 / 2 6.25

2(3.76) 2 3.76 18.75 70.5

C:

H:

S:

O:

N:


The molar (A/F) ratio can now be found ...

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8 18 2 2 2 2 2C H + 12.5(O + 3.76 N ) 8CO + 9H O + 47N

8 18 2 2 2 2 2 2C H + 18.75(O + 3.76 N ) 8CO + 9H O + 6.25O + 70.5N

0

,

1 3.76/

1mol actA F

0

,

1 3.76/

1mol stoichA F

/mol

A F 2 2moles of O + moles of Nmoles of air= =

moles of fuel moles of fuel

04.76 4.76 12.5 59.5

04.76 4.76 18.75 89.25


/ / airmass mol

fuel

MA F A F

M

The mass-based (A/F) ratio can be found knowing the molecular masses of the air and the fuel,

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Table C.13a

lbm28.97

lbmolairM

fuel C H S O NMW M M M M M

The molecular mass of the air is,

The molecular mass of the fuel is,

Table C.20 Table C.20

8 12.01115 lbm/lbmol 18 1.00797 lbm/lbmol 114.23 lbm/lbmolfuelMW

Example – Octane CombustionNow, the mass-based (A/F) can be found ...

19

,

,

28.97/ 59.5 15.09

114.23

28.97/ 89.25 22.63

114.23

mass stoich

mass act

A F

A F

Once the (A/F) ratios are determined, the equivalence ratio can be found,

/ 1/ /

/ 1/ /actual act

stoich stoich

F A A F

F A A F

1/ 89.250.667

1/ 59.5

1/ 22.630.667

1/15.09

(molar based)

(mass based)

Example – Octane CombustionThe dew point of the products is the temperature where the water vapor condenses,

Tdp = Tsat at Pw (partial pressure of the water vapor)

ww w w

Py P y P

P

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8 18 2 2 2 2 2 2C H + 18.75(O + 3.76 N ) 8CO + 9H O + 6.25O + 70.5N

2

1 2 3 4 5

90.096

8 9 0 6.25 70.5vy

0.096 1 atm 0.096 atm 1.411 psiaw wP y P

113.4 Fdp sat wT T P

Example – Problem 15.42

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Given: Combustion exhaust with 9.1% CO2, 8.9% CO, 82% N2, and no O2

Find: a) fuel model CnHm

b) mass percent of carbon and hydrogen in fuelc) molar air/fuel ratio and percent theoretical air (PTA)d) dew point temperature at .106 MPa


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STEP 1: Write balance equation using ORSAT data

CnHm + a(O2 + 3.76 N2)

9.1 CO2 + 8.9 CO + bH2O + 82 N2

STEP 2: Solve for unknowns and write fuel model CnHm

n = ? m = ? a=? b=?


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STEP 3: Compute molar mass of fuel & mass composition

Mfuel = 18lbmolC/lbmolfuel*(12lbm/lbmolC)

+ 33lbmolH/lbmolfuel*(1lbm/lbmolH)

= 249 lbm/lbmolfuel

C: 18*(12)/249 87%

H: 33*(1)/249 13%


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STEP 4: Calculate molar air/fuel ratio

21.8 * (1 + 3.76) /1 = 103.8 moles air / mole fuel

STEP 5: Write equation for stoichiometric combustion

C18H33 + 26.25(O2 + 3.76 N2) 18 CO2 + 16.5 H2O + 98.7 N2

STEP 6: Find theoretical air

%TA = (21.8 / 26.75) * 100 = 83%


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STEP 7: Find dew point of combustion products

# moles of H2O (in original equation) = 16.5

# moles of other combustion products = 100

# moles of all products (in original equation) = 116.5

Pw = .106 * (16.5/116.5) = .015 Mpa

Tsat (.015 MPa) = 54 C

Department of Mechanical Engineering ME 322 – Mechanical Engineering Thermodynamics Lecture 36 Combustion Reactions.

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