Department of Mechanical Engineering ME 322 Mechanical ...

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

Fuel

In 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

Oxidant

The oxidant must contain oxygen. The most abundant ‘free’

source is atmospheric air. By molar percent, atmospheric air

is considered to be ...

Nitrogen 79%

Oxygen 21%

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.

6


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

7

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 mol

mol

air

mass mol massfuel 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 (=1)

– CO2, H2O, N2

• Lean ( < 1 with T < 1800 R)

– CO2, H2O, N2, O2

• Rich ( > 1 with T < 1800 R)

– CO2, H2O, N2, O2, CO, H2

• Rich ( > 1 with T > 1800 R)

– CO2, H2O, N2, O2, CO, H2, H, O, OH, N, C(s), NO2, CH4

9

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 equations

5 unknowns

(0 through 4)

Atomic Balances

Stoichiometric Combustion of a General Fuel in Air

10

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 equations

6 unknowns

(0 through 5)

Requires a stoichiometric

balance first (to get 0)

Lean Combustion of a General Fuel in Excess Air

11

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


13

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 0

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

14

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

15

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 ...

17

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


/ / air

mass molfuel

MA F A F

M

The mass-based (A/F) ratio can be found knowing the

molecular masses of the air and the fuel,

18

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


Now, 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)


The 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

20

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

21

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 fuel

c) molar air/fuel ratio and percent theoretical air (PTA)

d) dew point temperature at .106 MPa


22

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=?


23

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%


24

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

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