Chemical Equilibrium

Estimation of Combustion Products under Equilibrium Conditions in Reciprocating

Under Supervision Of Prof. H.N. Gupta

Submitted by Ashish Kumar Agrawal Roll No. 08306EN017

Chemical equilibrium

In a chemical process, chemical equilibrium is the state in which the chemical activities or concentrations of the reactants and products have no net change over time.

Usually, this would be the state that results when the forward

To understand the concept of Chemical equilibrium one general chemical equation can be expressed as, aA + bB ↔ cC + dD Where, A and B are the reactants and, C and D are the products. a, b, c, d are stoichiometric coefficent

Equilibrium Constant Equilibrium constant expresses the ratio of the product of the concentrations of the reaction products (right side) to the product of the concentration of the reactants (left side). The equilibrium constant can be expressed as:

Thermodynamics of combustion The reaction of the hydrocarbon fuel and air is represented by the general equation: a(CnHmOr) + a/Ф(n+m/4 – r/2)[O2 + 78/21 N2 + 1/21 Ar] Σ xi (i=1 to12) Where, CnHmOr = General formula fuel a = Mole fraction of fuel

The following 12 species (q=12) , are considered to be present in the products in the cylinder and in the exhaust gases. They will be refer by the number shown against their names.   3.

 

H2O

4.

H2

5.

OH

6.

H

7.

N2

N CO2 CO O2

7. 8. 9. 10. 11.

In the from of Equilibrium constant   Kp1 = (x4/√x2) √P Kp2 = (x11/√x10) √P Kp3 = (x7/√x5) √P Kp4 = (x10/b^2)P Kp5 = (x3/(b√x2)) √P Kp6 = (bx9/x8) Kp7 = (x6/(b√x5)) √P   Where b = x1/x2   The equilibrium constants for this reactions are calculated using   ln Kp = [ Σ (υ*g(T)/R*T )R - Σ (υ*g(T)/R*T )P] – ∆Ho/RT

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There are 13 unknowns ; the 12 species fractions (xi) and the total number of moles of fuel (a) ; we have only 7 equation..

•

DOF = 13 – 7 = 6

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Hence, 6 more equations are needed for the solution, One of these is Σxi = 1

  •

The remainder are the atomic mass balance for Argon , Carbon , Hydrogen , Oxygen and Nitrogen.

After that we get 13 non –liner equations and this can be solve by some approximations    

Expression for a: 

When

ф>1,

a = 1.3 / [{n + 0.5m + 1.863(2n + 0.5m – r)/ ф}*exp(0.13T/1000) ]   and when ф<1,   a = 1.3 / [{0.25m + 2.363(2n + 0.5m – r)/ ф + 0.5r}*exp(0.13T/1000) ]     Expression for b :  30000/T];

when T< 3000 K, b = exp[-9.0 + 0.5*logP +

All the values of xi are calculated and below shown equations are checked for balance. Σxi= 1 (i= 1 to 12) Mass balance for oxygen, x1 + x3 + x6 + 2x8 + x9 + 2x10 + x11 = a*y Where,

y = 1/ ф (n+m/4 – r/2)*2 +r

If they do not balance within a stipulated accuracy, an adjustment is made to a and b using the Newton - Raphson technique and

Computer program in Turbo-c and MATLAB

Results and Discussion

Main curve Composition variation with Temperature at P = 50 atm

0

10

 

N2 CO2 ­1

10

­2

10

Mole Fraction

NO ­3

CO

10

­4

10

­5

10

N

­6

10   1800

2000

2200

2400 Temperature in Kelvin

2600

2800

3000

H20 H2 OH H N2 NO N CO2  CO O2 O Ar

Composition variation with Pressure and Temperature for NO 0.018

 

first line for lowest pressure of 1 atm with others for +10 atm increments as we move down

0.016

0.014

Composition for different Pressures

0.012

0.01

0.008

0.006

0.004

0.002

0  1000

1200

1400

1600

1800

2000 Temperature in Kelvin

2200

2400

2600

2800

3000

0.08

Composition variation with Pressure and Temperature for CO

 

first line for lowest pressure of 1 atm with others for +10 atm increments as we move down

0.07

0.06

Composition for different Pressures

0.05

0.04

0.03

0.02

0.01

0  1000

1200

1400

1600

1800

2000 Temperature in Kelvin

2200

2400

2600

2800

3000

Composition variation with Pressure and Temperature for CO2

0.13

 

0.12

0.11

0.1

Composition for different Pressures

0.09

0.08

0.07

0.06

0.05

0.04

0.03   1000

first line for lowest pressure of 1 atm with others for +10 atm increments as we move down 1200

1400

1600

1800

2000 Temperature in Kelvin

2200

2400

2600

2800

3000

Conclusion Spark-ignition and diesel engines are a major source of urban air pollution. The spark-ignition engine exhaust gases contain oxides of nitrogen (nitric oxide and nitrogen dioxide), carbon monoxide (CO) and organic compounds which are unburned or partially burned hydrocarbons. NOxemission is controlled from exhaust gas recirculation (EGR). Spark timing significantly affects NOx emission levels. NOx level is significantly reduce at the maximum brake torque (MBT). Carbon monoxide is reduce when the engine is working under lean condition.   A 3-way catalyst removes all the three pollutants

Thank you…