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Topic: Combustion of Propane  (Read 6456 times)

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Offline jturn14

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Combustion of Propane
« on: November 14, 2006, 05:24:23 PM »
hi, this might be a little involved, but i'd appreciate any help.

i'm working on a project involving the combustion of propane, resulting in soot. I'm having trouble balancing the equation. Here is the equation, as assigned by my professor:

C3H8 + 0.0.476O2 --> aCO + bCO2 + cH2 + dH2O + eC9H

The equations I'm using are as follows:

C/H:  3/8 = (a + 2b + 9e)/(2c+2d+e)
C/O: 3/.9524 = (a+b+9e)/(a+2b+d)
H/O: 8/.9524 = (2c+2d+e)/(a+2b+d)
a+b+c+d+e = 1

.228 = (ad)/(bc)  (equilibrium for water-gas shift)

those are my equation. I wrote a Matlab program to solve the nonlinear system, giving:

a = 0.1879, b=0.000021110, c=0.76714, d=0.0000196, e=0.0449

Are these values reasonable? They don't seem to balance when balancing the equation using mole fractions.

If anyone can help, that'd be great. I'm at school, and the security settings on these stupid computers won't let me post the Matlab code into this post. So i will post it when I get home.


For anyone with Matlab that would like to check my code, here it is:

clc
clear all;
x1 = .2; x2 = .2; x3 = .2; x4 = .2; x5 = .2;
d = 1;
while norm(d)>1e-12
f1 = -8*(x1)-8*(x2)+6*(x3)+6*(x4)-69*(x5);
f2 = 2.048*(x1)+5.048*(x2)+3*(x4)-8.57*(x5);
f3 = -8*(x1)-16*(x2)+1.904*(x3)+6.096*(x4)+.952*(x5);
f4 = (x1)+(x2)+(x3)+(x4)+(x5)-1;
f5 = .228*(x2)*(x3)-(x1)*(x4);
A = [-8 -8 6 6 -69;
    2.048 5.048 0 3 -8.57;
    -8 -16 1.904 6.096 .952;
    1 1 1 1 1;
    -(x1) .228*(x3) .228*(x2) -(x1) 0;];
b=[-f1 -f2 -f3 -f4 -f5]';
d=inv(A)*b;
x1=x1+d(1,1);
x2=x2+d(2,1);
x3=x3+d(3,1);
x4=x4+d(4,1);
x5=x5+d(5,1);
end
fprintf(1, 'x1 = %9.9f     x2 = %9.9f    x3 = %9.9f    x4 = %9.9f    x5= %9.9f', x1, x2, x3, x4, x5)

Offline Borek

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Re: Combustion of Propane
« Reply #1 on: November 15, 2006, 03:47:19 AM »
Are you sure it is an organic chemistry question?

Are these values reasonable? They don't seem to balance when balancing the equation using mole fractions.

They seem to be the solution of the set. Hard to say anything else.
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Offline billnotgatez

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Re: Combustion of Propane
« Reply #2 on: November 15, 2006, 04:37:29 AM »
Quote
C3H8 + 0.0.476O2 --> aCO + bCO2 + cH2 + dH2O + eC9H

I look at this and say, "where is the soot?" and "can hydrogen really exist after burning?”

Organic Chemistry is so confusing.

Offline Borek

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Re: Combustion of Propane
« Reply #3 on: November 15, 2006, 05:37:17 AM »
can hydrogen really exist after burning?

There is an equilibrium (used for town gas production):

C + H2O <-> H2 + CO

which can make it happen depending on conditions. It doesn't mean it happens here, just it is not that impossible ;)

You are right about soot, I suppose C9H is some approximate formula (soot doesn't have to be pure carbon - it may contain remains of hydrocarbons, at least that's what I would expect).
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Offline billnotgatez

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Re: Combustion of Propane
« Reply #4 on: November 15, 2006, 05:42:22 AM »
Oh Well, the professor gave him the equation so I guess we have to accept it.

Offline AWK

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Re: Combustion of Propane
« Reply #5 on: November 15, 2006, 06:12:38 AM »
hi, this might be a little involved, but i'd appreciate any help.

i'm working on a project involving the combustion of propane, resulting in soot. I'm having trouble balancing the equation. Here is the equation, as assigned by my professor:

C3H8 + 0.0.476O2 --> aCO + bCO2 + cH2 + dH2O + eC9H

The equations I'm using are as follows:

C/H:  3/8 = (a + 2b + 9e)/(2c+2d+e) Why 2
C/O: 3/.9524 = (a+b+9e)/(a+2b+d)
H/O: 8/.9524 = (2c+2d+e)/(a+2b+d)
a+b+c+d+e = 1 I doubt if this assumption was correct

.228 = (ad)/(bc)  (equilibrium for water-gas shift)

those are my equation. I wrote a Matlab program to solve the nonlinear system, giving:

a = 0.1879, b=0.000021110, c=0.76714, d=0.0000196, e=0.0449

Are these values reasonable? They don't seem to balance when balancing the equation using mole fractions.

If anyone can help, that'd be great. I'm at school, and the security settings on these stupid computers won't let me post the Matlab code into this post. So i will post it when I get home.


For anyone with Matlab that would like to check my code, here it is:

clc
clear all;
x1 = .2; x2 = .2; x3 = .2; x4 = .2; x5 = .2;
d = 1;
while norm(d)>1e-12
f1 = -8*(x1)-8*(x2)+6*(x3)+6*(x4)-69*(x5);
f2 = 2.048*(x1)+5.048*(x2)+3*(x4)-8.57*(x5);
f3 = -8*(x1)-16*(x2)+1.904*(x3)+6.096*(x4)+.952*(x5);
f4 = (x1)+(x2)+(x3)+(x4)+(x5)-1;
f5 = .228*(x2)*(x3)-(x1)*(x4);
A = [-8 -8 6 6 -69;
    2.048 5.048 0 3 -8.57;
    -8 -16 1.904 6.096 .952;
    1 1 1 1 1;
    -(x1) .228*(x3) .228*(x2) -(x1) 0;];
b=[-f1 -f2 -f3 -f4 -f5]';
d=inv(A)*b;
x1=x1+d(1,1);
x2=x2+d(2,1);
x3=x3+d(3,1);
x4=x4+d(4,1);
x5=x5+d(5,1);
end
fprintf(1, 'x1 = %9.9f     x2 = %9.9f    x3 = %9.9f    x4 = %9.9f    x5= %9.9f', x1, x2, x3, x4, x5)

This is a system of equation with two parameter variables.
3=a+b+9e
8=2c+2d+e
0.47602=a/2+b+d/2
Assume different values for e between 0 and 1/3 and for each of these values of e solve system of equation with one parameter variable. On the thermodynamic data (and at least heat of reaction=0) you can probably exclude some values of f.
AWK

Offline jturn14

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Re: Combustion of Propane
« Reply #6 on: November 15, 2006, 09:08:35 PM »
i figured it out

the a+b+c+d+e=1 assumption is correct, since i'm calculating mole fractions.

thanks for your input guys :)

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