00:01
In this question, the ethyl alcohol at a c2h5 gas is burned in a steady flow adiabatic combustion chamber with 90 % excess air.
00:21
The adiabatic flame temperature of the products is to be determined and the adiabatic flame temperature as a function of the present percent excess air is to be plotted here.
00:34
So when we analyze the question, the complete combustion reaction, in this case can be written as c2h5oh, which is gas plus 1 plus a02 plus 3 .76n2 gives, i'm writing it down, the product i'm writing it down, 2 co2 plus 3h2 plus o2 plus fn2.
01:30
So this is the complete combustion reaction in this case.
01:36
Now, we go slowly.
01:44
Okay, let's move down.
01:49
Now, where the eighth is, what is this a -t -h -given? is this tachymetric coefficient of air.
02:10
Now, the oxygen balance gives the oxygen balance that we are talking about gives 1 plus 1 plus excess of a.
02:31
2 is equals to 2 into 2 plus 3 plus 1 plus 1 plus x a .f into 2.
02:47
Now, the reaction equation with products in the equilibrium is, we are going to write down the reaction equation with products in the, products in equilibrium.
03:02
So it goes c2h5oh gas plus 1 plus 802, 3 .76n2 gives you a number.
03:23
Writing the product, the green color, acu, we are giving the variable here, then we'll balance it.
03:42
So when we see this, we understand, now we are going to, the coefficients are determined by the mass balance.
03:53
So that is what we are going to look here.
03:56
So the carbon balance is equals to two is equals to a plus b, now hydrogen balance, six is equal, to 2d that gives you d equals to 3.
04:13
Now the oxygen balance 1 plus 1 plus e x a into 2 equals to 2a plus b plus d plus 2 e and the nitrogen balance is 1 plus x x into 2 equals to 2a plus 2a plus 2a.
04:50
So these are the coiff mass balances that we have got.
04:55
Solving the above equation, we find the coefficients to be.
05:05
X is equals to 0 .9.
05:12
8 is equals to 3 .a equals to 2.
05:21
B equals to 0 .000000 8644 .4.
05:29
D is equals to 3, t equals to 2 .7, and f equals to 21 .43...