00:01
Gaseous fuel mixture of 60 % propane c3 hatchate and 40 % butane c4 h10 on a volume basis is burnt in air so that the air fuel ratio is 25k air per kg.
00:33
Fuel when the combustion process is complete.
00:36
Then determine the moles of nitrogen in air supplied to the combustion process in kilo -mole per kilo -mole fuel.
00:56
Part b asks the moles of water formed in the conversion process process in kilo -mole per kilo -mole of fuel.
01:12
Third, to find the moles of oxygen, the product gases in kilo -mole per kilo -mole of fuel.
01:25
To find the number of moles, we need to find first the stoichiometric coefficient for air, air and unknown number of moles of the gases in the growths by balancing the compounds x is equals to 3 into 0 .6 plus 4 into 0 .4 equals 3 and 8 into 0 .6 plus 10 into 0 .4 is equals to 2y from here the value of y will be y equals 4 .4 then twice of this trotometer coefficient is 2x plus y let this be the third equation this the second equation is this one and the first equation is for the equation of compound form which is 0 .6 c3h8 plus 0 .4 c4h10 plus atho2 plus a .tho2 gives x times co2 plus y times h2 plus z times n2.
03:14
This is our first equation.
03:17
Based on this we have calculated the value of x and now we have to find the value of a .t .h 3 .76 times a .t .h is equal to z.
03:35
This is our third equation.
03:39
Substituting the value of x and y in equation 2, we get ath equals 5 .6.
03:47
Substitute it again in third equation.
03:50
We get z value will be 21 .1.
03:53
0 .06.
03:55
Now substituting the value for x, y, z and a t h into our first equation, this will become 0 .6 c3h8 plus 0 .4 c4 h10 plus 5 .6 times o2 plus 3 .76 times n2 gives 3 .4 co2 gives 3 .4 co2 plus 4 .4 h2o plus 21 .06 n2.
04:38
Expressing the theoretical air fuel ratio is equals to mass of air by mass of fuel...