CH4 is established as the fuel used and we know that 7.5 moles of carbon and 29.83 moles of hydrogen is supplied. So, 7.5 CH4 will give 7.5 moles of carbon. Also, to give 29.83 moles of hydrogen from CH4, we will need to provide (29.83/4 = 7.45) moles of CH4.
Sir, I guess, we had to do 270+270 mol of water vapor because we have done partial and complete combustion on a water vapor right ? why did you take 270 mol of water vapor. ?
in considering the excess air you have taken per mol of c2h6 , 3.5 moles of oxygen is required but.. that is the case with coversion of c2h6 to co2 but you have not considered the conversion of c2h6 to co where per mol c2h6 ,2.5 moles of o2 is required
Hello harsh. In the previous lecture where he explained terminologies regarding combustion, he has mentioned that even if both CO and CO2 are formed in the reaction, theoretical air and excess air are calculated based on complete combustion only.
Dear teacher thank you so much.This videos are very important for me.And you are explaining fluently....
Amazing way of explanation. Really good 💯❤️
Best professor.
In example 1, why only moles of carbon is considered for calculating the moles of CH4 entering? What about Hydrogen?
My friend, hydrocarbon is burning, so 1 mole of CH4
CH4 is established as the fuel used and we know that 7.5 moles of carbon and 29.83 moles of hydrogen is supplied.
So, 7.5 CH4 will give 7.5 moles of carbon.
Also, to give 29.83 moles of hydrogen from CH4, we will need to provide (29.83/4 = 7.45) moles of CH4.
In example 2why sir has multipled with 1.5for air supplied?
50% excess air is supplied so,Air supplied = (150%) * theoretical air = (150/100) * theoretical air = 1.5 * theoretical air.
Sir, I guess, we had to do 270+270 mol of water vapor because we have done partial and complete combustion on a water vapor right ? why did you take 270 mol of water vapor. ?
in considering the excess air you have taken per mol of c2h6 , 3.5 moles of oxygen is required but.. that is the case with coversion of c2h6 to co2 but you have not considered the conversion of c2h6 to co where per mol c2h6 ,2.5 moles of o2 is required
Hello harsh. In the previous lecture where he explained terminologies regarding combustion, he has mentioned that even if both CO and CO2 are formed in the reaction, theoretical air and excess air are calculated based on complete combustion only.
Nice explaination
why we wrote 1.5 to calculate the air which is supplied ??
Hi there! 1.5 is an indication that air supplied is 150%( 100% + 50%(excess air)) of the theoretical air requirement.
how 7.5 moles of carbon meant that 7.5 moles of ch4 will also be there