Question

The % analysis by volume of product gas is H2 = 18.3%, CH2 = 3.4%, CO =25.4%, CO2 =5.1% and N2 = 47.8%. Calculate the volume of air required in m3 of the gas?​

Answer 1

A : 1.3643 m3 this is the answer

I hope it's help you

Answer 2

Given:

Volume of product gas is  $H_{2}$= 18.3%,  $CH_{2}$ = 3.4%, $CO$ =25.4%, $CO_{2}$ =5.1% and $N_{2}$ = 47.8%.

To Find:

Volume of air required in $m^{3}$ of the gas

Solution:

Let the Gaseous fuel be 100 mole  

$CO + \dfrac{1}{2} O_{2} = CO_{2}$

Volume of $O_{2}$ required for combustion CO = 1/2 × 25.4 = 12.7

$H_{2} + \dfrac{1}{2} O_{2} = H_{2}O$

Volume of $O_{2}$ required =  1/2 × 18.3 = 9.15

Volume of $O_{2}$ required in $CH_{2}$ =  2 × 3.4 = 6.8

Total moles of $O_{2}$  required = 28.65

So, Volume of air required  for 100 moles of fuel gas = 28.65 × (100/21) = 136.43 mole

Therefore air required per $m^{3}$ of gas = (136.43/100) = 1.3643  $m^{3}$