Question

n-Butane fuel (C4H10) is burned with the stoichiometric amount of air. Determine the mass fraction of each product. Also, calculate the mass of carbon dioxide in the products and the mass of air required per unit of fuel mass burned.

Answer 1

Answer:

1. 0.1852
2. 0.0947
3. 0.7201
4. 3.0345 kg CO$_{2}$ / Kg C$_{4}$ H$_{10}$
5. 15.3848 Kg air / kg C$_{4}$ H$_{10}$

Explanation:

Molar masses of each product are :

Butane = 58 kg /kmol

Oxygen = 32 kg/kmol

Nitrogen = 28 kg/kmol

water = 18 kg/kmol

1) Calculate the mass fraction of carbon dioxide

= ( 4 * 44 ) / ( (5 * 18) + (4 *44 )+ (24.44 * 28) )

= 176 / 950.32

= 0.1852

2) Calculate the mass fraction of water

= ( 5 * 18 ) /  (( 5* 18 ) + ( 4*44) + ( 24.44 * 28 ))

= 90 / 950.32

= 0.0947

3) Calculate the mass fraction of Nitrogen

= (24.44 * 28 ) / ((4 * 44 ) + ( 24.44 * 28 ) + ( 5 * 18 ))

= 684.32 / 950.32

= 0.7201

4) Calculate the mass of Carbon dioxide in the products

Mco2 = ( 4 * 44 ) / 58  = 3.0345 kg CO$_{2}$ / Kg C$_{4}$ H$_{10}$

5) Mass of Air required per unit of fuel mass burned

Mair = ( 6.5 * 32 + 24.44 *28 ) / 58  = 15.3848 Kg air / kg C$_{4}$ H$_{10}$