Answer:
- 0.1852
- 0.0947
- 0.7201
- 3.0345 kg CO[tex]_{2}[/tex] / Kg C[tex]_{4}[/tex] H[tex]_{10}[/tex]
- 15.3848 Kg air / kg C[tex]_{4}[/tex] H[tex]_{10}[/tex]
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[tex]_{2}[/tex] / Kg C[tex]_{4}[/tex] H[tex]_{10}[/tex]
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[tex]_{4}[/tex] H[tex]_{10}[/tex]
