written 5.2 years ago by
vibhavarikulkarni
• 480
|
modified 4.0 years ago
by
sanketshingote
• 100
|
$CH_4$ = 35%,$C_2 H_4$ = 5%, CO = 15%, $H_2$ = 40%, $N_2$ = 1 %, water vapours =4%, ( mol wt of air = 28.949)
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written 5.1 years ago by
vibhavarikulkarni
• 480
|
modified 8 months ago
by
adityakumarpradhan8694
• 30
|
| Constituent | %by weight | Wt of each per kg of fuel |
|---|---|---|
| $H_2$ | 40 | 0.4 x 0.5 = 0.20 |
| $CH_4$ | 35 | 0.35 x 2 = 0.7 |
| $N_2$ | 1 | Does not contribute |
| $CO$ | 15 | 0.11 |
| $C_2H_4$ | 3 | 0.05 x 3 = 0.15 |
$H_2 +1/2 O_2 \rightarrow H_2O$ = 0.20 $m^3$
$CH_4 +2O_2 \rightarrow CO_2 + 2H_2O$ = 0.7 $m^3$
$CO+1/2 O_2 \rightarrow CO_2$ = 0.11 $m^3$
$C_2H_4 +3O \rightarrow_2 2CO_2 + 2H_2O$ = 0.15 $m^3$
$=0.7 m^3 + 0.11m^3 + 0.2m^3 + 0.15 m^3$
Amount of oxygen required = $1.16 m^3$
Volume of air required = $1.16 m^3$
$1.16 m^3$ of air will weigh = $(1.16/22.4) \times 28.949$ = $1.499 Kg$
$\therefore$ Wt of air required = 1.499 Kg
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