Question

15kg / h of Propane should be burned with 15% excess air.

2.1.-Determine the air / fuel ratio to be used.
2.2.-Determine the amount of water, kg / h that produces the combustion.
2.3.-Determine the amount of burned gases, kg / h that combustion produces.

Answer 1

2.1) Molar mass Propane ,C3H8 = 44g

The balanced chemical equation for the combustion of propane is:

C3H8(g)+5O2(g)→3CO2(g)+4H2O(g).

Oxygen needed = 160 g

Since air contains about 23% oxygen hence air needed would be equal to (100/23)*160 = 695.65 g

Air needs to be 15 % excess so actual air = 695.65(1+0.15) = 800 grams

Air fuel ratio = 800 / 44 = 18.18

2.2 ) 44 grams is giving 72 gram water

1 gram will give 72/44 grams water

15kg/hr will give (72/44) * 15 = 24.54 kg water / hour

2.3)44 grams propane gives 132 grams burned gases

1 grams give 132/44 grams of burned gases

15kg/hr give (132/44)*15kg/hr = 45kg/hr burned gases

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