Question

A gas contains 75.0 wt% methane, 10.0% ethane, 5.0% ethylene, and the balance water. (a) Calculate the molar composition of this gas on both awet and a dry basis and the ratio (mol H2O/ mol dry gas). (b) If100kg/hofthisfuelistobeburnedwith30%excessair,whatistherequiredairfeedrate(kmol/ h)? How would the answer change if the combustion were only 75% complete?

Answer 1

a) molar composition of this gas on both a wet and a dry basis are

5.76 moles and 5.20 moles respectively.

Ratio of moles of water to the moles of dry gas =0.108 moles

b) Total air required = 68.51 kmoles/h

So, if combustion is 75% complete; then it is termed as incomplete combustion which require the same amount the same amount of air but varying product will be produced.

Step-by-step explanation:

Let assume we have 100 g of mixture of gas:

Given that :

Mass of methane =75 g

Mass of ethane = 10 g

Mass of ethylene = 5 g

∴ Mass of the balanced water: 100 g - (75 g + 10 g + 5 g)

Their molar composition can be calculated as follows:

Molar mass of methane
`CH_4}= 16 g/mol`

Molar mass of ethane
`C_2H_6= 30 g/mol`

Molar mass of ethylene
`C_2H_4 = 28 g/mol`

Molar mass of water
`H_2O=18g/mol`

number of moles =
`(mass)/(molar mass)`

Their molar composition can be calculated as follows:

`n_(CH_4)= (75)/(16)`

`n_(CH_4)=` 4.69 moles

`n_(C_2H_6) = (10)/(30)`

`n_(C_2H_6) =` 0.33 moles

`n_(C_2H_4) = (5)/(28)`

`n_(C_2H_4) =` 0.18 moles

`n_(H_2O)= (10)/(18)`

`n_(H_2O)=` 0.56 moles

Total moles of gases for wet basis = (4.69 + 0.33 + 0.18 + 0.56) moles

= 5.76 moles

Total moles of gas for dry basis = (5.76 - 0.56)moles

= 5.20 moles

Ratio of moles of water to the moles of dry gas =
`(n_(H_2O))/(n_(drygas))`

=
`(0.56)/(5.2)`

= 0.108 moles

b) If 100 kg/h of this fuel is burned with 30% excess air(combustion); then we have the following equations:

`CH_4 + 2O_2_((g)) ------> CO_2_((g)) +2H_2O`

4.69 2× 4.69

moles moles

`C_2H_6+ (7)/(2)O_2_((g)) ------> 2CO_2_((g)) + 3H_2O`

0.33 3.5 × 0.33

moles moles

`C_2H_4+3O_2_((g)) ----->2CO_2+2H_2O`

0.18 3× 0.18

moles moles

Mass flow rate = 100 kg/h

Their Molar Flow rate is as follows;

`CH_4 = 4.69 k moles/h\\C_2H_6 = 0.33 k moles/h\\C_2H_4=0.18kmoles/h`

Total moles of
`O_2` required = (2 × 4.69) + (3.5 × 0.33) + (3 × 0.18) k moles

= 11.075 k moles.

In 1 mole air = 0.21 moles
`O_2`

Thus, moles of air required =
`(1)/(0.21)*11.075`

= 52.7 k mole

30% excess air = 0.3 × 52.7 k moles

= 15.81 k moles

Total air required = (52.7 + 15.81 ) k moles/h

= 68.51 k moles/h

So, if combustion is 75% complete; then it is termed as incomplete combustion which require the same amount the same amount of air but varying product will be produced.