A gas contains 75.0 wt% methane, 10.0% ethane, 5.0% ethylene, and the balance water. (a)

Problem 4.84

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 a wet and a dry basis and the ratio (mol H2O/ mol dry gas). (b) If 100 kg/h of this fuel is to be burned with 30% excess air, what is the required air feed rate (kmol/h)? How would the answer change if the combustion were only 75% complete?

                                                               Step by Step Solution

Step 1 of 2

a)

The number of moles of substance can be calculated as follows:

Here, n is the number of moles of substance, w(kg) is the weight of substance in kilograms,

M(kg/kmol) is the molar mass of the substance in kilograms per kilo moles.

The wet basis includes the water with other substances. The dry basis only includes substances

not water

The air feed rate  can be calculated as follows:

Here,  is the moles of theoretical oxygen,  is the number of moles of air and  is the number of moles of oxygen

Let the total weight of gas (w) be 100 kg.

The weight percent of methane (w%) is 75.0%.

The weight of methane  can be calculated as follows:

Here, w% is the weight percent of methane and w is the total weight of the air.

Substitute the value of w% as 75.0%, w as 100 kg in the above equation as follows:

The molar mass of methane (M) is 16.0 g/mol.

So 16.0 g/mol = 16.0 kg/kmol.

The number of moles of methane  can be calculated as follows:

Substitute the value of  as 75 kg and M as 16.0 kg/kmol in the above equation as follows:

The weight percent of ethane (w %) is 10.0 %.

The weight of ethane  can be calculated as follows:

Here, w % is the weight percent of ethane and w is the total weight of the air.

Substitute the value of w % as 10.0%, w as 100 kg in the above equation as follows:

The molar mass of ethane (M) is 30.0 g / mol.

The number of moles of ethane  can be calculated as follows:

Substitute the value of  as 10 kg and M as 30.0 kg / kmol in the above equation as follows:

The weight percent of ethylene (w %) is 5.0 %

The weight of ethylene  can be calculated as follows:

Here, w % is the weight percent of ethylene and w is the total weight of the air.

Substitute the value of w % as 5.0%, w as 100 kg in the above equation as follows:

The molar mass of ethylene (M) is 28.0 g / mol.

The number of moles of ethylene  can be calculated as follows:

Substitute the value of  as 5kg and M as 28.0 kg / kmol in the above equation as follows:

The weight of water  can be calculated as follows:

Rearrange the above equation as follows:

Substitute the value of w as  as  as 10 kg and  as 5 kg in the above equation as follows:

The molar mass of water (M) is 18.0 g / mol.

The number of moles of water  can be calculated as follows:

Substitute the value of  as 10 kg and M as 18.0 kg / kmol in the above equation as follows:

The total number of moles of wet basis  can be calculated as follows:

Substitute the value of  as  as  as 0.18 kmol.  as 0.55 kmol in the above equation as follows:

For wet basis;

The mole percent of methane (n %) can be calculated as follows.

Substitute the value of  as 4.69 kmol and  as 5.75 kmol in the above equation as follows:

Therefore, the molar composition of methane for wet basis is 81.57 %

The mole percent of ethane (n %) can be calculated as follows:

Substitute the value of  as 0.33 kmol and  as 5.75 kmol in the above equation as follows.

Therefore, the molar composition of ethane for wet basis is 5.74%

The mole percent of ethylene (n %) can be calculated as follows:

Substitute the value of  as 0.18 kmol and  as 5.75 kmol in the above equation as follows:

Therefore, the molar composition of ethylene for wet basis is 3.13%

The mole percent of water (n %) can be calculated as follows:

Substitute the value of  as 0.55 kmol and  as 5.75 kmol in the above equation as follows:

Therefore, the molar composition of water for wet basis is 9.57%

The total number of moles of dry basis  can be calculated as follows:

the above equation as follows:

For dry basis;

The mole percent of methane (n %) can be calculated as follows:

Substitute the value of  as 4.69 kmol and  as 5.2 kmol in the above equation as follows:

Therefore, the molar composition of methane for dry basis is 90.19 %

The mole percent of ethane (n %)  can be calculated as follows:

Substitute the value of  as 0.33 kmol and  as 5.2 kmol in the above equation as follows:

Therefore, the molar composition of ethane for dry basis is 6.35%

The mole percent of ethylene (n %)  can be calculated as follows:

Substitute the value of  as 0.18 kmol and  as 5.2 kmol in the above equation as follows:

Therefore, the molar composition of ethylene for dry basis is

The ratio of moles of water over moles of dry basis  can be calculated as follows:

Substitute the value of  as 0.55 kmol and  as 5.2 kmol in the above equation as follows:

Therefore, the ratio  is 0.105.

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