Answer:
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.
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 [tex]CH_4}= 16 g/mol[/tex]
Molar mass of ethane [tex]C_2H_6= 30 g/mol[/tex]
Molar mass of ethylene [tex]C_2H_4 = 28 g/mol[/tex]
Molar mass of water [tex]H_2O=18g/mol[/tex]
number of moles = [tex]\frac{mass}{molar mass}[/tex]
Their molar composition can be calculated as follows:
[tex]n_{CH_4}= \frac{75}{16}[/tex]
[tex]n_{CH_4}=[/tex] 4.69 moles
[tex]n_{C_2H_6} = \frac{10}{30}[/tex]
[tex]n_{C_2H_6} =[/tex] 0.33 moles
[tex]n_{C_2H_4} = \frac{5}{28}[/tex]
[tex]n_{C_2H_4} =[/tex] 0.18 moles
[tex]n_{H_2O}= \frac{10}{18}[/tex]
[tex]n_{H_2O}=[/tex] 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 = [tex]\frac{n_{H_2O}}{n_{drygas}}[/tex]
= [tex]\frac{0.56}{5.2}[/tex]
= 0.108 moles
b) If 100 kg/h of this fuel is burned with 30% excess air(combustion); then we have the following equations:
[tex]CH_4 + 2O_2_{(g)} ------> CO_2_{(g)} +2H_2O[/tex]
4.69 2× 4.69
moles moles
[tex]C_2H_6+ \frac{7}{2}O_2_{(g)} ------> 2CO_2_{(g)} + 3H_2O[/tex]
0.33 3.5 × 0.33
moles moles
[tex]C_2H_4+3O_2_{(g)} ----->2CO_2+2H_2O[/tex]
0.18 3× 0.18
moles moles
Mass flow rate = 100 kg/h
Their Molar Flow rate is as follows;
[tex]CH_4 = 4.69 k moles/h\\C_2H_6 = 0.33 k moles/h\\C_2H_4=0.18kmoles/h[/tex]
Total moles of [tex]O_2[/tex] required = (2 × 4.69) + (3.5 × 0.33) + (3 × 0.18) k moles
= 11.075 k moles.
In 1 mole air = 0.21 moles [tex]O_2[/tex]
Thus, moles of air required = [tex]\frac{1}{0.21}*11.075[/tex]
= 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.
