Question

Propanol is burned with excess air to produce a mixture of comstock gases. The burner feed consists of 600 kh / h of propanol and 8700 kg / h of air. It is found that the total conversion of propanol is 90%. 95% of the propanol is burned through the complete reaction and the rest through incomplete reaction.

A. Determine the stoichiometry

CH3CH2CH2OH+ _O2---> _CO2+ H20

CH3CH2CH2OH+_O2 ---> _CO2 +_H20

B. LImitint reagent.

C. Excess reactive and %excess.

D. The molar composition on a humid base of the combustion gases% molar

E. The molar composition of the free gases of water vapor, % molar.

Answer 1

Given data :

Propanol Mass flow rate = 600 Kg/h

Propanol Molar flow rate = Mass flow rate/Molar mass = 600(kg) /60(kg/kmol) = 10 kmol/h

Mass flow rate of air = 8700 Kg

Molar flow rate of air = Mass flow rate / Molar mass = 8700(kg) / 29 (kg/kmol) = 300 Kmol /h

Molar flow rate of O₂ = Mole fraction of O₂ * Molar flow rate of air = 0.21*300 = 63 Kmol/h

Molar flow rate of N₂ = Mole fraction of N₂ * Molar flow rate of air = 0.79*300 = 237 Kmol/h

The reaction between Propanol and air (O₂) can be given as

CH₃CH₂CH₂OH + 4.5O₂ 3CO₂ + 4H₂O

So each mole of propanol requires about 4.5 mols of O₂ forcomplete combustion.

Molar rate of propanal in our system = 10 kmol/h

Molar rate of O₂ required according to stoichimery = 4.5 * Molar rate of propanal = 4.5*10 = 45 kmol/h

Molar rate of O₂ supplied = 63 kmol/h

B) From above calculations it is evident that "Propanol" is the limiting reactant. Excess reactant is O₂ .

C) Amount of excess reactant :

Excess flow rate of O₂ = Actual Flow rate - Theortically required flow rate = 63 - 45 = 18 kmol/h

Percentage excess = (Actual Flow rate - Theortically required flow rate /Theortically required flow rate)*100 = ((63 - 45) / 45) *100

Percentage excess = 40 %*

D) Products :

a) Moles of Propanol converted:

It is given that 90% of propannol got converted. Thus the amount of Propanol converted = 0.9 * 10 = 9 kmol/h

Propnaol exiting with the gases = 10 - 9 = 1 kmol/h

b) CO₂

From stoichiometry for each mole of 3 mol of CO₂ is produced. So for 9 kmol/h of propnaol, 3*9 = 27 kmol/h of CO₂ is produced.

c) H₂O

From stoichiometry for each mole of 4 mol of H₂O is produced. So for 9 kmol/h of propnaol, 4*9 = 36 kmol/h of H₂Ois produced.

d) O₂

From stoichiometry for each mole of 4.5 mol of O₂ is consumed. So for 9 kmol/h of propnaol, 4.5*9 = 40.5 kmol/h of CO₂ is produced.

CO₂ exiting with the gases = 63 - 40.5 = 22.5 kmol/h

e) N₂

N₂ leaving the system = N₂ entering the system = 237 kmol/h

The product stream :

Propanol = 1 kmol/h

O₂ = 22.5 kmol/h

N₂ = 237 kmol/h

CO₂ = 27 kmol/h

H₂O = 36 kmol/h

D) Molar composition humid based :

Total molar flow rate of gases = 1 + 22.5 + 237 + 27 + 36 = 323.5 Kmol/h

% Propanol = (Molar fflow rate of propanol / Total molar flow rate)*100 = (1/323.5) * 100 = 0.309 %

Similarly,

% O₂ = 22.5/323.5 * 100 = 6.955 %

% N₂ = 237/323.5 * 100 = 73.261 %

% CO₂ = 27/323.5 * 100 = 8.346 %

% H₂O = 36/323.5 * 100 = 11.128 %

E)

D) Molar composition on water vapour free based :

Total molar flow rate of gases excluding water vapour = 1 + 22.5 + 237 + 27 = 287.5 Kmol/h

% Propanol = (Molar flow rate of propanol / Total molar flow rate)*100 = (1/287.5)*100 = 0.347 %

Similarly,

% O₂ = 22.5/287.5 * 100 = 7.826 %

% N₂ = 237/287.5 * 100 = 82.434 %

% CO₂ = 27/287.5 * 100 = 9.391 %

% H₂O = 36/287.5 * 100 = 12.521 %