Question

Question 3 Consider the following combustion reaction of butane (C4H10): C,H,+ 13 02 2 → 4C0, +5H,0 a) Case I: 25 kmol/h of C4H10 and 1750 kmol/h of air is fed to the reactor. Determine the percentage excess air used. You may assume that the air contains 21% oxygen (O2) and 79% nitrogen (N2) (on a mole basis). [15 marks] b) Case II: 125 kmol/h of C4H10 and oxygen (O2) is fed to the reactor. The product stream leaving the reactor contains 3.05 mol% C4H10, 28.43 mol% carbon dioxide (CO2), 35.53 mol% water H20 and remainder oxygen (O2). Calculate the percentage conversion of C4H10. [15 marks]

Answer 1

Solution:

Chemical Reaction

• C₄H₁₀ + 6.5 O₂ --> 4CO₂ + 5H₂O

Case 1

Given data

• 25 kmol / h of C₄H₁₀
• 1750 kmol of Air having 21% oxygen & 79% Nitrogen.

First, We calculate oxygen required for complete combustion.

1 mole C₄H₁₀ required 6.5 mole of O₂

25 kmol C₄H₁₀ required = 25*6.5 = 162.5 kmol of O₂

Now, We calculate amount of air required.

Air contains 21% O₂ and 79% N₂ by mole percent.

Air required = 162.5 / 0.21 = 773.81 kmol of air

Now we calculate Percentage excess air.

% Excess air = [(Air feed - Air required) / Air required ] *100

= [(1750 - 773.81) / 773.81] * 100

= 126 %

Case 2

Given data

• 125 kmol / h of C₄H₁₀
• Product contain 3.05 % of C₄H₁₀ , 28.43 % of CO₂, 35.53 % of H₂O

We assume that total moles of product = X

Hence,

Moles of C₄H₁₀ in product = 0.0305X

Moles of CO₂ in Product = 0.2843X

First we calculate Moles of C₄H₁₀ Reacted.

Moles of C₄H₁₀ reacted = 125 - 0.0305X

Now, we calculate Moles of CO₂ produce.

1 mole C₄H₁₀ produce 4 mole of CO₂

(125 - 0.0305X) mole C₄H₁₀ produce = 4*(125 - 0.0305X) mole CO₂

Now we compare CO₂ generated and CO₂ in product.

0.2843X = 4*(125 - 0.0305X)

X = 1230.6178 Moles

Moles of C₄H₁₀ reacted = 125 - 0.0305X = 125 - 1230.6178*0.0305 = 87.47 mole

% Conversion of C₄H₁₀ = (87.47 / 125 )*100 = 69.97%