Chemical Engineering Calculations Problem Set 3

1. A 100 kmol/h stream that is 97 mole% carbon tetrachloride (CCl 4) and 3% carbon disulfide (CS2)
is to be recovered from the bottom of a distillation column. The feed to the column is 16 mole% CS 2
and 84% CCl4, and 2% of the CCl4 entering the column is contained in the overhead stream leaving
the top of the column.

(a) Draw and label a flowchart of the process and do the degree-of-freedom analysis.
(b) Calculate the mass and mole fractions of CCl4 in the overhead stream, and determine the
molar flow rates of CCl4 and CS2 in the overhead and feed streams.

Given:
Bottom stream: 100 kmol stream (97% mol CCl4, 3% mol CS2)
Feed stream: 16% mol CS2, 84% mol CCl4
Required: (a) (b)
Solution:
Overhead stream (O)
(xccl4% mol CCl4, xcs2% mol CS2)




Feed stream (F)
(84% mol CCl4, 16% mol CS2)
Distillation
Column




Bottom stream (B)
100 kmol/h (97% mol CCl4, 3% mol CS2)




Degree of Freedom Analysis:

𝑁𝑈 = 4
• Feed stream
• Overhead stream
• % mol of carbon tetrachloride
• % mol of carbon disulfide

𝑁𝐸 = 4
• Overall material balance
• Carbon tetrachloride or carbon disulfide balance
• Summation of mol fractions in the overhead stream = 1
• Process specification (2% of the carbon tetrachloride entering the column is contained in
the overhead stream)

𝐷𝐹 = 4 − 4 = 0

Basis: 100 kmol (1 hour operation)

Amount of CCl4 in the overhead stream: F(0.84)(0.02)

OMB:

, 𝐹 = 𝑂 + 𝐵; 𝐹 = 𝑂 + 100
CCl4 Balance:
𝐹 (0.84) = F(0.84)(0.02) + 100(0.97); 𝐹 = 117.83 𝑘𝑚𝑜𝑙; 𝑂 = 17.83 𝑘𝑚𝑜𝑙
CS2 Balance:
𝐹 (0.16) = 𝑂(𝑥𝑐𝑠2 ) + 100(0.03)
117.83(0.16) = 17.83(𝑥𝑐𝑠2 ) + 100(0.03)
𝑥𝑐𝑠2 = 0.889; 𝑥𝑐𝑐𝑙4 = 0.1109
Molar flowrates:
𝑘𝑚𝑜𝑙
𝑂(𝑥𝑐𝑠2 ) = 17.83(0.889) = 15.85 ℎ
𝑘𝑚𝑜𝑙
𝑂(𝑥𝑐𝑐𝑙4 ) = 17.83(0.1109) = 1.98 ℎ


Overhead stream Molar flowrate (kmol/h) Mass fraction Mol fraction
Carbon tetrachloride 1.98 0.2012 0.1109
Carbon disulfide 15.85 0.7988 0.889