Documentation:CHBE Exam Wiki/Module 1 - Process Basics

CHBE 241 Exam resources wiki
Chemical and Biological Engineering
Welcome to the CHBE Exam Resources Wiki! This wiki is intended to host past exams with fully worked-out hints and solutions
Past Exams
Final Exam 2016W
Midterm Exam 1 2016W
Midterm Exam 2 2016W
Problem Sets
Module 1 - Process Basics
Module 2 - Reactors
Module 3 - Separations 1
Module 4 - Separations 2
Module 5 - Non-reactive Energy Balances
Module 6 - Reactive Energy Balances

Question 1

Given a solution that is 20 wt% HCl with a specific gravity of 1.2, find the molarity of the solution.
The molecular weight of HCl is 36.5 g/mol.

Solution

${\begin{aligned}{\frac {\text{0.2g HCl}}{\text{g solution}}}\cdot {\frac {\text{1 mole HCl}}{\text{36.5g HCl}}}\cdot {\frac {\text{1.2 g solution}}{\text{mL solution}}}\cdot {\frac {\text{1000 mL}}{\text{1 L}}}&=6.57M\end{aligned}}$

Question 2

The molar composition of a gas mixture composed of species A, B and C are 30 mole%, 50 mol% and 20 mol% respectively. The molecular weights of species A, B and C are M_A = 12 g/mol, M_B = 32 g/mol and M_C = 50 g/mol respectively.

Question 2a

Calculate the average molecular weight of the gas mixture

Solution

${\begin{aligned}(0.3)({\text{12 g/mol}})+(0.5)({\text{32 g/mol}})+(0.2)({\text{50 g/mol}})&={\text{29.6 g/mol}}\end{aligned}}$

Question 2b

What are the weight fractions of species A, B and C? Using 100 moles as the basis

Solution

Number of moles (mol)	Molecular Weight (g/mol)	Mass (g)	X
30	12	360	0.1216
50	32	1600	0.5405
20	50	1000	0.3378
Total	-	2960	1

Question 3

A stream is 70% H₂ and 30% N₂ by mass. Will the mole fraction of H₂ be greater than, equal to or less than 70%? Explain.

Solution

Since H₂ has a smaller molecular weight, the number of moles and mole fraction of H₂ will be larger. Hence, the mole fraction will be larger than 70%.

Question 4

Blueberries contain about 85 wt% water and 15 wt% sugar (solids). To make blueberry jam, blueberries and pure sugar (solids) are mixed at a ratio of 60:40 by mass. Upon mixing, the mixture is then heated until the final water content of the mixture is one-third water by mass.

Question 4a

Draw and label a flowchart of this process.

Solution

Question 4b

Perform the degree of freedom analysis

Solution

Parameter	Values
Unknowns	3 (m1, m2, m3)
Species balances	2
Feed ratio	1
DOF	0

Question 4c

Calculate how many kilograms of blueberries are needed to make a kilogram of jam.

Solution

Based on the feed ratio given,
${\frac {m_{1}}{m_{2}}}={\frac {60}{40}}$
The sugar balance is
$0.15{m_{1}}+m_{2}=2/3$

By solving both the equations simultaneously,
$m_{1}=0.816kg$
$m_{2}=0.544kg$

Question 4d

The cost of sugar and blueberries are $0.15/lb and $2/lb respectively while the cost of operating the evaporator is $85 per day. If the blueberries are fed at 50 kg/hr, what is the cost to produce jam per day? Assume that the production period per day is 10 hours.

Solution

Cost of blueberries:
${\begin{aligned}2\$/lb\times 2.2046lb/kg\times 50kg/hr&=\$2205\end{aligned}}$

Cost of sugar:
${\begin{aligned}0.15\$/lb\times 2.2046lb/kg\times 33.33kg/hr&=\$110\end{aligned}}$

Total cost:
${\begin{aligned}{\text{cost of sugar}}+{\text{cost of blueberries}}+{\text{cost of operation}}=\$2400\end{aligned}}$