Documentation:CHBE Exam Wiki/Module 4 - Separations 2

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

Is it physically possible to have a case where the atmospheric pressure is 740 mm Hg and absolute pressure is 700 mm Hg? Explain.

Solution

It is possible if it has a vaccum of 40mm Hg

Question 2

What is the volume of 100 moles of O₂ at STP?

Solution

$\begin{aligned} (100 moles) (22.4 L/mol) & = 2240 L \end{aligned}$

Question 3

If atmospheric pressure is 760 mm Hg and the absolute pressure is 600 mm Hg, what is the gauge pressure?

Solution

$\begin{aligned} P_{g} & = P_{abs} - P_{atm} \\ = -160 mmHg or 160 mmHg of vacuum \end{aligned}$

Question 4

At 20°C, the solubility coefficient of CO₂ in water at is 0.0901 cm³ CO₂ (STP)/cm H₂O(l).

Question 4a

For the given solubility coefficient, determine the Henry's law constant [in atm/mole fraction of CO₂ in H₂O]

Solution

Determine the number of moles of water and CO₂ by using 1cm³ H₂O (l) as the basis,

Water
At STP, the specific gravity of water is 1,
$\begin{aligned} n_{H_{2} O} & = 1 g H_{2} O \cdot \frac{1 m o l}{18 g} \\ = 0.0555 m o l H_{2} O \end{aligned}$

CO₂
$\begin{aligned} n_{C O_{2}} & = 0.0901 c m^{3} \cdot \frac{1 m o l}{22400 c m^{3} (S T P)} \\ = 4.022 \cdot 1 0^{- 6} m o l C O_{2} \end{aligned}$

Determine the mole fraction of CO₂,
$\begin{aligned} x_{C O_{2}} & = \frac{4.022 \cdot 1 0^{- 6}}{4.022 \cdot 1 0^{- 6} + 0.0555} \\ = 7.246 \cdot 1 0^{- 5} \end{aligned}$

Determine the Henry's constant,
$\begin{aligned} H_{C O_{2}} & = \frac{1 a t m}{7.246 \cdot 1 0^{- 5}} \\ = 13800 a t m \end{aligned}$

Question 4b

How many grams of CO₂ can be dissolved in a 600mL of soda at 20°C if pure CO₂ is present in the space above the soda in the bottle at 3 atm.
Assume that the soda has similar properties as water.

Solution

Determine the mole fraction of CO₂ in the soda,
$\begin{aligned} x_{C O_{2}} & = \frac{p_{C O_{2}}}{H_{C O_{2}}} \\ = \frac{3 a t m}{13800 a t m} \\ = 2.174 \cdot 1 0^{- 4} \end{aligned}$

$\begin{aligned} n_{C O_{2}} & = 600 m L \cdot \frac{1 L}{1000 m L} \cdot \frac{1000 g H_{2} O}{1 L} \cdot \frac{1 m o l H_{2} O}{18 g H_{2} O} \cdot \frac{2.174 \cdot 1 0^{- 4} m o l C O_{2}}{1 m o l H_{2} O} \cdot \frac{44 g C O_{2}}{1 m o l C O_{2}} \\ = 0.319 g C O_{2} \end{aligned}$

Question 4c

If all this CO₂ gas was released from solution at body temperature, what would the volume of CO₂ gas be in litres?
Assume that the conditions in the body are: temperature = 37°C and pressure = 1 atm.

Solution

$\begin{aligned} V & = 0.319 g C O_{2} \cdot \frac{1 m o l C O_{2}}{44 g C O_{2}} \cdot \frac{22.4 L}{1 m o l} \cdot \frac{(273 + 37) K}{273 K} \\ = 0.184 L \end{aligned}$

Question 5

Estimate the vapor pressure of acetone (mm Hg) at 45°C

Question 5a

from data in and the Clausius–Clapeyron equation

Solution

By linearizing the Clausius-Clapeyron equation, $\begin{aligned} l n p^{*} & = \frac{a}{(273.15 + T [C^{\circ}])} + b \\ y & = a x + b \end{aligned}$

From Perry's handbook
When T is 39.5°C, p* is 400 mm Hg
When T is 56.5°C, p* is 760 mm Hg
Based on the values,
$\begin{aligned} x_{1} & = 3.1980 \cdot 1 0^{- 3} \\ y_{1} & = 5.99146 \\ x_{2} & = 3.0331 \cdot 1 0^{- 3} \\ y_{2} & = 6.63332 \end{aligned}$

Since the equation is linearized, we can interpolate the value of y when T is 45°C, x is 3.1432·10^-3

$\begin{aligned} y & = y_{1} + \frac{x - x_{1}}{x_{2} - x_{1}} \cdot (y_{2} - y_{1}) \\ = 6.204765 \\ p^{*} & = e x p (6.204765) \\ = 495 m m H g \end{aligned}$

Question 5b

from the Cox chart (Figure 6.1-4 - Felder: Elementary Principles of Chemical Processes), and

Solution

45°C equals to 113°F
Using the Cox chart, p* is approximately 10 psi.

$\begin{aligned} 10 p s i & = 10 p s i \cdot \frac{760 m m H g}{14.6 p s i} \\ = 520 m m H g \end{aligned}$

Question 5c

from the Antoine equation using parameters

Solution

The parameters for acetone are

$\begin{aligned} a & = 7.11714 \\ b & = 1210.595 \\ c & = 229.664 \end{aligned}$

$\begin{aligned} l o g p^{*} & = 7.11714 - \frac{1210.595}{45 + 229.664} \\ = 2.7096 \\ p^{*} & = 512 m m H g \end{aligned}$

Question 6

A stirred tank is fed with aqueous acetone solution at a rate of 35 lb /h along with a stream of pure methy isobutyl ketone (MIBK). The resulting mixture is sent to a settler operating at 25°C. One of the phases contains 75 wt% MIBK with a flow rate of 40lb/h.
Determine the flow rate and composition of the second product stream and the rate at which MIBK is fed to the unit.

Solution

Let F1 be the feed containing acetone solution and F2 be the stream of pure MIBK being fed into the tank Let 1 be Phase 1 and 2 be Phase 2

From the ternary phase diagram for water, MIBK and acetone,
Phase 1: x_M,1 - 0.75, x_W,1 - 0.05, x_A,1 - 0.20
Phase 2: x_M,2 - 0.03, x_W,2 - 0.85, x_A,2 - 0.12

Overall mass balance:
$\begin{aligned} \dot{m_{F 1}} + \dot{m_{F 2}} & = \dot{m_{1}} + \dot{m_{2}} \\ 32 + \dot{m_{F 2}} & = 40 + \dot{m_{2}} \end{aligned}$

MIBK balance:
$\begin{aligned} \dot{m_{F 2}} & = \dot{m_{1}} \cdot x_{M, 1} + \dot{m_{2}} \cdot x_{M, 2} \\ \dot{m_{F 2}} & = 40 \cdot (0.75) + \dot{m_{2}} \cdot (0.03) \end{aligned}$

Solving the simultaneous equations,
$\begin{aligned} \dot{m_{F 2}} & = 30.8 l b / h \\ \dot{m_{2}} & = 25.8 l b / h \end{aligned}$