Documentation:CHBE Exam Wiki/Module 3 - Separations 1

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

Convert the temperatures in parts (a) and (b) and temperature intervals in parts (c) and (d):
a) T=95°F to °R, °C,K

Solution

$\begin{aligned} T & = [95 + 459.7]^{\circ} R \\ = 55 4^{\circ} R \\ = \frac{554}{1.8} K \\ = 308 K \\ = (308 - 273)^{\circ} C \\ = 3 5^{\circ} C \end{aligned}$

b) T=-50°C to K, °F, °R

Solution

$\begin{aligned} T & = [- 50 + 273]^{\circ} K \\ = 22 3^{\circ} K \\ = [223 \cdot 1.8]^{\circ} R \\ = 40 1^{\circ} R \\ = (401 - 460)^{\circ} C \\ = - 58 . 6^{\circ} F \end{aligned}$

c) ΔT=50°C to K, °F, °R

Solution

$\begin{aligned} T & = 5 0^{\circ} C \cdot \frac{1^{\circ} K}{1^{\circ} C} \\ = 5 0^{\circ} K \\ = 5 0^{\circ} C \cdot \frac{1 . 8^{\circ} F}{1^{\circ} C} \\ = 9 0^{\circ} F \\ = 5 0^{\circ} C \cdot \frac{1 . 8^{\circ} R}{1^{\circ} C} \\ = 9 0^{\circ} R \end{aligned}$

d) ΔT=120°R to °F, °C,K

Solution

$\begin{aligned} T & = 12 0^{\circ} R \cdot \frac{1^{\circ} F}{1^{\circ} R} \\ = 12 0^{\circ} F \\ = 12 0^{\circ} R \cdot \frac{1^{\circ} K}{1 . 8^{\circ} R} \\ = 66 . 7^{\circ} K \\ = 12 0^{\circ} R \cdot \frac{1^{\circ} C}{1 . 8^{\circ} R} \\ = 66 . 7^{\circ} C \end{aligned}$

Question 2

To regulate the temperature of a sand bath, a thermostat control with dial markings from 0 to 100 is used. A straight line that passes through the points (30, 300°F) and (90, 660°F) represents a calibration plot on the logarithmic coordinates of temperature (°F) versus dial setting,D.

Question 2a

Determine the equation for T in terms of D

Solution

$\begin{aligned} l n T & = l n K + m \cdot l n R \\ T & = K R^{m} \end{aligned}$

Determine the slope and intercept,
$\begin{aligned} s l o p e & = \frac{l n (660 / 300)}{l n (90 / 30)} \\ = 0.718 \\ l n K & = l n 660 - 0.718 \cdot l n (90) \\ K & = 26.12 \end{aligned}$

Hence,

$\begin{aligned} T & = 26.12 \cdot D^{0.718} \end{aligned}$

Question 2b

What is the estimated dial setting to obtain 500°F?

Solution

$\begin{aligned} R & = {(\frac{500}{26.12})}^{1 / 0.718} \\ = 61.1 \end{aligned}$

Question 3

The following data was obtained from an experimental run,

x	0.5	1.4	11	35	54	72	84
y	2.1	4.2	4.9	5.4	5.8	6	6.25

Question 3a

Plot the data on logarithmic axes

Solution

Question 3b

Determine the coefficients of power law expression y =ax^b

Solution

$\begin{aligned} b & = 0.1701 \\ ln a & = 1.1033 \\ a & = 3.014 \end{aligned}$

Question 4

Calculate the pressure and gas-phase composition (mole fraction) in a system containing a liquid that is 0.5mol% N₂ and 99.5 mol% water that is in equilibrium with nitrogen gas and water vapor at 90°C

Solution

Use Henry's law for N₂
At 90°C, H_N₂ is 12.6 × 10⁴ atm/mol
The partial pressure of N₂ is then
$\begin{aligned} p_{N_{2}} & = x_{N_{2}} \cdot H_{N_{2}} \\ = (0.005) \cdot (12.6 \cdot 1 0^{4}) \\ = 630 a t m \end{aligned}$

Use Raoult's law for water
At 90°C, p_H₂O^* is 525.76 mm Hg [0.692 atm] The partial pressure of H₂O is then
$\begin{aligned} p_{H_{2} O} & = x_{H_{2} O} \cdot p_{H_{2} O}^{*} \\ = (0.995) \cdot (0.692) \\ = 0.688 a t m \end{aligned}$

The total pressure of the system is
$\begin{aligned} P & = p_{H_{2} O} + p_{N_{2}} \\ = 0.688 + 630 \\ = 630.7 a t m \end{aligned}$

The vapor mole fractions:
$\begin{aligned} y_{H_{2} O} & = 1.09 \cdot 1 0^{- 3} \\ y_{N_{2}} & = 0.999 \end{aligned}$

Question 5

A mixture in the gaseous phase at 1 atm consisting of 40 mole% benzene and 60% toluene is cooled isobarically from an initial temperature of 115°C

Question 5a

At what temperature does the first drop of condensate form?What is its composition?

Solution

Hint: Find the intersection of the vapor curve with y_B is 0.4
$\begin{aligned} T & = 10 2^{\circ} C \\ x_{B} & = 0.2 \\ x_{T} & = 0.8 \end{aligned}$

Question 5b

Determine the mole fractions of benzene in the vapor and liquid phases when the temperature of the system is at 98°C.
What is the ratio (total moles in vapor/total moles in liquid) at this point.

Solution

Hint: Find the intersection of the vapor and liquid curve when T is 98°C
$\begin{aligned} x_{B} & = 0.3 \\ y_{B} & = 0.53 \end{aligned}$

Assume that the number of moles in the system is 1 mol.
$\begin{aligned} 1 & = n_{V} + n_{L} \\ 0.4 & = 0.53 \cdot n_{V} + 0.3 \cdot n_{L} \end{aligned}$

Solving the 2 systems of equations,
$\begin{aligned} n_{V} & = 0.43 \\ n_{L} & = 0.57 \end{aligned}$

Ratio of vapor to liquid: 0.75

Question 5c

At what temperature does the last bubble of vapor condense?What is its composition?

Solution

Hint: Find the intersection of the liquid curve with x_B is 0.4
$\begin{aligned} T & = 9 5^{\circ} C \\ y_{B} & = 0.63 \\ y_{T} & = 0.37 \end{aligned}$

Question 6

A tank containing 50L of air-carbon tetrachloride gas mixture is at a temperature of 50°C, at an absolute pressure of 1.5 atm and a relative saturation of 40%.
To remove the CCl₄ through gas adsorption, activated carbon is added to the tank. The volume of activated carbon can be assumed as negligible.

Question 6a

Calculate p_CCl₄ at the moment the tank is sealed, assuming ideal gas behavior and neglecting adsorption that occurs prior to sealing.

Solution

Let CCl₄ be c

$\begin{aligned} 0.4 & = \frac{p_{c}}{p_{c}^{*} (5 0^{\circ} C)} \\ p_{c} & = 0.4 \cdot (300 m m H g) \\ = 120 m m H g \end{aligned}$

Question 6b

Calculate the total pressure in the tank and the partial pressure of carbon tetrachloride at a point when quarter of the CCl₄ initially in the tank has been adsorbed.

Solution

Determine the initial moles of gas in the tank
$\begin{aligned} n_{0} & = \frac{P_{0} V_{0}}{R T_{0}} \\ = \frac{1.5 a t m \cdot 50 L}{0.08206 L a t m / m o l K \cdot 323 K} \\ = 2.83 m o l \end{aligned}$

Initial number of moles of CCl₄ in the tank
$\begin{aligned} n_{c 0} & = \frac{p_{c 0}}{P_{0}} \cdot n_{0} \\ = \frac{120 m m H g}{1.5 a t m / m m H g \cdot 760 m m H g} \cdot 2.83 m o l \\ = 0.298 m o l \end{aligned}$

Total number of moles of CCl₄ adsorbed
$\begin{aligned} n_{c_{a d s}} & = 0.25 \cdot n_{c 0} \\ = 0.0744 m o l \end{aligned}$

Total number of moles in the tank
$\begin{aligned} n_{t o t} & = n_{0} - n_{c_{a d s}} \\ = 2.83 - 0.0744 \\ = 2.756 m o l \end{aligned}$

Pressure in the tank after adsorption [Assume T and V are constant]
$\begin{aligned} P & = \frac{n_{t o t} R T_{0}}{V_{0}} \\ = \frac{2.756 m o l \cdot 0.08206 L a t m / m o l K \cdot 323 K}{50 L} \\ = 1.461 a t m \\ = 1110 m m H g \end{aligned}$

Partial pressure of CCl₄
$\begin{aligned} p_{c} & = y_{c} \cdot P \\ = \frac{n_{c}}{n_{t o t}} \cdot P \\ = \frac{0.75 \cdot n_{c 0}}{2.756 m o l} \cdot 1.461 a t m \\ = 0.118 a t m \\ = 90.0 m m H g \end{aligned}$

Question 6c

How much activated carbon must be added to the tank to reduce the mole fraction of CCl₄ in the gas to 0.001? Assume that the following relationship is true for the conditions in the tank
$\begin{aligned} X_{C C l_{4}}^{*} = \frac{0.0762 \cdot p_{c}}{1 + 0.096 p_{c}} \end{aligned}$

Solution

Determine the number of moles of air in the tank

$\begin{aligned} n_{a i r} & = n_{0} - n_{c 0} \\ = 2.83 - 0.105 \\ = 2.725 m o l \end{aligned}$

Determine the number of moles of CCl₄ in the tank

$\begin{aligned} y_{c} & = 0.001 \\ = \frac{n_{c}}{n_{c} + 2.725} \\ n_{c} & = 2.725 \cdot 1 0^{- 3} m o l \end{aligned}$

Determine the number of moles in the tank
$\begin{aligned} n_{t o t} & = n_{c} + n_{a i r} \\ = 2.725 \cdot 1 0^{- 3} + 2.725 \\ = 2.728 m o l \end{aligned}$

Determine p_c in the tank
$\begin{aligned} p_{c} & = y_{c} \cdot P \\ = 0.001 \cdot \frac{n_{t o t} \cdot R \cdot T_{0}}{V_{0}} \\ = 0.001 \cdot \frac{2.728 m o l \cdot 0.08206 L a t m / m o l K \cdot 323 K}{50 L} \cdot \frac{760 m m H g}{1 a t m} \\ = 1.099 m m H g \end{aligned}$

Determine the mass ratio of adsorbed CCl₄ to carbon in the tank
$\begin{aligned} X_{C C l_{4}}^{*} & = \frac{0.0762 \cdot p_{c}}{1 + 0.096 p_{c}} \\ = 0.0758 m m H g \end{aligned}$

Determine the mass of CCl₄ adsorbed
$\begin{aligned} m_{a d s} & = (n_{c 0} - n_{c}) \cdot M W_{c} \\ = (0.105 - 2.728 \cdot 1 0^{- 3} m o l \cdot \frac{153.85 g}{1 m o l} \\ = 15.7 g \end{aligned}$

Determine the carbon required
$\begin{aligned} m_{c} & = \frac{m_{c_{a d s}}}{X_{C C l_{4}}^{*}} \\ = 207 g \end{aligned}$

Question 7

A solution containing 100 lb KNO₃ /100 lb H₂O is cooled in a cooling crystallizer operating at 30°C. The slurry of from the crystallizer contains solid crystals suspended in saturated solution which then fed to a filter to separate the solid and liquid components.

Question 7a

Determine the production rate of crystals (lb /lb feed).

Solution

At 30°C, the solubility of the salt in water is 48g KNO₃ /100 lb H₂O
$\begin{aligned} w_{K N O_{3}} & = \frac{48 g K N O_{3}}{48 g K N O_{3} + 100 g H_{2} O} \\ = 0.324 \end{aligned}$

Mass balance [using 1lb as basis for the feed solution]:
Let m₁ be the mass of the saturated solution leaving the crystalizer in terms of the mass flow rate of the feed stream and
m₂ be the mass of crystals in terms of the mass flow rate of the feed stream

$\begin{aligned} Overall mass balance \\ 1 & = m_{1} + m_{2} \\ Salt mass balance \\ 0.5 & = 0.324 m_{1} + m_{2} \\ m_{1} & = 0.74 \\ m_{2} & = 0.26 \end{aligned}$

Question 7b

Determine the solid-to-liquid mass ratio of the slurry leaving the crystallizer

Solution

$\begin{aligned} Solid-to-liquid mass ratio & = \frac{0.26}{0.74} \\ = 0.351 \end{aligned}$