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

showSolution

${\begin{aligned}T&=[95+459.7]^{\circ }R\\&=554^{\circ }R\\&={\frac {554}{1.8}}K\\&=308K\\&=(308-273)^{\circ }C\\&=35^{\circ }C\\\end{aligned}}$

b) T=-50°C to K, °F, °R

showSolution

${\begin{aligned}T&=[-50+273]^{\circ }K\\&=223^{\circ }K\\&=[223\cdot 1.8]^{\circ }R\\&=401^{\circ }R\\&=(401-460)^{\circ }C\\&=-58.6^{\circ }F\\\end{aligned}}$

c) ΔT=50°C to K, °F, °R

showSolution

${\begin{aligned}T&=50^{\circ }C\cdot {\frac {1^{\circ }K}{1^{\circ }C}}\\&=50^{\circ }K\\&=50^{\circ }C\cdot {\frac {1.8^{\circ }F}{1^{\circ }C}}\\&=90^{\circ }F\\&=50^{\circ }C\cdot {\frac {1.8^{\circ }R}{1^{\circ }C}}\\&=90^{\circ }R\\\end{aligned}}$

d) ΔT=120°R to °F, °C,K

showSolution

${\begin{aligned}T&=120^{\circ }R\cdot {\frac {1^{\circ }F}{1^{\circ }R}}\\&=120^{\circ }F\\&=120^{\circ }R\cdot {\frac {1^{\circ }K}{1.8^{\circ }R}}\\&=66.7^{\circ }K\\&=120^{\circ }R\cdot {\frac {1^{\circ }C}{1.8^{\circ }R}}\\&=66.7^{\circ }C\\\end{aligned}}$

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

showSolution

${\begin{aligned}lnT&=lnK+m\cdot lnR\\T&=KR^{m}\\\end{aligned}}$

Determine the slope and intercept,
${\begin{aligned}slope&={\frac {ln(660/300)}{ln(90/30)}}\\&=0.718\\lnK&=ln660-0.718\cdot ln(90)\\K&=26.12\end{aligned}}$

Hence,

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

Question 2b

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

showSolution

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

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

showSolution

Question 3b

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

showSolution

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

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

showSolution

Use Henry's law for N₂
At 90°C, H_N2 is 12.6 × 10⁴ atm/mol
The partial pressure of N₂ is then
${\begin{aligned}p_{N_{2}}&=x_{N_{2}}\cdot H_{N_{2}}\\&=(0.005)\cdot (12.6\cdot 10^{4})\\&=630atm\end{aligned}}$

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
${\begin{aligned}p_{H_{2}O}&=x_{H_{2}O}\cdot p_{H_{2}O}^{*}\\&=(0.995)\cdot (0.692)\\&=0.688atm\end{aligned}}$

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

The vapor mole fractions:
${\begin{aligned}y_{H_{2}O}&=1.09\cdot 10^{-3}\\y_{N_{2}}&=0.999\end{aligned}}$

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?

showSolution

Hint: Find the intersection of the vapor curve with y_B is 0.4
${\begin{aligned}T&=102^{\circ }C\\x_{B}&=0.2\\x_{T}&=0.8\\\end{aligned}}$

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.

showSolution

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?

showSolution

Hint: Find the intersection of the liquid curve with x_B is 0.4
${\begin{aligned}T&=95^{\circ }C\\y_{B}&=0.63\\y_{T}&=0.37\\\end{aligned}}$

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.

showSolution

Let CCl₄ be c

${\begin{aligned}0.4&={\frac {p_{c}}{p_{c}^{*}(50^{\circ }C)}}\\p_{c}&=0.4\cdot (300mmHg)\\&=120mmHg\end{aligned}}$

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.

showSolution

Determine the initial moles of gas in the tank
${\begin{aligned}n_{0}&={\frac {P_{0}V_{0}}{RT_{0}}}\\&={\frac {1.5atm\cdot 50L}{0.08206Latm/molK\cdot 323K}}\\&=2.83mol\end{aligned}}$

Initial number of moles of CCl₄ in the tank
${\begin{aligned}n_{c0}&={\frac {p_{c0}}{P_{0}}}\cdot n_{0}\\&={\frac {120mmHg}{1.5atm/mmHg\cdot 760mmHg}}\cdot 2.83mol\\&=0.298mol\end{aligned}}$

Total number of moles of CCl₄ adsorbed
${\begin{aligned}n_{c_{ads}}&=0.25\cdot n_{c0}\\&=0.0744mol\end{aligned}}$

Total number of moles in the tank
${\begin{aligned}n_{tot}&=n_{0}-n_{c_{ads}}\\&=2.83-0.0744\\&=2.756mol\end{aligned}}$

Pressure in the tank after adsorption [Assume T and V are constant]
${\begin{aligned}P&={\frac {n_{tot}RT_{0}}{V_{0}}}\\&={\frac {2.756mol\cdot 0.08206Latm/molK\cdot 323K}{50L}}\\&=1.461atm\\&=1110mmHg\\\end{aligned}}$

Partial pressure of CCl₄
${\begin{aligned}p_{c}&=y_{c}\cdot P\\&={\frac {n_{c}}{n_{tot}}}\cdot P\\&={\frac {0.75\cdot n_{c0}}{2.756mol}}\cdot 1.461atm\\&=0.118atm\\&=90.0mmHg\end{aligned}}$

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
${\begin{aligned}X_{CCl_{4}}^{*}={\frac {0.0762\cdot p_{c}}{1+0.096p_{c}}}\end{aligned}}$

showSolution

Determine the number of moles of air in the tank

${\begin{aligned}n_{air}&=n_{0}-n_{c0}\\&=2.83-0.105\\&=2.725mol\end{aligned}}$

Determine the number of moles of CCl₄ in the tank

${\begin{aligned}y_{c}&=0.001\\&={\frac {n_{c}}{n_{c}+2.725}}\\n_{c}&=2.725\cdot 10^{-3}mol\\\end{aligned}}$

Determine the number of moles in the tank
${\begin{aligned}n_{tot}&=n_{c}+n_{air}\\&=2.725\cdot 10^{-3}+2.725\\&=2.728mol\\\end{aligned}}$

Determine p_c in the tank
${\begin{aligned}p_{c}&=y_{c}\cdot P\\&=0.001\cdot {\frac {n_{tot}\cdot R\cdot T_{0}}{V_{0}}}\\&=0.001\cdot {\frac {2.728mol\cdot 0.08206Latm/molK\cdot 323K}{50L}}\cdot {\frac {760mmHg}{1atm}}\\&=1.099mmHg\end{aligned}}$

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

Determine the mass of CCl₄ adsorbed
${\begin{aligned}m_{ads}&=(n_{c0}-n_{c})\cdot MW_{c}\\&=(0.105-2.728\cdot 10^{-3}mol\cdot {\frac {153.85g}{1mol}}\\&=15.7g\end{aligned}}$

Determine the carbon required
${\begin{aligned}m_{c}&={\frac {m_{c_{ads}}}{X_{CCl_{4}}^{*}}}\\&=207g\\\end{aligned}}$

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).

showSolution

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

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

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

Question 7b

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

showSolution

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