Steps in solving the problems

Question 1

Gas Phase Concentration, based on Dalton’s Law

Answer 1

Pa=yPT
Y=y/(1-y)

Question 2

Liquid Phase Concentration

Answer 2

x=C/pm (pm: molar density of the solution, moles of solution per unit volume of solution)

Question 3

Overall Material Balance

Answer 3

V1 + L2 = V2 + L1

Question 4

Solute Balance

Answer 4

L’(X1 - X2) = V’(Y1 - Y2)
L’/V’ = (Y1 - Y2) / (X1 - X2)

Question 5

At any height Z of the absorber (the operating line equation)

Answer 5

Y = (L’/V’)X - (L’/V’)X1 + Y1

Question 6

Percentage Recovery

Answer 6

% recovery = (1 - Y2/Y1)*100

Question 7

Henry’s Law - for dilute solution

Answer 7

y = Pa/Pt
Pa = Hx
y = (H/Pt)x

Question 8

Raoult’s Law - for ideal solution

Answer 8

Pa = x(Pa)^0
y = [(Pa)^0 / Pt]*x

Question 9

Vaporization Constant

Answer 9

y = Kx where: K = vaporization constant

Question 10

Formula for (L’/V’)minimum

Answer 10

(L’/V’)minimum = (Y1 - Y2) / (X1^* - X2)

Question 11

Absorption Factor, not when A is large, the solute tends to be absorbed in the liquid phase

Answer 11

A = L’ / mV’

Question 12

The equilibrium data for a absorption may be represented by the equation Y = 5.2X^2, where: X = kmol of dissolved absorbate material per kmol of pure absorbing liquid and Y kmol of gaseous absorbable material per kmol inert gas. The unit is operated as continuous counter-current absorption. The entering gas contains 5 moles of absorbable material per 20 mols of inert gas. The absorbing liquid enters the column as a pure material. Determine the minimum ratio of liquid to gas if the gases leaving the tower contain 1 mol absorbable material per 50 mol of inert gas.

Answer 12

x2=0, Y2=1/50, Y1=5/20
X2=0, since x=0
get X1 from equilibrium eqn, thus X1=0.21926
(L’/V’)minimum = (Y1-Y2)/(X1*-X2) = 1.05

Question 13

Dry gas containing 75% mol air and 25% mol ammonia vapor enters the bottom of a cylindrical packed absorption tower that is 2ft. in diameter. Nozzles in the top distribute water over packing. A solution of ammonia in water is drawn from the bottom of the column and scrubbed gas leaves the top. The gas enters at 80 degrees fahrenheit and 730 mmHg pressure. It leaves at 600 degrees fahrenheit and 730 mmHg. The leaving gas contains, on the dry-basis 1% mol ammonia. If the entering gas flows through the empty bottom of the column at an average velocity of 1.5 ft/s, how many cubic feet of entering gas are treated per hour? how many pounds of ammonia are absorbed per hour?

Answer 13

(note: if stated as dry-basis matic capital Y and if it’s not dry-basis nor solute free then small y)
velocity –> volumetric flow rate –> use PV=nRT –> molar flow rate
R = 1545 ((lbf/ft^2)(ft^3))/(lbmol-R)
Convert 730 mmHg ((4.7 lb/in^2) / 760 mmHg) (144in^2 / 1 ft^2)
molar flow rate = 41.34381 lbmol/hr = V1

Do Ammonia balance, also V1 = V2

Answer is x1L1 = 170.44 lb/hr

Question 14

A gas mixture containing 1% acetone and 99% dry air by volume is contacted with pure water in a plate gas absorber in which 90% of the acetone is recovered. The inlet gas is 30 kmol/hr and inlet water flow rate is 90 kmol/hr. The absorber operates at 27 degrees Celsius and 1 atm pressure. What is the concentration of the strong liquor? What is the minimum liquid flow rate in kmol/hr? What is the number of theoretical plates?

Answer 14

get the equilibrium data
use solute balance to determine V’ (V’=V1(1-y1)=V2(1-y2))
use percent recovery to get y2, then Y2
use solute balance to determine L’ (L’=L1(1-x1)=V2(1-x2))
use solute free balance to solve for X1 (V’(Y1-Y2)=L’(X1-X2))
answer x1=2.99x10^-3

use V’(Y1-Y2)=L’minimum(X1-X2) to get L’minimum
use equilibrium eqn to get X1

Use tiller tour eqn to solve for number of stages

Question 15

A benzene-air mixture is to be scrubbed in a simple counter-current absorption tower using a non-volatile absorption oil as solvent. The inlet gas contains 5% benzene and the entering gas flow is 600 lbmol/hr. Solubility of benzene in oil follows Raoult’s law. The tower operates isothermally at 80 degrees fahrenheit and the average molecular weight of the oil is 200 and the tower pressure is 1 atm.
Required:
— minimum oil (lb/hr) needed to recover 90% of the entering benzene
— theoretical plates stages required if the oil rate is 1.5 times the minimum

Answer 15

note:
- Hanapin ng wala sa diagram
– Get the equilibrium eqn

Solution:
–> use percent recovery to solve for Y2. %recovery = 1- Y2/Y1
–> use equilibrium eqn to get x1, then X1.
–> solve for V’. (V’=V1(1-y1))
–> substitute the values to get L’minimum. (L’minimum = V’ (Y1-Y2/X1-X2))
–> V1 = lb/hr, it should be converted to lbmol/hr
–> compute L’actual. (L’actual = 1.5L’minimum)

–> to solve for theoretical plates, y1 and x2 is given
–> y2 is determined using %recovery
–> L1 is not given so to get X1, then x1, use solute-free balance. (L’(X1-X2)=V’(Y1-Y2))

Question 16

Engr. Mayor is being tasked to reduce the concentration of ammonia in the mixture of air and ammonia, from 2.5% to 0.05% by using water as a solvent in a counter-current gas absorption tower. The tower operates at 233 degrees fahrenheit and 1 atm. The gas is entering the bottom of the tower at a rate of 7777 cubic feet per hour. What is the minimum liquid to gas ratio? What is the concentration of the strong liquor? How many theoretical stages are required at twice the minimum water rate?
Required:
— minimum liquid to gas ratio
— x1
— theoretical stages

Question 17

A mixture of 5% butane and 95% air is absorbed in a bubble plate tower containing 8 ideal plates. The absorbing liquid is a heavy non-volatile oil having a molecular weight of 250 and a specific gravity of 0.90. The absorption takes place at 1 atm and 60 degrees fahrenheit. The butane is to be recovered to the extent of 95%. The vapor pressure of butane at 60 degrees fahrenheit is 28 psi and liquid butane has a density of 4.84 lb/gal at 60 degrees fahrenheit. Calculate the gallons of fresh absorbing oil per gallon of butane recovered.
Required:
L2/(x1L1) = L2/(RV’Y1)

Answer 17

1. Equilibrium Equation
2. Solve for y2 using percent recovery
3. solve for y1* using tiller-tour eqn since number of stages is given
4. solve for x1 using equilibrium eqn
5. solve for L’/V’
6. compute the final answer which is L’/RV’Y1
7. the units should be converted into gal oil/gal C4H10

note: sp=density/standard density which is usually 1000

Question 18

The purge gases from a synthetic ammonia plant contain 4% ammonia by volume and 96% hydrogen, nitrogen, argon and other inerts. This gas is to be scrubbed at 5 atmosphere pressure with water in a bubble plate tower maintained at 68 degrees fahrenhei by cooling coils. With water and inert gas rates of 0.88 and 4 lb/min-ft^2 of tower cross-section respectively, how many theoretical plates will be acquired to produce a solution containing 13% ammonia by weight? The average molecular weight of the inert gas is 21.

Data:
the following gives the equilibrium partial pressure of ammonia solutions at 68 degrees fahrenheit

Partial pressure
(mmHg) 12 19.2 31.7 50 69.6 166

gram NH3/ 100 gram
H2O 2 3 5 7.5 10 20

Answer 18

note:
% by volume = % by mole in ideal gas

Solution:
–> equilibrium data. (X=mol NH3/mol H20) and (Y=Pa/P-Pa)
–> operating line. (Y=(L’/V’)X - (L’/V’)1 + Y1)
convert lb/min-ft^2 to lbmol/min, use MW tower area is not given, so assume 1 ft^2
convert 13% ammonia by weight to mol NH3/mol H20
–> operating line data. X values are unchanged and Y values were solved using (Y=(L’/V’)X - (L’/V’)1 + Y1)
–> Plot the data

Question 19

A benzene-air mixture is to be scrubbed in a simple counter-current absorption tower using a non-volatile absorption oil as solvent. The inlet gas contains 5% benzene and the entering gas flow is 600 lbmol/hr. Solubility of benzene in oil follows Raoult’s law. The tower operates isothermally at 80 degrees fahrenheit and the average molecular weight of the oil is 200 and the tower pressure is 1 atm.
Required:
— minimum oil (lb/hr) needed to recover 90% of the entering benzene
— theoretical plates stages required if the oil rate is 1.5 times the minimum

Answer 19

–> compute the equilibrium data. use the capital Y (Y = Pa/P-PaX)
–> the range of values for X to be used is from 0.1 to 1 and use equilibrium eqn to solve for values of Y

–> for operating data (Y=(L’/V’)X - (L’/V’)1 + Y1)
first calculate V’. (V’=V1(1-y1))
calculate Y2 using the percent recovery. (%recovery = 1- Y2/Y1)
solve for X1* using the equilibrium equation
consequently solve for minimum liquid to gas ratio. (L’/V’)minimum = Y1-Y2/X1-X2
solve the L’minimum
get the L’actual. (L’actual = 1.5L’minimum)
solve for X1. (L’/V’)=Y1-Y2/X1-X2
substitue the values, the X values are unchanged and Y values were calculated using (L’/V’)=Y1-Y2/X1-X2

–> plot

Question 20

X-burner and Nishinoya are budding chemical engineers that were tasked to perform a pilot scale test for their gas absorption equipments with MLBB company. Upon trial, they found out that gas stream contains 4.0 mol% NH3 and its ammonia content is reduced to 0.5 mol% in a packed absorption tower at 293K and 1.013x10^5 Pa. The inlet pure water flow is 68 kmol/h and the total inlet gas flow is 57.8 kmol/h. The tower diameter is 0.747 m. The film mass transfer coefficients are k’ya=0.0739 kmol/s-m^3-mol fraction and k’xa=0.169 kmol/s-m^3-mol fraction. Using the design methods for dilute gas mixtures, they were tasked to:
— Calculate the tower height using k’ya
— Calculate the tower height using K’ya

Answer 20

The solution is dilute so use Henry’s Law
1. Plot the operating lin equation
proceed to step 2. determine L’,V’.V2,L1,V,L,x1,X1,y1,Y1,x2,X2,y2,Y2

Question 21

X-burner and Nishinoya are budding chemical engineers that were tasked to perform a pilot scale test for their gas absorption equipments with MLBB company. Upon trial, they found out that gas stream contains 4.0 mol% NH3 and its ammonia content is reduced to 0.5 mol% in a packed absorption tower at 293K and 1.013x10^5 Pa. The inlet pure water flow is 68 kmol/h and the total inlet gas flow is 57.8 kmol/h. The tower diameter is 0.747 m. The film mass transfer coefficients are k’ya=0.0739 kmol/s-m^3-mol fraction and k’xa=0.169 kmol/s-m^3-mol fraction. Using the design methods for dilute gas mixtures, they were tasked to:
— Calculate the tower height using k’ya
— Calculate the tower height using K’ya

Answer 21

The solution is dilute so use Henry’s Law
1. Plot the operating lin equation
proceed to step 2. determine L’,V’.V2,L1,V,L,x1,X1,y1,Y1,x2,X2,y2,Y2