# Separation Process Compilation of Problem Set

September 5, 2017 | Author: Kaye Cabrera Fabros | Category: Solution, Solubility, Concentration, Chemical Substances, Chemistry

#### Description

1

By extracting kerosene, 2 tons of waxed paper is to be dewaxed in a continuous countercurrent extraction system that contains a number of ideal stages. The waxed paper contains, by weight, 25% paraffin wax and 75% paper pulp. The extracted pulp is put through a dryer to evaporate the kerosene. The pulp, which retains the unextracted wax after evaporation, must not contain over 0.2 kg of wax per 100 kg of wax free pulp. The kerosene used for the extraction contains 0.05 kg of wax paper per 100 kg of wax free kerosene. Experiments show that the pulp retains 2.0 kg of kerosene per kg of kerosene and wax free pulp as it is transferred from cell to cell. The extract from the battery is to contain 5 kg of wax per 100 kg of wax free kerosene. a Find the overflow stream. b Find the underflow stream c Kg wax / kg kerosene in the underflow d No. of stages

Given: Vbyb

V1:Overflow

S: Solvent

5 kg wax Vaya 100 kg kerosene

V2y2

Lbxb

Feed : 4 Tons Wax paper 25% wax 75% pulp

5 kg wax L1xkerosene 1 100 kg

Required: Overflow and Underflow streamsN stages. Solutions: OMB :

F+ S=V 1 + L1

Material balance for the wax 4 Ton x

1000 kg =4000 kg 1Ton

4000 kg x .25=1000 kg

Wax balance

0.05 kg wax 100 kg kerosene

4000 kg x .75=3000 kg

2 kg kerosene 1 kg pulp

L1: Underflow

0.2 kg wax 100 kg free pulp

First Sta

1000 kg+

0.05 kg wax 5 kg wax 0.2 kg wax V b= V a+ ( 0.75 x 4000 kg ) 100 kg kerosene 100 kg kerosene 100 kg pulp

Solvent Balance V b=V a+

2 kg kerosene ( 0.75 x 4000 kg ) =V a +6000 kg 1 kg pulp

Solving simultaneously we get the streams V a=20141 . 4141 kg Kerosene

V b=V 2 =20081. 4141 kg kerosene

undeflow=6000 kg kerosene 0.2 kg wax x 0.75 x 4000 kg pulp 100 kg pulp kg wax x b= =0 .001 6000 kg kerosene kg kerosene

Wax balance: 20081.4141 y 2 +1000=

5 ( 20141.4141) +6000 x 1 100

equilibrium condition x1 = y a= y 2=0.05

kg wax kg kerosene

No. of ideal stages 0.001−0.0005 ( 0.05−0.01529 ) N= +1=4 .5398 ≈5 0.01529−0.0005 ln ( 0.05−0.001 ) ln

y 2=0.01529

2

In a single step solid-liquid extraction soybean oil has to be extracted from soybean flakes using hexane as solvent. 100 kg of the flakes with an oil content of 20 wt% are contacted with 100 kg fresh hexane. 1.5 kg of inert material hold back a constant value of 1 kg solution.

extract (overflow)

solvent

V1

V2 extraction step

L0

L1 feed

Total balance:

L0 + V2 = M = L1 + V1 = 100 + 100 = 200 kg

underflow

Balance for compound A: L0 wA,L0 + V2 wA,V2 = M wA,M with the feed concentration wA,L0 = 0.8 and the suggestion, that no solid particles are included in the overflow, so wA,V2 = 0 follows: 100 * 0.8 + 100 * 0 = 200 * wA,M wA,M = 0.4 Balance for compound B: L0 wB,L0 + V2 wB,V2 = M wB,M with the feed concentration wB,L0 = 0.2 and with the knowledge, that pure hexane is used as solvent, wB,V2 = 0, follows 100 * 0.2 + 100 * 0 = 200 * wB,M wB,M = 0.1 The concentration of compound C (solvent) in the mixing point M can be determined either by a mass balance for compound C L0 wC,L0 + V2 wC,V2 = M wC,M with wC,L0 = 0, because no solvent is included in the feed, and with wC,V2 = 1, pure hexane, follows 100 * 0 + 100 * 1 = 200 * wC,M wC,M = 0.5 or by the rule, that the sum of the mass percent of each compound in the point M has to be 1. wA,M + wB,M + wC.M = 1 0.4 + 0.1 + wC.M = 1 wC.M = 0.5

With these concentrations the mixing point M can be drawn in the diagram, which has to be on the connection line of feed point F and solvent C.

It is given, that 1 kg inert material retains 1.5 kg solution (extractable substance + solvent = miscella = overflow). Therefore the concentration of the underflow is

inert m

w

A,Underflow =

inert material+extractable substa 1.5 w

A,Underflow =w A,L1

= 1.5 + 1

The amount of the leaving flows L1 and V1 can be calculated from the mass balance for compound A M wA,M = V1 wA,V1 + L1 wA,L1 with wA,V1 = 0 (no solid material in the overflow) and wA,L1 = 0.6 (underflow)

L1=M

W A ,M 0.4 =200 W A , L1 .0 .6 L1=133.333 kg

With the total balance M = L1+V1

follows V1 = M - L1 = 200 - 133.333 V1 = 66.666 kg The concentrations of B and C in the overflow V1 are calculated with the suggestion that no inert material A is included in the overflow. W B ,V 1=

B 20 = A+ B+C 0+ 20+100

W C , V 1=

C 100 = A +B+ C 0+20+100

B ,V 1=0.1667 ¿ W¿ W C , V 1=0.8333 The composition of the underflow can be calculated by mass balances for compound B and C. L1 wB,L1 + V1, wB,V1 = L0 wB,L0 + V2 wB,V2 W B , L1=

LO × W B , Lo−V 1 ×W B ,V 1 100 ×0.2−66.666 ×0.1667 = L1 133.333

W B , L1=0.067 W A ,L 1+ W B , L 1+W C , L1=1 W C , L1 =1−0.6−0.67 W C , L1 =0.333

Total mass (kg)

Wt% A

Wt% B

Wt%C

Feed LO

100

80

20

0

Solvent V2

100

0

0

100

Overflow V1

66.666

0

16.667

83.333

Underflow L1

133.333

60

6.7

33.3

Situation for problems no. 23-28 By extraction with kerosene with 0.05 lb wax per 100 lb kerosene, 2 tons of waxed paper per day is to be dewaxed in a continuous countercurrent extraction system that contains a number of ideal stages. The waxed paper contains, by weight, 25 percent paraffin wax and 75 percent paper pulp. The extracted pulp is put through a dryer to evaporate the kerosene. The pulp, which retains the unextracted wax after evaporation, must not contain over 0.2 lbs of wax per 100 lbs of waxfree kerosene-free pulp. Experiment show that the pulp retains 2.0 lb of kerosene per lb of kerosene and wax-free pulp as it is transferred from cell to cell. The extract from the battery is to contain 5 lb of wax paper per 100 lb of wax-free kerosene. Per 100 lb of wax-free kerosene-free pulp,

23. The kerosene in the exhausted pulp is equal to a. 150 lb c. 117 lb b. 200 lb d. 212 lb 24. The kerosene in the strong solution is equal to a. 561 lb c. 761 b. 651 lb d. 671 lb 25. The wax in the strong solution is equal to a. 35.35 lb c. 55.33 lb b. 33.55 lb d. 53.53 lb 26. The wax in the underflow to unit 2 is equal to a. 8 lb c. 12 lb b. 10 lb d. 14 lb 27. The wax in the overflow from the second cell to the first is a. 10.22 lb c. 12.11 lb b. 11.12 lb d. 13.19 lb 28. The total number of ideal stages is equal to a. 3 c. 5 b. 4 d. 6 Given: Solute = Wax Solvent = Kerosene Inert = Pulp

Y1

Y2

1

YN+1

2

N XN

X1 F= 2 Tons 25% Solute 75% Inert

Solution: In Feed: Inert = 100 lb 100 lbinert =133.3333lb Feed = 0.75 Solute = (0.25)(133.3333 lb) = 33.3333 lb In Final Underflow: Inert = 100 lb 0.2lb solute x 100 lb inert =0.2 lb Solute = 100 lb inert Solvent =

2 lb solvent x 100 lb inert=200lb lb inert

Overall Solvent Balance 0 + VN+1 = V1 + 200

Equation 1

Overall Solute Balance 0.05 lb solute 33.3333 + ( 100 lb solvent

5 lb solute x VN+1) = ( 100 lb solvet

x V1) + 0.2

Equation 2

From Equation 1: VN+1 – V1 = 200 From Equation 2: 5 x 10−4

VN+1 – 0.05V1 = -33.1333

Equate Equation 1 and Equation 2, solve for VN+1 and V1: VN+1 = 871.3798 lb V1 = 671.3798 lb Solute in V1: 5 lb solute 100 lb solvent

x 671.3798 = 33.5690 lb

Solvent Balance in Stage 1: 0 + V2 = 671.3798 +200 V2 = 871.3798 lb Solute Balance in Stage 2: 33.3333

+

solute

in

V2

=

5 lb solute x 671.3798lb solvent ( 100lb solvent ) Solute in V2 = 10.2357 lb Solute in Y2 =

5 lb solute x 200 lb solvent =10 lb 00lb solvent

(

5 lb solute x 200 lb solvent ) 100 lb solvent

+

Solving for Number of Stages: Y N+1 −X N ] Y 2−X 1 N=1+ Y −Y 2 ln ⁡[ N +1 ] XN− X1 ln ⁡[

where: 0.05 lb solute −4 =5 x 10 100 lb solvent

YN+1 =

0.2 lb solute −3 =1 x 10 200 lb solvent

XN =

10.2357 lb solute =0.0117 Y2= 871.3798lb solvent X1 =

10 lb solute =0.05 200 lb solvent

5 x 10−4 −1 x 10−3 ] 0.0117−0.05 5 x 10−4 −0.0117 ln ⁡[ ] 1 x 10−3−0.05

ln ⁡[ N=1+

N = 3.9396 = 4 stages

Problems no. 29-31. 100 kg of solid containing 50% of a soluble material were treated with 200 kg of a solvent containing the same solute at 3% concentration in a vessel under the constant agitation. After a long time, pressing separated in the solution and the solid. The solid analyzed 0.75 kg of solvent per kg of inert solid. 29. The amount of solute in the final underflow is approximately equal to a. 10.82 kg

c. 2.78 kg

b. 8.54 kg

d. 7.16 kg

30. The amount of solvent in the extract is approximately equal to a. 106.2 kg

c. 178.3 kg

b. 216.0 kg

d. 156.5 kg

31. How much extract was collected? a. 201.68 kg

c. 216.08 kg

b. 106.21 kg

d. 192.86 kg

Given: Overflow, V1

Vo = 200 kg

(Extract)

3% solute 97% solvent

Feed, F = 100 kg

Underflow, L1

50% solute

solid =

0.75 kg solvent kg inert solid

50% solid

Required: 29.) amount of solute in final underflow 30.) amount of solvent in the extract 31.) V1

Solution: In the Feed, F:

solute=( 0.5 )( 100 kg )=50 kg

inert solid =( 0.5 ) (100 kg )=50 kg

In Vo:

solute=( 0.03 )( 200 kg )=6 kg solvent =( 0.97 )( 200 kg )=194 kg

In overflow, V1.:

solute=a solvent =Vi−a

Solute balance:

Continuation... Inerts balance: inerts∈F=inerts∈L1=50 kg

Solution balance: solution ∈F +Vo=solution ∈ L1+V 1

( 50+0 ) +200=93.5−a+V 1 156.5+a=Vi

At Equilibrium: solute solute ( solution ) =( solution ) V1

L1

a 56−a = ( 156.5+ a ) ( 37.5+56−a ) V1

L1

a=45.1753 kg

29.) Amount of solute in underflow, L1: solute=56−a=56−45.1753=10.8247 kg 30.) Amount of solvent in Extract, Vi: solvent =V 1−a=156.5+ a−a=1 56.5 kg 31.) V1: V 1=solvent V + soluteV =156.5+45.1753=201.6753 kg 1

1

Situation for problems no. 32-34 A solid B, contains a soluble component, A, of mass fractions

x A=0.25

,

x B=0.75

and is

to be recover A by a solvent extraction with C. Solid B and solvent C are mutually totally insoluble. The extracted solid is to be screw passed to a 0.75 kg of solution/kg of B underflow. The entrainment of B in the overflow can be neglected. Per kg of feed and to obtain 85% of A in the extract overflow. 32. The composition of the solution in the underflow is a. 0.04 c. 0.01 b. 0.07 d. 0.10 33. The amount of solvent in the underflow is a. 0.44 b. 0.53

c. 0.88 d. 1.33

34. How much solvent C (A free) must be fed? a. 3.5000 kg b. 2.5712 kg

c. 1.7000 kg d. 5.2311 kg

GIVEN:

REQUIRED: 32. x1 in L1 33. solvent in L1 34. solvent C

SOLUTION: In Feed, F: F = 1 kg Solute (A) : (0.25)(1) = 0.25 kg Inerts (B) : (0.75)(1) = 0.75 kg

In Underflow , L1 : Inerts (B) : 0.75 kg Solution (A + C) : (0.75)(0.75) = 0.5625 kg Solute : (1 - 0.85)(0.25) = 0.0375 kg Solvent : (0.5625 - 0.0375) = 0.5250 kg x 1=

solute 0.0375 = =0.0667 solution 0.5625

In Overflow , V1 : Solute (A) : (0.85)(0.25) = 0.2125 kg V1 = ? ? ?

Solution Balance: (0.25 + 0) + C = V1 + 0.5625 V1 = C - 0.3125

@ equilibrium: solute solute ( solution ) =( solution ) V1

L1

0.2125 ( C−0.3125 ) =( 0.0375 0.5625 ) V1

C = 3.5000 kg

L1

Situation for problems no. 35-38 Seeds containing 30% weight oil are extracted in a countercurrent plant and 88% of the oil is recovered in a solution containing 55% by weight of oil. The seeds are extracted with fresh solvent and 1 kg of solution is removed in the underflow in association with every 1.5 kg of insoluble material. 35. The amount of solvent in final extract is approximately equal to a. 26.4 kg b. 21.6 kg

c. 46.67 kg d. 43.07 kg

36. The amount of solvent in final underflow is approximately equal to a. 26.4 kg b. 21.6 kg

c. 46.67 kg d. 43.07 kg

37. The concentration of oil in the solvent stream for stage 1 is approximately equal to a. 0.55 b. 0.08

c. 0.18 d. 0.34

38. How many ideal stages are needed to attain the desired separation? a. 4 b. 6

c. 8 d. 10

GIVEN:

REQUIRED: 35. Solvent in V1 36. Solvent in LN 37.Concentration of oil V2 38.N SOLUTION: Basis: 100 kg of Feed In Feed, F: Insoluble = 0.70(100 kg) = 70 kg Oil = 0.30(100 kg) = 30 kg In final Overflow, V1: Oil = 0.88 (30 kg) = 26.4 kg 45 Solvent = 26.4 kg 55 = 21.6 kg

( )

y1=x1=0.55

In final Underflow, LN=L1=46.6667 kg Oil= 0.12(30 kg) = 3.6 kg Insoluble = 70 kg

Solution = 70 kg

( 1.51 kkgg solution insoluble )

= 46.6667 kg

Solvent = 46.6667 kg– 3.6 kg= 43.0667 kg 3.6 =0.0771 xN= 46.6667 In Fresh Solvent, VN+1: Solvent = 21.6 kg + 43. 0667 kg = 64.6667 kg Solute = 0 yN+1 = 0 Solute Balance around Stage 1: 30 kg + 64.6667 kg (y2) = 26.4 kg + 46.6667 kg(0.55) y2= 0.3412 = x2 0−0.0771 ( 0.3412−0.55 ) 1+ 0−0.3412 ln ( 0.0771−0.55 ) ln

N=

= 4.05

Situation for problems 39-42 Calcium-carbonate precipitate can be produced by the reaction of an aqueous solution of sodium carbonate and calcium oxide. The by-product is aqueous sodium hydroxide. Following decantation, the slurry leaving the precipitation tank is 5 wt% calcium carbonate, 0.1 wt% sodium hydroxide, and the balance water. One hundred thousand lb/h of this slurry is fed to a two-stage, continuous, countercurrent washing system to be washed with 20,000 lb/h of fresh water. The underflow from each thickener will contain 20 wt% solids. 39. The amount of extract 40. The amount of sodium hydroxide in final extract 41. The amount of sodium hydroxide in final underflow 42. The percent recovery of sodium hydroxide in the extract Given: 20,000 lb/h (V1)

1

(V2)

(V3)

2 (F) 100,000 lb/h

(L1)

5 wt% Calcium Carbonate 0.1 wt% Sodium Hydroxide

Solution: In Feed: Calcium Carbonate=0.05(100,000)=5000 lb/h Sodium Hydroxide=0.001(100,000)=100lb/h

Water=0.949(100,000)=94900 lb/h

(L2 ) 20wt% solid

Solid Balance: Solid ∈F=Solid ∈L 2 5000

lb =0.20 ( L2 ) h

L2=25000

lb h

OMB: F+V 3=V 1+ L 2

100,000+20,000=V 1+25,000

V 1=95000

lb h

Stage 1 (@ equilibrium) (

Solute Solute ) =( ) Solution (V 1) Solution ( L1)

(

Solute Solute ) =( ) 95,000 (V 1 ) 20,000 (L 1)

Solute ∈V 1=4.75 Solute∈L 1−−−eqn(1) Stage 2 (@ equilibrium) (

Solute Solute ) =( ) Solution (V 2) Solution ( L2)

(

Solute Solute ) =( ) 20,000 (V 2) 20,000 (L2 )

Solute ∈V 2=Solute∈L 2−−−eqn(2) Solute Balance in Stage 2 Solute L 1+ SoluteV 3=Solute V 2 +Solute L 2 Solute L 1+0=SoluteV 2 +Solute L2 Solute L 1=Solute L 2+ Solute L2 Solute L 1=2 Solute L2−−−eqn( 3) Overall Solute Balance Solute F + Solute V 3=SoluteV 1 + Solute L2 100+0=Solute V 1+ Solute L2 −−−eqn (4) Substitute eqn (1) to eqn (4) 100=4.75 Solute L1 + Solute L2−−−eqn(5) Substitute eqn (3) to eqn (5) 100=4.75 ( 2 ) Solute L2 +Solute L 2 Solute L 2=9.52

lb h

Using eqn 4: 100+0=Solute V 1+ 9.52 Solute V 1=90.48

lb h

Percent Recovery ( R )=

SoluteV 1−SoluteV 3 Solute F

R=

( 90.48−0 ) x 100 100

R=90.48

PROBLEM 43-46 Ground roasted coffee contains 8% soluble solids, 2% water, and 90% inert insoluble solids. In order to obtain an extract with high soluble solids content without having to concentrate it for spray drying, a countercurrent extraction process is to be used to prepare the extract. It is desired that the final extract contain 0.15kg soluble/kg water and that the soluble of the spent coffee grounds not to exceed 0.008 kg/kg dry inert solids. The coffee grounds carry 1 kg water/kg of soluble-free inert solids and this quantity is constant with the solute concentration in the extract.

REQD: 43) The amount of final extract is approximately equal to a. 55.81 kg

b. 48.54 kg

c. 72.8 kg

d. 28.1 kg

44) The concentration of the solution adhering to the extracted solids is approximately equal to a. 0.0936

b. 0.0079

c. 0.1304

d. 0.0032

45) The water/coffee ratio to be used in the extraction is a. 1.37

b. 2.88

c. 0.98

d. 1.87

46) The number of extraction stages needed for this process is a. 5

b. 6

c. 7

d. 8

SOLUTION:

Overflow, V1 0.15 kg solute kg H2O

V2

V3

V4

Y2

Y3

Y4

Yn+1 1

2 R1

Feed, F 8% Solute

In

2% H2O

Solute:

L1

Solvent: 2kg Inerts: 0.9(100)=90kg

In final underflow: Inerts= inerts in F=90kg Solute=0.008(90)=0.72kg Solvent=90(1)=90 kg XN= 0.008

Solvent, Vn+1

Solution:

3

R2

the

L2

0.08(100)= R= 1 kg H2O/kg Inerts

N R3

feed:

L3

Final Underflow, Ln

basis(100

Xn

kg)

Solute/Inerts = 0.008

8kg 0.02(100)=

Solute ( Solution )

in LN

0.72 = 90+0.72 =0.0079 (#44)

In final overflow, V1: Solute balance: 8+0=0.72+Solute in V1 Solute =0.15 Y1= X1= Solvent

Final Overflow(extract)=solute+solvent = 7.28+48.5333kg=V1 = 55. 8133 kg(#43)

In solvent stream, Vn+1 Yn+1=0(pure water) Solvent= Vn+1=? Overall solvent bal: 2+Vn+1=90+48.5333 Vn+1=136.5333

Ratio: Vn+1 136.5333 = =1.3653 F 100

(#45)

Solute for Y2 using Solute balance around stage 1 8+ V2Y2=L1X1+7.28 V2= Vn+1=136.5333 Y2=? L1=LN=90

X1=Y1=0.15 8+136.5333=90(0.15)+7.28 Y2=0.0936

Solve for N:

0.008 ( 0.0926−0.15 ) 0−0.0936 ln ( 0.008−0.15 )

ln N=1+

=5.69=6 stages(#46)

Problem 47 Given: V1

V2 YN+1

Y1

Feed= 50 tons/hr

1

Y2

2

L1

LN

X1`

XN

48% H2O 40% Pulp

N

R=

3 tons H 20 tons Pulp

12%Sugar Required: N = ? Solutions: tons In Feed: H2O = 0.48(50) = 24 hr

Pulp = 0.40(50) = 20

tons hr

Sugar = 0.12(50) = 6

tons hr

In Final Overflow: Sugar = 0.97(6) = 5.82

Solution =

5.82 0.15

tons hr

= 38.8

tons hr

H2O = V1 = 38.8 – 5.82 = 32.98

Y1 =

5.82 32.98

tons hr

= 0.1765

X1 = Y1 = 0.1765

at equilibrium

In Final Underflow: Sugar = 0.03(6) = 0.18

H2O = LN = 20(3) = 60

tons hr tons hr

XN =

0.18 60

= 0.003

In Fresh Solvent: OMB (Solvent): LN + V1 – Lo VN+1 = 60 + 32.92 – 24 H20 = VN+1 = 68.98

tons hr

YN+1 = 0 (pure solvent) Sugar Balance Around Stage 1: Sugar in F + V2Y2 = L1X1 + V1Y1 tons V2 = V3 = V4 = ….. = VN+1 = 68.98 hr tons L1 = L2 = L3 = L4 = ….. = LN = 60 hr 6 + 68.98Y2 = 60(0.1765) + 5.82 Y2 = 0.1509 Solving for N: 0−0.003 0.1509−0.1765 0−0.1509 ln 0.003−0.1765

ln ⁡ N=

N = 16.36 = 17

+1

Problem 48 Constant Solution Retention: L8V – solution flowrates : X8Y – solute/solution In Feed: H2O = 0.48(50) = 24 tons/hr Pulp = 0.40(50) = 20 tons/hr Sugar = 0.12(50) = 6 tons/hr In Final Overflow: Sugar = 0.97(6) = 5.82 tons/hr Y1 = 0.15 Solution = V1 =

5.82 0.15

X1 = Y1 = 0.15

= 38.8 tons/hr

(@ equilibrium)

In Final Underflow: R=

3 tons solution ton dry pulp

Sugar = 0.03(6) = 0.18 tons/hr Solution = LN = 3(20) = 60 tons/hr XR =

0,18 60

= 0.003

In Fresh Solvent H2O = VN+1 = LN + V1 – L0

(overall solution balance)

H2O = VN+1 = 60 + 38.8 – (24 + 6) VN+1 = 68.8 tons/hr

YN+1 = 0

(pure solvent)

Sugar Balance Around Stage 1: Sugar in F + V2Y2 = L1X1 + V1Y1 V2 = V3 = V4 = . . . = VN+1 = 68.8 tons/hr L1 + L2 + L3 = . . . = LN = 60 tons/hr 6 + 68.8Y2 = 60(0.15) + 38.8(0.15) Y2 = 0.1282 Solving for N: 0−0.003 ] 0.1282−0.15 0−0.1282 ln[ ] 0.003−0.15

ln [ N=

N = 15.49 = 16

Situation for problems no. 49-52 A seashore sand contains 85% insoluble sand, 12% salt and 3% water. 1000 lb/hr of this mixture is to be extracted in a countercurrent washing system with 2000 lb/hr of pure water so that after drying it will contain only 0.2% salt. The sand retains 0.5 lb of water per pound of insoluble sand. 49. The mass of salt in the final underflow is equal to a. 1.7 lb/hr c. 2.3 lb/hr b. 1.2 lb/hr d. 2.5 lb/hr 50. The concentration of salt in the final overflow is equal to a. 0.03 c. 0.07 b. 0.05 d. 0.09 51. The concentration of salt in the solvent stream for stage 1 is approximately equal to a. 0.023 c. 0.07 b. 0.015 d. 0.19 52. The number of washing is approximately equal to a. 3 c. 5 b. 4 d. 6 Given:

2000 lb/hr Y1

Y2

1

YN+1

2 L1

F= 1000 lb/hr 12% Solute 85% Inert 3% Solvent

Solution: In Feed : Inert = 1000 lb/hr (0.85) = 850 lb/hr Solute = 1000 lb/hr (0.12) = 120 lb/hr Solvent = 1000 lb/hr (0.03) = 30 lb/hr In Final Underflow : Inert = 850 lb/hr 0.5lb solvent ×850 lbinert =425lb solvent Solvent = lbinert

Solute : solute ∈LN =

0.2 ( inert + solute∈ LN ) 100

solute ∈ LN =

0.2 0.2 inert + solute ∈ LN 100 100

solute ∈LN =

0.2 lb 0.2 (850 )+ solute∈ LN 100 hr 100

solute ∈LN =1.7034 lb/hr

N LN

after drying = 0.2% sa

Final underflow = inert +solvent + solute ¿ ( 850+425+1.7034 )

lb hr

FinalUnderflow =1276.7034 lb/hr

Overall Material Balance (OMB) : F+V N +1=LN +V 1

1000

lb lb lb +2000 =1276.7034 +V 1 hr hr hr

V 1=1723.2966

lb hr

Overall Solute Balance : F solute +V N +1solute =V 1Solute + L Nsolute

120

lb lb +0=V 1solute +1.7034 hr hr

V 1solute=118.2966

lb hr

concentration of solute∈V 1=

118.2966 1723.2966

concentration of solute∈V 1=0.06865≈ 0.07

solvent ∈V 1 =( 1723.2966−118.2966 )

lb lb =1605 hr hr

In Stage I : V1=1605 lb/hr

V2 = 2000 lb/hr

LN= 425 lb/hr

F= 30 lb/hr

solute solvent ¿ ¿ ¿ lb hr solute ∈L1 = lb lb 1605 425 hr hr

118.2966

solute ∈L1 =31.3246

lb hr

Solute Balance in Stage I : F solute +V 2solute =V 1solute + L1solute

120

lb lb lb +V 2solute =118.2966 +31.3246 hr hr hr

solute ∈V 2=29.6212

lb hr

concentration of solute∈V 2=

29.6212 2000

concentration of solute∈V 1=0.01481≈ 0.015

Solving for Number of Stages: Y N+1 −X N ] Y 2−X 1 N=1+ Y −Y 2 ln ⁡[ N +1 ] XN− X1 ln ⁡[

where:

YN+1 = 0 1.7034 lb solute =4.008 x 10−3 425 lb solvent

XN =

Y2=

X1 =

29.6212 lb solute =0.01481 2000 lb solvent 31.3246 lb solute =0.07370 425lb solvent

0−4.008 x 10−3 ] 0.01481−0.0737 0.01481−0 ln ⁡[ ] 0.07370−4.008 x 10−3 ln ⁡[

N=1+

N = 2.7352 = 3 stages

55. A slurry of flaked soybeans weighing 100 kg contains 75 kg inert solids and 25 kg of solution 10 weight % oil and 90 weight % solvent hexane. This slurry is contacted with 100 kg pure hexane in a single stage so that the value of retention for the outlet underflow is 1.5 kg on insoluble solid per kg solvent in the adhering solution. The composition of underflow leaving the extraction stage in percent by weight oil is GIVEN: V1

V0 = 100 kg hexane

y1

y0

F = 100 kg

L1 x1

Inert = 75 kg Sol’n = 25 kg

REQUIRED: The composition of underflow leaving the extraction SOLUTION: In Feed: F = 100 kg Inert = 75 kg Sol’n = 25 kg Oil (solute) = .10(25 kg) = 2.5 kg Inert balance: Inert in feed = Inert in L1 Inert in L1= 75 kg In Underfeed (L1): Inert = 75 kg Solvent = ? = 50 kg solvent =

75 kg inert kg inert 1.5 kg solvent

Solute = ? Solute balance: Solute in F + Solute inV0 = Solute in V1 + Solute in L1 2.5 kg + 0 = Solute in V1 + Solute in L1 Eq. 1 solute ∈V 1=2.5−solute∈L1

Solvent balance: Solvent in F + Solvent in V0 = Solvent in V1 + Solvent in L1 22.25 kg + 100 kg = solvent in V1 + 50 kg Solvent in V1 = 72.5 kg At Equilibrium: solute ∈V 1 solute∈ L1 = solution∈V 1 solution∈L1 Eq. 2 solute∈V 1 solute ∈L2 = solute∈V 1 + solvent ∈V 1 solute∈ L2+ solvent ∈L2 Subs. Eq. 1 to Eq. 2: 2.5−solute ∈ L1 solute∈L1 = ( 2.5−solute∈ L1 ) +solvent ∈V 1 solute ∈L1 + solvent ∈ L1 Solute in L1 = 1.0204 kg Subs to Eq. 1 Solute in V1 = 1.4795 Composition on underflow leaving: V ¿ 1 L1

¿

1.4795 kg 1.0204 kg = 1.45

56. Tung meal containing 55% oil is to be extracted at a rate of 4000 kg per hour using nhexane containing 5% wt oil as solvent. A counter current multiple stage extraction system is to be used. The meal retains 2 kg of solvent per kg of oil free meal while the residual charge contains 0.11 kg oil per kg oil free meal while the product is composed of 15 weight percent oil. The theoretical number of ideal stages is (A) 3

(C) 5

(B) 4

(D) 6

Given: V1 V n+1 15% oil 5% oil F=4000

1

2

3

kg hr

Ln 55% oil 0.11 kg oil kg oil free meal R= Required: Theoretical number of ideal stages Solution: Basis: 1 hr

2 kgsolvent kg free meal

n

In the Feed, kg oil : 0.55 x 4000=2200 kg kgmeal :0.45 x 4000=1800 kg In the final underflow, 0.11 kg oil kgoil : x 1800 kg=198 kg kg oil free meal kg solvent :

2 kg solvent x 1800 kg=3600 kg kg free meal

kg meal :1800 kg 2200+V n+ 1=V 1+ 3600+198

Overall Solution Balance: V n+1 =V 1+1598

 eq. 1

Overall Solute Balance:

2200+0.05 V n+1 =198+0.15 V 1

eq 2

V 1=20819 kg V n+1 =22417 kg At equilibrium condition, solute solute V 1= L solution solution 1

(

) (

)

x 20819 kgoil∈ L1 ( 0.1520819 )=( 3600+kg oil∈L 1 ) Kg oil in L1= 635.29 kg Solute balance in stage 1:

2200+kg oil∈V 2=0.15 x 20819+635.29

Kg oil in V2= 1558.14 kg Solvent balance in stage 1: 0+kg solvent ∈V 2=3600+ 0.85 x 20819 Kg solvent in V2= 21296.15 kg y2 :

1558.14 =0.0682 1558.14+21296.15

xn :

198 =0.0521 3600+198

x1 :

635.29 =0.15 635.29+3600

At constant underflow, y n+1 −x log y 2−x 1 N−1= y n+1− y 2 log x n−x 1 n

0.05−0.0521 0.0682−0.15 N−1= 0.05−0.0682 log 0.0521−0.15 log

N= 3.1665 ≈ 4 stages

57. Coconut oil is to be produced from dry copra in two stages. First, through expellers to squeeze out part of the coconut oil and then through a counter current multi stage solvent extraction process. After expelling, the dry copra cake contains 20% residual oil. In the solvent extraction operation, 90% of the residual oil in the expeller cake is extracted as a solution containing 50% by weight oil. If fresh solvent is used and on kg of solution with every 2 kg of insoluble cake is removed with the underflow, the number of ideal stages is (A) 4

(C) 6

(B) 5

(D) 7

Given:

V1 V n+1 90% recovery

1

2

3

n

V n+1

50% oil

Ln

F Copra R=

20% oil

Required: Number of Ideal Stages Solution: Basis: 100 kg Copra In the Feed, F= 100 kg Kg oil: 0.20 x 100= 20 kg Kg inert: 0.80 x 100= 80 kg In the Solvent, V(n+1) y n+1=

solute =0 solution

In the Final Underflow, Ln, Kg inert= 80 kg 1 kg solution x 80 kg inert =40 kg Kg solution: 2 kg inert Ln :80 kg+40 kg=120 kg solute :0.10 x 20=2kg solute 2 : =0.05 solution 40 In the Final Overflow, V1 y 1=0.50 kg solute :0.90 x 20=18 kg

1 kg solution 2 kg cake

V 1:

18 kg =36 kg 0.50

In the first undeflow, L1 solute = y 1=0.50 solution Kg inert: 80 kg Kg solution: 40 kg L1 :80 kg +40 kg=120 kg kg solute :0.50 x 40 kg=20 kg

In the Ovreflow 2, V2, OMB on stage 1, F+V 2 =L1+V 1 100+V 2=120+36 V 2=56 kg Solute balance on stage 1, solute ∈F + solute∈V 2=Solute∈ L1+ solute∈V 1 20+ y 2 x 56=20+18 y 2=0.3214 For constant underflow, y n+1 −x log y 2−x 1 N−1= y n+1− y 2 log x n−x 1 n

0−0.05 0.3214−0.50 N−1= 0−0.3214 log 0.05−0.50 log

N= 5 stages

58. Roasted copper ore containing the copper as CuSO4 is to be extracted in countercurrent stage extractor. Each hour, a charge consisting of 10 tons gangue, 1.2 tons CuSO4 and 0.5 ton water is to be treated. The strong solution produced is to consist of 90% wt. water and 10% wt. CuSO4. The recovery of CuSO4 is to be 98% of that in the ore. Pure water is to be used as fresh solvent. After each stage, one ton inert gangue retained 2 tons of water plus the copper sulfate dissolve in that water. Equilibrium is attained in each stage. The number of stages required is.

Given: OverFlow water) 90% water, 10% CuSO4

Solvent(Pure

Feed 10 tons gangue 1.2 tons CuSO4 0.5 tons water Solution: Inert in feed = inert in underflow Amount of solution in underflow 2ton s olution 10 ton inert x =20 ton solution 1 toninert Amount of overflow

Underflow R=

1 ton gangue 2 tonsolution

1.2 tons x 0.98 =11.76 tons 0.10 Overall balance of solute and solvent 11.76 + 20 = 0.5 + solvent stream Solvent stream = 30.06 tons Composition of final underflow Solute in underflow = 1.2 – 1.2x0.98 =0.024 tons 0.024 −3 =1.2 x 1 0 %wt. = 20

For stage 1 11.76 tons

30.06 tons

1.2 tons CuSO4 20 tons 0.5 tons Water At equilibrium solute s olute ( ) =( ) =0.10 solution overflow solution underflow Solute balance ; let x = fraction of solute at solvent stream 1.2 = 30.06 x =1.176 + 2 X = 0.065735 0−1.2 x 1 0−3 ] 0.065735−.10 0−0.065735 ln ⁡[ ] 1.2 x 1 0−3 −.10 ln ⁡[

Number of stages

=1+

= 9.226 = 10 stages

Situation for Problems 59-63 Oil is to be extracted from meal by means of benzene using a continuous countercurrent extractor. The unit is to be treat 1000 kg of meal (based on completely exhausted solid) per hour. The untreated meal contains 400 kg of oil and is contaminated with 25 kg of benzene. The fresh solvent mixture contains 10 kg of oil and 655 kg of benzene. The exhausted solids are to contain 60 kg of unextracted oil. Experiments carried out under conditions identical with those of the projected battery show that the solution retained depends on the concentration of the solution, as shown in table below. All quantities are given in an hourly basis. Concentration, kg oil/kg solution 0.0 0.1 0.2 0.3

Solution retained, kg/kg solid 0.500 0.505 0.515 0.530

Concentration, kg oil/kg solution 0.4 0.5 0.6 0.7

Solution retained, kg/kg solid 0.550 0.571 0.595 0.620

59. The concentration of the strong solution or extract is approximately equal to a. 0.56 b. 0.58

c. 0.60 d. 0.62

60. The concentration of the solution adhering to the extracted solids is approximately equal to a. 0.193 b. 0.218

c. 0.021 d. 0.118

61. The mass of the solution leaving with the extracted meal is approximately equal to

a. 507 kg/h b. 306 kg/h

c. 418 kg/h d. 621 kg/h

62. The mass of the extract is approximately equal to a. 583 kg/h b. 512 kg/h

c. 536 kg/h d. 571 kg/h

63. The number of stages required is a. 3 b. 4

c. 6 d. 7

Given:

Final Overflow, Vi

V2, 1

y2

VN, 2

yN

Solvent, VN+1 N

10 kg oil 655 kg benzene

Feed, F = 1000 kg meal/hr

L1,

L2,

400 kg oil

x1

x2

LN 60 kg unextracted oil

25 kg benzene 575 kg solid

Solution: In the feed: F = 1000 kg meal/hr Solute: 400 kg oil

In the Solvent: VN+1 = 10+655 = 665 kg

Solvent: 25 kg benzene

Solute: 10 kg oil

Inert Solid: 1000 – (400+25) = 575 kg

Solvent: 655 kg benzene

Solution: 400 + 25 = 425 kg/h solution af =

400 425

= 0.941

In the Final Underflow: LN Solute: 60 kg unextracted oil

Benzene: Ln – 60

OMB Solute: Feed + Solvent = Final (Underflow + Overflow) Oil: 400 + 10 = 60 + Final Overflow

Let: a = mass fraction of oil in final underflow b = mass fraction of oil in final overflow bVn+1 =

10 665

Final Overflow Solute: 350 kg/h OMB Solvent: Feed + Solvent = Final (Underflow + Overflow) Benzene: 25 + 655 = 447 + Final Overflow

= 0.015

Final Overflow Solvent: 233 kg/h Vi = 350 + 233 = 583 kg/h extracted By trial and error, b=

Assume aVn+1 = 0.1,

350 583

= 0.60

from table, Solution in Ln = 0.505 kg kg solid

At equilibrium:

LN = 0.505 (1000) = 505 kg/hr

from table, Solution = 0.595

avn+1 =

60 505

= 0.119

MB: Feed + V2 = V1 + L1 Solution

Ln

60 507

=

0.507

425 + V2 = 583 + 595 V2 = 953 kg Oil Balance:

Ln = 0.507(1000) = 507 kg/h avn+1 =

kg kg solid

At stage 1:

@ avn+1 = 0.119, from table, kg kg solid

a = bvi = 0.60,

595 (0.60) + 583 (0.6) = 425 (0.) + 753Y2 y2 = 0.408

= 0.118

0.015−0.118 ] 0.408−0.6 0.015−0.408 ln ⁡[ ] 0.118−0.6 ln [

@ Final Underflow, Ln: Benzene: 507 – 60 = 447 kg/h @ Final Overflow

N=1+

= 4.05 = 4

64. An oil-sand mixture that is 25% (by mass) oil and 75% (by mass) sand is to be extracted or leached with 75 tons/day of naphtha in a countercurrent extractor. The feed consists of 100 tons/day of mixture. The final extract (overflow) produced contains 35% (by mass) oil and 65% (by mass) naphtha, and the underflow from each unit consists of 32% (by mass) oil and 68% (by mass) sand. The overall efficiency of the extraction is 80% (by mass). Assume the solvent is miscible with the oil in all portions and the extractor has reached equilibrium conditions in each stage. Assume there is no sand in the overflow. The number stages required to effect the desired separation of oil from sand is a. 3

c. 5

b. 4

d. 6

Given: 75 tons/day of

OverFlow Yoil = 0.35 Ynaphtha = 0.65 Feed, F 100 tons/day

Xsolution = 0.32

Xoil =0.25

Xsand =

Xsand = 0.75 Overall efficiency = 80%

Required: Number of stages Solution: Assume: 1 day

*in the feed* (100

tons day

*amount of raffinate* of mixture)(1 day) = 100 tons of mixture

75 tons of sand 0.68

110.29 Sand = 100(0.75) = 75 Oil = 100(0.25) = 25 *OMB* Feed + naphtha = raffinate + extract 100 + 75 = 110.29 + extract Extract = 64.71 *Naphtha balance* Amount of naphtha entering = amount of naphtha extract + amount of naphtha raffinate Let X = mass fraction of naphtha in raffinate 75 = (64.71) (0.65) + (110.29) (X) X= 0.2986

Mass fraction of oil in raffinate = 1- 0.2986 – 0.68 = 0.0214

The further solution will be subjected to a graphical method

=

65. A copper ore containing 10.3% by mass copper sulfate, 85.4% by mass inert and 4.3 % by mass water is to be extracted with pure water in a counter current extractor. The daily feed consist of 281 tons. The final extract produced contains 10% by mass copper sulfate and 90% by mass water. The underflow from each stage consist of 66.7% by mass solution and 33.3% by mass inert. The process is to recover 92% of the copper sulfate from the ore. Assume the extractor has reached equilibrium conditions in each stage the minimum number of stages required to effect the desired separation of copper sulfate from the inert. Given: Overflow 10% CuSO4, 90% water Solvent %recovery = 92

Feed 281 tons 10.3 % CuSO4 solution 85.4 % inert 4.3 % water Solution: Basis: 281 tons feed .103(281) = 28.943 kg CuSO4 .854(281) = 239.974 kg inert .043(2810 = 12.083 kg water Inert in feed = inert in underflow Amount of solution in underflow 239.974 ( .667 ) =480.6686 tons solution .333

66.7 % 33.3 % inerts

Amount of overflow 0.92(28.943) =266.2756 0.10 %solute in underflow 28.943−(28.943 x 0.92) =4.817 x 1 0−3 480.6636 Solvent balance 12.083 + solvent stream = 0.90(266.2756) + 480.6636 – 2.31544 Solvent stream =705.9182

At stage 1

266.2756 tons solution

28.943 tonsCuSO4 12.083 tons water

705.9182 tons solution

480.6686 tons solution

At equilibrium (

solute solute ) =( ) =0.10 solution overflow solution underflow

Solute balance; let x be fraction of solute at solvent stream .10(266.2756) + .10(480.6686) = 28.943 + 705.9182x X = 0.0648

−3

0−4.817 x 1 0 ] 0.0648−.10 0−0.0648 ln ⁡[ ] 4.817 x 1 0−3−.10 ln ⁡[

Number of stages

=1+

= 6.1727 = 7 stages

1. 60 tons per day oil sand (25 wt% oil and 75 wt % sand) is to be extracted with 40 tons per day of naphthalene in a counter current extraction battery. The final extract from the battery is to contain 40 wt% oil and 60 wt% naphthalene and the underflow from each unit is expected to consists of 35 wt% solution and 65 wt% sand. If the overall efficiency of the battery is 50%, how many stages will be required?

GIVEN: Final Vo

Vn+1

F

Final Ln

Final Vo ( Final Overflow)

Final Ln ( Final Underflow)

X oil= 0.40

X naphthalene+ X oil= 0.35

X naphthalene= 0.60

X sand= 0.65

Feed= 60 tons/day

Vn+1= 40 tons/day

X oil= 0.25 X sand= 0.75

Required: N (Number of Stages) =?

X naphthalene= 1

Detailed Solution: Let A= Oil (Solute) B= Sand (Insoluble Solid) C= Naphthalene (Solvent)

In the Feed F= 60 tons/day A= 0.25(60) = 15 tons/day B= 0.75(60) = 45 tons/day

Overall Insoluble Solid Balance: (B)FEED=(B)UNDERFLOW (B)UNDERFLOW= 45 tons/day

In the underflow (B)UNDERFLOW= 45 tons/day=0.65 (underflow) Underflow= 69.53 tons/day Ln=(69.23)(0.35)= 24.23 tons/day

Liquid Balance: ( Solute+Solvent) 15+40=24.23+Vo Vo= 30.67 tons/day

Solvent Balance: 40= (C)UNDERFLOW + (0.60)(30.67) (C)UNDERFLOW= 21.598 tons/day

In the Underflow: Xc = ((C)UNDERFLOW/ Ln)= 21.598/24.23= 0.89 Xn= Xa= 1-0.89= 0.11

No. of Stages: N THEO= 1+ ln( (Yn+1- XN)/ (Y2-X1))/ln(((Yn+1- Y2)/ (XN-X1)))

Balance at Stage 1: 0.25(60)+ Y2(40)= 0.40( 30.77)+0.40(24.23) Y2= 0.175

Substitute: Yn+1= 0 XN= 0.11

Y2= 0.175 X1= 0.40 N THEO = 1+ ln( (Yn+1- XN)/ (Y2-X1))/ln(((Yn+1- Y2)/ (XN-X1))) N THEO= 2.42 stages N ACTUAL= N THEO/EFFECIENCY= 2.42/ 0.50 N ACTUAL= 4.84= 5 STAGES

(Principles of Mass Transfer and Separation Processes by Binay K. Dutta) A solid feed containing 22% of solute, 3% water and 75% inerts (insoluble) is to be leached a rate of 1 ton per hour with water in a countercurrent leaching cascade. The strong leachate leaving the unit should have 16% of the solute in it. Desired recovery of the solute in the feed is 99%. The overflow does not have any entrained inert in it, and the amount of solution retained in the sludge is 0.45 kg solution per kg inert. Analytically determine the number of stages required for the separation.

Given:

Final V1

Vn+1

F

Final Ln

Solution:

Basis: 1 hour operation 1 ton = 1000 kg

In Feed Solute: 1000(0.22) = 220 kg Water: 1000(0.03) = 30 kg Inert: 1000(0.75) = 750 kg

In Underflow: 0.45 kg solution x 750 kg inert =337.5 kg kg inert LN = 337.5 kg solution

Mass of solute leaving with the sludge (99% recovery) = (220)(0.01) = 2.2 kg Solute = 2.2 kg Solvent = 335.3 kg xN =

2.2 =0.00652 337.5

In Overflow: Solvent balance: Solvent in F + VN+1 = LN + V1 30 + VN+1 = 335.3 + V1 V1 = VN+1 – 305.3 Solute balance : Solute in F + VN+1 = LN + V1 220 + 0 = 2.2 + (VN+1 – 305.3) (0.16/0.84)

VN+1 = 1448.75 kg V1 = 1143.45 kg Solute in V1 = 182.95 kg Solvent in V1 = 960.5 kg

Solute Balance at Stage 1: VN+1 = 1448.75 = V2 X1 = Y1 = 0.16 YN+1 = 0 Pure solvent 220 + V2y2 = L1x1 + 182.95 220 + 1448.75y2 = 337.5 (0.16) + 182.95 Y2 = 0.0117 x N =0.00652

Using the equation: y N +1−x N ) y 2−x 1 N−1= y N +1− y 2 ln ⁡( ) x N −x1 ln ⁡(

0−0.00652 ) 0.0117−0.16 N−1= 0−0.0117 ln ⁡( ) 0.00652−0.16 ln ⁡(

N= 2.2

N= 3 STAGES

N = 3 stages