Separation Process Compilation of Problem Set

Fir st S ta ge

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:

Required: Overflow and Underflow streamsN stages. Solutions:

OMB : F+S=V1+L1

Material balance for the wax

4 Ton x1000 kg 1Ton =4000 kg 4000 kg x .25=1000 kg 4000 kg x .75=3000 kg Wax balance Vbyb V1:Overflow 5 kg wax 100 kg kerosene S: Solvent 0.05 kg wax 100 kg kerosene L1: Underflow 0.2 kg wax 100 kg free pulp Feed : 4 Tons Wax paper

25% wax 75% pulp 5 kg wax 100 kg kerosene Vaya Lbxb V2y2 2 kg kerosene 1 kg pulp L1x1

1000 kg+ 0.05 kg wax 100 kg keroseneVb= 5 kg wax 100 kg keroseneVa+ 0.2 kg wax 100 kg pulp(0.75 x 4000 kg ) Solvent Balance V_b=V_a+2 kg kerosene 1 kg pulp (0.75 x 4000 kg)=Va+6000 kg

Solving simultaneously we get the streams

V_a=20141 . 4141 kg Kerosene V_b=V₂=20081. 4141 kg kerosene undeflow=6000 kg kerosene x_b= 0.2 kg wax 100 kg pulpx 0.75 x 4000 kg pulp 6000 kg kerosene =0 .001 kg wax kg kerosene Wax balance: 20081.4141 y₂+1000= 5 100(20141.4141)+6000 x1

equilibrium condition x₁=y_a=y₂=0.05 kg wax

kg kerosene y2=0.01529

No. of ideal stages

N= ln

(

0.001−0.0005 0.05−0.01529

)

ln

(

0.01529−0.0005 0.05−0.001

)

+1=4 .5398 ≈5

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 L0 step L1 feed underflow Total balance: L0 + V2 = M = L1 + V1 = 100 + 100 = 200 kg

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

w _A,Underflow =

inert material

inert material+extractable substance+solvent

w _{A,Underflow =}w _{A,L1 =}

1.5

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 WA , M WA , L1

=200 0.4 .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 A+B+C= 20 0+20+100 W_{C , V 1}= C A +B+C= 100 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× WB , Lo−V1×WB ,V 1 L1 =100 ×0.2−66.666 ×0.1667 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

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 wax-free kerosene-wax-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

1

2

N

X1 Y2 Y1 XN YN+1 F= 2 Tons 25% Solute 75% Inert Solution: In Feed: Inert = 100 lb Feed = 100 lbinert_0.75 =133.3333lb Solute = (0.25)(133.3333 lb) = 33.3333 lb In Final Underflow: Inert = 100 lb

Solute = 0.2lb solute_{100 lbinert} x 100 lbinert =0.2 lb

Solvent = 2 lb solvent_lbinert x 100 lbinert=200lb

Overall Solvent Balance

0 + VN+1 = V1 + 200 Equation 1

Overall Solute Balance

33.3333 + ( _{100 lb solvent x V}0.05 lb solute N+1) = (

5 lb solute

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 100 lb solvent x 200 lb solvent) +

(

5 lb solute 100lb solvent

)

x 671.3798lb solvent Solute in V2 = 10.2357 lb Solute in Y2 = 5 lb solute 00lb solvent x 200 lb solvent =10 lb

Solving for Number of Stages: N=1+ ln ⁡[YN+1−XN Y2−X1 ] ln ⁡[YN +1−Y2 XN−X1 ] where: YN+1 = 0.05 lb solute 100 lb solvent=5 x 10 −4 XN = 0.2 lb solute 200 lb solvent=1 x 10 −3 Y2= 10.2357 lb solute 871.3798lb solvent=0.0117 X1 = 10 lb solute 200 lb solvent=0.05 N = 1 + ln ⁡[5 x 10−4−1 x 10−3 0.0117−0.05 ] ln ⁡[5 x 10 −4_−0.0117 1 x 10−3_−0.05 ] 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

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+V₁ (50+0)+200=93.5−a+V1 156.5+a=Vi At Equilibrium:

(

solute solution

)

V1 =

(

solute solution

)

L1

(

a 156.5+a

)

V1 =

(

56−a 37.5+56−a

)

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=V₁−a=156.5+a−a=1 56.5 kg

31.) V1:

Situation for problems no. 32-34

A solid B, contains a soluble component, A, of mass fractions xA=0.25 , xB=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

b. 0.07 c. 0.01d. 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 x₁= solute solution= 0.0375 0.5625=0.0667 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 solution

)

V1 =

(

solute solution

)

L1

(

0.2125 C−0.3125

)

V1 =

(

0.0375 0.5625

)

L1 C = 3.5000 kg

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 kgd. 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 Solvent = 26.4 kg

(

45₅₅

)

= 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

(

_{1.5 kg insoluble}1 k g solution

)

= 46.6667 kg Solvent = 46.6667 kg– 3.6 kg= 43.0667 kg xN= 3.6 46.6667=0.0771 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 N= 1+ ln

(

0−0.0771 0.3412−0.55

)

ln

(

0−0.3412 0.0771−0.55

)

= 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: 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 20,000 lb/h (V1) (V2) (V3)

1

2

5 wt% Calcium Carbonate 0.1 wt% Sodium Hydroxide (L2 ) (L1) (F) 100,000 lb/h 20wt% solid

Solid Balance: Solid ∈F=Solid ∈L 2 5000lb h=0.20 ( L2) L2=25000lb h OMB: F+V 3=V 1+L 2 100,000+20,000=V 1+25,000 V 1=95000lb h Stage 1 (@ equilibrium) ( Solute Solution)(V 1) =( Solute Solution)(L1) (Solute 95,000)(V 1 ) =(Solute 20,000)(L 1)

Solute ∈V 1=4.75 Solute∈L 1−−−eqn(1)

Stage 2 (@ equilibrium) ( Solute Solution)(V 2) =( Solute Solution)(L2) (Solute 20,000)(V 2) =(Solute 20,000)(L2 )

Solute ∈V 2=Solute∈L 2−−−eqn(2)

Solute Balance in Stage 2

Solute_{L 1}+Solute_{V 3}=Solute_{V 2}+Solute_{L 2}

Solute_{L 1}+0=Solute_{V 2}+Solute_L2

Solute_{L 1}=Solute_{L 2}+Solute_L2

SoluteL 1=2 SoluteL2−−−eqn(3)

Overall Solute Balance

Solute_F+Solute_{V 3}=Solute_{V 1}+Solute_L2

100+0=SoluteV 1+SoluteL2−−−eqn (4)

Substitute eqn (1) to eqn (4)

100=4.75 SoluteL1+SoluteL2−−−eqn(5) Substitute eqn (3) to eqn (5)

100=4.75 (2) Solute_L2+Solute_{L 2} SoluteL 2=9.52 lb h Using eqn 4: 100+0=Solute_{V 1}+9.52 Solute_{V 1}=90.48lb h

Percent Recovery ( R)=SoluteV 1−SoluteV 3 SoluteF

R=(90.48−0) 100 x 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:

Solution:

In the feed: basis(100 kg)

Solute: 0.08(100)= 8kg Solvent: 0.02(100)= 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 R= 1 kg H2O/kg Inerts Solute/Inerts = 0.008 R3 L3 Xn R1 L1 R2 L2 N 3 2 1 Y4 Y2 Y3 Final Underflow, Ln Solvent, Vn+1 Yn+1 V2 V3 V4 0.15 kg solute kg H2O 8% Solute 2% H2O Overflow, V1 Feed, F

(

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

Y1= X1= Solute Solvent=0.15 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 F = 136.5333 100 =1.3653 (#45)

Solute for Y2 using Solute balance around stage 1

8+ V2Y2=L1X1+7.28

V2= Vn+1=136.5333

Y2=?

X1=Y1=0.15 8+136.5333=90(0.15)+7.28 Y2=0.0936 Solve for N: N=1+ ln

(

0.008 0.0926−0.15

)

ln

(

0−0.0936 0.008−0.15

)

=5.69=6 stages(#46) Problem 47 Given: V1 V2 YN+1 Y1 Y2 L1 LN Feed= 50 tons/hr X1` XN 48% H2O

40% Pulp R = 3 tons H 20_{tons Pulp}

2 N

12%Sugar

Required: N = ? Solutions:

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 Solution = 5.82_{0.15 = 38.8} tons_hr H2O = V1 = 38.8 – 5.82 = 32.98 tons_hr Y1 = _{32.98 = 0.1765}5.82 X1 = Y1 = 0.1765 at equilibrium In Final Underflow: Sugar = 0.03(6) = 0.18 tons_hr H2O = LN = 20(3) = 60 tons_hr

XN = 0.18₆₀ = 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 V2 = V3 = V4 = ….. = VN+1 = 68.98 tons_hr L1 = L2 = L3 = L4 = ….. = LN = 60 tons_hr 6 + 68.98Y2 = 60(0.1765) + 5.82 Y2 = 0.1509 Solving for N: N = ln ⁡ 0−0.003 0.1509−0.1765 ln 0−0.1509 0.003−0.1765 +1 N = 16.36 = 17

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 = 38.8 tons/hr X1 = Y1 = 0.15 (@ 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)

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: N = ln[ 0−0.003 0.1282−0.15] ln[ 0−0.1282 0.003−0.15] 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

1

2

N

L1 Y2 Y1 LN YN+1 F= 1000 lb/hr 12% Solute 85% Inert

3% Solvent after drying = 0.2% salt

2000 lb/hr 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

Solvent = 0.5lb solvent_lbinert ×850 lbinert =425lb solvent

Solute : solute ∈L_N=0.2 100

(

inert +solute∈LN

)

solute ∈L_N=0.2 100inert + 0.2 100solute ∈LN solute ∈L_N=0.2 100(850 lb hr)+ 0.2 100solute∈LN solute ∈L_N=1.7034 lb/hr

Final underflow = inert +solvent + solute

¿(850+425+1.7034)_hrlb

FinalUnderflow=1276.7034 lb/hr Overall Material Balance (OMB) :

F+V_{N +1}=L_N+V₁ 1000lb hr+2000 lb hr=1276.7034 lb hr+V1 V₁=1723.2966 lb hr Overall Solute Balance :

F_solute+V_{N +1solute}=V_1Solute+L_Nsolute

120lb hr+0=V1solute+1.7034 lb hr V_1solute=118.2966lb hr concentration of solute∈V₁= 118.2966 1723.2966 concentration of solute∈V₁=0.06865≈ 0.07 solvent ∈V₁=(1723.2966−118.2966 )lb hr=1605 lb hr

In Stage I : solute solvent ¿ ¿ ¿ 118.2966lb hr 1605 lb hr =solute ∈L1 425lb hr solute ∈L1=31.3246 lb hr

Solute Balance in Stage I : F_solute+V_2solute=V_1solute+L_1solute

120lb hr+V2solute=118.2966 lb hr+31.3246 lb hr solute ∈V₂=29.6212lb hr V2 = 2000 lb/hr V1=1605 lb/hr LN= 425 lb/hr F= 30 lb/hr

concentration of solute∈V₂=29.6212 2000

concentration of solute∈V₁=0.01481≈ 0.015

Solving for Number of Stages:

N=1+ ln ⁡[YN+1−XN Y2−X1 ] ln ⁡[YN +1−Y2 XN−X1 ] where: YN+1 = 0 XN = 1.7034 lb solute 425 lb solvent =4.008 x 10 −3 Y2= 29.6212 lb solute 2000 lb solvent =0.01481 X1 = 31.3246 lb solute 425lb solvent =0.07370 N = 1 + ln ⁡[ 0−4.008 x 10 −3 0.01481−0.0737] ln ⁡[ 0.01481−0 0.07370−4.008 x 10−3] N = 2.7352 = 3 stages

V1 y1 V0 = 100 kg hexane y0 L1 x1 F = 100 kg Inert = 75 kg Sol’n = 25 kg

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:

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 1.5 kg inert 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 ∈V₁=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 ∈V₁ solution∈V1 = solute∈L1 solution∈L1 Eq. 2 solute∈V₁ solute∈V1+solvent∈V1 = solute ∈L2 solute∈L2+solvent ∈L2

Subs. Eq. 1 to Eq. 2: 2.5−solute∈L₁

(

2.5−solute∈L₁

)

+solvent ∈V₁= solute∈L₁ solute ∈L1+solvent∈L1 Solute in L1 = 1.0204 kg Subs to Eq. 1 Solute in V1 = 1.4795

Composition on underflow leaving:

¿V1

¿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 n-hexane 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=4000kg hr L_n 55% oil 0.11kg oil kg oil free meal

R=2 kgsolvent kg free meal Required:

Theoretical number of ideal stages Solution: Basis: 1 hr n 3 2 1

In the Feed,

kg oil : 0.55 x 4000=2200 kg

kgmeal :0.45 x 4000=1800 kg In the final underflow,

kgoil : 0.11 kg oil

kg oil free meal x 1800 kg=198 kg

kg solvent :2 kg solvent

kg free mealx 1800 kg=3600 kg kg meal :1800 kg

Overall Solution Balance: 2200+Vn+ 1=V1+3600+198

V_n+1=V₁+_{1598  eq. 1}

Overall Solute Balance: 2200+0.05 Vn+1=198+0.15 V1 _{eq 2}

V₁=20819 kg V_n+1=22417 kg At equilibrium condition,

(

solute solution

)

V1=

(

solute solution

)

L1

(

0.15 x 20819 20819

)

=

(

kgoil∈L1 3600+kg oil∈L 1

)

Kg oil in L1= 635.29 kg

Solute balance in stage 1: 2200+kg oil∈V2=0.15 x 20819+635.29

Kg oil in V2= 1558.14 kg Solvent balance in stage 1:

0+kg solvent∈V₂=3600+0.85 x 20819 Kg solvent in V2= 21296.15 kg

y₂: 1558.14

x_n: 198 3600+198=0.0521 x₁: 635.29 635.29+3600=0.15 At constant underflow, N−1= log yn+1 −xn y₂−x₁ log yn+1−y2 x_n−x₁ N−1= log0.05−0.0521 0.0682−0.15 log0.05−0.0682 0.0521−0.15

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 50% oil Vn+1 F Ln Copra

20% oil R=1 kg solution_{2 kg cake}

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 solution=0

In the Final Underflow, Ln, Kg inert= 80 kg

Kg solution: 1 kg solution_{2 kg inert} x 80 kg inert =40 kg L_n:80 kg+40 kg=120 kg solute :0.10 x 20=2kg solute solution: 2 40=0.05 In the Final Overflow, V1

y₁=0.50 kg solute :0.90 x 20=18 kg n 3 2 1

V₁:18 kg

0.50=36 kg In the first undeflow, L1

solute solution=y1=0.50 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+V₂=L₁+V₁ 100+V₂=120+36 V₂=56 kg

Solute balance on stage 1,

solute ∈F +solute∈V₂=Solute∈L₁+solute∈V₁ 20+ y₂x 56=20+18

y₂=0.3214

For constant underflow,

N−1= log yn+1 −xn y₂−x₁ log yn+1−y2 x_n−x₁ N−1= log 0−0.05 0.3214−0.50 log 0−0.3214 0.05−0.50

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 Solvent(Pure

water)

90% water, 10% CuSO4

Feed Underflow

10 tons gangue R=_{2 tonsolution}1 ton gangue 1.2 tons CuSO4

0.5 tons water Solution:

Inert in feed = inert in underflow Amount of solution in underflow

10 toninert x2ton s olution

1 toninert =20 ton solution Amount of overflow

1.2 tons x 0.98

0.10 =11.76tons 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

%wt. = 0.024₂₀ =1.2 x 1 0−3

For stage 1

11.76 tons 30.06 tons

1.2 tons CuSO4 20 tons

0.5 tons Water At equilibrium ( solute solution)overflow =( s olute solution)underflow =_0.10

Solute balance ; let x = fraction of solute at solvent stream 1.2 = 30.06 x =1.176 + 2 X = 0.065735 Number of stages = 1 + ln ⁡[ 0−1.2 x 1 0 −3 0.065735−.10] ln ⁡[ 0−0.065735 1.2 x 1 0−3−.10] = 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 Solution retained, kg/kg solid Concentration, kg oil/kg solution Solution retained, kg/kg solid 0.0 0.500 0.4 0.550 0.1 0.505 0.5 0.571 0.2 0.515 0.6 0.595 0.3 0.530 0.7 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

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, VN, Solvent, VN+1

1 y2 2 yN N 10 kg oil

655 kg benzene Feed, F = 1000 kg meal/hr L1, L2, LN

400 kg oil x1 x2 60 kg unextracted oil

25 kg benzene 575 kg solid

Solution:

In the feed: F = 1000 kg meal/hr Solute: 400 kg oil Solvent: 25 kg benzene Inert Solid: 1000 – (400+25) = 575 kg Solution: 400 + 25 = 425 kg/h solution af = 400 425 = 0.941 In the Solvent: VN+1 = 10+655 = 665 kg Solute: 10 kg oil Solvent: 655 kg benzene

In the Final Underflow: LN

Benzene: Ln – 60

Let:

a = mass fraction of oil in final underflow b = mass fraction of oil in final overflow

bVn+1 =

10

665 = 0.015

By trial and error, Assume aVn+1 = 0.1,

from table, Solution in Ln = 0.505

kg kg solid LN = 0.505 (1000) = 505 kg/hr avn+1 = 60 505 = 0.119 @ avn+1 = 0.119,

from table, Solution Ln = 0.507

kg kg solid Ln = 0.507(1000) = 507 kg/h avn+1 = 60 507 = 0.118 @ Final Underflow, Ln: Benzene: 507 – 60 = 447 kg/h @ Final Overflow

OMB Solute: Feed + Solvent = Final (Underflow + Overflow)

Oil: 400 + 10 = 60 + Final Overflow Final Overflow Solute: 350 kg/h

OMB Solvent: Feed + Solvent = Final (Underflow + Overflow)

Benzene: 25 + 655 = 447 + Final Overflow Final Overflow Solvent: 233 kg/h

Vi = 350 + 233 = 583 kg/h extracted

b = 350_{583 = 0.60}

At equilibrium: a = bvi = 0.60,

from table, Solution = 0.595 _{kg solid}kg

At stage 1: MB: Feed + V2 = V1 + L1 425 + V2 = 583 + 595 V2 = 953 kg Oil Balance: 595 (0.60) + 583 (0.6) = 425 (0.) + 753Y2 y2 = 0.408 N = 1 + ln [0.015−0.118 0.408−0.6 ] ln ⁡[0.015−0.408 0.118−0.6 ] = 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: Required: Number of stages Solution: Assume: 1 day OverFlow Yoil = 0.35 Ynaphtha = 0.65 75 tons/day of Xsolution = 0.32 Xsand = Feed, F 100 tons/day Xoil =0.25 Xsand = 0.75 Overall efficiency = 80%

*in the feed* *amount of raffinate* (100 tons_{day 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

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 66.7 % solution 85.4 % inert 33.3 % inerts 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

Amount of overflow 0.92(28.943) 0.10 =266.2756 %solute in underflow 28.943−(28.943 x 0.92) 480.6636 =4.817 x 1 0 −3 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 705.9182 tons solution

28.943 tonsCuSO4 480.6686 tons solution 12.083 tons water At equilibrium ( solute solution)overflow =( solute solution)underflow =0.10

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

Number of stages = 1 + ln ⁡[0−4.817 x 1 0 −3 0.0648−.10 ] ln ⁡[ 0−0.0648 4.817 x 1 0−3−.10] = 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 naphthalene= 1

X sand= 0.75

Required:

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 kg inert x 750 kg inert=337.5 kg 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 x_N= 2.2 337.5=0.00652 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:

N−1= ln ⁡( yN +1−xN y2−x1 ) ln ⁡(yN +1−y2 xN−x1 ) N−1= ln ⁡( 0−0.00652 0.0117−0.16) ln ⁡( 0−0.0117 0.00652−0.16) N= 2.2 N= 3 STAGES