Mass Transfer Part (11)

Worked Examples

1) A 5%(by weight) weight) soluti solution on of acetal acetaldehy dehyde de in toluen toluenee is extract extracted ed with water water in a 3 stage cross current unit. If 100 kgs of water is used per stage for 500 kgs of feed, calculate (using graphical method) the percentage extraction of acetaldehyde and the weights of final raffinate and mixed extract. The equilibrium relationship is given by the equation, Y = 2.3 X where Y = kg acetaldehyde/kg Water and X = kg acetaldehyde/kg toluene. Assume that toluene and water are immiscible with each other. Solution:

A: toluene,

B : water,

C: acetaldehyde,

F = 500 kg,

xF = 0.05,

Y = 2.3 x,

B = 100 kg water/stage

Three stage cross – current operation Assume solvent to be pure i.e. ys’ = 0 F = 500 kg, A = 475 kg, and C = 25 kg Slope = (– A/B) So (– A/B) for each stage = (– 475/100) = (– 4.75) Draw the operating line with a slope of – 4.75 for each stage X F

=

x F

(1 − x F )

=

0.05 1 − 0.95

X (kg acetaldehyde/ kg toluene) Y (kg acetaldehyde/ kg Water)

= 0.0526 0

0.01

0.02

0.03

0. 0 4

0.05

0.0 6

0

0.02 3

0.046

0.06 9

0.09 2

0.115

0.138

Since Since syste system m is immisc immiscibl ible, e, the whole of solvent solvent goes in extrac extract. t. The feed feed introduced in 1st stage stage just just passes passes through through all stages stages and comes out as final final raffinate: A plot between X and Y is drawn. The operating line is drawn with a slope of – 4.75 for each of the three stages. Weight of A in final raffinate = A = 475 kg Final raffinate contains X3 = 0.0161 kg C/kg A (from graph) Amount of C in raffinate = 475 × 0.016 = 7.6 kg

100

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Total weight of raffinate = 475 + 7.6 = 482.6 kg Total C extracted = (Y1 + Y2+ Y3) ×100 = 100 × (0.082 +0.055 + 0.037) = 17.4 kg In extract, the amount of B = 100 kg (in each stage) Y3 = 0.037 kg C/kg B (from graph) Amount of C in final stage extract = 0.037 × 100 = 3.7 kg Total weight of extract = 300 + 17.4 = 317.4 kg % Extraction = (17.4/25) × 100= 69.6%

Fig. 10.25 Example 1

2) 100 Kg of a soluti solution on containing containing acetic acetic acid acid and water contai containing ning 25% acid acid by weight weight is to be extr extrac acte ted d with with isopr isopropy opyll ether ether at 20°C 20°C.. The The tota totall solv solven entt used used for for extraction is 100kg. Determine the compositions and quantities of various streams if,









Equilibrium data:

Water layer (wt %) Acid (x) Water (A) 0.69 9 8 .1 1.41 9 7 .1 2 .9 9 5 .5 6.42 9 1 .7 1 3.3 8 4 .4 2 5.5 7 1 .1 3 6.7 5 8 .9 4 4.3 4 5 .1 4 6.4 3 7 .1 Solution:

.

A → water, water,

Ether layer (wt %) Acid (y) Water (A) 0.18 0 .5 0.37 0 .7 0.79 0.8 1.93 1 .0 4.82 1 .9 1 1 .4 3 .9 2 1 .6 6 .9 3 1 .1 10.8 3 6 .2 15.1

B → isopro isopropyl pyl ether, ether, C → Acetic Acetic acid, acid,

F = 100 kg,

A = 75 kg,

and C = 25 kg,

xF = 0.25

Total solvent used = 100 kg = B B

0.012 0.014 0.016 1 9 x 0.006 0.014 0.029 9 1 B 0.993 0.989 0.984 2 3 1 y 0.001 0.003 0.007 8 7 9 i) Single stage operation:

0.018 8 0.064 2 0.970 7 0.019 3

0.023 0.133 0.932 8 0.048 2

0 .0 3 4 0 .2 5 5 0.84 7 0.11 4

0.04 4 0.36 7 0.71 5 0.21 6

0.10 6 0.44 3 0.58 1 0.31 1

0.165 0.464 0.487 0.362

By total and component material balances, F + S = M1 100 + 100 = M1 = 200 kg x M 1

=

Fx F + sy s F + S

=

100 × 0.25 + 100 × 0 100 + 100

= 0.125

Locate M1 on the Fs line corresponding to xM1. By trial and error, a tie line is drawn which passes through through M1. The co–ordinates (x1,y1) obtained are (0.18, 0.075)









R 1 × 0.18 + 0.075 E1 = 200 × 0.125 R 1 + E1 = 200


Solving we get, E 1

 x − x  = M 1  M 1 1    y1 − x1 

Quantities of product streams are E1 = 104.76 kg R 1 = 95.24 kg (ii) Two stage operation: F = 100 kg,

S = 50 kg

S + F = M1 x M 12

=

Fx F + sy s F + S

=

100 × 0.25 + 50 × 0 100 + 50

= 0.167

M1 = 50 + 100 = 150 kg Locate M1,2 on the Fs line corresponding to xM1,2. By trial and error, a tie line is









and solving we get, E 12

 x − x   0.167 − 0.215  150 = M 12  M 12 12  =   =57.6 kgs  y x 0 . 09 0 . 215 − −    12 12 

R 12 12 = 150 - 57.6 = 92.4 kgs Similarly for II stage, xM22= 0.1395, M2 = 92.4+50 = 142.4 kg x2 = 0.175 and y2 = 0.07 (from tie line) E2 = 48.14 kg. R 2 = 94.26 kg. Percentage recovery = 3)

(25 − 94.26 × 0.175) 25

× 100 = 34.02%

1000 Kg/hr of an acetone-wate acetone-waterr mixture mixture containing containing 20% by weight of acetone acetone is to be counter-currently extracted with trichloroethane. The recovered solvent to be used is free from acetone. The water and trichloroethane are insoluble. If 90% recovery of acetone is desired estimate the number of stages required if 1.5 times the minimum solvent is used. The equilibrium relationship is given by y =1.65x, where where x and y are weight fractio fractions ns of aceton acetonee in water water and trichlor trichloroet oethane hane respectively. XF = 0.2/ (1– 0.2) = 0.25 X NP = 0.25 × 0.1 = 0.025 y1= 1.65 × 0.2=0.33 Y1 = 0.33/0.67 = 0.49 (the same value is got from plot also) A Bmin

800 Bmin

= =

− Y s X F − X Np Y 1

0.49 − 0 s 0.25 − 0.025

Bmin = 367.35 kg Bact = 1.5 × Bmin = 1.5 × 367.35 = 551.025 kg A Bact

=

Y 1,act − Y s X F

− X Np









An operating operating line with a slope of 1.452 is drawn and by step-wise step-wise construction construction the number of stages is determined as 5


4.

Water – dioxane so solution is is to be be se separated by by extraction pr process using be benzene as solvent. At 25°C the equilibrium distribution of dioxane between water and benzene is as follows:











At these concentrations concentrations water and benzene are substantia substantially lly insoluble. insoluble. 1000 kg of a 25% dioxane water solution is to be extracted to remove 95% of dioxane. The benzene is dioxane free. (i) (i)

Calc Calcul ulat atee the the benz benzene ene requi require reme ment nt for for a sing single le batc batch h oper operat atio ion. n.

(ii) (ii)

Calc Calcul ulat atee the the benz benzen enee requ requir irem emen entt for for a five five stag stagee cros crosss curre current nt

operation with 600 kgs of solvent used in each stage x y X =x/(1-x) Y =y/(1-y)

0.051 0.052 0.054 0.05485

0.189 0.225 0.233 0.29

raffinate in each stage (B)

(F)

or

(R)

F = 1000 kg (A = 750 kg, C = 250 kg) xF = 0.25,

XF = 0.25/0.75 = 0.333

XRNp = 0.05 × 0.333 = 0.01665 Y in = 0 Y1 = 0.0175 (From plot) A

Y 1

− Y s

0.252 0.32 0.337 0.471

Solution:

Solvent amount

of

feed

= or









Amount of solvent used is 600 kgs A B

=

750 600

=1.25

Draw Draw oper operat atin ing g line liness with with a slop slopee of -1.2 -1.25 5 and and dete determ rmin inee the the raff raffin inat atee concentration. X final = 0.0175 (0.333 − 0.0175) × 100 % recovery = 0.333

= 94.75%









5.

1000 1000 kil kilog ogra rams ms per per hou hourr of of a solu soluti tion on of of C in A con conta tain inin ing g 20%C 20%C by wei weight ght is to be counter currently extracted with 400 kilograms per hour of solvent B. The components A and B are insoluble. The equilibrium distribution of component C between A and B are as follows; Wt. of C/Wt. of A

0.05

0.20

0.30

0 .4 5

0.50

0.54

Wt. of C/Wt. of B

0.25

0.40

0.50

0 .6 5

0.70

0.74

How many theoretical stages will be required to reduce the concentration of C to 5% in effluent? Solution:

F = 1000 kg/hr, xF = 0.2,

(A = 800 kg/hr C = 200 kg/hr)

xRNp = 0.05

Assume solvent to be pure Countercurr urrent ent extraction

ys = Ys = 0

Solvent = B = 400 kg/hr A and B are insoluble XF = 0.2/ (1 – 0.2) = 0.25, A B

=

− Y s X F − X Np Y 1

Slope = A

A B Y 1

=

800 400

−0

=2

X R,NP = 0.05/ (1 – 0.05) = 0.0526










6.

Water – dioxane dioxane solution solution is to be separated separated by extraction extraction process process using benzene









Solution:

Benzene: B

Water: A

Dioxane: C

F = 1000 kg (A = 750 kg, C = 250 kg), x y X =x/(1-x) Y =y/(1-y)

0.051 0.052 0.054 0.05485

0.189 0.225 0.233 0.29

0.252 0.32 0.337 0.471

xF = 0.25, XF = 0.25/0.75=0.333 XRNp = 0.05 × 0.333 = 0.01665 A

=

Bmin

− Y 1 X NP − X F Y NP +1

A Bmin

=

Y NP +1 X NP

− Y 1

− X F

=

0 − 0.36 365 5 0.01665

Bmin = 650 kgs Bact = 1.5 × 650 =975 kgs A Bact

750 975

= =

− Y 1 , act X NP − X F

Y NP +1

0 − Y 1 , act 0.01665 − 0.333

Y1,act = 0.243

− 0.33 333 3

= 1.1154



















X F

xF = 0.01

= X 0 =

0.01 (1 − 01)

= 0.0101

F = 1000 kg, (C = 10 kg, A = 990 kg),

−

A Bn

990 1500

=

− Y s ) ( X n − X n −1 )

=

−0 0.0101 − X 1

B = 1500 kg

(Y n

Y 1

A line with a slope of -0.66 is drawn from (0.0101,0) to obtain X1 and Y1 Y1 = 0.66 [(0.0101) – X1] Y1= 0.0037 (From graph) X1= 0.0045 Amount of nicotine in extract = 0.0037 × 1500 = 5.55 kg % extraction = (5.55/10) × 100 = 55.5% For 3 stages (− A/B) = − 990/500 = − 1.98. 3 Lines with a slope of -1.98, each, are drawn staring from (0.0101,0)

















Water: A

Dioxane: C

x y X =x/(1-x) Y =y/(1-y) F = 1000 kg/hr, xF = 0.2,

0.051 0.052 0.054 0.05485

Benzene: B

0.189 0.225 0.233 0.29

XF=X0=0.2/0.8=0.25

Countercurrent extraction X Np= 0.2 × 0.25 = 0.05 A Bmin

800 Bmin

= =

− Y 1 X NP − X F Y NP +1

0 − 0.3075 0.05 − 0.25

Bmin = 520.33 Kgs

(From Graph)

0.252 0.32 0.337 0.471



















3)

1000 Kg Kg/hr of an ac acetone-water mixture co containing 10 10% by weight of acetone is is to be counter-currently extracted with trichloroethane. The recovered solvent to be used is free from acetone. The water and trichloroethane are insoluble. If 95% recovery of acetone is desired estimate the number of stages required if 1.5 times the minimum solvent is used. The equilibrium relationship is given by y=1.65x, where x and y are weight fractions of acetone in water and a nd trichloroethane respectively.

4)

Repeat pr problem 3 for a 4-stage cr crosscurrent op operation us using 30 300 Kg Kg/hr of of so solvent in each stage and determine the % recovery.









ii)

The ex extraction is ca carried ou out in in tw two st stages wi with 10 100kgs kgs of of so solvent

in each stage. Equilibrium data: Water layer (wt %) Acid Water 0.69 98.1 1.41 97.1 2 .9 95.5 6.42 91.7 1 3.3 84.4 2 5.5 71.1 3 6.7 58.9

Ether layer (wt %) Acid Water 0.18 0.5 0.37 0.7 0.79 0. 8 1.93 1 4.82 1.9 11.4 3.9 21.6 6.9









11)

1000 ki kilograms pe per ho hour of of a solution of of C in A containing 10 10%C by by we weight is is to to be counter currently extracted with 500 kilograms per hour of solvent B. The components A and B are insoluble. The equilibrium distribution of component C between A and B are as follows; Wt. of C/Wt. of A

0.05

0.20

0.30

0 .4 5

0.50

0.54

Wt. of C/Wt. of B

0.25

0.40

0.50

0 .6 5

0.70

0.74

How many theoretical stages will be required to reduce the concentration of C in A to 2%? 12)

Acetone is is to to be be re recovered fr from di dilute aqu aqueous so solutions by li liquid – liq liquid









i) Determine the number of theoretical stages if the solvent rate is 1.5 times the minimum. ii) If the the same operation is done in a 3-stage crosscurrent crosscurrent battery with 60 kgs of solvent in each stage, estimate the number of stages needed?

Mass Transfer Part (11)

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