Seminar 11 (winter 2017 ) with solutions PDF

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DEPARTMENT OF CHEMICAL & MATERIALS ENGINEERING

ChE 416

Equilibrium Stage Processes Seminar 11 (April 3)

Winter 2017

Q1. Single stage liquid-liquid extraction Isopropyl ether (E) is used to separate acetic acid (A) from water (W). One hundred kilograms of 30 wt% A-W solution is contacted with 120 kg of ether (E). What are the compositions and weight of the resulting extract and raffinate? What would the concentration of acid in the (ether rich) extract be if all ether were removed? Equilibrium data are given below. Draw an equilateral triangular diagram and another right triangular diagram with at least four tie-lines (tie-lines data are in bold font) and use those diagram for you calculations.

Wt% A 1.41 2.89 6.42 13.30 25.50 36.70 45.30 46.40

Water-Rich Layer Wt% W 97.1 95.5 91.7 84.4 71.1 58.9 45.1 37.1

Wt% E 1.49 1.61 1.88 2.3 3.4 4.4 9.6 16.5

Wt% A 0.37 0.79 1.93 4.82 11.4 21.6 31.1 36.2

Ether-Rich Layer Wt% W 0.73 0.81 0.97 1.88 3.9 6.9 10.8 15.1

Wt% E 98.9 98.4 97.1 93.3 84.7 71.5 58.1 48.7

(Note: Both diagrams can also be found below) Q2. Multistage liquid-liquid extraction 8000 kg/h of a 26 wt. % solution of acetic acid in water solution is extracted in a continuous countercurrent system with pure (isopropyl) ether to obtain a raffinate concentration of 5 wt.% of acetic acid. a) Determine the minimum flow rate of solvent. b) Find the number of stages required for (S/F) = 1.5 (Smin/F) For this problem use both equilateral triangular diagram and right triangular diagram shown below. The necessary tie-lines can be drawn from the data given in Q1.

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DEPARTMENT OF CHEMICAL & MATERIALS ENGINEERING

ChE 416

Equilibrium Stage Processes Seminar 11 (April 3)

Winter 2017

Figure 1: Acetic acid-water-isopropyl ether equilibrium data at 101 kPa and 20 C (Equilateral triangular diagram)

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DEPARTMENT OF CHEMICAL & MATERIALS ENGINEERING

ChE 416

Equilibrium Stage Processes Seminar 11 (April 3)

Winter 2017

Figure 2: Acetic acid-water-isopropyl ether equilibrium data at 101 kPa and 20 C (right triangular diagram)

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DEPARTMENT OF CHEMICAL & MATERIALS ENGINEERING

ChE 416

Equilibrium Stage Processes Seminar 11 (April 3)

Winter 2017

Solutions: Q1.

Use a ternary diagram for equilibrium data to plot immiscibility (or miscibility) curve (or use the diagram provided) Draw the tie lines.

Similarly, draw the tie-lines in a right-triangular ternary diagram. 4

DEPARTMENT OF CHEMICAL & MATERIALS ENGINEERING

ChE 416

Equilibrium Stage Processes Seminar 11 (April 3)

Winter 2017

Given: Feed, F = 100 kg x(F)a = 0.3 x(F)w = 0.7 Solvent (Isopropyl ether), S = 120 kg Let the hypothetical mixing point is M. So, from the mass balance: F+S = M = E+R  M =F+S From the solute balance, M xA,M = F xA,F + S yA,S Or, (F+S) xA,M = F xA,F + S yA,S ----------1 Thus, point S, M and F lie on a straight line. By using the inverse lever-arm rule: ---------- 2

Locate the feed point F (xA,F = 0.3, xD,F = 0.7) in the diagram. Draw a this line.

line. M must be somewhere in

We can locate the M by two different methods. Method 1(using mass and component balance): Use equation 1 to get the composition of the solvent at mixing point (M) ----------1 

[Since, yA,S = 0 (pure solvent) ] 5

DEPARTMENT OF CHEMICAL & MATERIALS ENGINEERING

ChE 416

Equilibrium Stage Processes Seminar 11 (April 3)

Winter 2017

By inverting, 

=

    So,

xA,M = 0.136 Locate the point in the graph on

line and find out the composition of others.

xW,M = 0.319 The composition of xE,M (solvent composition) can be obtained directly by reading from equilateral triangular diagram. We cannot read it from right triangular diagram. However, in either case, we can use the following formula in order to get the composition of the solvent. xE,M = [1- (xA,M + xW,M) = [1 – (0.136 + 0.319)] = 0.545 Method 2(using inverse-lever-arm rule): From the inverse-lever-arm rule, we get, From the lever arm rule, we get,



+1

 6

DEPARTMENT OF CHEMICAL & MATERIALS ENGINEERING

ChE 416

Equilibrium Stage Processes Seminar 11 (April 3)

Measure the length of

line. Let the length is L cm.

Now calculate the length of [For example, if

line which is

= 10.8 cm,

So, locate point M on

Winter 2017

cm]

line using a measuring scale and find out the compositions of M.

The compositions should be the same as we calculated by using method 1 Extract and Raffinate compositions: • Draw a tie line ER passing through M • Points E and R on miscibility curve represent the composition of extract and raffinate • Find the amount of E & R using lever-arm rule or mass balance From the diagram, the Extract compositions: yA,E = 0.09 yW,E = 0.04 However, yi = 1 So, yE,E = 0.87 From the diagram, the Raffinate compositions: xA,R = 0.2 xW,R = 0.755 However, xi = 1 xE,R = 0.045 For the amount of E and R From overall mass balance: M= E+R = F+S  E+R= 100+120=220 So, R = (220-E) From solute balance, M xA,M = E yA,E + R xA,R 220×0.136 = E ×0.09+ (220-E) × 0.2 E = 128 kg R = (220-128) = 92 kg

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DEPARTMENT OF CHEMICAL & MATERIALS ENGINEERING

ChE 416

Equilibrium Stage Processes Seminar 11 (April 3)

Winter 2017

Alternatively, E and R can be obtained by the inverse=lever-arm rule as follows: =

/

Use a scale to measure and and from there get the ratio of E/R. We have another equation of E+R = 220. Solving these two equations will give the value of E and R. [Consider the following calculation true for one diagram (the length may not be same for all diagram due to different size of the diagram] = 3.8 cm and = 6 cm So, E/R = 6/3.8 E = 1.0328 R E = 1.579 R E = 1.579 (220-E) E = 134.7 kg R = (220-134.7) = 85.3 kg

For ether free composition in the extract: xA = 0.09/(0.09+0.04) = 0.692 and xW= (1-0.692) = 0.308

Q2. a) Minimum flow rate of solvent: Given: F = 8000 kg/h, xAF = 0.26 and xRN = 0.05 xAS = 0 (because of pure solvent) Step 1: Locate F, S and RN Step 2: Join S and F Step 3: Join RN and S and extend to the extract (left) side (as the tie lines slope towards the extract). 8

DEPARTMENT OF CHEMICAL & MATERIALS ENGINEERING

ChE 416

Equilibrium Stage Processes Seminar 11 (April 3)

Winter 2017

Step 4: Locate several tie lines (shown in the diagram) Step 5: Extend tie lines to the SRN mixing line. Step 6: Tie line closest point to S corresponds to P min (it would be furthest if SRN were on the raffinate side). (Usually, tie line through F usually corresponds to Pmin, however, it is not always obvious) Step 7: Mixing line from Pmin through F locates E1 Step 8: Connect RN and E1 in order to complete the mass balance (overall mass balance) Step 9: M is located at the intersection of SF and RNE1 Step 10: Get Smin/F by using any of the followings: 1. Using inverse liver arm rule: 2. Using solute (A) balance: F.xAF + SminxAS = (F+Smin)xAM  Smin /F = (xAF –xAM)/(xAM – xAS) = (xAF/xAM – 1) for pure solvent

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DEPARTMENT OF CHEMICAL & MATERIALS ENGINEERING

ChE 416

Equilibrium Stage Processes Seminar 11 (April 3)

Winter 2017

10

DEPARTMENT OF CHEMICAL & MATERIALS ENGINEERING

ChE 416

Equilibrium Stage Processes Seminar 11 (April 3)

Winter 2017

From the above diagrams, We have got, xAM = 0.09 So, Smin /F = (xAF/xAM – 1) = (0.26/0.09 -1) = 1.89

So, minimum flow rate of solvent (Smin) = 1.89 × F = 1.89 × 8000 = 15120 kg/h b) S/F = 1.5 (Smin/F) So, S/F = 1.5× 1.89 = 2.835 S = 2.835×2000 = 22680 kg/h

Step 1: Get the mixing point (M) 11

DEPARTMENT OF CHEMICAL & MATERIALS ENGINEERING

ChE 416

Equilibrium Stage Processes Seminar 11 (April 3)

Winter 2017

We know, MxAM = FxAF + SxAS M=F+S So, We know already, F = 8000 kg, xAF = 0.26 S = 22680 kg/h, xAS = 0 So, xm = 0.068 Step 2: Get y1 as xN is given (Let x be the concentration in raffinate and y in the extract (solvent) of the species to be extracted by the solvent). xN = 0.05 (given) Draw a line from xN (RN) passing through the mixing point M (xM) to get y1 (E1) on the miscibility curve.

Figure: A counter-current N-stage L-L extraction system Step 3: Get E1 Step 4: Get P by extrapolating E1F and SRN [E1 and F are passing streams; hence they are on operating lines. Similarly, S and RN are also on operating lines] 12

DEPARTMENT OF CHEMICAL & MATERIALS ENGINEERING

ChE 416

Equilibrium Stage Processes Seminar 11 (April 3)

Winter 2017

P is the focus of all the operating lines because of the following balance:

[Note that E1 and R1 are leaving streams, so y1 and x1 must be on same equilibrium/tie lines. However, R1 and E2 are passing stream, so x1 and y2 must be one same operating line (connected to R1 and P through E2)] Step 5: Get the x1 that is in equilibrium with y1. x1 can be obtained in the equilibrium curve from the tie line drawing from y1. Step 6: Get the operating line passing through x1 in order to get y2 An operating line drawn from P and passing through x1 will intersect equilibrium curve at y2. Step 7: Repeat step 5 and 6 until the tie line goes below to any tie line passing through xN. Count the number of tie lines for number of stages. Number of stages = 5 13

DEPARTMENT OF CHEMICAL & MATERIALS ENGINEERING

ChE 416

Equilibrium Stage Processes Seminar 11 (April 3)

Winter 2017

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DEPARTMENT OF CHEMICAL & MATERIALS ENGINEERING

ChE 416

Equilibrium Stage Processes Seminar 11 (April 3)

Winter 2017

15...