Assignment 3 (Winter 2017 ) with solutions PDF

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DEPARTMENT OF CHEMICAL & MATERIALS ENGINEERING

ChE 416

Equilibrium Stage Processes

Winter 2017

Assignment 3 (Due 4:00pm, Feb. 15, 2017) Q1. Murphree Efficiency and Binary Distillation using McCabe Thiele method (25 marks) Use enclosed y-x diagram for your solution [20+5] A distillation column with a total condenser and a partial reboiler is separating methanol and water. Feed is a super-heated vapor where 3 moles of feed vaporize 2 mole of liquid. Feed rate is 100 mol/h, and a feed composition is 60% methanol. The column separates this feed into a liquid distillate product of 96 mol% methanol and liquid bottom product of 96 mole% water. The column operates at 1 atm. The y-x diagram for the system is enclosed (Figure 1). L/V is equal to 0.75. Determine the number of trays and optimal feed tray location for the actual column at 75% Murphree vapor -tray efficiency by the McCabeThiele method. Q2. Optimal side-stream location [2+2+6+4+2+4+5] Following figure shows a distillation column assuming that methanol (M) and ethanol (E) form an ideal solution with a constant relative volatility of 1.745. Please see Figure 2 for equilibrium data.

a) Calculate D and B b) Determine reflux ratio (R) c) Calculate the following flow rates of all three sections: Top section: V, L Intermediate section: V, L Bottom section: and

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DEPARTMENT OF CHEMICAL & MATERIALS ENGINEERING

ChE 416

Equilibrium Stage Processes

Winter 2017

d) From the mass and/or component balance show the intermediate operating line in terms of y = mx + c, where m is the slope of the operating line and c is the intercept. Find the value of m and c e) From the balance around reboiler show the bottom operating line in terms of y = mx + c. Find the value of m and c f) Determine the optimal feed and side stream locations g) Determine the number of equilibrium stages

Q3. Multiple feed streams using McCabe Thiele method [6+2+6+4+5+2] There are two feed streams A and B at their corresponding bubble points available in a plant. The streams consist of binary mixtures of methanol and water whose characteristics are given in Table1. You are given an assignment to design a distillation column, which is expected to produce a top product of 95% methanol and a bottom product of 95% water. There is a high premium on distillation equipment cost and you are expected to keep the number of plates the minimum. The equilibrium data is given in Figure 1. Your supervisor fixes the reflux ratio L/D = 2.0 and asks you to explore the following two conditions: (i) Case 1. Mix the two streams A and B, bring them to their bubble point and then have one feed to the column. (ii) Case 2. Feed the two streams at two different places in the column. Table 1: Characteristic of the two streams. Stream Flow rate [mol/h] Composition A 100 x = 0.5 B 100 x = 0.4 a) For case 1, calculate D, B and determine the number of stages b) For case 2, find the followings: i) Calculate D and B ii) Calculate all the flow rates in all three sections: V, L, V, L, and iii) From the mass and/or component balance show the intermediate operating line in terms y = mx + c, where m is the slope of the operating line and c is the intercept. Find the value of m and c iv) Determine the number of stages c) Which of the above two cases will give you the minimum number of trays? Support your answers with the proper McCabe-Thiele diagrams.

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DEPARTMENT OF CHEMICAL & MATERIALS ENGINEERING

ChE 416

Equilibrium Stage Processes

Winter 2017

Q4. Open steam using McCabe Thiele method [4+12+4+5] We are separating a mixture containing methanol and water in a distillation column using open steam heating and a partial condenser. Feed is a super-heated vapor where 2 moles of feed vaporize 1 mole of liquid. Feed rate is 250 mol/h, and a feed composition is 80% methanol. Distillate composition is yD=0.96 and bottoms composition is xB=0.1. Use a reflux ratio, L/D = 1.84. The equilibrium data are given in Figure 1. a) b) c) d)

Find the optimum feed plate location. Determine the total number of equilibrium stages needed. Find the composition of the liquid leaving the 4th stage below the partial condenser. Determine steam rate in kg/h and distillate rate in kmol/h

Figure 1. Equilibrium data for methanol and water at 1 atm 3

DEPARTMENT OF CHEMICAL & MATERIALS ENGINEERING

ChE 416

Equilibrium Stage Processes

Winter 2017

Figure 2. Equilibrium data for methanol and ethanol at 1 atm

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DEPARTMENT OF CHEMICAL & MATERIALS ENGINEERING

ChE 416

Equilibrium Stage Processes

Winter 2017

Solutions: Solutions of Q1. Given: Feed: F = 100 mol/h zF = 0.60 Feed condition: Super-heated vapor Distillate: yD = 0.96 Partial condenser acts as a stage. Bottoms: yB = 0.04 Total reboiler.

Feed is a super-heated vapor where 3 moles of feed vaporize 2 mole of liquid. Amount vaporized = v = (2/3) F

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DEPARTMENT OF CHEMICAL & MATERIALS ENGINEERING

ChE 416

Equilibrium Stage Processes

Winter 2017

We know,

Then, So, Slope of q-line : Intersection: y = x = zF = 0.60 q-line can be drawn from the intersection point with a slope of 0.4. Alternatively, Equation of q-line is given as

At x = 0, y = 0.36 Draw q-line using the points y = 0.4 at x = 0 and x =z F= 0.6 on diagonal line that intersects the equilibrium curve. R-line: R-line can be obtained by component mass balance over rectifying section:

Slope = L/V = R/(R+1) = 0.75 So, R = 3

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DEPARTMENT OF CHEMICAL & MATERIALS ENGINEERING

ChE 416

Equilibrium Stage Processes

Winter 2017

In terms of D (distillate) and R (Reflux), R-line:

So, the slope of R-line is L/V or R/(R+1). It intersects y-axis (x = 0) at y = xD/(R+1)

Slope = L/V = R/(R+1) = 0.75 Intercept, xD/(R+1) = 0.24

R- line passes through the point x = xD = 0.96 on the 45° line and the point of intersection of y-axis (x = 0) at y = xD/(R+1) = 0.24. S-line: S-line intersects the point of intersection of q-line and R-line. We can draw S-line that passes through the point x= xB = 0.04 on the 45° line and the point of intersection of q-line and R-line. Stages for Murphree vapor efficiency The stages are stepped off from an efficiency line, which for a Murphree vapor efficiency of 0.8 is positioned 70% of the vertical distance from the operating line to the equilibrium curve, as governed by the following equation:

where yn+1 is the location on the operating line, yn is the location on the efficiency line, and yn* is the location on the equilibrium line. As seen in the plot, 8 stages are required including partial reboiler. Optimal feed stage is the 6th stage. Partial reboiler is assumed to have 100% efficiency. So the first step is between the original equilibrium curve and the S-line.

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DEPARTMENT OF CHEMICAL & MATERIALS ENGINEERING

ChE 416

Equilibrium Stage Processes

Winter 2017

[Note that partial condenser is also assumed to have 100% efficiency and in that case the first step is between the original equilibrium curve and the R-line. However, counting can be started from top or from bottom, it will not hamper the number of stages]

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DEPARTMENT OF CHEMICAL & MATERIALS ENGINEERING

ChE 416

Equilibrium Stage Processes

Winter 2017

Solution of Q2. Given: Light key: methanol Feed: F = 100 kmol/h zF = 0.75 Feed condition: Two phase mixture that is 20% vapor Distillate: xD = 0.96 R = 1.2 (L/D)min Side Stream: S = 15 kmol/h xS = 0.2 Bottoms: xB = 0.05

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DEPARTMENT OF CHEMICAL & MATERIALS ENGINEERING

ChE 416

Equilibrium Stage Processes

Winter 2017

a) We can get the value of D and B using overall mass balance and component balance. Over mass balance: F=D+S+B 100 = D + 15 + B B + D = 85 ----- (1) From over all component balance: FzF = DxD + SxS + BxB 1000.75 = D0.96 + 150.2 + B×0.05 0.96D + 0.05B = 72 --------- (2) Solving equations (1) and (2), we get, D = 74.45 kmol/h B = 10.55 kmol/h b) In order to get R value for R = 1.2 (L/D)min , we have to draw q-line and Rmin line first. q-line: Equation of q-line is given as

The feed is 25 mol% vaporized. i.e., ψ = 0.25, so, q = 0.75 Slope of q-line: q/(q-1) = 0.75/(0.75-1) = -3 So, the q-line passes through x = 0.75 and has a slope of -3 [since, zF = 0.75] Rmin-line:

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DEPARTMENT OF CHEMICAL & MATERIALS ENGINEERING

ChE 416

Equilibrium Stage Processes

Winter 2017

Rmin-line passes through the intersection of q-line (x = 0.75) and the equilibrium curve from x = y = xD = 0.96 and meets y-axis (x = 0) at y = xD/(Rmin+1) Rmin-line cuts y axis (at x = 0) at y = 0.41 (from the graph) So, xD/(Rmin + 1) = 0.40 Rmin = 1.4 So, R = 1.2×Rmin = 1.2 × 1.4 = 1.68 c) Calculation of flow rates in all three sections Top Section: Reflux ratio, L/D = 1.68 So, L = 1.68D L = 1.68×74.45 = 125.076 mol/h From the condenser balance, V=L+D V = 125.076 + 74.45 = 199.526 mol/h Intermediate section: Consider the following two phase feed diagram for balancing liquid and vapor in the feed tray:

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DEPARTMENT OF CHEMICAL & MATERIALS ENGINEERING

ChE 416

Equilibrium Stage Processes

Winter 2017

L = L + LF = 125.076 + 75 = 200.076 V = V + VF V = V – VF = 199.526 – 25= 174.526 [Since F is 25% vapor, 75% of the feed is liquid]. So, LF = 0.75×100 = 75 mol and VF = 0.25×100 = 25] Bottom section: = L' – S = 200.076 – 15 = 185.076 mol/h As the side stream is saturated liquid. It has no vapor part. So the vapor steam entering and leaving the side stream withdrawal tray will not change. = V' = 174.526 d) Lets do a component balance around D and F of the distillation column. Vʹy + FzF = Lʹx +DxD So, intermediate operating line: We need to determine L and V for calculating slope

and intercept

.

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DEPARTMENT OF CHEMICAL & MATERIALS ENGINEERING

ChE 416

Equilibrium Stage Processes

Winter 2017

Let’s calculate the slope and the intercept: Slope,

=

= 1.146

Intercept, = = - 0.02 So the equation of intermediate operating line: y = 1.146 x - 0.02 [ Note that the intermediate operating line can also be obtained by doing a balance around side stream and bottom. The operating line would be :

]

e)

Lets do a component balance around reboiler for the bottom operating line.

y + BxB = /x Rearranging, y = / x + BxB/

Where slope , / = 185.076 /174.526 = 1.06 And intercept, BxB/ = 10.55×0.05/174.526 = 0.003 So the bottom operating line is: y = 1.03x+0.003 f) Optimal feed location: 10th stage from the top (see the graph) Side stream location: 18th stage from the top (see the graph) g) Number of equilibrium stages: 20 stage + one reboiler that acts as an equilibrium stage.

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DEPARTMENT OF CHEMICAL & MATERIALS ENGINEERING

ChE 416

Equilibrium Stage Processes

Winter 2017

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DEPARTMENT OF CHEMICAL & MATERIALS ENGINEERING

ChE 416

Equilibrium Stage Processes

Winter 2017

Solution of Q3 Let us consider a total condenser and a partial reboiler for the cases stated in the problem. a) Let use first consider case 1 with one feed stream. Case 1 Mixing two steams give a new feed and composition.

From a mass and component balance we get, F = 200 mol/h (F = F1 + F2 = 200) and z FF = z 1F 1 + z 2F 2 zF = 0.45 Now consider this is input as feed of one single steam. First, find D and B Overall mass balance: F = D + B ---------------- (1) 200 = D + B Component balance: zFF = xDD + xBB ----------------- (2) 0.45×200 = 0.95D + 0.05B 90 = 0.95D + 0.05B Solving simultaneous equations

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DEPARTMENT OF CHEMICAL & MATERIALS ENGINEERING

ChE 416

Equilibrium Stage Processes

Winter 2017

we get, D = 88.89 mol/h B = 111.11 mol/h Given that R=L/D = 2 Because of saturated liquid feed, q = 1 and q-line is vertical at y= x= zF = 0.45 R- line:

The slope of R-line is L/V or R/(R+1). It intersects y-axis (x = 0) at y = xD/(R+1) Intercept, xD/(R+1) = 0.95/(2+1) = 0.316 So, we can draw top operating line from point y = x = xD = 0.95 and using slope or the y-intercept, 0.316. This line will cut q-line and S-line can be drawn by adding the point of intersection of R-line and q-line and the point y = x= xB = 0.05.

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DEPARTMENT OF CHEMICAL & MATERIALS ENGINEERING

ChE 416

Equilibrium Stage Processes

Winter 2017

Total number of stages = 5.5 stages + 1 stage for partial reboiler acting as an equilibrium stage b) Now, let us consider case 2 with two feed streams. i) Find D and B Overall mass balance: F1 + F2 = D + B ---------------- (1) 100 + 100 = D + B D + B = 200

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DEPARTMENT OF CHEMICAL & MATERIALS ENGINEERING

ChE 416

Equilibrium Stage Processes

Winter 2017

Component balance: z1F1 + z2F2 = xDD + xBB ----------------- (2) 0.5×100+ 0.4×100 = 0.95D + 0.05B 90 = 0.95D + 0.05B Solving simultaneous equations we get, D = 88.89 mol/h B = 111.11 mol/h ii) Calculation of flow rates in all three sections Top Section: Reflux ratio, L/D = 2 So, L = 2D L = 2×88.89 = 177.78 mol/h From the condenser balance, V=L+D V = 177.78 + 88.89 = 266.67 mol/h Intermediate section: Consider the following diagram for balancing liquid and vapor in the top feed tray:

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DEPARTMENT OF CHEMICAL & MATERIALS ENGINEERING

ChE 416

Equilibrium Stage Processes

Winter 2017

L = L + LF1 = 177.78 + 100 = 277.78 V = V + VF1 V = V – VF1 = 266.67 – 0= 266.67 [Since F1 is saturated liquid, there is no vapor part in the feed. So, LF1 = F1 = 100 and VF1 = 0] Bottom section: = 277.78 +100 = 377.78 = 266.67 - 0 = 266.67 [Since F2 is also saturated liquid, there is no vapor part in the feed. So, LF2 = F2 = 100 and VF2 = 0] iii) Lets do a component balance around D and F1 of the distillation column.

Vʹy + F1zF1 = Lʹx +DxD So, intermediate operating line: We need to determine L and V for calculating slope

and intercept

.

Let’s calculate the slope and the intercept:

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DEPARTMENT OF CHEMICAL & MATERIALS ENGINEERING

ChE 416

Slope, =

Equilibrium Stage Processes

Winter 2017

= 1.04

= = 0.13 So the equation of intermediate operating line:

Intercept

iv) Construction of q-lines and all other operating lines: q-lines: 1st feed goes through y = x = zF = 0. 5 Feed 1 (saturated liquid, so q = 1) has a vertical q1-line. 2nd feed goes through y=x=xF = 0.4 Feed 2 (saturated liquid, so q = 1) has a vertical q2- line. Top Line (R-Line) Given, xD = 0.95 R = L/D = 2 Top operating line:

The slope of top operating line is L/V or R/(R+1). It intersects y-axis (x = 0) at y = xD/(R+1) Intercept, xD/(R+1) = 0.95/(2+1) = 0.316 So, we can draw top operating line from point y = x = xD = 0.95 and using slope or the y-intercept, 0.316. I-Line: The equation of intermediate operating line:

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DEPARTMENT OF CHEMICAL & MATERIALS ENGINEERING

ChE 416

Equilibrium Stage Processes

Winter 2017

Operating line passes through the intersection point of the top-line and q-line for F1 intersection and the point of intersection of y-axis (x=0) at y = = 0.13. The line intersects the horizontal q-line of F2. Bottom-Line (S-Line): So, bottom-line intersects the point of intersection of q-line for F2 and intermediate op. line. We can draw bottom-line that passes through the point x= xB = 0.05 on the 45° line and the point of intersection of q-line for F2 and the intermediate operating line.

Total number of stages = 5.5 stage + 1 stage for partial reboiler acting as an equilibrium stage c) Comparing two cases, we see that there is no change in the top operating line (since R is same for both cases) and for the given problem the required numbers of stages seems to be same for each case.

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DEPARTMENT OF CHEMICAL & MATERIALS ENGINEERING

ChE 416

Equilibrium Stage Processes

Winter 2017

So, considering only number of stages, it is difficult to make any comment. However, from the argument of entropy, feed with two streams is preferable.

Solution of Q4 Given: Feed: F = 250 mol/h zF = 0.80 Feed condition: Super-heated vapor Distillate: yD = 0.96 Partial condenser acts as a stage. Bottoms: xB = 0.1 Open steam. So, no reboiler.

Feed is a super-heated vapor where 2 moles of feed vaporize 1 mole of liquid. Amount vaporized = v = (1/2) F

We know, Then, So, Slope of q-line : Intersection: y = x = zF = 0.80

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DEPARTMENT OF CHEMICAL & MATERIALS ENGINEERING

ChE 416

Equilibrium Stage Processes

Winter 2017

q-line can be drawn from the intersection point with a slope of 0.33. Alternatively, Equation of q-line is given as

At x = 0, y = 0.533 Draw q-line using the points y = 0.533 at x = 0 and x =z F= 0.8 on diagonal line that intersects the equilibrium curve. For partial condenser, R-line: Given: R = 1.84 We can get the y intercept, yD/(R+1) and draw R line using this intercept and the point x = y = yD = 0.96. yD/(R+1) = 0.96/(1+1.84) = 0.338 R-line has been depicted in the diagram below. Intersection of R-line and q-line give a point for S-line and the other point is at x = x B = 0.1 at y = 0. Draw S-line using these two points. Once operating lines are fixed, steps can be drawn (see the diagram)

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DEPARTMENT OF CHEMICAL & MATERIALS ENGINEERING

ChE 416

Equilibrium Stage Processes

Winter 2017

From the McCabe-Thiele diagram, a) Optimum feed stage: Stage 3 below the partial condenser. b) Number of equilibrium stages: 4.8 stages + 1 Partial condenser acting as an equilibrium stage c) Find the liquid compositions leaving 4th stage: For a binary distillation column, leaving streams from an equilibrium stage are in equilibrium. So the streams leaving form 3rd stages must be on equilibrium curve. From the diagram above, we find: Composition of liquid stream from the 4th stage: 0.305 d) Determine steam rate in kg/h and distillate rate in kmol/h

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DEPARTMENT OF CHEMICAL & MATERIALS ENGINEERING

ChE 416

Equilibrium Stage Processes

Winter 2017

From overall mass balance: F+S = D+B 250 + S = D + B D + B - S = 250 -------------------- (1) From component balance: FzF + Sys = DxD +BxB 250×0.8 + 0 = 0.96D + 0.1B [since steam is pure so, ys = 0] 0.96D + 0.1B = 200 -----------------(2) From the diagram, Slope of bottom operating line: / = 1.42 However, for open steam,

= S and

=B

So, B/S = 1.42 --------------------------- (3) Solving the above 3 equations we get, D = 185.75 mol/h = 0.18575 kmol/h and S = 152.53 mol/h = 0.15253 kmol/h = 0.15253 kmol/h ×18 kg/kmol = 2.74554 kg/h [Note that instead of 3rd equation from the graph another equation can be obtained by simply doing an internal mole balance as follows: V = L+D = (R+1)D

= V-VF = (...