Assignment 2 (2017 winter) with solutions PDF

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DEPARTMENT OF CHEMICAL & MATERIALS ENGINEERING

ChE 416

Equilibrium Stage Processes

Winter 2017

Assignment 2 (Due 4:00pm, Feb. 01, 2017) Q1. Multicomponent Flash (12+13) A liquid mixture consisting of 100 kmol of 60 mol% benzene, 25 mol% toluene, and 15 mol% o-xylene is flashed at following conditions: i) 100 C and 1 atm ii) 100 °C and 2 atm iii) 105 °C and 0.1 atm iv) 150 °C and 1 atm Assume ideal solutions and use saturation pressure for these components from the following Table. Saturated Vapor Pressure, Pis (atm) Temperature 100°C 105°C 150°C

Benzene

Toluene

o-xylene

1.79 2.04 5.79

0.74 0.86 2.76

0.26 0.31 1.17

a) Identify the state of the feed mixture after flashing (i.e., subcooled, two phase or superheated) at each conditions mentioned above. b) Compute the amounts of liquid and vapor products and their compositions for the feed that formed two phase at the above conditions.

Q2. Multicomponent flash distillation (25) A feed that is 45.0 mol% n-butane, 35.0 mol% n-pentane, and 20.0 mol% n-hexane is fed at a rate of 1.0 kmol/min to a flash drum operating at 200 kPa. What is the highest temperature at which the flash drum can operate (still have vapor and liquid present)? Use the DePriester chart for equilibrium data.

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DEPARTMENT OF CHEMICAL & MATERIALS ENGINEERING

ChE 416

Equilibrium Stage Processes

Winter 2017

Figure: DePriester chart for equilibrium data (Wankat, 2007)

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DEPARTMENT OF CHEMICAL & MATERIALS ENGINEERING

ChE 416

Equilibrium Stage Processes

Winter 2017

Q3. Gas Liquid System (25) Plot the solubility data into solute free fractions in gas and liquid phases. A vapor mixture having equal volumes of NH3 and N2 is to be contacted at 20°C and 1 atm with water to absorb a portion of the NH3. If 14 m3 of this mixture is brought into contact with 10 m3 of water and if equilibrium is attainted, calculate the percent of the ammonia originally in the gas that will be absorbed. Both temperature and total pressure will be maintained constant during the absorption. The partial pressure of NH3 over water at 20 °C is as follows:

Q4. Gas Liquid System (15+10) Air is used to strip off some organic pollutant from liquid stream containing 1 mol of solute per 100 mole water. The air stream used for stripping is free of the solute. a) In a real operation, 15 moles of pure air is used per mole solute free water. What will be the mole fraction of the organic pollutant in the effluent water and gas stream? (G-L equilibrium data for this system is given in the following chart.) What % of organic pollutant is stripped off in this operation? b) How many moles of air will be needed per mole of solute free liquid to reduce the concentration of the organic component to water to 5% of the original concentration in water? In this part please use the equilibrium compositions given by Y = 0.1X

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DEPARTMENT OF CHEMICAL & MATERIALS ENGINEERING

ChE 416

Equilibrium Stage Processes

Winter 2017

Solutions: Solution of Q1. a) Note that for the existence of two phase we have to meet the following conditions: Essential conditions: Some Ki < 1 and other Ki > 1 Sufficient conditions: ∑(ziKi) > 1 (to be above bubble point ) and ∑(zi/Ki) > 1 (to be below dew point) Let’s find the K values for each condition and check if the essential and sufficient conditions are satisfied. Results are tabulated below:

Flash conditions i) T = 100 °C, P = 1 atm ii) T = 100 °C, P = 2 atm iii) T = 105°C, P = 0.1 atm iv) T = 150 °C, P = 1 atm

State of flashed feed K2 K3 ziKi zi/Ki 1.79 0.74 0.26 1.298 1.24996 Two phases Subcooled 0.895 0.37 0.13 0.649 2.49991 (∑ziKi < 1) Superheated 20.4 8.6 3.1 14.855 0.10687 (∑zi/Ki < 1) Superheated 5.79 2.76 1.17 4.3395 0.32241 (∑zi/Ki < 1)

K1

From the above table, it can be seen that only the first flash condition (T = 100 °C, P = 1 atm ) satisfied both essential and sufficient conditions in order to form two phase. b) Let’s compute the amounts of liquid and vapor products and their compositions for the feed at T = 100 °C, P = 1 atm

Calculate f(ψ) from the Rachford-Rice (R-R) equation: -------- (1) We are looking for a value of V/F for which f(ψ) = 0 or close to zero. A continuous guess of ψ is required, however, a higher number of trials might be required if the first guess is very poor.

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DEPARTMENT OF CHEMICAL & MATERIALS ENGINEERING

ChE 416

Equilibrium Stage Processes

Winter 2017

After a several guess of ψ, we got the following values of f(ψ). 0.4 0.6 0.7 0.8

-0.13 -0.045 0.0045 0.051

From the plot of ψ vs f(ψ), let’s estimate the root.

From the above figure, we get ψ = 0.69 for which f(ψ) = 0 or close to zero. Now we can calculate xi and yi from the following equations:

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DEPARTMENT OF CHEMICAL & MATERIALS ENGINEERING

ChE 416

Equilibrium Stage Processes

Winter 2017

Sample calculations:

Since F = 100 kmol V = 100 × 0.693 = 69.3 kmol L = (100 – 69) = 30.7 kmol

Check xi and yi if for the accuracy. Here we got, xi = 1.0005 and yi = 0.9997 The values are close to unity.

Solution of Q2.

Highest temperature that two phase exist is the dew point. We know, Rachford-Rice (R-R) equation:

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DEPARTMENT OF CHEMICAL & MATERIALS ENGINEERING

ChE 416

Equilibrium Stage Processes

Winter 2017

=0 So, at dew point: ψ = V/F = 1; f(1) = 0; so, Σ (zi/Ki) = 1 In order to find Tdew: Step 1: Guest T Step 2: Read Ki from DePriester chart at TGuess and P = 200 kPa Step 3: Check Σ (zi/Ki) = 1. Yes  record the dew point temperature No  Follow step 1 by guessing a new T until we get, Σ (zi/Ki) = 1. Initial guess: TGuess = Σ zi Tiboil For Tiboil (boiling T of pure i)  yi = 1 and xi = 1, So Ki = yi/xi = 1 By using DePriester chart at P = 200 kPa and Ki, we get TnC4boil = 19 C TnC5boil = 58 C TnC6boil = 94 C TGuess, initial : 0.45 ×19 + 0.35×58 + 0.20×94 = 48 C From DePriester chart at P = 200 kPa and TGuess: T 48 60 59

∑zi/Ki KnC4 KnC5 KnC6 2.3 0.76 0.29 1.35 2.9 1.05 0.42 0.96 2.9 1 0.4 1.005

So, Tdew = 59 C

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DEPARTMENT OF CHEMICAL & MATERIALS ENGINEERING

ChE 416

Equilibrium Stage Processes

Winter 2017

Solution of Q3.

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DEPARTMENT OF CHEMICAL & MATERIALS ENGINEERING

ChE 416

Equilibrium Stage Processes

Winter 2017

Solutions of Q4

Entering Polluted Liquid: Liquid stream is containing 1 mole of solute per 100 mole water. So, Xo = 0.01moles solute/mole liquid

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DEPARTMENT OF CHEMICAL & MATERIALS ENGINEERING

ChE 416

Equilibrium Stage Processes

Winter 2017

Entering Stripping Gas: Pure air is using for refining the liquid. So, y0 = Y0 = 0 a) Xo = 0.01 and Y0 = 0 Slope: It is stated that 1 mole of solute free liquid requires 15 moles of air. So, Slope = - Lʹ/Gʹ = -1/15 = - 0.06667

From component balance: LXo + GYo = LX1 + GY1

Slope = - L/G = -1/15 Graphical solution:

We can draw the operating line, using the feed point (Xo,Yo) and the slope (-1/15)

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DEPARTMENT OF CHEMICAL & MATERIALS ENGINEERING

ChE 416

Equilibrium Stage Processes

Winter 2017

The operating line FA intersects the equilibrium curve at A. The coordinates of A (X 1, Y1) represent the compositions of the resulting phases. From the graph, Y1 = 0.0004 and X1 = 0.004 So the mole fraction of the organic pollutant in the effluent water: x1= X1/(1+X1) = 0.004/(1+0.004) = 0.00398 and the mole fraction of the organic pollutant in the gas stream y1= Y1/(1+Y1) = 0.0004/(1+0.0004) = 0.0003998 Percentage removal of organic pollutant = (Xo – X1)/Xo ×100% = (0.01 – 0.004)/0.01 ×100% = 60%

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DEPARTMENT OF CHEMICAL & MATERIALS ENGINEERING

ChE 416

Equilibrium Stage Processes

Winter 2017

b) Xo = 0.01 mole solute /mole H2O and Y0 = 0 X1 = 0.05× Xo = 0.05× 0.01 = 0.0005 mole solute /mole H2O The equation of equilibrium curve is, Y= 0.1X. So, we can get, Y1 = 0.1X1 = 0.1× 0.0005 = 0.00005 mole solute/mole air From the mass balance, we got the following equation.

If we can find out the slope = - L/G, we can get the moles of air will be needed per mole of solute free liquid to reduce the concentration of the organic component.

So, 190 moles of pure air is required per mole solute free liquid (water).

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