Assignment 1 (2017 winter) with solutions PDF

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DEPARTMENT OF CHEMICAL & MATERIALS ENGINEERING

ChE 416

Equilibrium Stage Processes

Winter 2017

Assignment 1 (Due 4:00pm, Jan. 20, 2017) Q1 (4+9+2=15) Equilibrium ratios, K Experimental measurements for the propane-isopentane system at T = 75 °C and P =10.135 bar show liquid phase propane mole fraction to be x1 = 0.29, while the vapor phase mole fraction is y1 = 0.665. (a) Determine the experimental K-values and AB of the binary system. (b) Estimate the K-value and A,B i) Using Raoult’s law. The Antoine’s equation and the values of the constants for propane and isopentane are given below.

Table 1: Antoine’s constant for n-heptane and toluene, Psat is in [bar] and T is in [°K] Propane

isopentane

A

4.53678

3.97183

B

1149.36

1021.864

C

24.906

-43.231

ii) Using nomograph shown in Wankat 2007 (Figure can also be found below) iii) Using nomograph shown in Henely and Seader (Figure can also be found below) (c) Compare results in (a) and (b-(i)). Q2 (20+10 = 30) Ideal Gas-Liquid System: The Antoine’s equation and the values of the constants for n-heptane and toluene are given below.

Table 1: Antoine’s constant for n-heptane and toluene, Psat is in [mmHg] and T is in [°C] n-heptane

Toluene

A

6.90253

6.95805

B

1267.828

1346.773

C

216.823

219.693

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DEPARTMENT OF CHEMICAL & MATERIALS ENGINEERING

ChE 416

Equilibrium Stage Processes

Winter 2017

Consider a two-phase binary system of n-heptane and toluene. Assume that the both liquid and vapor phases behave ideally, assuming Raoult’s law is valid. a) Prepare and plot y-x, and T-x-y diagrams at 101.3 kPa, where x and y are mole fractions of the more volatile component. For vapor pressure of pure component, use Antoine equation b) From the following experimental data, plot the experimental y-x, and T-x-y diagrams on the same figure as part (a). Then compare the experimental y-x and T-x-y diagrams with those from Raoult’s law in part (a), and comment on the accuracy of the Raoult’s law for n-heptane and toluene system.

Table 2: VLE experimental data for n-heptane + toluene mixture at 1 atm. x & y are mole fractions. Ref: Steinhauser and White [Ind. Eng. Chem., 41, 2912 (1949)] xn-heptane

y n-heptane

T [°C]

0.03

0.05

110.75

0.13

0.21

106.80

0.25

0.35

104.50

0.35

0.45

102.95

0.50

0.58

101.35

0.69

0.74

99.73

0.84

0.86

98.90

0.94

0.95

98.50

0.99

0.99

98.35

Q3 (8+8+4=20) Using Henry’s constant Henry's law constant for the solubility of CO in water at 25 °C is 6.66×104 atm/mole fraction. Saturation pressure of H2O at 25 °C is 0.031316 atm. At 2 atm, determine the followings: (a) The equilibrium ratio of H2O, CO. (b) The values of xH2O, xCO, yH2O and yCO (c) Is it a viable option to remove CO from flue gas by using water? Comment briefly.

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DEPARTMENT OF CHEMICAL & MATERIALS ENGINEERING

ChE 416

Equilibrium Stage Processes

Winter 2017

Q4 (5×5 + 10= 35) Inverse Liver Rule i) Using T-y-x diagrams in Figure 1, determine the temperature, amounts and compositions of the vapor and liquid phases at 101 kPa for the following conditions with a 100 kmol mixture of nC6 (H) and nC8 (O). (a) zH = 0.52,  = V/F = 0.8; (b) zH = 0.3, yH = 0.56; (c) z H = 0.4, =0; (d) zH=0.4, =1.0; and (e) zH = 0.46, T = 205 °F [Note: 4i (a) might require trial and error] ii) For a feed composition z H= 0.4, describe the states of this mixture as temperature is changed from 170 ºF to 270 ºF. Also please describe the condition as it crosses the bubble and dew point curves.

Figure 1. T-x-y phase equilibrium diagram for the n-hexane-n-octane sytem at 1 atm

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DEPARTMENT OF CHEMICAL & MATERIALS ENGINEERING

ChE 416

Equilibrium Stage Processes

Winter 2017

4

DEPARTMENT OF CHEMICAL & MATERIALS ENGINEERING

ChE 416

Equilibrium Stage Processes

Winter 2017

K-Values for Hydrocarbons (Henely & Seader Fig 2.4) by Hadden & Grayson 1961

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DEPARTMENT OF CHEMICAL & MATERIALS ENGINEERING

ChE 416

Equilibrium Stage Processes

Winter 2017

Solution of # 1 a) Experimental values of equilibrium ratios: K1 = y1/x1 = 0.665/0.29 = 2.293 K2 = y2/x2 = (1-0.6650)/ (1-0.29) = 0.335/0.71 = 0.472 α12 = K1/K2 = 2.293/0.472 = 4.86 b) From Raoult’s law: i) Given: T = 75 °C and P =10.135 bar

From Antoine equation, for pure propane

Psat = 28.566 bar

for pure isopentane:

Psat = 4.174 bar K1 = P1s/P = 28.566/11.51 = 2.482 K2= P2s/P = 4.174/11.51 = 0.363 α1,2 = K1/K2 = 2.786/0.399 = 6.837 ii) Given: T = 75 °C and P =10.135 bar P = 10.135 bar = 1013.5 kPa

From the graph (Wankat 2007, Fig 2-11): K1 = 2.4 K2 = 0.455 α1,2 = K1/K2 = 2.4/0.455 = 5.27

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DEPARTMENT OF CHEMICAL & MATERIALS ENGINEERING

ChE 416

Equilibrium Stage Processes

Winter 2017

iii) Given: T = 75 °C and P =10.135 bar T = 75 °C = 167 °F P =10.135 bar = 147 psia From the graph (Henely & Seader Fig 2.4): K1 = 2.35 K2 = 0.44 α1,2 = K1/K2 = 2.35/0.44 = 5.34 7

DEPARTMENT OF CHEMICAL & MATERIALS ENGINEERING

ChE 416

Equilibrium Stage Processes

Winter 2017

8

DEPARTMENT OF CHEMICAL & MATERIALS ENGINEERING

ChE 416

Equilibrium Stage Processes

Winter 2017

(c) Comparison of experimental value of α1,2 with that from Raoult’s law shows a disagreement of ~41%. Hence, ideal behavior can often lead to highly erroneous results. Use of Process Simulator will increase the accuracy for such a system.

Solution of # 2 Let n-heptane and Toluene be component 1 and 2, respectively. (a) For T-x-y, consider a temperature range starting from the boiling point of pure n-heptane (more volatile component) up to the boiling point of the less volatile component. To find the boiling point of pure component at P=101.3kpa (760 mmHg), substitute Psat with 760 mmHg, and calculate T from Antoine equation, with the corresponding Antoine coefficient for each component. From Antoine equation, for pure n-heptane (eq 1) for pure Toluene: (eq 2)

Again, by rearranging, for n-heptane: (eq 3) For Toluene: (eq 4) Within range of 98.4 to 110.6 C, choose some arbitrary temperatures. For each T, calculate P1sat and P2sat from Antoine equation (eq 1 and 2, respectively). To obtain xi and yi, start with Raoult’s Law ( ) for component 1 and 2. Add the equations together. Hence: x1Ps1 + x2Ps2 = (y1 + y2) P

(eq 5)

Also x1+x2=1, and y1+y2=1, combined with (eq 5): x1Ps1 + (1-x1)Ps2 = P

(eq 6)

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DEPARTMENT OF CHEMICAL & MATERIALS ENGINEERING

ChE 416

Equilibrium Stage Processes

Rearranging that to obtain x1: x1 = (P- Ps2)/(Ps1- Ps2)

Winter 2017

(eq 7)

For y1, according to Raoult’s law: y1 = x1Ps1/P

(eq 8)

Look at the following table for the results for various temperatures: Table. Calculated x1 and y1 for each T T (°C) 98.42 99 100 101 102 103 104 105 106 107 108 109 110 110.6

Ps1 (Kpa) 101.29 103.03 106.08 109.20 112.39 115.66 118.99 122.41 125.90 129.46 133.10 136.83 140.63 ------

Ps1 (Kpa) -----71.93 74.15 76.43 78.76 81.15 83.60 86.10 88.66 91.29 93.97 96.71 99.52 101.24

x1 1.00 0.94 0.85 0.76 0.67 0.58 0.50 0.42 0.34 0.26 0.19 0.11 0.04 0.00

y1 1.00 0.96 0.89 0.82 0.74 0.67 0.59 0.51 0.42 0.34 0.25 0.15 0.06 0.00

Figure. T-x-y diagram for problem 2

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DEPARTMENT OF CHEMICAL & MATERIALS ENGINEERING

ChE 416

Equilibrium Stage Processes

Winter 2017

Figure. x-y diagram for problem 2

b)

Figure. T-x-y experimental data versus Raoult’s law

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DEPARTMENT OF CHEMICAL & MATERIALS ENGINEERING

ChE 416

Equilibrium Stage Processes

Winter 2017

Figure. x-y experimental data versus Raoult’s law Although Raoult’s law can predict the composition of liquid phase and vapor phase, it results in error when predicting the temperature of the vapor-liquid equilibria. This tells us that we should always look into the T-x-y diagram to confirm Raoult's law's applicability. Solution of # 3 Henry's law constant for the solubility of CO in water at 25 °C is 6.66×104 atm/mole fraction. Saturation pressure of H2O at 25 °C is 0.031316 atm. At 2 atm, determine the followings: (d) The equilibrium ratio of H2O, CO. (e) The values of xH2O, xCO, yH2O and yCO Given that P = 2 atm a) At 40 C KH2O = PsH2O /P = 0.031316 /2= 0.015658 KCO = HCO/P = 6.66×104 /2=33300 b) At 40 C xCO = (1- KH2O)/( KCO- KH2O) = (1- 0.015658)/( 33300 - 0.015658) = 0.000029559 xH2O = (1- xCO) = (1- 0.000029559) = 0.99997044 y CO = KCO× x CO = 33300 × 0.000029559 = 0.9843147 yH2O = (1- yCO) = (1- 00.9843147) = 0.0156853

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DEPARTMENT OF CHEMICAL & MATERIALS ENGINEERING

ChE 416

Equilibrium Stage Processes

Winter 2017

c) The liquid fraction of CO shows that the CO is barely soluble to water at the given temperature and pressure. So it is not viable option to remove CO from flue gas by using water.

Solution of # 4

Given: T-x-y diagram in figure 1 and 100 kmol mixture. So, F = 100 kmol. Let zH = mole fractions of n-hexane in the feed and ψ = V/F. Using inverse lever-arm rule as displayed by Line DEF in Fig 1, Fig 2 is depicted. The results for parts (a) through (e) are as follows:

Given a) zH = 0.52, ψ = 0.8 b) zH = 0.3, yH = 0.56 c) zH = 0.4, ψ = 0 d) zH = 0.4, ψ = 1 e) zH =0.46 , T =205 °F

T, F 219 225 198 137.5 205

V, kmol 80 33.33 0 100 31.03

yH 0.62 0.56 0.786 0.4 0.74

xH 0.22 0.17 0.4 0.90 0.334

Note that (a) need trial and error in order to match ψ = V/F = 0.8 = 4/5. We have to find a tie line through zH = 0.52 so that we get, DE/DF = 4/5

Figure 1. T-x-y phase equilibrium diagram for the n-hexane-n-octane sytem at 1 atm

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DEPARTMENT OF CHEMICAL & MATERIALS ENGINEERING

ChE 416

Equilibrium Stage Processes

Winter 2017

Figure 2. T-x-y phase equilibrium diagram for the n-hexane-n-octane sytem at 1 atm

b) From 170 °F as we move up till we reach bubble point curve the state of the mix is sub-cooled liquid. At liquid curve it is saturated liquid. Between saturated liquid curve and saturated vapor curve it is a mix of vapor and liquid, which separates into vapor phase and liquid phase in equilibrium as given by a tie-line. At vapor curve it is saturated vapor. From vapor curve as we move upwards the mixture is superheated vapor.

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