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Quiz 11 April 9, 2020 Chemical Engineering Thermodynamics

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Quiz 11 April 9, 2020

Chemical Engineering Thermodynamics

Vacuum distillation of ethanol from an ethanol (1)/water (2) mixture can lead to a lower energy load and a higher concentration of ethanol compared to distillation at atmospheric pressure. Data for the equilibrium concentrations of vapor and liquid ethanol at 190 mmHg are given in the attached excel sheet. [Beebe A.H.; Coulter K.E.; Lindsay R.A.; Baker E.M. Equilibria in Ethanol-Water System at Pressures Less Than Atmospheric. Ind.Eng.Chem. Ind.Ed. 34 1501- 1504 (1942).]

a) Use this data to obtain the one-parameter Margules coefficient by adding to the table columns for Psat,1, Psat,2, g1, g2, calculated ycalc,1 for the bubble point, and the calculated P values, the calculated (Pcalc-P)2, and a cell containing the sum of the (Pcalc-P)2 values.

b) Using solver find the minimum of the sum of (Pcalc-P)2 by varying the Margules coefficient, A12 (P is 190mmHg). (This is the least-squares method.)

After solving for A12 make a plot of ycalc,1 and y1 versus x1. Why do y1calc and y1 disagree?

c) Use the Margules coefficient to calculate the dew pressure, Pdew, for T = 50.5°C;

y1 = 0.6925.

d) Does an azeotrope exist at T = 50.5°C for this system (use the Margulis model)?

At the azeotrope x1 = y1, and x2 = y2.

Using expressions for y1 in terms of x1, for y2 in terms of x2, and x1 + x2 =1, solve for x1,azeotrope at the azeotrope for T = 50.5°C.

Use this value of x1,azeotrope to calculate Pazeotrope at the azeotrope.

Calculate y1,azeotrope at the azeotrope.

Is this a maximum boiling or a minimum boiling temperature azeotrope (remember P vs x1 and T vs x1 plots are different)?

e) Make a scatter plot of T versus x1 and T versus y1 on the same chart with the x/y range 0 to 1 and the T range from 45 to 65 °C. On the same plot add your T versus ycalc,1. Does this plot support your prediction of an Azeotrope?

3.49

77.2 mmHg

0.422

173 mmHg

0.422

Minimum Boiling Temperature

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ANSWERS a)

b)

0.6 0.7 0.8 0.9 1

y vs x

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y1calc and y1 disagree because this is optimized for pressure not y. If we had used (y-ycalc)2 for the least squares optimization this plot would look better but P would be wrong.

c)

d) x1 = y1 P/(Psat1g1) and x1 = y1 at the azeotrope so, P = Psat1g1 = Psat2g2

g1 = exp(x22A12)

Psat1/Psat2 = g2/ g1 = exp((x22-x12)A12) = exp((1-2x12)A12) Solve for x1; calculate P; calculate y1

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e)

45 47 49 51 53 55 57 59 61 63 65

0 0.2 0.4 0.6 0.8 1

x1 y1 y1calc

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