# Che425: Problem set #3

Download 44.83 Kb.
 Date 15.05.2016 Size 44.83 Kb.
CHE425: Problem set #3
1. Consider the equilibrium stage with C components shown below. Conduct a degree-of-freedom analysis by performing the following steps: (a) list and count the variables; (b) write and count the equations relating the variables; and (c) calculate the degree of freedom. Analysis: (a) Number of variables = 7C + 22
(b) The equations are:

Total number of equations = NE = 3C + 12

(c) Degrees of freedom = 4C +10

2. Consider an adiabatic equilibrium flash shown below with all the indicated variables. (a) Determine the number of variables. (b) Write all the independent equations that relate the variables. (c) Determine the number of equations. (d) Determine the number of degrees of freedom. (e) What variables would you prefer to specify in order to solve an adiabatic-flash problem? Analysis: a) Variables are those appearing in Figure 4.10a

NV = 3C + 9

(b) C Component material balances

(c) NE = 2C + 6

(d) ND = C + 3

(e) Specify the feed completely ( feed rate, temperature, pressure and C - 1 mole fractions) plus exiting pressure.

3) Determine the number of degrees of freedom for a nonadiabatic equilibrium flash for the liquid feed and three products shown below. Number of degrees of freedom = C + 4

4. For the seven-phase equilibrium system shown below, assume air consists of N2, O2, and argon. What is the number of degrees of freedom? What variables might be specified? Note: all species exist in all phases. Number of degrees of freedom = 4

Specify T, P, and mole fractions of argon and oxygen in the air.

5.2 A liquid mixture containing 25 mol% benzene and 75 mol% ethyl alcohol, in which components are miscible in all proportions, is heated at a constant pressure of 1 atm from 60oC to 90oC. Using the spline command in Matlab to plot the following T-x-y experimental data:

 y, x = mole fraction of benzene in vapor and liquid phase, respectively T,oC 78.4 77.5 75 72.5 70 68.5 67.7 68.5 72.5 75 77.5 80.1 y 0 0.075 0.28 0.42 0.54 0.60 0.68 0.73 0.82 0.88 0.95 1.0 x 0 0.015 0.05 0.12 0.22 0.31 0.68 0.81 0.91 0.95 0.98 1.0

Note: the following Matlab codes can plot the data using spline command:

T=[78.4 77.5 75 72.5 70 68.5 67.7 68.5 72.5 75 77.5 80.1];

y=[0 0.075 0.28 0.42 0.54 0.60 0.68 0.73 0.82 0.88 0.95 1.0];

x=[0 0.015 0.05 0.12 0.22 0.31 0.68 0.81 0.91 0.95 0.98 1.0];

ppx=spline(x,T);

ppy=spline(y,T);

xp=0:0.02:1;yp=xp;

Tx=ppval(ppx,xp);

Ty=ppval(ppy,yp);

plot(xp,Tx,yp,Ty)

grid on

Title('Txy data for benzene-ethyl alcohol at 1 atm')

xlabel('x,y: mole fraction benzene')

ylabel('T(^oC)')

Determine: (a) the temperature where vaporization begins; (b) the composition of the first bubble of vapor; (c) the composition of the residual liquid when 25 mol% has evaporated, assuming that all vapor formed is retained in the apparatus and is in equilibrium with the residual liquid. (d) Repeat part (c) for 90 mol% vaporized. (e) Repeat part (d) if, after 25 mol% is vaporized as in part (c), the vapor formed is removed and an additional 35 mol% is vaporized by the same technique used in part (c).

Analysis: See plot of T-x-y data on next page, as drawn with a spreadsheet. Curved (instead of straight) lines connecting the points would be a good improvement.

(a) For a benzene mole fraction of 0.25, a vertical line from M intersects the liquid line at N at 69.4oC, which is the bubble point.

(b) The benzene mole fraction in the vapor at 69.4oC, obtained from the left-most vapor line at P, is 0.56. (c) To find the benzene mole fraction in the liquid at 25 mol% vaporization, extend a dashed, vertical line upward from the bubble point at N, as shown in the figure on the next page, until point B is reached. At this point, using the inverse lever-arm rule, the ratio of the AB line length to the BC line length is 25/75. The benzene mole fraction in the equilibrium liquid at point A is 0.175 at a temperature of 71.2oC.

(d) To find the benzene mole fraction in the liquid at 90 mol% vaporization, extend a dashed, vertical line upward from the bubble point, as shown in the figure on the next page, until point E is reached. At this point, using the inverse lever-arm rule, the ratio of the DE line length

to the EF line length is 90/10. The benzene mole fraction in the equilibrium liquid at point D is 0.045 at a temperature of 75.1oC.

(e) To find the benzene mole fraction in the liquid when the liquid from part (c) is removed from the vapor and further vaporized, proceed as follows. If we start with 100 moles, then the liquid at 25 mol% vaporized is 75 moles with a benzene mole fraction of 0.17. If an additional 35 mol% is vaporized, we will have 35 moles of vapor in equilibrium with 40 moles of liquid. Therefore, extend a dashed, vertical line upward from point A, as shown in the figure on the next page, until point H is reached. At this point, using the inverse lever-arm rule, the ratio of the GH line length to the HI line length is 35/40. The benzene mole fraction in the liquid at G is 0.05 at a temperature of 74.9oC. 6.2 Stearic acid is steam distilled at 200oC in a direct-fired still. Steam is introduced into the molten acid in small bubbles, and the acid in the vapor phase leaving the still has a partial pressure equal to 70% of the vapor pressure of pure stearic acid at 200oC. Use Matlab to plot the kg acid distilled per kg steam added as a function of total pressure from 101.3 kPa to 3.3 kPa at 200oC. Label the plot with your name using the Title command. The vapor pressure of stearic acid at 200oC is 0.40 kPa.

 P, kPa kg A/kg B 101.3 0.0438 75 0.0592 50 0.0890 25 0.1790 15 0.3006 10 0.4553 5 0.9376 3.3 1.4650 7. The relative volatility, , of benzene to toluene at 1 atm is 2.5. Construct x-y and T- x-y diagrams for this system at 1 atm. On the same diagrams, plot the data obtained using the following vapor pressure data and ideal solution:

 TA, TB = temperature of benzene and toluene at the given vapor pressure, respectively P, torr 20 40 60 100 200 400 760 1,520 TA, OC  2.6 7.6 15.4 26.1 42.2 60.6 80.1 TB, OC 18.4 31.8 40.3 51.9 69.5 89.5 110.6 136

Use the diagram for the following: (a) A liquid containing 70 mol% benzene and 30 mol% toluene is heated in a container at 1 atm until 25 mol% of the original liquid is evaporated. Determine the temperature. The phases are then separated mechanically, and the vapors condensed. Determine the composition of the condensed vapor and the liquid residue. (b) Calculate and plot the K-values as a function of temperature at 1 atm.

Note: Use Matlab to plot the diagrams and label the figures with your name using the Title command. T, oC Ps of A, torr Ps of B, torr KA KB xA yA 80.1 759.9 290.0 0.9998 0.3816 1.000 1.000 82.5 817.4 314.9 1.0755 0.4144 0.886 0.953 85.0 880.8 342.7 1.1590 0.4510 0.775 0.899 87.5 948.0 372.5 1.2474 0.4901 0.673 0.840 90.0 1019.1 404.4 1.3409 0.5321 0.579 0.776 92.5 1094.1 438.5 1.4396 0.5769 0.490 0.706 95.0 1173.4 474.9 1.5439 0.6249 0.408 0.630 97.5 1256.9 513.7 1.6539 0.6760 0.331 0.548 100.0 1345.0 555.2 1.7697 0.7305 0.259 0.459 102.5 1437.6 599.3 1.8916 0.7885 0.192 0.363 105.0 1535.0 646.2 2.0198 0.8503 0.128 0.259 107.5 1637.3 696.1 2.1544 0.9159 0.068 0.146 110.0 1744.7 749.1 2.2957 0.9856 0.011 0.025 110.5 1766.8 760.1 2.3248 1.0001 0.000 0.000

(a) To find the temperature at 25 mol% vaporization, starting with a liquid mixture of 70 mol% benzene and 30 mol% toluene, extend a dashed, vertical line upward from point M on the T-y-x diagram on the previous page until point B is reached. At this point, using the inverse lever-arm rule, the ratio of the AB line length to the BC line length is 25/75. The temperature is 88oC. The benzene mole fraction of the equilibrium vapor when condensed is the same as the equilibrium vapor at point C or 0.88. The benzene mole fraction in the residue liquid is the same as the equilibrium liquid at point A or 0.65.

(b) The Raoult's law K-values are included in the above table, and are plotted

below. 8. Saturated-liquid feed of F = 40 mol/h, containing 50 mol% A and B, is supplied to the apparatus in the following figure. The condensate is split so that reflux/condensate = 1. (a) If heat is supplied such that W = 30 mol/h and relative volatility = 2, what will be the composition of the overhead and the bottoms product. (b) If the operation is changed so that no condensate is returned to the still pot and W = 3D, compute the product compositions.

xA = 0.4575, xB = 0.5425 for the bottoms
yA = 0.6275, yB = 0.3725 for the distillate
(b) Note that the solution to Part (a) was independent of the reflux ratio. Accordingly, the solution to Part (b) is the same as for Part (a)
Ref: J. D. Seader and E. J. Henley, Separation Process Principles , Wiley, 2011

Share with your friends:

The database is protected by copyright ©essaydocs.org 2020
send message

Main page