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Minimum number of actual plates (total reflux).
Number of actual plates for R = 1.5 Rmin.
Kilograms per hour of products for a feed of 907.3 kg/h.
Kg/h of saturated steam at 273.7 kPa for reboiler heat duty using given enthalpy data.
Rigorous enthalpy balance around the reboiler. Analysis: McCabe-Thiele plots are made in terms of benzene mole fractions, since benzene is
the more volatile component. The equilibrium curve is plotted from the data in Exercise 7.13.
(a) For a saturated liquid feed, minimum reflux corresponds to a pinch point located at
the intersection of a vertical q-line passing through xF = 0.5 and the equilibrium curve as shown
in the McCabe-Thiele diagram below. From the equilibrium data, this intersection is at y = 0.72
and x = 0.5. Then, the slope of the rectifying section operating line, (L/V)min is (0.95 - 0.72)/(0.95
- 0.50) = 0.511. From a rearrangement of Eq. (7-7), Rmin = (L/V)min /[1 - (L/V)min ] = 0.511/(1 0.511) = 1.045.
(b) The McCabe-Thiele plot for minimum stages at total reflux is shown below. The
operating lines are coincident with the 45o line. Equilibrium stages are stepped off starting from
xB = 0.05 to xD =0.95. It is seen that just less than 7 equilibrium stages are needed. Call it Nt = 7.
From Eq. (6-21), for an overall plate efficiency of 65%, i.e. Eo = 0.65, the actual minimum
number of plates = Na = Nt / Eo = 7/0.65 = 10.8.
(c) Operating reflux ratio = R = 1.5 Rmin = 1.5(1.045) = 1.57. From Eq. (7-7), the slope
of the operating line for the rectifying section = L/V = R/(1 + R) = 1.57(1 + 1.57) = 0.611. On
the McCabe-Thiele diagram on the next page, the rectifying section operating line has this slope
and passes through the point, y=0.95, x=0.95. the stripping section operating line passes through
the point, y=0.05, x=0.05 and intersects the vertical q-line at the point where the rectifying
section operating line intersects the q-line. As seen, the equilibrium stages are stepped off
starting at the top, with a switch from the rectifying section to the stripping section to minimize
the number of stages and, thus, locating the optimal feed stage. The result is just over 10
equilibrium stages plus a partial reboiler. Call it 11 equilibrium stages plus a partial reboiler.
Applying Eq. (6-21), Na = 11/0.65 = 16.9 or 17 actual plates plus the partial reboiler. Analysis: (continued) Exercise 7.28 (continued) Analysis: (continued) Exercise 7.28 (continued) Analysis: (continued) Exercise 7.28 (continued) Exercise 7.28 (continued) Analysis: (continued)
(d) MW of benzene = 78.11. MW of toluene = 92.14.
Let F = kmol/h of feed. Then by mass material balance with an equimolar feed,
0.5F (78.11) + 0.5F(92.14) = 907.3
Solving, F = 10.66 kmol/h. For the equimolar feed, the component flow rates in the feed are:
5.33 kmol/h each for benzene and toluene
Next calculate the distillate and bottoms flow rates from,
overall total mole balance: F = nF = 10.66 = D + B
overall benzene mole balance:
FxF = 5.33 = 0.95D + 0.05B
Solving Eqs. (1) and (2)...
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- Spring '11
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