Unformatted text preview: , D = 5.33 kmol/h and B = 5.33 kmol/h
Therefore in terms of mass flow rates, total total distillate rate is,
mD = 0.95(5.33)(78.11) + 0.05(5.33)(92.14) = 420.1 kg/h
Therefore the bottoms rate = mB = 907.3  420.1 = 487.2 kg/h
(e) First compute the kmol/h of vapor leaving the reboiler, using the assumption of
constant molar overflow. From part (c), the reflux ratio = 1.57. Therefore, the reflux rate =
1.57(5.33) = 8.37 kmol/h. Below the feed stage, the liquid rate = 8.37 + 10.66 = 19.03 kmol/h.
The vapor rate leaving the reboiler = 19.03  5.33 = 13.70 kmol/h. From the plot above, the
composition of the reboiler vapor = 12 mol% benzene. Neglecting the sensible heat and using
the enthalpy data given, after converting from Btu/lbmol to kJ/kmol, the reboiler heat duty is,
QR = 2.324[0.12(13.7)(18,130  4,900) + 0.88(13.7)(21,830  8,080)] = 436,000 kJ/kmol.
From Perry's Handbook, latent heat of vaporization of steam at 273.7 kPa (404 K) = 2,172 kJ/kg
Therefore, we need 436,000/2,172 = 200.7 kmol/h or 3,616 kg/h.
(f) A rigorous enthalpy balance around the reboiler takes into account the sensible heat
effect since the temperature of the liquid entering the reboiler is not the same as the temperatures
of the equilibrium liquid and vapor leaving the reboiler. Let N = conditions leaving the reboiler
and N1 be the conditions leaving the stage above the reboiler. Then,
QR = VN HVN + LN H LN − LN −1 H LN −1
Note that in the simplified enthalpy balance of part (e), the following equation was
applied,
QR = VN HVN − H LN
Since, VN = LN − LN −1 , this is equivalent to assuming H LN = H LN −1 Exercise 7.29
Subject: Preliminary design for the distillation of a mixture of ethanol and water at 1 atm. Given: Bubblepoint feed containing 20 mol% ethanol in water. Unit consisting of a
perforatedtray column, partial reboiler, and total condenser. Distillate to contain 85 mol%
alcohol and a 97% recovery of alcohol. Vaporliquid equilibrium data.
Assumptions: Constant molar overflow.
Find: (a)
(b)
(c)
(d) Molar concentrations in the bottoms product.
Minimum values of L/V, L/D, and VB/B.
Minimum number of equilibrium stages and actual plates for Eo = 0.55.
Number of actual plates for L/V = 0.80. Analysis: From the vaporliquid equilibrium data for 1 atm., it is seen that ethanol is more
volatile than water for ethanol mole fractions in the liquid from 0 to 0.8943, which is the
azeotrope concentration. The distillate composition is within this region.
(a) Take a basis of F = 100 kmol/h.
(1)
Overall total material balance: F = 100 = D + B
(2)
Ethanol recovery: 0.97FxF = 0.97(100)(0.20) = 19.4 = DxD = 0.85D
Solving Eq. (2), D = 22.82 kmol/h. From Eq. (1), B = 100  22.82 = 77.18 kmol/h
Ethanol in bottoms = 20  19.4 = 0.6 kmol/h
Therefore, ethanol mole fraction in bottoms = 0.6/77.18 = 0.00777
Water mole fraction in bottoms = 1.0  0.00777 = 0.99223
(b) In the McCabeThiele diagram on the next page, the given equilibrium data are
plotted. For a bubblepoint liquid feed, the qline is vertical at xF = 0.2. The minimum reflux in...
View
Full Document
 Spring '11
 Levicky
 The Land

Click to edit the document details