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Unformatted text preview: min = 1/ [( L / V )  1] = 1/(4.96  1) = 0.253. (c) The boilup ratio = VB = 1.2(0.253) = 0.3036. From Eq. (712), the slope of the stripping section operating line = L / V = (VB + 1)/VB = (0.3036 + 1)/0.3036 = 4.294. This line intersects the vertical qline (xF = 0.05) at 0.002 + 4.294(0.05  0.002) = 0.2081. Therefore, the slope of the rectifying line = L/V = (0.35  0.2081)/(0.35  0.05) = 0.4730. From a rearrangement of Eq. (78), R = (L/V)/[1  (L/V)] = 0.473/(1  0.473) = 0.8975. The equation for the rectifying section operating line, using Eq. ((79), with a modification for a partial condenser as determined from Fig. 7.18, is,
y n +1 = R 1 xn + y D = 0.473xn + 0.184 5 R +1 R +1 (2) Analysis: (c) (continued) Exercise 7.22 (continued) The equation for the stripping section operating line, using Eq. (712) is, 1 V +1 y m +1 = B xm  x B = 4.294 xm  0.00659 VB VB (3) We can now calculate stage by stage down from the top, starting from yD = 0.35, alternating between the equilibrium curve, Eq. (1) and the appropriate o...
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 Spring '06
 selebi
 Distillation, pH, Mass Transfer

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