# 771077 1 09150 analysis continued this line passes

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Unformatted text preview: line. The equation for this line is given by Eq. (7-9), L 1 1 (2) x+ x D = 0.915x + (0.98) = 0.915x + 0.08326 V R +1 11.77 + 1 Below the feed stage, with 10 mol% vaporization of the feed, L / V = 10535. From a . rearrangement of Eq. (7-12), the boilup ratio, VB, is 18.6916. The equation of the stripping section operating line is given by Eq. 7-14), L 1 1 y= x- x D = 10535x - . (0.04) = 10535x - 0.00214 . (3) V VB 18.6916 y= The equation for the q-line is given by Eq. (7-26), y= q 1 0.9 1 x- zF = x- (0.62) = -9 x + 6.2 q -1 q -1 0.9 - 1 0.9 - 1 (4) Exercise 7.39 (continued) Based on Eqs. (1) to (4), the McCabe-Thiele diagram in terms of P, the more volatile component, is drawn below for three regions: (2) x = 0.2 to 0.4, (3) x = 0.4 to 0.6, and (4) x = 0.6 to 0.8, in order to gain accuracy. In these three regions, 28 stages are stepped off in the rectifying section up to x = 0.8, and 34.3 stages are stepped off in the stripping section down to x = 0.2. Let region (1) extend from x = 0.8 to 0.98 (i.e. xD). Apply the Kremser equation...
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