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Separation Process Principles- 2n - Seader &amp; Henley - Solutions Manual

# Take the result as 33000 kmolh or 33000 18 594000 kgh

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Unformatted text preview: ms Cl2/L 1.773 2.27 2.74 y 0.132 0.197 0.263 x 0.000450 0.000576 0.000695 These equilibrium data and those given in the exercise are now converted to mole ratios using, Y = y/(1 - y) and X = x/(1 - x) y 0.006 0.012 0.024 0.040 0.060 0.132 0.197 0.263 x 0.000100 0.000150 0.000200 0.000250 0.000300 0.000450 0.000576 0.000695 Y 0.00604 0.01215 0.02459 0.04167 0.06383 0.15207 0.24533 0.35685 X 0.000100 0.000150 0.000200 0.000250 0.000300 0.000450 0.000576 0.000695 (a) To determine the minimum water rate, it is best to do this graphically because of the high degree of curvature of the equilibrium curve. The pinch region of infinite stages at minimum absorbent rate occurs at the bottom of the column, such that the leaving liquid is in equilibrium with the entering gas, with a Y = 0.25. With mole-ratio coordinates, the operating line is straight and passes through the following point at the top of the column {X = 0.0, Y = 0.0101}. Exercise 6.35 (continued) Analysis: (a) (continued) The point at the bottom of the column must pass through the point {X* in equilibrium with Y = 0.25}. This is shown in the following plot, which is similar to Fig. 6.9. From the solute material balance, Eq. (6-6), the slope of the operating line is given by L'/V'. From the above plot, Xbottom = 0.000581, and, thus, the slope of that line is, L'/V' = (0.25 - 0.0101)/(0.000581 - 0.0) = 413. Since V' = 80 kmol/h, L' = 80(413) = 33,000 kmol/h. Alternatively, since the Cl2 absorption rate = V'(Ybottom - Ytop) = 80(0.25 - 0.0101) = 19.19 kmol/h, then L' = 19.19/ Xbottom = 19.19/0.000581 = 33,000 kmol/h. Take the result as 33,000 kmol/h or 33,000 (18) = 594,000 kg/h. (b) Twice the minimum water rate = 2(33,000) = 66,000 kmol/h. This gives a mole ratio of chlorine in the leaving water of Xbottom =19.19/66,000 = 0.000291 moles Cl2 / mole water. This operating line is shown in the plot below. To obtain NOG , using mole ratios, the integral in Eq. (6-138) or Table 6.7 can be used, Analysis: (b) (continued) Exercise 6.35 (continued) N OG = Y = 0.25 dY (Y − Y *) Y = 0.0101 (1) 1 (Y − Y *) as a function of Y, where for a given value of X, the value of Y is obtained from the operating line and Y* is obtained from the equilibrium curve, with values as follows. The values of X are taken from the above table, where the values of Y are the corresponding values of Y*. For given values of X, values of Y for the operating line are obtained from the following equation of the straight operating line passing through the points { X = 0.0, Y = 0.0101} and { X = 0.000291, Y = 0.25}, L′ 66, 000 Y = 0.0101 + X = 0.0101 + X = 0.0101 + 825 X V′ 80 Eq. (1) can be solved graphically or numerically. Both require a table of values of Note that several values of X are added by interpolation to increase the accuracy. Analysis: (b) (continued) X 0.000000 0.0000125 0.000025 0.000050 0.000075 0.000100 0.000150 0.000200 0.000250 0.000291 Y* 0.00000 0.00075 0.00150 0.00301 0.00452 0.00604 0.01215 0.02459 0.04167 0.06030 Exercise 6.35 (continued) Y 0.0101 0.0204 0.0307 0.0514 0.0720 0.0926 0.1339 0.1751 0.2164 0.2500 1/(Y-Y*) 99.01 50.86 34.22 20.69 14.82 11.55 8.22 6.64 5.72 5.27 ∆Y Avg 1/(...
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