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It is, of course, impossible to treat every mass transfer operat ion in existence-there are too many variations. And new techniques of separation are being developed daily. But by means of the typical examples cited in these last two chapters, the reader should be able to fit the basic principles of control to his own operation. Do not try t o apply t he cquations given herein as all-encompassing formulas. They are intended to demonstrate a point: the evolution of a cont rol system out of mass and heat balances. Writing these balances on your own process should always be the first step in designing its control system.
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1. Treyhal, R. E.: Mass Transfer Operations, p. 204, McGraw-Hill Book Company, New York, 1955. 2. Ibid., p. 158. 3. The Foxboro Company, Foxhoro I>ew Point Recording System, Using the Dewcel Element, Technical Information Sheet 1%30a. 4. Moore, J. G., and W. E. IIeslcr: Evaporation of Heat Sensitive Materials, Chem. Eng. Prog., February, 1963.
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5. Treybal, op. cit., p. 568.
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1P.l What arc the upper and lower limits of !/ and z in the streams leaving the absorpt,ion tower \l hat will make u approach its lower limit What will happen to z under these conditions 12.9 Saturated air at 60 F is being heated t.o 72 F. \l hat is the relative humiditv after heating 1 2 . 3 The three effects of a triljlc-effect evaporator are operated at 12, 7, and 2 psia, reslxxtively. How man\. pounds of water will 1 lb of 20 psia saturated steam evaporate, excluding losses How nlany ljounds of solution can be concentrated from 35 to 70 weight percent solids from 1 lb of steam lg.4 Feed to a crystallizer contains 44 weight percent solute. It is chilled to a controlled temperature where saturation represrnts only 17 weight percent solute. Khat is the ratio of uiot,her liquor to feed flow (weight basis) needed to control the discharged slurp- at 70 weight percent crystals 1 9 . 5 -1 column is fed a ternary mixture of 27 percent -1, 53 percent B, and 20 percent C. Product .I is to leave thr bottom of the tower essentially pure. Components B and C forin a heterogeneous azeotropc whose composition is 80 percent B and 20 percent C. It separates upon condensation into two layers: the heavy layer is comprised of 90 ljercent C and the light laptr of 10 percent C. X11 the heavy layer is withdrawn as distillate, together with part of the light layer, the balance being returned as reflus. Calculate, per mole of feed, the flows of vapor, reflus, and both distillate streams necessary to deliver pure bottoms product. 19.6 Design a system to control the amount of both B and C in the bottom
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APPENDIX
1.1 T, = 1.33 min; R = O.-Q min. This setting is likely to be conservative
because there will be mne spreading of the solids on the belt, which acts as capacity. 1.2 7, = 1.0 niin; Ez = 0.092 mill; P = 4000/,. 1.3 The steady-state gain is 1 /k at a I)hase angle of zero; the dynamic gain is a vertical vector of magnitude T,,/~TT~/I at, -90 . The resultant, however, is not thr sum of these comlmncnts, but is the recil)rocal of the resultant shown in Fig. 1.17; its lnagnitude is given in IQ. (1.22) and its I)hase in Eq. (1.24); it appears in the fourth quadrant. 1.4 The I)rolmrtional vector has a magnitude of 100/P aud zero phase; the derivative vector is vertical, with a magnitude of 2OOnD/Pr, and a phase of $90 . The resultant is a sum of the cornlwnents, having a magnitude of and a l)hase angle of tan-l 27rD/r,. 100 2/l + (2 TD/7 )2/P, 1 . 5 G1 = 0.5; rd/rl = 1.21, which falls on the curve of Fig. 1.26. 1 . 6 \TYth prolmrtiorml l)lus derivative, T, = 2.67 min P = 2.8 ; o, II = 0.425 m i n . Kith l)rol)ortional 111~s reset, T, = 8.0 min, I =)11.9(x, R = 1.27 min.
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