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

01 jm2 assumptions spherical crystals temperature of

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Unformatted text preview: oC. Final solution at 30oC with 90% crystallization of the urea in the anhydrous form. Assumptions: Equilibrium in the crystallizer. Find: Fraction of water that must be evaporated. Analysis: From Table 17.7, solubility of urea in water at 100oC = 730 g/100 g water, and solubility of urea in water at 30oC = 135 g/100 g water Take a basis of 1,000 kg of feed. Therefore, have 0.90(730) = 657 g urea/100 g water Therefore, feed is: [657/(657 + 100)]1,000 = 868 kg urea 1,000 – 868 = 132 kg water Urea crystals formed at 30oC = 0.90(868) = 781 kg Urea remaining in solution = 868 – 781 = 87 kg Therefore, water in solution = [100/135]87 = 64.4 kg By overall mass balance, water evaporated = 132 – 64.4 = 67.6 kg Fraction of water evaporated = 67.6/132 = 0.512 Exercise 17.13 Subject: Heat addition to the feed of a crystallizer Given: Data from Examples 17.3 and 17.5. Crystallizer feed is 4,466 lb/h of 37.75 wt% MgSO4 in water at 170oF and 20 psia. Instead of adding 246,000 Btu/h of heat to recirculating magma in the vacuum, evaporating crystallizer, that heat is to be added to the feed. Assumptions: Linear extrapolation of the enthalpy chart of Figure 17.10. Find: Exit temperature from the feed heat exchanger. Analysis: Use the enthalpy-concentration chart for aqueous solutions of MgSO4 in Figure 17.10. Hfeed in = -20 Btu/lb for 37.75 wt% MgSO4 at 170oF By an enthalpy balance on the heat exchanger, Hfeed out = -20 + 246,000/4,466 = 35.1 Btu/lb For 37.75 wt% MgSO4, this enthalpy is off the chart of Figure 17.10. Therefore, make a linear extrapolation, assuming a constant specific heat for the solution. For the temperature range of 170oF to 230oF, the enthalpy change is -20 to 18 Btu/lb. Therefore, the average specific heat = [18 – (-20)]/(230 – 170) = 0.63 Btu/lb-oF. Using this specific heat, Tfeed out = 170 + 246,000/[4,466(0.63)] = 257oF. This is a much higher temperature than the temperature of 85oF in the crystallizer. Therefore, a much higher heating-medium temperature would be required. Furthermore, the pressure of the feed would have to be increased to suppress vaporization in the heat exchanger. The conclusion is that a heat exchanger on the recirculating magma is preferred over a heat exchanger on the feed. Exercise 17.14 Subject: Rate of heat addition to a crystallizer Given: Data from Exercise 17.11. Feed of 5,870 lb/h of 35 wt% MgSO4 in water at 180oF and 25 psia. Recycle of 10,500 lb/h of saturated MgSO4 at 80oF and 25 psia. From calculations made in Exercise 17.11, three phases leave the crystallizer at 85oF and 0.58 psia in the vapor space: 1,490 lb/h of steam; 3,720 lb/h of crystals of MgSO 4 7H 2 O ; and 11,160 lb/h of a saturated solution of 28 wt% MgSO4. Assumptions: No heat losses from the crystallizer Find: Rate of heat addition to the crystallizer. Analysis: Use Figure 17.10 for enthalpies of liquid and crystals. Use steam table for steam. An enthalpy balance around the crystallizer gives, mfeed H feed + mrecycle H recycle + Qin = msteam H steam + mli...
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This document was uploaded on 02/24/2014 for the course CBE 2124 at NYU Poly.

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