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

01 jm2 assumptions spherical crystals temperature of

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

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...
View Full Document

{[ snackBarMessage ]}

Ask a homework question - tutors are online