Question 27.5: A continuous vacuum crystallizer is fed with a 31 percent Mg......

A continuous vacuum crystallizer is fed with a 31 percent MgSO_4 solution. The equilibrium temperature of the magma in the crystallizer is 86°F (30°C), and the boiling-point elevation of the solution is 2°F (1.11°C). A product magma containing 5 tons (4536 kg) of MgSO_4 ~⋅~ 7H_2O per hour is obtained. The volume ratio of solid to magma is 0.15; the densities of the crystals and mother liquor are 105 and 82.5 lb/ft³ , respectively. What is the temperature of the feed, the feed rate, and the rate of evaporation?

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Figure 27.13 shows the graphical solution of the problem. The vapor leaves the crysta11izer at the pressure corresponding to 84°F and carries 2°F superheat, which may be neglected. From steam tables, the enthalpy of the vapor is that of saturated steam at 0.5771 \rm{lb_f /in.^2} , and the coordinates of point a are H = 1098 Btu/lb and c = 0. The enthalpy and average concentration of the product magma are calculated from data given by Fig. 27.4. The straight line fd is the 86°F isotherm in the area bcih of Fig. 27.4. The coordinates for its terminals are, for point f, H = -43 Btu/lb and c = 0.285, and for point d, H = -149 Btu/lb, c = 0.488. The mass ratio of crystals to mother liquor is

\frac{0.15 \times 105}{0.85 \times 82.5}=0.224

The rate of production of mother liquor is 10,000/0.224 = 44,520 lb/h, and the total magma produced is 10,000 + 44,520 = 54,520 lb/h. The average concentration of MgSO_4 in the magma is

\frac{0.224 \times 0.488+0.285}{1.224}=0.322

The enthalpy of the magma is

\frac{0.224(-149)+(-43)}{1.224}=-62.4~ Btu / lb

These are the coordinates of point e. The point for the feed must lie on the straight line ae. Since the feed concentration is 0.31, the enthalpy of the feed is the ordinate of point b, or -21 Btu/lb. Point b is on the 130°F (94.4°C) isotherm, so this temperature is that of the feed. By the center-of-gravity principle, the evaporation rate is

54,520 \frac{-21-(-62.4)}{1098-(-21)}=2,017 ~\rm{lb / h~ (915 ~kg / h )}

The total feed rate is 54,520 + 2017 = 56,537 lb/h (25,645 kg/h).

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