Evaporation in Double-Effect Reverse-Feed Evaporators. A feed containing 2 wt % dissolved organic solids in water is fed to a double-effect evaporator with reverse feed. The feed enters at 100°F and is concentrated to 25% solids. The boiling-point rise can be considered negligible as well as the heat of solution. Each evaporator has a 1000-ft2 surface area and the heat-transfer coefficients are Uj = 500 and U2 = 700 btu/h-ft'°F. The feed enters evaporator number 2 and steam at 100 psia is fed to number 1. The pressure in the vapor space of evaporator number 2 is 0.98 psia. Assume that the heat capacity of all liquid solutions is that of liquid water. Calculate the feed rate F and the product rate L1 of a solution containing 25% solids. (Hint: Assume feed rate of, say, F = 1000 lbm/h. Calculate the area. Then calculate the actual feed rate by multiplying 1000 by 1000/calculated area.) a F 133 800 lbm/h (60 691 kg/h), L = 10700 lbm /h (4853 kg/h).

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As per the given data Solids in feed 2 Solids in concentrate 25 HTA1 HTA2 1000 ft2 U1 500 Btu h ft2 ... View the full answer