An aqueous slurry at 30C containing 20.0 wt% solids is fed to an evaporator in w

Question

An aqueous slurry at 30C containing 20.0 wt% solids is fed to an evaporator in which enough water is vaporized at 1 atm to produce a product slurry containing 35.0 wt% solids. Heat is supplied to the evaporator by feeding saturated steam at 2.6bar absolute into a coil immersed in the liquid. The steam condenses in the coil, and the slurry boils at the normal boiling point of pure water. The heat capacity of the solids may be taken to be half that of liquid water.

(a) Calculate the required steam feed rate (kg/h) for a slurry feed rate of 1.00 X 10^3 kg/h.

(b) Vapor recompression is often used in the operation of an eveporator. Suppose that the vapor generated in the evaporator described above is compressed to 2.6 bar and simultaneously heated to the satuaration temperature at 2.6 bar , so that no condensation occurs. The compressed steam and additional saturated steam at 2.6 bar are then then fed to the evaporator coil, in which isobaric condensation occurrs. How much addtional steams are required

Explanation / Answer

Basis : 1000 kg/hr of slurry. It contains 20 weight % solids , solid flow rate =1000*20/100 =200 kg/hr

Slurry in the product= 35%. let P= Product. Hence P*0.35= 200, P= 200/0.35=571.43 kg/hr

water removed in the evaporator = 1000-571.43= 428.57 kg/hr, enthalpy of vaporization =2676 Kj/Kg

hence heat required to be removed = 428.57*2676 Kj/hr=1146853 Kj/hr

steam is avaiable at 2.6 bar and its enthalpy is =2718 Kj/kg

Hence steam required = 1146853/2718= 422 kg/hr

b) vapor generated = 428.57 kg/hr this has been compressed to 2.6 bar

change in enthalpy = enthalpy of vapor- enthalpy of liquid = 2177 Kj/kg

hence enthalpy of the evaporated steam at 2.6 bar= 2177*428.57 =932997 Kj

additional enthalpy required =1146583-932997=231856.4 Kj

enthalpy change of saturated steam supplied = 2177

Additional steam required = 231856.4/2177=106.5 kg/hr

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