A singleeffect evaporator is to be used to concentrate a foo

A single-effect evaporator is to be used to concentrate a food solution containing 15% (by mass) solids to 50% solids. The feed stream enters the evaporator at 291K at a rate of 1.0 kg/s. Steam is available at a pressure of 2.4 bar. It remains at saturation conditions through the evaporator. An absolute pressure of 0.07 bar is maintained in the evaporator (that is, above the evaporating food solution). Assuming that the properties of the feed solution are the same as water, and taking the overall heat transfer coefficient to be 2300 W/m^2 middot K, calculate the rate of steam consumption and the necessary heat transfer surface area. You will need steam tables to solve this problem; interpolate as needed. Remember, the overall temperature driving force in the evaporator is the difference between the steam temperature and the temperature of the saturated liquid \"feed\" in the evaporator.

Solution

Single Effect Evaporator Calculations Food Solution Solids water Total Kg/s Kg/s Kg/s Feed rate 1 Intial Foold Solution is 15% Solids 0.15 0.85 1 Final Food Solution is 50% Solids 0.15 0.15 0.3 Evaporation required   0.7 Kg/s Feed Temperature 291 K Taking the feed as pure water, entalpy at 291K 75.5 kJ/Kg From Steam Tables Enthalpy of Steam hs at 2.4 bar 2715 kJ kg1 Enthalpy of Condensate hc 530 kJ kg1 Ts Temparature of Steam at 2.4 bar 126.1 oC Enthalpy of Vapor in Evaporator at 0.07 Bar Vacuum 2572 kj kg-1 Enthalpy of liquid The Enthalpies of are functions of the pressure within the evaporator Now Write the Heat Balance as below, S = steam flow rate Kg/s S (2715-530) = (0.70x2572)+(0.30x163)-(1.0x75.5) From the above equation We get S 0.812 Kg/s Q Heat Q = S*( hs-hc) 1774.22 kW Te Temerature of Saturated Liquid at evaporator pressure 0.07bar 39 oC U Heat Transfer Coefficient 2300 W/m2/K 2.3 kW/m2/K From the Rate of Evaporation, the Heat Trafer Area is calculated. Evaporator Area = A = Q/U(Ts-Te) Evaporator Area 8.856487 m2