A single effect evaporator has a thermal load, Qe of 2355 kJ/s and the heating steam temperature is 115 C. The seawater temperature, Tew, is 30 C and the feed seawater temperature, Tr, is less than the boiling temperature, T, (102C), by 10 C. The condensed vapor temperature is less than the boiling temperature by 0.985 C. We assume that the If the distillate product flow rate and the Boiling Point elevation are equals respectively to 1 kg/s and 0.994 C. lleating Steam Evaporator U=1.9695+1.2057x10-2T-8.5989x10-5(T)2+2.5651x10-7(Th) Ue = 1.7194+3.2063x10-Ty+1.5971x10-5(Ty)-1.9918x10-7(Tv) Reject Brine My Th Condensed fram MT, Feed Seawater Cooling Seawater Condenser Distillate Product Max Ta Calculate: 1- The heating steam flow rate, assuming a steam latent heat equal to 2216.73 KJ/kg, 2- The Single effect Performance Ratio, 3- The overall heat transfer coefficient in the evaporator and condenser, 4- The specific heat transfer areas in the evaporator and condenser, Specific heat C = 4.2 KJ/kg C, X = 70000 ppm while X, = 42000 ppm. 5- The specific cooling water flow rate. AU and Ue can be given by the following correlations: Ay = 2501.897149-2.407064037 Ty+1.192217x10-3 (T)2-1.5863x10-5 (T,)3 Product Vapor Ma Te Feed and Cooling Seawater MeMe Tew