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A stream of air at 650.0 C and 835 torr with a dew point of 50.0 C flowing at a rate of 1915 L/s is to be cooled in a spray cooler. A fine mist of liquid water at 15.0C is sprayed into the hot air at a rate of 110.0 g/s and evaporates completely. The cooled air emerges at 1 atm. Physical Property Tables Material Balances For the flow diagram below, use two material balances and three constraints to determine all unknowns except for the outlet temperature T3. A g H2O(l)/s n (mol H2O(l)/s) B C CL air/s n (mol/s) HE *1 (mol H2O (V) /mol) (1-x4) (mol DA/mol) DC, E torr Tdp = C n (mol/s) X3 (mol H2O(v)/mol) (1-x) (mol DA/mol) T: (C), 1 atm Assume A = 110.0 g, B = 15.0 C, C = 1915 L, D = 650.0 C, E = 835 torr, F = 50.0C. ni = 27.78 mol/s n2 = 6.11 mol/s n3 = i 33.81 mol/s X1 = i mol H20(V)/mol X3 = i mol H2O(v)/mol Outlet Temperature Calculate the final temperature of the emerging air stream, assuming that the process is adiabatic. Suggestion: Derive expressions for the enthalpies of dry air and water at the outlet air temperature, substitute them into the energy balance, and use an iterative or automated method to solve the resulting fourth-order polynomial equation. Outlet Temperature: i C Save for Later Attempts: 0 of 3 used Submit Answer Heat Transferred At what rate (a positive number) is heat transferred from the hot air feed stream in the spray cooler? i kW