As shown in the figure, two kilograms of water undergo a process from an initial state where the water is saturated vapor at 120C, the velocity is 30 m s-1, and the elevation is 6 m, to a final state where the water is saturated liquid at 10C, the velocity is 25 m s-, and the elevation is 3 m. Determine in kJ, (a) the exergy at the initial state, (b) the exergy at the final state, and (c) the change in exergy. Let T = 25C, p = 1 atm, and g = 9.8 m s-. Answer: (a) E = = 1000.1 kJ (b) E = 3.9 kJ (c) AE = -996.2 kJ 6 m - Saturated vapor at 120C 30 m/s 3 m Saturated liquid at 10C 25 m/s Po = 1 atm, To = 25C, 8=9.8 m/s