A nuclear power plant has a core temperature of 5000 K, and a cooling tower temp

Question

A nuclear power plant has a core temperature of 5000 K, and a cooling tower temperature of 373 K, and is able to produce 1.21 gigawatts of power. Assume that the plant operates at its maximum efficiency [Carnot efficiency]. giga = 10^9 1 watt = 1 J/s the Latent Heat of Vaporization (LV) of water is 2.26 10^6J/kg How much water is evaporated per second at the cooling towers? mass of water evaporated per unit time =

Explanation / Answer

maximum efficiency (ETA) = [T1 - T2]/T1 only a reversible carnot engine can have max theoretical efficiency working between source (T1 = 5000 K) and sink (300 K} (ETA) = [4700]/5000 = 0.94 --------------------------------------… ETA = output power/rate of heat flow = 1.21*10^9/Q-hot Q_hot = 1.21*10^9/0.94 = 1.287*10^9 J/s ------------------------------ change in entropy (dS) = change in entropy of hot source + change in entropy of sink dS = +Q-hot/5000 + (- Q-sink)/300 where>>. Q-sink = rate of heat rejected to water = Q-hot - 1.21*10^9 (rate of work done) W Q-sink = 1.287*10^9 - 1.21*10^9 {J/s} -------------------------------------- work output rate = m-dot * L = 1.21*10^9 m-dot = rate of mass evaporated = kg/s L = latent heat just work out!!!

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