- Aug 10, 2001
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It has been proposed to use the thermal gradients of the ocean to drive a heat engine. Suppose that a certain location the water termperature is 295K at the ocean surface and 277K at the ocean floor. If the engine is to produce 1 GW of eletrical power, what minimum volume of water must be processes to suck out the heat in every second?
maximum efficiency = 1 - T(cold)/T(hot) = 1 - (277K/295K) = .061
maximum efficiency = work done per unit time/heat absorbed per unit time = .061 = (1x10^9 J/S)/Q(hot)
Q(hot) = 1.64x10^10 J/S
minimum volume of water processed = Q(hot)/(heat capacity of water*change in water temperature)
= (1.64X10^10 J/S)/[4186 J/(K*kg)*18K] = 217657 kg/s = 217.657 m^3/s