A question about heat engines (i.e., a physics problem)

[Random Variable]

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

 

[Gunslinger08]

I'm pretty sure that 1m^3 saltwater != 1kg saltwater

Saltwater is heavier than fresh water... so it should be > 1kg per m^3.

 

[Random Variable]

Originally posted by: joshsquall
I'm pretty sure that 1m^3 saltwater != 1kg saltwater

Saltwater is heavier than fresh water... so it should be > 1kg per m^3.

Other than that is it there something wrong with what I did?

 

[Gunslinger08]

No, otherwise, it seems okay. Well, the heat capacity for salt water is different, as well.

Beyond that, seems valid.

 

[Random Variable]

Originally posted by: joshsquall
No, otherwise, it seems okay. Well, the heat capacity for salt water is different, as well.

Beyond that, seems valid.

I thought that maybe I should be dividing by T(hot) and not the change in T.

 

[Generalen]

I'm sure 1 m^3 weight alot more then 1 kg regardless if salt or freshwater. More like 1000 kg.

 

[Random Variable]

Originally posted by: Generalen
I'm sure 1 m^3 weight alot more then 1 kg regardless if salt or freshwater. More like 1000 kg.