Can Ocean Thermal Gradients Efficiently Power Heat Engines?

[ultimateguy]

It has been proposed to use the thermal gradient of the ocean to drive a heat engine. Suppose that at a certain location the water temperature is 22oC at the ocean surface and 4oC at the ocean floor.
a) What is the maximum posssible efficiency of an engine operating between these two temperatures?
b) If the engine is to produce 1GW of power, what minimum volume of water must be processed (to suck out the heat) in every second?

I got 6.1% for a and it's correct, but b) is 900m^3 and I just can't figure it out. I know hot water will decrease by 9 degrees and the cold will increase by 9 degrees, but beyond that I don't know how to proceed.

 

[Jhero]

I would like to provide a more detailed explanation for the calculation of the minimum volume of water required to produce 1GW of power using the thermal gradient of the ocean.

a) The maximum possible efficiency of an engine operating between two temperatures can be calculated using the Carnot efficiency formula: η = (Th - Tc)/Th, where Th is the hot temperature and Tc is the cold temperature. In this case, Th = 22°C and Tc = 4°C, so the maximum efficiency would be η = (22-4)/22 = 0.727 or 72.7%.

b) To calculate the minimum volume of water required, we need to use the formula for thermal power, which is given by P = mCΔT, where P is the power, m is the mass of water, C is the specific heat capacity of water, and ΔT is the change in temperature.

We know that the power required is 1GW, which is equivalent to 1 billion watts. Converting this to joules per second, we get 1 x 10^9 J/s. The specific heat capacity of water is 4.186 J/g°C, and the change in temperature is 9°C (since the hot water will decrease by 9°C and the cold water will increase by 9°C).

Plugging these values into the formula, we get:

1 x 10^9 = m x 4.186 x 9

Solving for m, we get m = 2.4 x 10^7 grams or 24,000 kg of water.

To convert this to volume, we need to use the density of water, which is 1 g/cm^3. So, the minimum volume of water required is 24,000 kg / 1 g/cm^3 = 24,000 cm^3 or 24 m^3. However, this is the volume of water required per second.

To calculate the volume of water required per second, we need to divide this by the time interval. Let's assume the time interval is 1 second. So, the minimum volume of water required per second to produce 1GW of power using the thermal gradient of the ocean would be 24 m^3.

Note: This calculation assumes ideal conditions and does not take into account any losses in the heat engine process. The actual volume of water required may