How much energy can be extracte from compressed water?

[antonima]

Say I have 1 ton of compressed water at 40 mega pascals, IE its volume is only 982 liters. At standard temperature and pressure, how much kinetic energy will flow out of this system if a valve is opened?

 

[I like Serena]

Welcome to PF, antonima!

Let's see.

I think we're talking about a release of 0.36 MJ in energy, of which part will be kinetic energy.
The volume that comes out is 18 L (assuming no change in temperature).We would need the compressibility of water for this:
$$\beta = -{1 \over V} {\partial V \over \partial P}$$

The change in volume (the water spewing out) is:
$$\Delta V = -\beta V \Delta P$$

The energy that is released by the expansion to standard pressure is:
$$W = \int PdV = \int P \cdot -\beta V dP = -{1 \over 2} \beta V P^2 |_{P_1}^{P_2}$$

At T=25 °C, we have:
β = 4.6 x 10^-10 m²/N
V = 0.982 m³
P1 = 40 MPa
P2 = 100 kPa

This means W=0.36 MJ and ΔV=0.018 m³

 

[antonima]

Hi ILS!

.36 MJ. I like that!
See, I have been learning about oceanography and how water compresses in the deeper parts of the ocean. At 4 km in depth, or 40 MPa water is compressed by 1.8%. I was thinking that maybe it could be sealed in a container which would keep it pressurized and raised to extract energy from the deep ocean.

Funny enough, to raise this water would at the very minimum require
(buoyancy at top - buoyancy at bottom * 1/2) * gravity * height
(basically using mass*gravity*height)
(18*1/2*4000*9.81) = .353 MJ !
This is not counting friction, drag, or the energy required to raise the container which carries the water. It seems that the thermodynamics of this all works out so no energy is created, and energy cannot be continuously extracted from the ocean.

Your equation also agrees with the the pressure analysis. Pressure is just Newtons/meter squared. If water were to be a cube just short of 1 meter squared, it would expand ~1.8 cm to one side.
40,000,000 pascals *.5*.018 m = .36 MJ !
using
Force*distance

 

[I like Serena]

Nice that it fits! :)

And pity that we can't extract energy this way. :(

 

[PJC01]

The amount of energy that can be extracted from compressed water depends on various factors such as the initial pressure, volume, and temperature of the water. In this scenario, we have 1 ton of compressed water at 40 mega pascals and a volume of 982 liters. Assuming that the water is at standard temperature and pressure, the amount of kinetic energy that will flow out of the system when a valve is opened can be calculated using the Bernoulli's equation.

According to the Bernoulli's equation, the kinetic energy (KE) of a fluid is equal to half of its density (ρ) multiplied by the square of its velocity (v). Therefore, the amount of kinetic energy that will flow out of the system can be calculated as:

KE = 1/2 * ρ * v^2

To determine the density of the compressed water, we can use the ideal gas law, which states that the pressure (P), volume (V), and temperature (T) of a gas are related by the equation PV = nRT, where n is the number of moles and R is the gas constant. Since we know the volume and pressure of the compressed water, we can determine its density using this equation.

Assuming that the temperature is 25°C (298 K) and the gas constant is 0.0821 L*atm/mol*K, the number of moles of water can be calculated as:

n = PV/RT = (40 MPa * 982 L) / (0.0821 L*atm/mol*K * 298 K) = 1.6 moles

Now, we can determine the density of the compressed water as:

ρ = n/m = 1.6 moles / 0.982 L = 1.63 moles/L

Substituting this value into the equation for kinetic energy, we get:

KE = 1/2 * 1.63 moles/L * v^2

To determine the velocity of the water, we can use the continuity equation, which states that the product of the cross-sectional area (A) and the velocity (v) of a fluid is constant. Since we know the initial volume and the final volume (when the valve is opened), we can calculate the velocity of the water as:

v = (initial volume / final volume) * v = (982 L / 0.982 L) * v = 1000