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Pressure and potential energy per unit volume are related by Bernoulli's equation, but they are not the same thing.If someone says it is potential energy per unit volume...

[itex]P + 1/2*\rho*v^2 + \rho*g*h = constant[/itex]

Where:

[itex]P[/itex] is the pressure (energy/volume)

[itex]1/2*\rho*v^2[/itex] is the kinetic energy per unit volume

[itex]\rho*g*h[/itex] is the potential energy per unit volume

Have you dove to the bottom of a pool before? If so, you've experienced the exchange of potential energy for pressure energy. The water at the bottom of the pool doesn't have the same potential energy as the water at the top, but it is at a higher pressure.

The head form of Bernoulli's equation simply divides through by density * force of gravity. This leaves the equation in units of length. For why that's called the head form, I'll defer to someone more knowledgeable.

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Pressure energy is simply pressure. Force per unit area, or energy per unit volume are just two ways to express the same thing.

Pressure and potential energy per unit volume are related by Bernoulli's equation, but they are not the same thing.

[itex]P + 1/2*\rho*v^2 + \rho*g*h = constant[/itex]

Where:

[itex]P[/itex] is the pressure (energy/volume)

[itex]1/2*\rho*v^2[/itex] is the kinetic energy per unit volume

[itex]\rho*g*h[/itex] is the potential energy per unit volume

Have you dove to the bottom of a pool before? If so, you've experienced the exchange of potential energy for pressure energy. The water at the bottom of the pool doesn't have the same potential energy as the water at the top, but it is at a higher pressure.

The head form of Bernoulli's equation simply divides through by density * force of gravity. This leaves the equation in units of length. For why that's called the head form, I'll defer to someone more knowledgeable.

Pressure energy is simply pressure. Force per unit area, or energy per unit volume are just two ways to express the same thing.

Pressure and potential energy per unit volume are related by Bernoulli's equation, but they are not the same thing.

[itex]P + 1/2*\rho*v^2 + \rho*g*h = constant[/itex]

Where:

[itex]P[/itex] is the pressure (energy/volume)

[itex]1/2*\rho*v^2[/itex] is the kinetic energy per unit volume

[itex]\rho*g*h[/itex] is the potential energy per unit volume

Have you dove to the bottom of a pool before? If so, you've experienced the exchange of potential energy for pressure energy. The water at the bottom of the pool doesn't have the same potential energy as the water at the top, but it is at a higher pressure.

The head form of Bernoulli's equation simply divides through by density * force of gravity. This leaves the equation in units of length. For why that's called the head form, I'll defer to someone more knowledgeable.

U can see that in the equation u have written...the pressure energy (P) and potential energy (ro*g*h) are the same...isnt it?...really confused

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please elaborate more...with the spring as eg.

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Are thermodynamic pressure and hydraulic pressure different??

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please elaborate more...with the spring as eg.

I am not really sure how much more elaborate I can make this. Bernoulli's equation is a conservation of energy statement. The ##\rho v^2/2## term (dynamic pressure) is kinetic energy, the ##\rho g z## term (hydrostatic pressure) is the gravitational potential energy, and ##p## (static pressure) is sort of like spring potential energy in the sense that it comes from a bunch of moving molecules being in some space together and as you relax how much it is compressed, the pressure goes down and is transferred into some other sort of energy. In the case of a flow that is subject to Bernoulli's equation, that other form of energy is either kinetic or gravitational depending on the system.

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I understand your confusion. No, they are not the same. However, they are closely related.U can see that in the equation u have written...the pressure energy (P) and potential energy (ro*g*h) are the same...isnt it?...really confused

Imagine a column of water 10 meters tall open to the atmosphere. Assume zero velocity and incompressible (water is very nearly incompressible.)

With no velocity, the kinetic energy portion will drop out and leave:

[itex]P + \rho*g*h = constant[/itex]

Now image two points in the water; point one at the top and point two at the bottom. You can use the reduced Bernoulli's equation to describe their relationship.

[itex]P1 + \rho*g*h1 = P2 + \rho*g*h2[/itex]

For the h terms, h1 is 10m and h2 is 0m (at the bottom of the column.)

This leaves you with:

[itex]P1 + \rho*g*10 m = P2[/itex]

Now, because point one is open to the atmosphere we know that P1 (the static pressure at point one) is equal to about 101 kPa (atmospheric pressure) or more usefully, zero gauge. Taking the second option (0 gauge pressure) leaves this:

[itex]\rho*g*10 m = P2[/itex]

So you can see that the static pressure at the

For the sake of completeness:

[itex]P2 = \rho*g*10m = 1000 kg/m^3 * 9.81 m/s^2 * 10 m = 100kPa (gauge) [/itex]

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Cht

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