# Question on Bernoulli's equation

1. Jan 1, 2010

Hi guys, we use bernoulli's equation for inviscid flows, like water and gas.
But for gases, we also have the ideal gas law, let's say if i want to calculate energy within a gasous system, like a gas in a fixed space (e.g. a room; a tank etc...)

Looking at ideal gas law, PV=mRT, PV will give me the energy.

But if i use Bernoulli's equation, P/rho*g + v^2/2g +h, i can rearrange the equation to
P + v^2*rho/2 + rho*g*h, and if i times volume into it, it becomes PV + v^2*rho*V/2 +rho*g*h*V which gives me energy. For a gas 'sitting' in a container (without gas moving in or out of the container), I can take away the kinetic energy part right? And the potential energy part will stay because of gas particles occupying a constant space, is this a correct assumption?

The reason i am asking is because if the potential energy part also falls away, then the ideal gas law correspond to Bernoulli's equation (Even though i believe the potential energy term should stay). But if the potential energy do stay, then Bernoulli's equation will always have a higher energy value than the ideal gas law, which raise the question of when would you use ideal gas law and when would you use Bernoulli's equation to calculate energy in a gas system?

Thanks!

2. Jan 1, 2010

### rcgldr

The ideal gas law isn't taking GPE (gravitational potential energy) into account. Include GPE with the ideal gas law and you end up with Bernoulli's equation x mass (ignoring the speed related component).

Pressure energy = pressure x volume.
Bernoulli equation for ideal fluid: pressure/density + g h + 1/2 v2 = constant

Multiply this by mass and you get:
(pressure x volume) + m g h + 1/2 m v2 = total mechanical energy = constant

Usually an inviscid and incompressable fluid is used for Bernoulli examples, but you can't readily extract the pressure energy from an incompressable fluid. In the real world most forces coexist with some form of deformation (compression, expansion), in which case the pressure energy can be extracted. (One exception would be the force on an object from constant gravitational field).

3. Jan 2, 2010

Thank you Jeff.
But is there a reason why the ideal gas law don't take GPE into consideration? Because to me, in the container example in my 1st post, surely there are potential energy in the system beside the pressure energy? (let me get off topic abit and ask this: gravitational potential energy for gas would be calculated the same way as for liquid right? meaning that m would be total mass of the gas in that container and h would be the full height of the container since gas will occupy all the space?)

So are you saying that if i wanted to calculate the energy of a container with gas, i should use ideal gas law because bernoulli's is only for incompressible fluids?

And i am not sure as to what you mean by you can't readily extract pressure energy from an incompressible fluid? Do you maybe have an example?

Thank you! I really appreciate your help!

4. Jan 2, 2010

### rcgldr

It was just defined that way, ignoring the effect of gravity.
Each molecule of a gas or fluid has a different GPE depending on it's height.
There's an alternate form of Bernoulli that takes compressable flow into account:
http://en.wikipedia.org/wiki/Bernoulli's_principle#Compressible_flow_equation

Genernally Bernoulli is used for flows, while ideal gas law is for a non-moving gas.
Imagine a horizontal cylinder filled with with an incompressable fluid and a piston on one end. Regardless of the fluid's initial pressure, it will not move the piston. If the fluid was compressable+expandable, then a higher than external pressure fluid could expand, pushing the piston outwards, and lower than external pressure could compress, pulling the piston inwards.

5. Jan 3, 2010

Thanks for the answer, i think i have a much better grasp on this now!
But let us move onto a comparison between container with liquid and container with gas if you don't mind.

I have often been told that a system containing gas under pressure is much more dangerous (in terms of explosion due to over pressure, a hole in the container etc...) than a system containing liquid under pressure. And the reason was that gas holds contains more energy than liquid.

Assuming gas is steam and liquid is water, both contain in a container of equal size, operating under the same pressure, which one is more danagerous? According to the equations, energy in gas container (since there are no gas coming in/out, i will use ideal gas law) equals to PV. And the energy in water container (also no flow in/out) equals to PV + V*rho*g*h. So looking at these two equations, wouldnt the container with liquid actually be more danagerous than the gas container?

I am think it has got to do with the mass, if both water and steam has the same mass in the same size container and operate at the same temperature, then pressure in steam would be A LOT higher than the one in water, so i am thinking it is not correct to say that a system automatically poses more threat if it has gas in it, gas system is only more dangerous if the mass of gas and liquid is the same, does my reasoning make sense?

Thanks!!

6. Jan 3, 2010

### rcgldr

If the pressure inside the container was close to it's limits, than a slight disturbance (accleration) of the container with liquid would be more likely to burst the container than a gas, assuming the liquid was much denser and had much more mass.

Both the gas and liquid have a total mechanical energy equal to PV + V*rho*g*h. If the pressure was the same, then rho for the gas would be much less.

What makes the system more dangerous is the pressure energy (pressure times volume), and the pressure would be higher with a gas versus a liquid, if the mass was the same.

Last edited: Jan 3, 2010
7. Jan 3, 2010

### gmax137

Is this right? It seems to contradict experience... Or is it just that all real fluids are in fact compressible to some extent. So that by 'incompressible fluid' in this example we should imagine the behavior of a solid?

8. Jan 3, 2010

### rcgldr

All real solids are also compressable. All forces involving contact coexist with some form of deformation. Fluids are nearly incompressable, making them good for applications like hydraulics, where forces can be transmitted through the fluid with little excess motion required to compensate for compression factors.

Work done is still related to force times distance, and increasing the pressure energy of an otherwise non-flowing fluid involves compressing that fluid by applying a force over some finite distance. An incompressable fluid poses a problem because the distance involved is zero, so how can any mechanical work be peformed on a non-flowing incompressable fluid?

Last edited: Jan 3, 2010
9. Jan 4, 2010

Thanks for the explanation Jeff, i just have one point that i am a bit confused, you said in post #4 that
'Genernally Bernoulli is used for flows, while ideal gas law is for a non-moving gas.'

But in post #6 you mentioned
'Both the gas and liquid have a total mechanical energy equal to PV + V*rho*g*h. If the pressure was the same, then rho for the gas would be much less.'

But since it is a container with no gas coming in/out of the container, then wouldn't the total mechanical energy for gas just be the ideal gas las which = PV?

Thanks!

10. Jan 4, 2010

### rcgldr

Due to gravity, there will be a pressure differential within the container, lower at the top, higher at the bottom, and this pressure differential will generate a net downforce on the container that is exactly equal to the weight of the gas (or fluid) within the container.

11. Jan 8, 2010

### rcgldr

Assume that a liquid and gas are at the same pressure, but that the gas has much less mass. If there is a rupture, in the case of the much denser liquid, like water, much of the pressure energy goes into accelerating the fluid, with a relatively small amount of expansion, less than 2% (1.02 times) expansion at 1000 psi. While air at 1000 psi will expand by over 3300% (33 times). This means the air will peform more work on what ever it containing it if there's a rupture, resulting in much higher velocites of particles of the container.

Last edited: Jan 8, 2010
12. Jan 13, 2010

Hmm, why would the fluid expand when there is a rupture? I can understand that the fluid will want to rush out due to the pressure differences between inside the pressure and the atmosphere pressure, so the velocity of the fluid will increase, but what makes them expand?

Thanks!

13. Jan 13, 2010

### rcgldr

Because it's slightly compressable + expandable

http://www.engineeringtoolbox.com/fluid-density-temperature-pressure-d_309.html

In my example of 1000 psi = 68.95 bar = 6.895 x 10^6 N/m^2. Atmospheric pressure would be 14.696 psi = 1.01325 x 10^5 N/m^2. Using the supplied equation and graph for water at 21C:

density_at_1000psi = 998 / (1 - ((6.895 x 10^6-1.0135 x 10^5) / (2.15 x 10^9))) = 1001.21 kg / m^3.

The water compressed to .9968 of it's volume at 14.696psi, and the density increased by 1.0032 or 0.32%.
The fluid only rushes out if there is expansion. If the fluid was incompressable, the pressure would decrease with no expansion or flow of fluid (unless the rupture occurred where gravity would drain the fluid without involving expansion).

Last edited: Jan 13, 2010