Concept Question (Delta P or Differential Pressure) 2

[Poppapetro]

The question I have relates to basic fundamentals of fluid dynamics. Here is my way of thinking: The driving force for fluid movement is Delta P.

If you have two tanks with an equal volume of 20 cubic meters absolutely full of water and tank 1 is at 500 kPag and tank 2 is 100 kPag. Furthermore, there is a line connecting both tanks with a valve at the direct midpoint of this pipe from T1 to T2. If I open the valve, what is the direction of flow if both tanks have NO vent lines to atmosphere. Fundamentally, one would think flow would be from high pressure to low, but what replaces the liquid. Could some explain this concept at a conceptual/fundamental level?

 

[Oldliar]

I think that you actually have no flow. As you have two completely full tanks of water (assumed non-compressible in this case), you system is completly filled and you're not able to get any movement of mass.
Is this a question raised in some IQ test or am I completely off the track??

 

[zdas04]

First off, while dP defines a direction of flow, it's magnitude is a result of a combination of the flow and the total resistance to flow. I know this seems pedantic, but right now I'm involved in a heated argument with someone who feels that for a given flow rate, a higher dP is somehow "better" than a lower dP.

With two liquid-full pressure vessels at the same pressure, if you cross-connect them, Archimedes Principle would say that the pressure will equalize without any mass flow. The liquid doesn't have to "go" anywhere for the pressure to change in both vessels.

Think about a liquid-full vessel at a stable temperature. If you add heat to that vessel (through the wall) to raise the temperature 1F, then the pressure will increase about 100 psi. This is an example of a pressure change with no mass transfer. Your cross-connected vessels are another.

David

 

[Oldliar]

Okay, but back to the flow; as there is no mass transfer, there is no flow. So the answer must be that there is no flow (ie no movement of mass) between the tanks. Or?

 

[Compositepro]

If there is a pressure difference then there will be flow until the pressure equalizes. It may only be a fraction of a liter but there will be flow. The water will expand and the tank will shrink slightly.

 

[25362]

Assuming the containers are rigid, T1, at 6 bara, would decompress to about 4 bara, while T2, originally at 2 bara, would increase its pressure to the same level.

Considering a small change in specific volume, the transfer T1[→]T2, at 20[sup]o[/sup]C, would be about 1.8 kg.

 

[dcasto]

Zdas04, So higher energy cost for higher dP is better? Sounds like you are argueing with an account that likes simple rule.

Think about how two tanks of equal size temperature got different pressures. They added more molecules with a pump to the one with higher pressure. When you open the valve, the same number of molecules will need to be in each tank.

 

[Poppapetro]

Thanks for your help!

zdas04 (Mechanical)..."With two liquid-full pressure vessels at the same pressure, if you cross-connect them, Archimedes Principle would say that the pressure will equalize without any mass flow... Do you mean different pressures?... How does Archimedes Principle state this...Mathematically?

25362 (Chemical). How did you calculates the above?

 

[zdas04]

Dcasto,
Actually, the guy is a Chem E with a Masters who has spent his whole life in Artificial Lift with what is now a Major Oil Company. He keeps saying that if q=c(Pr^2-Pbh^2)^n then a bigger dP is more flow, so it follows that more dP is better for any flow. In a meeting last week I was reduced to yelling epithets at him because I couldn't get him to see that there is a certain resistance to flow built into the "c" term, and if resistance increases (through scale, or multi-phase flow) then for a constant "q", dP must be higher. I may start screaming again.

Poppapetro,
I haven't seen the mathmatical representation that the a force applied to a liquid will be transmitted throughtout the liquid in a very long time (first semester physics in the '70s?). I don't even know where to start looking for it. Some of the other posts have been correct that a measurable mass will transfer from one vessel to the other, since no liquid is truly incompressible.

David

 

[dcasto]

So, then he could approach the speed of light given enough dP with that logic.

 

[zdas04]

More like the volume of the universe traveling at the speed of light, not sure what that volume flow rate works out to (at standard conditions).

 

[quark]

The assumption of non compressibility is wrong here and infact you should use that concept to solve the problem. The bulk modulus of water is 2.2x10[sup]9[/sup]Pa. The equilibrium pressure will be 500+100 = 300 kPag.

There is 200 kPa increase in pressure, so the volumetric increment will be (200 kPa/2.2x10[sup]6[/sup]kPa)x20000kgs = 1.8kgs (I would prefer it to be in liters but there may be some reason why 25362 mentioned it in kgs)

 

[25362]

To quark, I did simply because I used a table of densities in kg/m[sup]3[/sup] from NIST.

 

[quark]

Ah, simple yet perfect. Now I know why you were considering pressure in absolute terms. Get the star.

 

[25362]

quark,

your approach is, at least, more apodictic and valid than mine, thus you certainly deserve a star.

 

[BigInch]

If you have two tanks "absolutely full of water" (also assumed at the same elevations), but at different pressures, the water in the high pressure tank is more compressed (yes water is compressible) than the water in the low pressure tank. When you open a connecting valve, the water will flow from the high pressure tank to the low pressure tank. Only enough water will flow (expand is a better concept) from the high pressure tank to the low pressure tank until the density of the water in both tanks is exactly equal, which also equalizes all pressures.

http://virtualpipeline.spaces.msn.com

 

[BigInch]

BTW, the fluid usually oscillates with back and forth flow in the connecting pipe around three or four times, until it settles down. (If you open the valve quickly.)

http://virtualpipeline.spaces.msn.com

 

[sailoday28]

With regard to final equilibrium pressure--
The mass inventory in each tank should be readily calculated from the initial conditions.

After the valve is opened and the pressures equalize, the final pressure may be calculated assuming internal volume of tanks remain constant. Since system liquid mass is constant, system specific volume is constant and known.
For ISOTHERMAL conditions and complete mixing, the known specific volume and temp will yield final equilib pressure.

Similar to the above, if the process is ADIABATIC, use the known initial total suste, internal energy to balance the final total system internal energy. When this occurs,the equilibrium press is obtained.
Regards

 

[sailoday28]

BigInch (Petroleum)Without a knowledge of lengths between the interconnecting piping to the quick opening valve nor the diameter, how can you justify the 3-4 oscillations?
Regards

 

[vzeos]

zdas04,
Apparently you and your colleague are talking past each other. Your colleague is correct. The equation that you wrote, q=c(Pr^2-Pbh^2)^n, appears to be a variant of the pipeline equation for high pressure gas. Pipeliners usually write (P[sub]1a[/sub][sup]2[/sup] - P[sub]2a[/sub][sup]2[/sup]) =kQ[sup]2[/sup] where P[sub]1[/sub] is the source pressure P[sub]2[/sub] is the terminal pressure, the a subscript signifies the absolute pressure, k stands for all of the other parameters in the equation that are being held constant and Q is the flow rate. The term (P[sub]1a[/sub][sup]2[/sup] - P[sub]2a[/sub][sup]2[/sup]) is commonly referred to in conversation as K Q Squared. The relationship is nonlinear and operating the pipeline at a higher source pressure will give you a greater flow rate under the same pressure drop. Look at the example below. Operating the pipeline with a source pressure of 300 psig will give you 27% more flow than if you operated it with a source of 200 psig while in both cases the pressure drop is 100 psid.
[tt]
P[sub]1[/sub] P[sub]2[/sub] P[sub]1[/sub]-P[sub]2[/sub] P[sub]1a[/sub] P[sub]2a[/sub] (P[sub]1a[/sub][sup]2[/sup]-P[sub]2a[/sub][sup]2[/sup]) (P[sub]1a[/sub][sup]2[/sup]-P[sub]2a[/sub][sup]2[/sup])[sup]0.5[/sup]
200 100 100 214.7 114.7 32940 181.49
300 200 100 314.7 214.7 52940 230.08

230.08/181.49 = 1.27
[/tt]