Static Pressure Is Lower Than Vapor Pressure.

As I understand-

The water flowing through the pipeline will have the following condition '

pressure at inlet = (outlet pressure - pressure drop in the pipe due to flow) ie

P_in = P_out - δ_p

As you reduce the pipe size the δ_p increases and that means that the P_in reduces for the same P_out

In your case if we assume that the P_out is 1 bar, the P_in is going below the water vapour pressure at the pipe inlet (and may be a bit inside too) which would result in cavitation.

UD15

There's something wrong here. -0.9044 MPa = -9.044 bar. You can't have a static pressure below (approx) -1 barg (= 0 bara).

Cheers......Codey

This is puzzling. If the tanks are both open to atmospheric pressure, I don't see how one could get a local pressure of -0,9044 MPa. (Maybe -0,9044 kPa, but even then I don't think it's below the vapor pressure of 21°C water. But I don't usually work in SI, so I might be missing something.)

Vapour pressure of water at 21°C is about 25 mbar = 2.5 kPa. But that's absolute pressure, of course, and positive. OP needs to explain the problem a lot better if we're to help him.

Cheers..........Codey

Vapour pressure of water at 21⁰C is about 0.025 bar. _{(1-0.025 = 0.975)}

BTW: The pressure of course will be below atmosphere in the inlet of the pipe even if both the tanks are open to atmosphere (refer bernoulli's principle) and can be calculated easily by calculating the heads.

For this one must know the flow and that is calculate-able since we know the head (2 mt) and then for different pipe sizes the flow for the corresponding pressure (head) drop may be calculated - once one knows the pipe routings - pipe type, number of bends,...

Then at the inlet of the pipe the head can be re-calculated to correspond to the flow. Only there will be a slight complexity there due to the possiblity of vortex formation .

UD15

Just guessing:

Since the Vapor Pressure at 21degC is~2.5m it is higher that the 2m differential between the 2 tanks. Therefore, if a big enough diam pipe is used, there could be formation of vapor in big qty to form a plug at some bend in the pipe(?) preventing flow therefore, the levels of the tanks will not become equal.

{I am not familiar with 'Fathom-AFT' warnings}

No, vapour pressure 25 mbar = 0.25 m water, not 2.5 m.

Cheers............Codey

Sorry, I was just reading the other inputs and picked it as is.

Now I understand that the Op is using a program and is having these results... Well I think that there is some error in the data specified to the program since there should be no problem of cavitation for a pipe flow and no vapor pressure issue since he has a very high water head above the pipe inlet and outle: assuming he is talking about a transfer from the bottom of the 1st tank to the other bottom..

Hmmm some curiosities here

Definitely there is an issue with the units (I really dont like MPa, kPa or Pa)

-0.9044 MPa (presumably gauge) equates to -904 kPag or -9barg ie -8barA which is a theoretical pressure.

However the vapour pressure of water at 21°C is 25 mbarA, -0.975barg, -97.5kPag, -0.098 Mpag.

I do love infinite reservoirs they would solve so many problems. Anyway we have two reservoirs with a 2m head between them. As soon as the line is opened water will start to flow from the reservoir wth the higher level to the one with the lower level. I assume the reservoirs are inifinite in lateral dimension so that no matter how much water flows from reservoir A to reservoir B there is still 2m head between them.

I thinkthe issue is the performance with different line sizes.

If you use different line sizes the frictional losses are reduced with larger line sizes for the same flow and so with a fixed head as you would expect the flow increases as the line size increases.

There is a pressure drop in entering the pipe depeding on the shape of the inlet to the pipe. For a typical smoothinlet this is K=0.5 velocity heads. There is also a loss of 1 velocity head for the fluid in the reservoir being at rest and then being accelerated to flowing velocity in the pipe

Head Loss (m) = 1.5 * v^2 / 2 g (g = 9.81 m/2^2)

Although the volumetric flow increases with line size I would expect the velocity to be similar although slowly increasing with line size at a max 2m/s

Head = 0.3m

Pressure Drop = m *SG /10.2 SG = 1.00 for water

Pressure Drop = 0.03 bar = 3 kPa

So in the first few mm of the pipe pressure is 98 kPa(a) 0r -3 kPa

But where is the connecting pipe relative to the water surface - if it were 10m below the surface of the higher reservoir and 8 m below the surface of the lower reservoir then there is 1bar (101kPa) of extra pressure from the depth of water

I think the problem is that somehow the programme thinks that there has been a pressure drop of 0.97 bar which brings the static pressure close to the vapour pressure which would cause cavitation.

The only way I can see this happening is if the connecting pipework goes UP. Now in theory you could siphon between the two reservoirs but it might be impractical in reality but to a PC it might be acceptable but I think that would be the first thing to check

I think it is related to the NPSH (Net Positive Suction Head), available. This is calculated from the following:

NPSHa = Habs +/- Hz - Hvp - Hl

Hz= Static head

Hvp= Vapor pressure head

Habs= Atmospheric plus manometric pressure head at the top level of the liquid

Hl= losses in the line due to pipe and fittings

If NPSH of the system is lower than what is required by a pump, than the pump will cavitate. Now since there is no pump there will be no cavitation.