Pressure drop in a pipe (ie) Bernoulli 5

[mechengdude]

Forgive me if this is a bonehead question...I have someone telling me one thing that I thought knew what they were talking about but now not so sure...Anywho

If you have a pipe with liquid in it (incompressible) and you know the Pressure, temperature, flow rate, velocity etc. It branches into two pipes of equal size then the pressure does not change correct?! This is assuming no loss due to any fittings, pumps, valves etc.
thanks

 

[BigInch]

Without energy being expended, there can be NO flow, incuding the assumption of zero friction. You are contradicting the 2nd law of thermodynamics.

Your example, "Given a very low friction environment, the water flowing through a canal into a lake has a lower top water level than the water in a lake. The water moves up hill as the kinetic energy is converted to potential energy."
.. . is not correct.

If such a thing occurred (and it can and does), it would be due to the velocity head being converted to static head, which technically is known as a hydraulic jump. If an open channel led into a reservoir, one of three things can happen,

1. the velocity of the channel is subcritical and the liquid level from channel to reservoir would fall,

2. the velocity of the channel is at critical velocity, in which case there would be no change in liquid level from the channel to the lake

3. the velocity of the channel is supercritical, in which case the level of the liquid would increase going from channel to reservoir

In ALL CASES the TOTAL ENERGY UPSTREAM OF ANY FLOW is greater than TOTAL ENERGY DOWNSTREAM OF ANY FLOW.. OTHERWISE THERE CAN BE NO FLOW!

Bernoulli describes total energy Z potential due to gravity, P potential due to pressure, V kinetic energy, HL friction. If you don't want to consider friction for some reason, make HL zero as I did above.

http://virtualpipeline.spaces.msn.com

 

[katmar]

Pressure is only one of the forms of energy that is considered in the Bernoulli Equation. As rcooper has pointed out the total energy remains the same because some kinetic energy is converted to pressure. If there were no flow there would be no increase in pressure.

An analogous situation is that in a vertical column of water the pressure is lower at the top than at the bottom, but the water has no problem in flowing from the zone of low pressure to that of high pressure. The potential energy term in Bernoulli takes care of this effect.

Katmar Software
Engineering & Risk Analysis Software

http://katmarsoftware.com

 

[BigInch]

Yes energy cannot be created or destroyed. Given that, it must be that some is being converted, if motion is occurring.

Niagra falls, pressure is 0 at the top and the Lake Eire level and at Lake Ontario. Agreed, but IMO trivial and irrelavent to the OP.

Katmar. Not a good analogy. There is no flow in a barrel of water without some component of total energy being converted to drive it, yet you don't explain what it might be.

In every case of the OP branch problem, pressure increases in the branch outlets and flow is reversed OR there is NO flow, hence no pressure change, assuming potential z1 = z2.

http://virtualpipeline.spaces.msn.com

 

[Feric]

To All:

Let me try to go back to the very beginning of the problem.

If I got it right, initially we do know inlet pressure, temperature, cross sectional area of the pipeline and velocity. Friction losses can be ignored.

Therefore, the volumetric flow rate is:

Q = v1*A1

At all the time, at any point of the pipeline, the volumetric flow rate is constant -- Q = constant.

The pipeline cross sectional area is fully filled with the flow resulting in A = A1 = A1wetted.

At a certain point, the pipeline splits in two branches and the cross sectional area stays the same.

A = A2 and A = A3 and A2 = A3

If I am not mistaken, v1 = v2 and v2 = v3 and v2 = v3

As much volumetric flow rate goes in, as much volumetric flow rate goes out ...

The continuity equation needs to hold resulting in

Q = v1*A1 = v2*A2wetted + v3*A3wetted

Assuming that A2wetted = A3wetted = Awetted

v1*A1 = v1*Awetted + v1*Awetted

v1*A1 = 2*v1*Awetted

Awetted = A1/2 = A2/2 = A3/2

In my opinion this is what is happening, the wetted cross sectional area in branches is half size the cross sectional area of the pipeline.

In general, there is no pressure los due to friction, and both continuity and Bernoulli equations are satisfied.

The flow is from the left to the right ...

Once we can agree that v1 = v2 = v3, then everybody is right.

I just do not know how it turned out to be that v2 = v3 = v1/2 ... -- this is against physics and common sense

I would appreciate if somebody can check my understanding of the problem and my outcome ...

Thanks,

Gordan Feric, PE
Engineering Software

http://members.aol.com/engware

 

[25362]

It is a common error to assume that an incompressible fluid fluid moves in a horizontal pipe (without friction) only from higher to lower pressure heads.

It does so from higher to lower "total" heads. This is what the Bernoulli equation tells us.

It is quite common to read "higher gage pressures" downstream when a pipe diameter is enlarged, as a result of a drop in "velocity head".

 

[BigInch]

Gordan Feric, PE

I think you've got it very well indeed.

--------------------------------------
25362

I don't think its a common "error"; its simply the typical beginning of the construction of a common generic pipe flow problem. In a single horizontal pipe, most people's initial thought of the problem is, "OK I'll assume constant diameter until somebody says its not, as well as, "flowing full" until somebody says its not, as well as no holes in the pipe wall between inlet and outlet, no partially open valves, the outlet end of the pipe is open, etc. Most people know that water will flow in a vertical pipe, if you put it in the top and the bottom of the pipe is not closed, so I think a lot of time was wasted above discussing things that (I thought) were extraneous to the OP.

IMO, in solving for velocities in the tee of the OP, its pretty much a waste of time to discuss the zero velocity solution, as well as ask for all the confirming information to arrive at what should be the more or less direct solution of interest, immediately discounting all the possible trivial variations.

---------------------------------------------
mechengdude,

Can you confirm that the tee is flowing full in all branches?

Can you confirm that the inlet is at the same elevation as the outlets?

Can you confirm that the velocities are nonzero?
Can you confirm that the inlets and outlets are all fully open?

We know that the fluid is incompressible and there is no friction, but can you confirm that the fluid does not expand?

Since you know all the velocities, why not list them (or just one of them) and I'll fill in the rest.

.......

http://virtualpipeline.spaces.msn.com

 

[BigInch]

Although I do agree, and admit that in the above I assumed without apparent reason , that the pressure term alone adequately described the total energy grade line at any point.

http://virtualpipeline.spaces.msn.com

 

[mechengdude]

As evident in the many posts, what I thought was a "slam dunk" really starts the ball rolling so to speak. To back up, my initial question stemmed from a discussion with an electrical engineer who on a schematic stated that if flow came in one pipe at 100 psi and then split at a tee the resulting pressure would be 50 psi. I looked at it and said "no don't think so". I think he was confusing it with some Ohms law stuff. I then set off to consult my text books and the internet to confirm my opinion that there will be some pressure drop but it should be a fucntion of "mechanical losses, friction". Anywho the assumptions were;
- no change in elevation
- no change in pipe diameter
- all liquid flow, pipe full
- no losses due to friction

Out of curiosity, I'm sticking a quick model into COSMOSFloWorks to see what really happens and will post the results.

thanks

 

[Feric]

To All:

Guys, you are good -- your credentials indicate so.

This is a good engineering discussion.

It is my pleasure to be able to make my contribution and be a part of it.

Engineering should be practiced and discussed like this.

As it has already been pointed out, it is tricky to quickly figure out what is in and what is out of the problem.

I am looking forward to having more discussions and analysis with you guys.

Thanks,

Gordan Feric, PE
Engineering Software

http://members.aol.com/engware

 

[rcooper]

Your electric engineer is confusing himself. I know I am going to get shot down by my more learned colleagues here as the analogy is not perfect, but it is close.

Voltage is analogous to pressure (Pa)
Current is analogous to flow rate (m3/s)

Current is a measure of the flow of electrons i.e. a flow rate.

Voltage is what pushes the electrons through the wires. Pressure pushes the fluid through the pipe.

So ask your electrical engineering colleague if the voltage at a junction varies just in the branches assuming there is low resistance in the conductors.

 

[mechengdude]

Now we're cooking ... Thanks for all your input. I think the rcooper post summed up what I was thinking. I've pretty much convinced myself that the pressure does not "split". I ran a CFD on it that confirmed this. I'd like to post the results but it seems the cut and paste doesn't work and I'm not ready to go through the trouble of figuring out how to upload the pics to a remote site and link them.

If anyone has a easy way to post pics here let me know.

 

[fxcostac]

I`m Fully agree with your statement rcooper but now in a practical case it would be as having a flow from a less diameter pipe to a larger diameter pipe. And to be more specific from a D diameter pipe to a 2D diameter pipe.

We can solve that and see what happend with pressure, velocity and maybe other things in our flow.

I know that there are other effects here but we have almost neglected all of them.

 

[BigInch]

Bernoulli will tell you, just not the direction of flow, that is until you include the HL friction term.

http://virtualpipeline.spaces.msn.com

 

[sailoday28]

BigInch
If "downstream" pressure is higher than upstream pressure, what usually happens?

I don't know of any flow that goes from a low pressure point to high pressure point. That's one of the few times I can use "ALWAYS" flow is from high pressure point to low pressure point.

Consider flow into a venturi?
What happens in a venturi, from the throat to the outlet?
Does the flow reverse?

 

[GregLamberson]

sailoday28

Not sure I understand your analogy. Pressure on the "downstream" side of a venturi is the delta-P between "upstream" pressure and the pressure at the venturi. That would mean that pressure on the "downstream" side would be less, if it wasn't it would be the upstream side.

Flow would only reverse if the pressures were reversed. The point holds true, flow will always go from high pressure to low pressure.

Greg Lamberson, BS, MBA
Consultant - Upstream Energy
Website:

http://www.oil-gas-consulting.com

 

[BigInch]

Sailo very good, but I think I already acknowledged that Bernoulli tells nothing about flow direction until friction is included. Did you not read the thread? Please see above.

As I should have remembered, in a frictionless environment, Bernoulli says nothing at all about flow. In fact in the Bernoulli equation, it is only the friction term that has anything at all to say about flow, so if friction is neglected, there's not much about flow that anybody can infer at all. Without friction, its just an energy balance equation.

http://virtualpipeline.spaces.msn.com

 

[BigInch]

No Greg, he's talking only about the flow from the throat to the outlet. There pressure usually increases as velocity slows down and the fluid expands, so "in a sense", flow goes from low pressure to high pressure, because pressure, as it usually is, in this case is not entirely descriptive of total energy difference between those two points.

http://virtualpipeline.spaces.msn.com

 

[BigInch]

I think flow over a wing section would have been a better analogy though.

http://virtualpipeline.spaces.msn.com

 

[25362]

A constant area short dividing manifold could be another interesting example. In some geometries as flow is bled off the (static) pressure raises towards the end of the manifold resulting in a larger flow rate out of the end branches.

 

[BigInch]

25362 A most excellent and relavent example(*). It would seem that to assure balanced flows in parallel meter tubes, parallel process lines, parallel pumps, header and outlet size, and the relative flows pulled from each outlet can be quite critical.

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