# Exit velocity of a water pipe.

• posted
Assume a horizontal water pipe. Then Bernoulli's equation reads:
Pi + 1/2(density)(Vi)^2 = Pe + 1/2(density)(Ve)^2 , where P and V are
the pressure and velocity and the i and e subscripts indicate initial
and exit values. There is no height term since the pipe is horizontal.
Let's say the inital pressure Pi is very high say thousands of bars. I
want to find the velocity on exit from the pipe when it empties to the
ambient pressure, Pe = 1 bar.
In addition to the Bernoulli principle an equation used to calculate
the velocity of a fluid stream is the continuity equation AeVe = AiVi,
where A is the cross-sectional area of the pipe, with e and i again
meaning initial and exit values. This equation holds since assuming an
incompressible liquid the total volume flowing in must equal the volume
flowing out.
Therefore if the pipe is at constant diameter the velocity should stay
the same. But that means in the Bernoulli equation above the velocity
cancels out so the equation becomes Pi = Pe = 1 bar, which is
incorrect. What's the resolution of this?
Secondly let's say the pipe is of smaller diameter on exit. Then Vi =
(Ae/Ai)Ve. Substituting this into Bernoulli gives:
Pi + 1/2(density)(AeVe/Ai)^2 = Pe + 1/2(density)(Ve)^2.
Then Pi-Pe = 1/2(density)(1-(Ae/Ai)^2)Ve^2. And solving for Ve this
becomes:
Ve = [2(Pi-Pe)/(density)(1-(Ae/Ai)^2)]^.5
The curious thing about this equation is if we make the exit
cross-section arbitrarily close to the initial cross-section then we
can make the exit velocity arbitrarily large.
Is this correct?
Bob Clark
• posted
Dear rgregoryclark:
Which means density is not constant. Likely (unless the pipe is very long), the Mach number will come into play.
Bad assumption for "thousands of bars". OK for lower pressures, however.
The resolution is that above a certain pressure, water no longer has defined solid/liquid/gas states. This is the same for any material.
Velocity in = velocity out, yes. *IF* the fluid is incompressible.
David A. Smith
• posted
Bob--
Is your intention to pump liquid hydrogen under high pressure?
• posted
Or one.
Bob
• posted
I think you are not setting up the problem correctly. Rearranging, we have DP = 1/2 p (ve^2-vi^2), where DP is the pressure drop along the tube, and 'p' is density. If you have a tube of constant diameter, it's not clear how you generate a pressure drop in the first place. So, let's make a pressure head by the usual way (i.e. a bag of fluid suspended a height z above the tube, make the pressure drop across the tube equal to pgz). Then, because the bag is a reservior, the initial fluid velocity *at the bag* is zero. Then your equation works.
No- look at the second equation. As the area ratio approaches 1, the pressure drop approaches zero as well. So in the third equation, the ratio is undefined.
At least that's how I would resolve the difficulties.
• posted
I'm confused---As far as I know, the only time there is zero pressure drop in a pipe is when the flow is zero. One of the problems about being too analytical is that it never solves any real world problems. MLD
• posted
The mistake is in the first line. Bernoulli's equation does not apply in a pipe because there is always friction from the walls. With a constant cross-sectional area the velocity is constant to satisfy continuity for an incompressible fluid. Look in any Fluid Mechanics book at the chapter on pipe flow and the Darcy-Weisbach equation for the pressure gradient in a pipe.
• posted
Again the Navier Stokes equation comes to the rescue. (It explains gravity in Dual Space theory). In any case, it says dP/dx = -rho*acc If you apply instant pressure pi at time T = 0, then dP/dx = (Pe - Pi)/L = -rho*acc The column of water will accelerater at the rate of: dv/dt = -(Pe - Pi)/L*rho As velocity accumulates, the pressure difference will be diminished dv/dt = (Pe - Pi - Bv)/Lrho assuming B is a linear viscous coefficient of drag.
John Polasek
• posted
Nope. You've missed out pipe friction, which is the most important cause of pressure loss in pipes.
• posted
You have had a number of replies, most of which are correct but looking at different aspects. I'll add a pragmatic point of view.
The equation is correct. Any real pump has limitations. If you run it without anything on the outlet, it will produce a finite rate of flow, not infinite. For a pipe whose length is much shorter than the diameter, friction can be neglected as you did, but then there is nothing to produce a pressure difference so Pi = Pe.
_____ \ Flow \_____ ---> _____ / ^ _____/ | ^ | | | inlet exit
On the left you have a large area and low speed. For an incompressible fluid, the volume rate is the same on both sides so the speed on the right must be higher. In order to increase the speed, there must be a force acting so there must be a pressure difference, a drop from inlet to exit.
The higher the pressure difference, the greater will be the flow rate and vice versa:
Flow ^ | / | / | / | / | / | / |/ +-------> Pressure difference
Going back to our real-world pump, we can draw a similar graph. With no back pressure at the outlet, you get maximum flow. If you block the outlet (or in some cases at some low rate), the pump will produce its maximum pressure. In between, increasing the back pressure will reduce the flow.
Flow ^ | |__ | \ | \ | \ | \ | \ +------+--> Back Pressure
Plot the two graphs together and add the outlet pressure to the nozzle pressure difference (plus any frictional losses) to find the back pressure at the pump and you can predict how your system will perform:
Flow ^ | / cross-section arbitrarily close to the initial cross-section then we
As you make the constriction arbitrarily small, the resistance of the nozzle goes to zero and if you could maintain the pressure difference, the speed would go to infinity. In the real world, the flow rate reaches the maximum your pump is capable of delivering and the pressure falls. In this regime, the pump determines the flow rate while the constriction then determines the pressure.
Conversely, if you take the outlet area towards zero for a given flow rate, you can make the pressure go to infinity. In reality the pump reaches its maximum pressure and then the constriction controls the flow rate. Your kitchen tap should work in this regime.
Notice that in one case you have fixed flow rate and diminishing pressure while in the other you have fixed pressure and a small flow. Neither gives much thrust since that depends on the rate of change of momentum so contains the product of flow rate and pressure difference. For your purposes, the trick is to find the optimum constriction to maximise the thrust. The NASA page you quoted earlier has a page on nozzle design that explores that aspect.
George
• posted
...
Minor correction, of course friction alters the slope of the nozzle line, not the intercept.
George
• posted
Nice work George-- As per your explanation, I have had to use the graphical method many times to resolve flow problems primarily because it was way too cumbersome to try and do it analytically. Most of the time besides a simple line or pipe there was usually parallel paths and one or more orifices involved. You mentioned "real world" a couple of times and therein lies the big difference between the scientist and engineer. I have great respect for those that can mathematically (and analytically) define the boundaries of a problems such as, in this case, the flow in a pipe. However, and I speak from my experience only, I more often than not had to tell these guys that what they are proposing and describing just won't work "in the real world". Pumps, orifices and nozzles etc. just don't always follow the pure mathematical approach. BTW, the pump characteristics that you mention follow the lines of a centrifugal pump. The positive displacement pump, such as a gear pump, would react differently in the same setup--Flow as a f(Speed)--some flow drop off as the discharge pressure increases. Limiting factor here is pump discharge pressure limited by a pressure relief valve. MLD
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and then there is the interesting problem of designing the pipe nozzle system for maximum nozzle jet power.
Design Result:
2/3 head loss in the nozzle 1/3 head loss in the pipe
Richard Saam
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Does that solution include the assumption of a a constant diameter pipe?
• posted
Yes

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