# Water speed in the closed system

Does water speed in the pipes having the same diameter is strictly the same in any point of a closed (heating) system (with pump working) ?
On the contrary, although liquids are "incompressible" in practice, however, does the piping still allows it to be so : the more far the water from the pump the slower it is moving ?

etudiant
Gold Member
The pumps have to work to overcome the pipe/fluid friction, so in a closed system, the throughput flow is the input flow to the pump.
In a pipeline, this means the longer the distance between pumps, the slower the flow and the lower the throughput.

I suppose, not only the flow slows more down the more far the liquid flows from the pump, but, there is a point at which its molecules stop for a moment - to accelerate after that towards the pump. It is at this point, I suppose, where the SP of the system stays always the same, independently on whether the system runs or stands still.
Am I right ?

etudiant
Gold Member
In an ideal world, where pressure pulses travel instantly, that should not be possible.
In practice, the pipe will always have friction and gradients to exert back pressure, so eventually the pump will stall. It seems plausible that the static pressure in the system at that point would be the same as if there were no pump.

russ_watters
Mentor
I don't think you answered the right question:

The velocity of water is the same everywhere in a pipe of constant diameter.

etudiant
Gold Member
I don't think you answered the right question:

The velocity of water is the same everywhere in a pipe of constant diameter.

It seemed to me though the OP was looking at a limit case, where the pump could not overcome the fluid friction in the system.

russ_watters
Mentor
Sounded to me like he was asking if water flowed at different velocities in different parts of the same system, based on distance from the pump.

jim hardy
Gold Member
Dearly Missed
I'm with Russ:

Does water speed in the pipes having the same diameter is strictly the same in any point of a closed (heating) system (with pump working) ?

Does not (area of pipe) X (velocity) = volume flow rate?
If you're not losing a volume of water anyplace how can it slow down?
Think about people on an escalator - the folks in the middle cannot slow down they'd get run over from behind.

On the contrary, although liquids are "incompressible" in practice, however, does the piping still allows it to be so : the more far the water from the pump the slower it is moving ?

Does the system branch out to feed several rooms?
If so, each room only gets a fraction of the total flow so it will move slowly within the individual branches.

Have faith in your formula : (Volume Flow Rate) = Area X Velocity ..
If you double the area by branching into two rooms(with same size pipe), you'll halve the velocity in each room.
In that sense, the piping arrangement might allow it.

let's say there is one longish line system with no branches, for instance. Two equal forces are excerted on the liquid by the pump. One force (decreasing along the line) pushes out and another force (increasing along the line) pulls the liquid back into the running pump. At some point in the system, where the two forces are equal, friction losses to = friction losses from. Standing pump system's SP there = running pump system' SP. As we are told "total pressure is constant along each streamline", how to convince oneself that the DP at that point is greater than the DP directly at the pump discharge (or directly at the suction point)?

jim hardy
Gold Member
Dearly Missed
how to convince oneself that the DP at that point is greater than the DP directly at the pump discharge (or directly at the suction point)?

by DP you mean dynamic pressure?

How could it be different? dynamic pressure is $1/2$ρV2 and in your unbranched line V is constant.

http://www.grc.nasa.gov/WWW/k-12/airplane/dynpress.html

One force (decreasing along the line) pushes out and another force (increasing along the line) pulls the liquid back into the running pump. At some point in the system, where the two forces are equal, friction losses to = friction losses from. Standing pump system's SP there = running pump system' SP.

Am I right here at least - or not ?

etudiant
Gold Member
I think you idealize forces somewhat.
There is viscosity and turbulence, plus pipe friction, so as the system gradually reaches the point where the pump is unable to push through the fluid on a sustained basis, things hiccup rather than acting uniformly. The pump may push forward a slug of water or it may just choke or cavitate, unable to keep up the flow. Smooth flow requires a reasonably large pressure gradient. Absent such, the models fail.

The major issue though is that you introduce a force that 'pulls the liquid back' into the running pump. That force does not exist.
In a circuit open to the air, atmospheric pressure may push fluid to fill in a void created by a pump, but in a closed circuit, atmospheric pressure is excluded. Fluid moves in response to pressure, not pull, but in a closed circuit, the pressure is set by the fluid intake of the pump.
So if friction starves the pump inflow, the system flow will decline accordingly. Nothing gets pulled, everything gets pushed and if it is not pushed hard enough, it does not flow.

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jim hardy
Gold Member
Dearly Missed
I'm not a mechanical engineer or thermodynamicist,
just an instrument guy who learned enough about fluid flow to understand its measurement with venturi type meters.
I wondered why as an undergrad in EE we took a course in fluid flow - but it turned out quite useful.
.........................................................................>>

In your simple system you described the fluid as incompressible and pipe diameter constant.
Therefore the velocity has to be the same everywhere as mentioned earlier.

There may well be some point in your system where the static pressure with flow is same as static pressure without flow. (Without flow the system is static - there is no dynamic pressure.)

Read up on Bernoulli's principle here - the energy in a flowing fluid is constant....
http://www.princeton.edu/~asmits/Bicycle_web/Bernoulli.html
Observe their assumption of no friction.

One can keep refining the equation by including more energy terms.
You have addressed only static pressure and dynamic pressure
the former being p ,
and the latter being $\frac{1}{2}$ρV2 .
You have not included a term for friction loss, or for work taken out of the system.
Were there no friction you wouldn't need the pump.

You can include in Bernoulli more terms - elevation, temperature, depending on how precise you want to be.

The friction losses in a real piping system eat away at that static pressure term.
Presumably that energy of pressure is changed into into heat.
It is replaced by the pump.
Since the dynamic pressure term is defined as velocity^2 and velocity is constant , heat could come from noplace else but pressure.

In a converging-diverging venturi flowmeter , downstream pressure will be slightly lower than upstream.
So the exchange between energies of velocity and pressure in a venturi is not quite perfect. But to my recollection it's better than 99% in a good one.
That's why we instrument guys prefer our flow venturis mounted horizontal, so we don't have to include an elevation term in our formula for flow.

Perhaps a genuine mechanical engineer can improve or correct my thoughts here.

old jim

Chestermiller and joh_eng
Chestermiller
Mentor
As an engineer with lots of practical fluid mechanics experience, I can tell you that I fully support Russ and Jim's assessment. If the fluid is incompressible, the mass flow rate is constant, and the pipe diameter is constant, the cross-section average velocity must be constant. I have no idea what you other guys are talking about.

joh_eng
Thanks for your answers ! I am very sorry that my last post has appeared - I did not want it to. I wrote, and re-wrote a rather long post but, as writing improve thinking I finally gave it up, but, somehow, that something popped up. Sorry to have taken your time answering it.

Only, Etudiant, in refrigeration there are both discharge and suction pressure technical terms. Are they correct because there takes place the different process - compression - not pumping (that is, positive displacement, although, there are PD pumps too)?

etudiant
Gold Member
In refrigeration, the fluid is not water, but usually a low boiling liquid that undergoes a phase change as part of the refrigeration cycle, picking up heat and then dumping it. That also involves pressure changes within the overall cycle, so there is room for all manner of surprising behavior.
I think your initial use of the work 'water' perhaps narrowed the focus of the discussion more than you had intended.

In refrigeration, the fluid is not water, but usually a low boiling liquid that undergoes a phase change as part of the refrigeration cycle, picking up heat and then dumping it. That also involves pressure changes within the overall cycle, so there is room for all manner of surprising behavior.
I think your initial use of the work 'water' perhaps narrowed the focus of the discussion more than you had intended.
BTW, water is used as refrigerant too.