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A basic fluid mechanics concept that's puzzled me for years

  1. Dec 4, 2014 #1
    And it seems like it's so straightforward for everyone else, which is incredibly frustrating. It's about the relationship between pressure and flow rate of a fluid moving in a pipe or closed system.

    Often what I've heard is, they have an inverse relationship...but then there are cases where they're directly proportional...I think the problem may be that I'm conflating different types of pressures...maybe the best way to ask it is this...when we speak about discharge pressure of a pump for example...what does that physically mean?

    For example, a C-Pump with an open impeller will produce fluid with high flow rate and low discharge pressure...ok, I get that it's low pressure in the sense that, there is less resistance to flow, i.e less pressure which then leads to high flow rate....but at the same time, a fluid moving faster, exerts more pressure across a fixed cross-sectional area than a fluid moving slower correct? So, which kind of pressure is the one usually referred to when speaking about "the pressure of the fluid" and why is that particular pressure significant?
     
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  3. Dec 4, 2014 #2
    It seems like you are talking about 2 different situations. One is laminar and the other is turbulent.

    The pump will cause turbulence at its exit. This is why most pumps are designed to push fluid instead of pull so the flow can straighten out over the pipe. The only exception I can think of is a hand water pump. This pulls water because it's just not a good idea to put a pump underground and if you do dig that far you might as well dig a well.

    The other case is laminar and applies to nozzles.

    The diverence is viscocity in the equations of conservation.
     
  4. Dec 4, 2014 #3
    You should be careful what you mean when you discuss pressures regarding pumps. A centrifugal pump will not develop "pressure", it develops flow and overcomes head. When the fluid is flowing, you'll be able to read pressure at the discharge of the pump as the fluid travels from a high energy state to a low energy state at the exit of the pipe system, but if you were to close off the discharge pipe, the pressure would fall off as the pump dead-heads (basically it will just slosh water around inside the volute).

    I don't know where you read about the high flow and low pressure though. I constantly work with pumps which can discharge huge volumes of water up hundreds of feet of elevation through miles of pipe. There's a lot of backpressure (in the form of friction loss and elevation change) on a system like that, so it's not unusual to see pumps which are capable of overcoming 1000+ psi ---- while the liquid is flowing.
     
  5. Dec 5, 2014 #4
    Yes I know this happens, but why does it happen? Can you explain it in more detail, I've forgotten some of the most basic principles tbh :/

    Also, I'm not fussed about the type of pump, it could be a positive displacement pump, my main concern is, when we speak about discharge pressure of a pump for example...what does that physically mean? I mean is it the pressure of the fluid against the pipe walls? If it is, then why is it said that a lower flow rate will give a higher pressure, and vice versa...wouldn't a fluid moving faster exert more pressure against the pipe?

    I feel like I might be missing something...
     
  6. Dec 5, 2014 #5

    Astronuc

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    Pumps, that is the impellers, displace the water at some velocity, and the work imposed by the pump changes the momentum of the water. Pressure on a pipe system usually comes from a standpipe or elevated storage tank or pressurizer (head). There is a static pressure and a pressure drop. In a closed loop, e.g.,. in a pressurized cooling system, one can have a pressure of 2000 psia and a differential pressure (pressure drop of 50 psid), so the pressure at the inlet of the pump might be 1950, and the discharge pressure would be 2000 psia.

    Similarly, in a high voltage (high tension) line, the voltage of the line at one location might be 345 kV, and hundreds of miles away, might be 328 kV, so the voltage drop along the line is 17 kV.
     
  7. Dec 5, 2014 #6
    Well, a pump per se does not produce a pressure but flow.
    The pressure at the pump is a result of the system the pump is connected to.
    With a system that has a very low resistance such as a large diameter pipe or a short pipe for example, the pump can basically pump the fluid at a maximum possible rate that the particular pump is capable of doing. Continuing with the example, if the same pump is now connected to a system that has more resistance to flow, such as a narrower or longer pipe, the pressure that the pump sees is now greater. With a positive displacement pump where the flow is ideally constant, the pressure that the pump sees can increase dramatically as the system resistance to flow increases. With a centrifugal pump, one will see the flow decrease and the pump pressure increase, as the restriction increases. But see travis King's post about centfugal pumps when the restriction to flow is too great.
     
  8. Dec 5, 2014 #7

    jack action

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    There is no conflict here. The fluid contains energy, mainly on 3 forms: potential, kinematic, and static. Let's forget about potential energy (difference in level) and concentrate on the other two which are related to the fluid velocity (½mv²) and the fluid pressure (PV). As long as you don't add (with a pump for example) or remove (with a piston for example) energy from the fluid, the following will be true (the energy is constant):

    PV + ½mv² = constant

    or, if we divide by V:

    P + ½ρv²= constant

    Which is the famous Bernoulli[/PLAIN] [Broken] equation.

    When the fluid is in motion, the velocity v is high, which means that the pressure P must be low. When you say:

    This high pressure across a fixed cross-sectional will appear only when the fluid will be decelerated (thus, v goes down), usually when the fluid hit something. The faster the initial fluid velocity, the higher the potential pressure when the fluid is fully decelerated. Just imagine a car crashing into a wall: The faster it goes, the bigger the damages.
     
    Last edited by a moderator: May 7, 2017
  9. Dec 5, 2014 #8
    I guess that makes more sense, but what if the fluid hit nothing....if you had a flowrate of 50bbl/min does the fluid exert more or less pressure on the walls of the pipe of the same fluid flowing at 20bbl/min?
     
    Last edited by a moderator: May 7, 2017
  10. Dec 5, 2014 #9

    jack action

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    Assuming they contain the same amount of energy, the answer is less.

    That is the whole concept of flight:

    Faster fluid velocity on top of the wing compare to bottom -> Lower pressure on top compare to bottom -> wing goes up!

    But it would be possible to create a higher flow rate without dropping the static pressure, though only if someone increase the energy input of the pump.
     
  11. Dec 5, 2014 #10
    OP, just a point, it really does matter whether it's a centrifugal pump or a PD or what have you. They all operate differently. A diaphragm pump can provide a whole range of flows at a given pressure (based on it's displacement and the period between strokes); whereas a centrifugal pump operates along a curve which is determined by the geometry of the impeller and volute. Basically, as the other folks here have mentioned, the pump is doing work on the water by spinning it and throwing it to the edges of the volute. This work imparts energy into the system. The system wants to be at a low energy state and will work to balance itself. It does this naturally by balancing whether the energy is lost do to useful work (lifting the volume of fluid to a higher elevation) or friction losses.

    Since pipe systems are typically static (they don't change much) they have specific properties. A pipe system has a static head (elevation change) and from its geometry one can determine the head loss ("pressure" loss) due to friction (from pipe walls due to velocity, from valves, from elbows, etc) for various flow rates. This is called the system curve and will increase rapidly as fluid volume increases. As you may know, centrifugal pumps have curves as well, and where these curve intersect is where the pump will operate (and thus, how much flow you'll see in the system). If you have a valve in the pipeline and you start to close it, you are increasing the resistance in that line and you will start to make that system curve steeper and steeper. This will in turn make the point of intersection ride leftward up the pump curve (as the pump must overcome more head) and as a result there is less fluid being pumped. This is a good quick summary: http://www.engineeringtoolbox.com/pump-system-curves-d_635.html

    That statement about pumps having higher pressure at lower flows is sort of a misunderstanding. They develop lower flows at higher back-pressures.
     
  12. Dec 5, 2014 #11

    russ_watters

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    It doesn't seem to me that anyobe is actually answering the OP's main question, which is actually a pretty simple one: the discharge or differential pressure of a pump, read from a curve or rating sheet is typically the STATIC pressure.

    So the reason why low pressure and high flow go together (as seen on the pump curve) is conservation of energy in Bernoulli's equation: for high flow, the energy of the impeller stays in flow/velocity pressure and isn't converted to static pressure.
     
  13. Dec 5, 2014 #12
    Yes! Thank you I was getting frustrated I didn't know how else to articulate my question..,

    and high flow = higher dynamic pressure, correct? If so, what is the significance of mentioning "low pressure" in reference to static pressure....why is static pressure significant? To be honest, I find the whole concept of static and dynamic a little confusing...what's the point of having two separate terms for pressure "static" and "dynamic" .. at the end of the day it's pressure due to the molecules of the fluid right?

    When we use pressure sensors to measure pressure, is the reading we get dynamic pressure? or static? or Total?

    Why do we need to know that static pressure is low at a higher rate...what are the physical consequences of that...?

    If I pumped fluid through an infinite pipe at 100bbl/min vs 20bbl/min which would make the pipe shake more?
     
  14. Dec 6, 2014 #13

    jack action

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    The static pressure is the pressure due to PV energy term I mentioned earlier. You can measure it with a pressure gauge. Th dynamic pressure is the pressure due to the ½mv² energy term I mentioned earlier. You cannot measure that with a pressure gauge unless you stop the fluid (v = 0). Usually you do so by putting the pressure gauge directly against the flow such that when the fluid hits the gauge sensor, it completely stops. You now measure the total pressure which is the static pressure + the dynamic pressure.

    To use your image, the static pressure is the pressure due to the immobile molecules and total pressure is the pressure due to the moving molecules. The dynamic pressure is the difference between the two.

    Like I said earlier, all it does tell you is how much energy is in the fluid. Example:

    Pressure is 1 000 000 Pa, density is 1 000 kg/m³ and velocity is 20 m/s. How much energy is in the fluid?

    1 000 000 Pa + ½(1 000 kg/m³)(20 m/s)² = 1 200 000 J/m³

    If we were to reduce the velocity of that fluid to 0, what would be its pressure?

    1 200 000 J/m³ - ½(1 000 kg/m³)(0 m/s)² = 1 200 000 Pa

    If we were to increase its velocity, how fast could we go? (Hint: You cannot have a negative static pressure so minimum static pressure is 0)

    sqrt((1 200 000 J/m³ - 0 Pa) / (½ (1 000 kg/m³))) = 49 m/s

    What you should notice now is that the energy per unit volume have exactly the same units as pressure (1 Pa = 1 J/m³). So when when we talk about total pressure (static pressure + dynamic pressure), we are in fact talking about the total energy content of the fluid.

    Another example: Which state of a fluid with a 1 000 kg/m³ density will require the largest energy input from a pump, one that its pressure is increased by 1 000 000 Pa and its velocity increased by 20 m/s or one that its pressure is increased by 1 100 000 Pa and its velocity increased by 10 m/s?

    1 000 000 + ½(1 000)(20)² = 1 200 000 J/m³
    1 100 000 + ½(1 000)(10)² = 1 150 000 J/m³

    So the first one would require a more powerful pump to deliver the same volume of fluid.

    Pipe shaking is a whole other thing ...
     
    Last edited: Dec 6, 2014
  15. Dec 6, 2014 #14

    Doug Huffman

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    I find the intersection of the system and pump performance curves illustrative.
     
  16. Dec 7, 2014 #15

    russ_watters

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    Yes.
    Static pressure is what pushes the fluid through the pipe and what you lose due to friction, but if you want to know about the energy being produced/consumed by the pump, you need all forms of energy (in Bernoulli's equation), so you have to make sure you include both in one form or another.
    Depends on the type and purpose of the sensor. Most pressure gauges you see attached to piping systems are measuring static pressure. But some flow meters use pito-static tubes, which measure dynamic (via total - static) pressure.
    Conservation of energy at the pump:

    http://www.jmpcoblog.com/storage/pump%20curve.jpg?__SQUARESPACE_CACHEVERSION=1368800250171 [Broken]

    You can see that for a real pump, flow rate goes up as pressure goes down, which keeps the pump power fairly constant as you run-out on the curve. That relationship breaks down near the left side of the curve as the pump impeller stalls.
    Shake? Velocity causes shaking (and noise). So the 100 bbl/min. Not sure what that has to do with the other part of the discussion though...
     
    Last edited by a moderator: May 7, 2017
  17. Jan 31, 2015 #16
    Interesant Thanks!
     
  18. Jan 31, 2015 #17
    Not to insert confusion here, but I can reduce head from a pump (into the system) and flow from it at the same time with a constant load. (variable speed drive)

    You have to think about what the downstream load is doing to understand. If you are pumping water thru a heat exchanger (or any load for that matter) on the discharge side of the pump, and the load on the pump is reduced via a flow control valve closing (throttling flow) on the inlet to the heat exchanger (reducing flow from the pump into the heat exchanger) the reduction in flow will correspond to an increase in pressure on the discharge of the pump.
    An increase in flow will decrease the pressure reading on the discharge of the pump. A decrease in flow will increase pressure on a pump.
    Think about a pressure gauge on a garden hose with the nozzle closed. Your pressure will be "x". and you have no flow. Open the nozzle and the pressure
    "X" will drop, and flow (thru the nozzle) will increase.

    I assume we are talking constant speed motor driven pumps.
    We run variable speed motors and turbine driven pumps where we can drop pressure and flow thru the pump by varying speed of the drive..

    Centrifugal pumps (impeller) produce velocity energy, the volute designed into the pump housing converts the velocity energy into pressure energy.
    Some centrifugals have no volutes, they have diaphragms to do the conversions.
     
  19. Jan 31, 2015 #18
    The physical consequences is effect on load, head, ect. On the receiving end of the pump.

    Is hammering the focus of that last part? If so it'd depend on more than flow and pressure and be a result of shockwaves. That is a whole other discussion entirely.
     
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