Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

I Trouble understanding energy aspect of Bernoulli's Equation

  1. Apr 26, 2017 #1
    9bda17fafb707c2e31e6bbc893d00b02be8dc740.png
    So the Bernoulli's EQ comes from conservation of energy. From the figure, I see that if Force 1 is greater than Force 2, the water will move to the right.

    The distance Force 1 travels gives work of F1d1, and the distance Force 2 is pushed back gives work F2d2, and net work on the system is F1d1 - F2d2.

    But then you have to equate the net work to the PE and KE change of the system, to get F1d1 - F2d2 = (0.5mv22 + mgh2) - (0.5mv12 + mgh1).

    But that seems to imply the volume of water displaced at the left suddenly appears at the displaced water on the right.

    I understand that both volumes of water are equal, but it seems as the though the water on the right side BEFORE it was pushed was already in the same pipe at the same height. So after pushing, the water at the right only changed horizontal position. Same for the left side, too. So shouldn't the differences in mgh be 0?
     
  2. jcsd
  3. Apr 26, 2017 #2

    sophiecentaur

    User Avatar
    Science Advisor
    Gold Member

    I think this is the crux of the question. If you consider an infinitesimal movement of water, each element of water in the sloping section will move upwards and gain GPE (and KE). The shaded portions in your diagram have no change in GPE (nor KE). The change in Energy is all in the rising, tapered bit. The increase in GPE is
    ρg∫A(h)dh (h from 0 to height H), which has to come to your mgh and the details don't matter too much - my Maths could be sloppy but . . . . My point is that the energy change is all in the sloping section. on the way up / faster.
     
  4. Apr 26, 2017 #3

    russ_watters

    User Avatar

    Staff: Mentor

    The two volumes of water are not different volumes of water being examined at the same time, they are the same volume of water examined at different times.
     
  5. Apr 26, 2017 #4
    If you envision a piston pushing the fluid below to the right and another piston being pushed by the fluid above to the right, then, during a small time interval, the potential energy of the volume of fluid contained between the two pistons (a closed system) increases.
     
  6. Apr 26, 2017 #5
    Can you explain why the KE in the flat sections doesn't change?

    I agree, but when you push the water, the left side water goes from the left pipe and into the beginning of the sloped section and the same volume exits the sloped section from the top and into the right side pipe. How can we compare the energies of the sections of water when they came from different places? It is not as though the water from the left pipe teleported itself to the right side.

    Shouldn't we be examining the KE and PE changes of just the water of one section? Like look at the KE and PE change of left side as it goes from the flat section to the entry of the sloped section?
     
  7. Apr 26, 2017 #6

    boneh3ad

    User Avatar
    Science Advisor
    Gold Member

    Maybe I ought to make a PF Insight about the Reynolds Transport Theorem. It's a lot easier, in my opinion, to approach Bernoulli's equation from the conservation of energy angle by starting with that.
     
  8. Apr 26, 2017 #7
    Sure, that would be great!
     
  9. Apr 26, 2017 #8
    You definitely need to include the KE and PE changes also. If you follow the approach I outlined in post #4, then you are dealing with a closed system that is experiencing a transient change, and work is being done on the fluid at both moving boundaries as well as the potential energy and kinetic energy of the fluid between the moving boundaries changing. The expanded version of the first law of thermodynamics captures all this if one neglects the change in internal energy (temperature of the fluid).

    Your course is trying to get you to transition from the closed system version of the first law of thermodynamics to the open system (control volume) version, which involves an energy balance over a control volume, with fluid entering and leaving. In such a case, the development usually starts out considering steady state flow, where none of the parameters are changing with time at all fixed spatial location. In this framework, a parcel of fluid entering at the bottom is imagined to instantly exit at the top, because nothing is changing with time in-between.

    The two developments I described are totally equivalent, and it is important to understand intuitively how the two are related. The Reynolds Transport theorem alluded to by @boneh3ad captures all this mathematically.
     
  10. Apr 26, 2017 #9

    sophiecentaur

    User Avatar
    Science Advisor
    Gold Member

    How is the KE changing if the water in the parallel sections is not accelerating? It did all its speeding up in the tapered section. (Ditto for the GPE)
     
  11. Apr 26, 2017 #10
    It looks to me like the diameter at 2 is less than the diameter at 1. So there is an increase in velocity and kinetic energy.
     
  12. Apr 26, 2017 #11

    boneh3ad

    User Avatar
    Science Advisor
    Gold Member

    Unfortunately, after a meeting at work today, it looks like I probably won't have time to actually do this for at least several more weeks, but at least the thread has inspired me to consider it for my to-do list.
     
  13. Apr 27, 2017 #12

    sophiecentaur

    User Avatar
    Science Advisor
    Gold Member

    Isn't the appropriate word "was"? If there is a change between states A and B, we don't normally look for an on-going change in state B if conditions don't change. All change has happened in the taper.
    I feel I must have missed something here but I have read and re-read the thread and I can't find it. Work 'was' done in the tapered section if the fluid is going faster. The pressure difference is another issue, surely and, as in analysing simple electrical circuits, the situation assumes steady state.
    Edit: I thought the OP was about this but , if you want to add a dynamic change to the situation (piston with pressure suddenly applied) then it isn't as simple and, with an incompressible fluid, there is, indeed, an instantaneous change in pressure between A and B. With a real fluid, a wave is involved.
     
    Last edited: Apr 27, 2017
  14. Apr 27, 2017 #13
    I guess I didn't explain very well.

    The beauty of the Bernoulli equation (which is nothing more than a steady state version of the open system form of the first law of thermodynamics, involving a single entering- and exit stream, with no heat transfer or shaft work, for an inviscid fluid) is that it allows us to treat the control volume as a black box, and to focus exclusively on the conditions at the inlet and outlet cross sections only.

    Anyway, no work is done in the tapered section because the walls are rigid. All the work is done at the inlet and outlet cross sections.

    I also didn't do a very good job about explaining with regard to my "pistons" example. I was trying to offer a different perspective on all this by reverting to a closed system transient version of the first law of thermodynamics in which the fluid behind the large slug of fluid we are considering (i.e., between the inlet and outlet cross sections) is acting as a piston to do work on our slug at cross section A, and the fluid ahead of the large slug of fluid we are considering is acting like a piston to receive work from our slug at cross section B. So there is no dynamic change associated with these fictitious pistons and no waves involved. The closed system version of the first law I am alluding to here is the expanded version, which includes changes in kinetic energy and potential energy (when appropriate):
    $$\frac{dU}{dt}+\frac{d(KE)}{dt}+\frac{d(PE)}{dt}=\dot{Q}-\dot{W}$$where, in this system,$$\frac{dU}{dt}=0$$$$\dot{Q}=0$$and the rate of doing work on the system is given by $$\dot{W}=\dot{m}(P_Bv_B-P_Av_A)$$with ##\dot{m}## representing the mass rate of flow, P representing pressure, and v representing specific volume.
     
  15. Apr 27, 2017 #14

    sophiecentaur

    User Avatar
    Science Advisor
    Gold Member

    If we were dealing with trucks of coal that were going from a dual carriageway to a single carriageway, each truck would need to be accelerated during the transition. That would involve a force and a distance, which is work doe on each truck. In the OP, both gpe and ke are increased in the taper and I don't see how that cannot involve work. There seems to be some sort of paradox - and there cannot be, so wassup? Is it just that your accounting is being done at a different place from mine?
     
  16. Apr 27, 2017 #15
    Like I said, all the work is being done at the inlet and outlet cross sections. Imagine your trucks without engines that are connected together by bars. You are pushing from behind and pulling from in front. That's all the work that is being done.
     
  17. Apr 27, 2017 #16

    sophiecentaur

    User Avatar
    Science Advisor
    Gold Member

    I appreciate that the fluid will only flow if power is supplied to the system so what you say is true because the work is all done by the pump. But how does that contradict my statement that work is done on the individual parts of the fluid that are actually being lifted or accelerated? Let some small sample of the fluid out from the narrow section and you could measure the KE and PE of the sample. An observer would see the acceleration in the taper and conclude that 'something' is providing energy to make it happen. I take your point that there are no little engines at work within the fluid so the ultimate source and sink of power is being supplied via the ends of the tube.
    If there were another taper, bringing the area back to the original area and height and there were no losses in the flow, the fluid in the narrow bit would have more energy but that energy would be returned to the system at the widening section. That could be a conceptual problem, I suppose; a sort of virtual energy. :eek:
    From what you say, it's as if we are 'forbidden' to look at it my way (??) which is only to look at what happens in one individual element in a circuit. (Yes, I accept that no energy escapes through the walls as it would in a resistor)
     
  18. Apr 27, 2017 #17
    There is nothing wrong with looking at smaller subsystems of a larger system. In that case, work is being done at the boundaries of the smaller subsystem, but, again, not within the subsystem.

    Work being done on smaller subsystems that are part of a larger system cancel out with one another, except at the boundaries of the larger system. Whenever we do an energy balance on a deformable system, we only include the work done at the boundary. So, once you define what you are calling your system, only the work done at the boundary is relevant.
     
    Last edited: Apr 27, 2017
  19. Apr 28, 2017 #18
    Can you explain why we are allowed to imagine that fluid entering at the bottom immediately exits the top? It doesn't seem like that's what happens in real life, so how does such a model work?

    In real life, the fluid that enters the pipe was just immediately left of the entrance, and the fluid leaving was just at the left of the exit. The two parcels of water aren't the same.
     
  20. Apr 28, 2017 #19
    They are not the same. But, if the system is at steady state, with nothing changing inside the control volume, it just works out mathematically to be the same as if the same parcel that entered the bottom exited immediately at the top.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: Trouble understanding energy aspect of Bernoulli's Equation
  1. Bernoulli's Equation (Replies: 12)

  2. Bernoulli's equation (Replies: 3)

Loading...