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Bernoulli Explained

  1. May 21, 2007 #1
    Hello, I am studying fluid mechanics and we have worked on Bernoullis theory a lot over the past couple of years. I am really struggling to grasp the concept of the theory and what it is about. Bernoulli states that in a system the pressure will always be constant (hence Pt = Ps + Pv + Ph) at any point.

    Can any one explain this to me. My understanding is that the more head in a system over a particular outlet the higher the potential (Ph) and the static head (Ps), so how can the outlets with a lower head equal the same pressure?

    Would static pressure be calculated as the potential head (p.g.h) + atmospheric pressure, then the total pressure would be the sum of the Ph + Ps + the dynamic velocity which would be given in m/s? Again my understanding is the more head involved the higher the velocity pressures, so how does a tap with only 3m of head have the same total pressure as a tap with 10m of head?

    If there's any examples of this to show the thoery or simple table calculations this would be very useful.
  2. jcsd
  3. May 21, 2007 #2
    Along a streamline

    rgh + rv²/2 is constant (if there is no losses, by friction for example)

    this says that if gravity is pushing a fluid over an height "h", its kinetic energy will increase.

    This conservation law above is the energy conservation: the work done by the gravity on a fluid is not lost but transformed in kinetic energy, or vice-versa. Obviously, this conservation law can be derived from the laws of motion of a (perfect) fluid: the Navier-Stokes equation without viscosisty losses.

    When there are losse, these must be introduced in the Bernouilli equation.
  4. May 21, 2007 #3
    Right so can you take a look at the following for me:

    (using 1000kg/m3 - density, 9.81m/s - gravity and 101325Pa Atmos)

    Tap 1 - 5m of head

    Potential (p.g.h) = 1000 x 9.81 x 5 = 49050Pa
    Static = 49050Pa + 101325Pa Atmos
    Dynamic (0.5pv2) = 49005Pa

    Total for tap 1 = 199380

    Tap 2 - 10m of head

    Potential (p.g.h) = 1000 x 9.81 x 10 = 98100Pa
    Static = 98100Pa + 101325Pa Atmos
    Dynamic (0.5pv2) = 98000Pa

    Total for tap 2 = 297425Pa

    Now am i right in saying the totals should add to the same amount, or have i done part of the calculation wrong? I have added potential on to the static, and added the sum of those to the dynamic (using 0.5pv2).

    Am i missing the important part of this theory? If so could you correct?
  5. May 21, 2007 #4


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    Why did both the static and dynamic pressure increase for the second tap? What is the background to the problem you are referring to?
  6. May 21, 2007 #5
    I am trying to understand how total pressure is calculated and how all system outlets will equal the same pressure (constant) within a system as bernoulli states.

    To answer your question the static pressure increased as the potential increased, and to calculate the static pressure i just added 101,325 pascals onto the potential. As for the dynamic pressure, using 0.5.p.v2, the velocity obvisouly increased as there was more head available.

    My problems relate to example questions i have been recentely calculating. But as of yet i have struggled to make all total pressures the same?

    Can this be done in table format? with static, potentials and dynamic pressures? If so is the static pressure, just the atmospheric pressure + the potential or am i doing this wrong.
  7. May 21, 2007 #6

    You concentrate too much on calculating something.
    You should start by stating a real physical problem.

    Do you know how to measure static and dynamic pressure, do you know what a Pitot tube is, have you made some measurement that you try to analyse?

    If you are dealing with paper exercices, you should try to understand their meaning first, in second the principles, in third the "how to do something".

    Sorry, but I don't understand anything to your tap problem.
    Could you try to explain us the purpose of this problem, and its meaning?
  8. May 21, 2007 #7
    The question i am relating it to is from the below. To calculate static pressure is this the potential pressure + atmospheric pressure = static?

    We were also asked to tabulate our results with 4 columns: Static, Potential, Dynamic and Total pressures, and were told all totals should be the same for each tap. (see below)

    An open tank of water is sited on high ground, with its base 50 m above datum level. It is filled with water to a depth of 2.5m. From the base of the tank, a straight pipe leads out, sloping downwards at a gradient of 1:100, terminating at 0.0 m above datum. At intervals of 1000m along the pipe, tap connections are provided.

    Calculate the static, potential, and dynamic pressures at each of the 5 taps when all the taps are turned off. Tabulate your results, your totals should be all constant.

    This is my problem, i know how to calculate each individual pressure but the totals dont add up? Should they?
  9. May 22, 2007 #8
    Ok, that's clear now.
    To make my answer shorter, I give you pressure in special units: meters of water.
    Again to be shorter, I express the pressure with respect to the atmospheric pressure, this is I assume at the top of the tank pstatic=0.

    At the base of the tank you have:
    static pressure = 2.5 m
    potential = 50 m
    dynamic pressure = 0 m
    total = 52.5

    At the first tap, 10 m below you have:
    static pressure = 12.5 m
    potential = 40m
    dynamic pressure = 0 m
    total = 52.5

    etc ...

    When you go lower along the pipe, the static pressure increases by the weight of the water above, but the potential decreases by the same quantity.
    Last edited: May 22, 2007
  10. May 22, 2007 #9
    Ok thank you, so you potential for the 10m tap would be total head 52.5 - the water head and the 10ms?!?!

    So how would this be expressed in pascals as apposed to meters??!
  11. May 23, 2007 #10
    1 m of water procuces a pressure of 1000 kg/m³ * 9.81 m/s² * 1 m = 9810 Pa

    additional comment:
    the atmospheric pressure compresses the whole system by 1 atm,
    but this play no role since the atm pressure does not change a lot on a heigth of 50 m and since water is not compressible
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