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How area affects flow velocity

  1. Nov 4, 2008 #1
    I am a mechanical engineering graduate. working in Chennai.

    My question is regarding fluid flow from a tank.

    I consider a water tank of diameter 'D' meters with 'h' meters of water column and with a orifice of diameter 'd' mm.

    The Bernoulli equation says sum of pressure energy, kinetic energy and potential energy is constant for a fluid at any point in the flow field.

    Rearranging the energy terms a equation for velocity can be derived in terms of pressure energy and potential energy. The size of the orifice does not feature in this equation.

    So irrespective of the orifice diameter will the water flow out of tank with same velocity?
    This is my question. Can anybody please help me?
  2. jcsd
  3. Nov 4, 2008 #2


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    Let's see your derivation. The orifice area directly affects the flow. I know I have looked at a DEQ for draining a tank before. Let me see if I can find the post.
  4. Nov 4, 2008 #3


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  5. Nov 6, 2008 #4
    yes velocity of the jet would remain constant. But as the diameter of the orifice increases, flow rate ll increase, & you ll need to replenish the lost fluid to maintain a constant head 'h'

    Also, if orifice diameter gets too big, velocity would vary across it
  6. Nov 6, 2008 #5


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    Brain fart there, Fred...

    The one issue that may be worth considering here is that due to viscosity effects, the efficiency of your orifice will be lower the smaller the orifice is. Besides that, yes, the flow velocity (not flow rate) is constant regardless of the area of the orifice.
  7. Nov 7, 2008 #6


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    In the general solution, no, the velocity will not be the same. You've forgotten to account for the conservation of mass (continuity equation).

    The exit velocity at the tank would be:

    [tex]v_2 = \sqrt{{\frac{2}{1- \left( \frac{A_2}{A_1}\right)^2}} \cdot \left(\frac{P_1 - P_2}{\rho} + g \cdot h \right) [/tex]

    P1 is the pressure at the top of the tank.
    P2 is the pressure at the bottom of the tank.

    If the tank is open to atmosphere and discharges to atmosphere the equation will obviously reduce. If not, use P1 and P2 as applicable.

    h is the height from the top of the fluid in the tank to the centerline of the orifice.

    A1 is the area of the tank.
    A2 is the area of the orifice (outlet).

    g is gravitational acceleration.

    [itex] \rho [/itex] is the fluid's density.

    Plainly the exit velocity is a function of the orifice area.

  8. Nov 7, 2008 #7


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    The OP was non-specific, stewartcs, but due to the choice of units, you can assume that the orifice diameter is to or three orders of magnitude smaller than the diameter of the tank. So the tank empties slowly and the velocity decreases slowly.

    Also, your point is not using a steady state (and the OP didn't really specify). In other words, the initial velocity is basically independent of orifice diameter, but the velocity some amount of time later when the tank is partly empty will depend on it. But change in velocity over time as the tank empties is not something that was spelled out in the OP, so I wouldn't assume he was asking that.

    Ther is no reason to insist on such complications to what is probably an ideal case and a question that was probably intended to be basic.
  9. Nov 7, 2008 #8


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    The OP asked this:

    If one assumes that the areas are significantly different, then the equation reduces such that the velocity is essentially independent of the orifice area. However, that's not what the OP ask. The OP asked about the derivation of the Bernoulli equation as applied to a tank draining problem and specifically asked if the orifice area would affect the exit velocity. To say that the orifice area does not affect the exit velocity is wrong in regards to the derivation of it. After he derives it, and makes the assumption that the orifice area is significantly smaller, then he may reduce the general solution to one that doesn't include the ratio of areas...but not before.

    The equation I gave is for the initial exit velocity and the orifice area does affect the initial exit velocity. Run a few hand calcs with that equation with the only difference being the orifice area and you'll come up with different initial velocities.

    The only thing I've insisted on is that the OP get the correct answer to his question. Including the viscosity effects as you suggested is the only unnecessary complication that I can see.

  10. Nov 10, 2008 #9
    Hello guys I am glad that I got so many good responses. I was out of town for a week on a official tour. Saw all your replies only today .

    After reading the replies I got another question in my mind which I have expressed as a picture.
    Consider a tank of diameter 100cm and head of water column is 75cm. The tank is fitted with a short pipe of diameter 10cm and length 25cm and is fitted with a open/close valve at its free end.

    Now will the water flow velocity from the pipe be same,for various positions of the valve, by considering this to be analogues to the flow from the orifice.
  11. Nov 10, 2008 #10
    Ideally yes.

    If friction is considered, then no. effective head ll be reduced, & hence velocity.
  12. Nov 10, 2008 #11
    I just read your first post again. You said you are a graduate mechanical engineer. I guess fluid mechanics course covers such novice concepts.
  13. Nov 12, 2008 #12
    Well our education system is not so strong and it focusses only in the examination point of view.

    Now my work is related to compressible and incompressible flow and I feel a strong need to understand the science behind these flow phenomenon.

    Well coming back to my previous post, in reality as the valve is opened more the water strikes the ground farther away from the pipe which only means that velocity is increasing as the flow area is increasing.
    This looks different from what happens in a nozzle where the velocity increase as the area decreases.
    How this happens?
  14. Nov 12, 2008 #13
    Don't say "our", because i am also a part of it, its your fault that you are concerned about exams only & not the subject. U know, I ll most probably get something from the mods here, but heck, it is people like you in this country who destroy everything people like me want to have. people here have dishonored education itself.

    Coming back to your question, as the valve is opened, pressure drop(head loss actually) across it reduces, effective head increases, & hence veloctiy increases.
    Heff = Htotal - hloss

    In case of nozzle you are talking about the velocity increase at the minimum cross section, in case of valve, you are talking about the flow downstream.

    Ever heard of pressure recovery??
  15. Nov 12, 2008 #14


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    I'm an ME with only an undergrad degree and this is basic fluid dynamics. I'm a little shocked by this thread, so no grief from me...
  16. Nov 12, 2008 #15
    Well Ank_gl I was speaking of education in a state called TAMILNADU in INDIA. I bet both you and RUSS WATTERS are not from my place.
  17. Nov 12, 2008 #16
    In any case guys my intention is not to point fingers at any system or person.

    It's just my feeling that my system didn't give me enough opportunity to prepare myself in a better way.

    At this point of time in my life I am able to recognise the strong correlation between theory and practice.

    Ank can you explain what you said in detail please....
  18. Nov 12, 2008 #17


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    For anyone interested, this is why the velocity of approach in a flow calculation looks like it does:

    For a tank with station 1 at the top and station 2 at the orifice:

    [tex]p_1+\frac{1}{2} \rho V_1^2 + \gamma z_1 = p_2+\frac{1}{2} \rho V_2^2 + \gamma z_2[/tex]

    If we say that:
    P1 = P2 = z2 = 0 and
    z1 = h
    D = diameter of the tank
    d = diameter of the orifice

    then the Bernoulli equation reduces to:

    [tex]\frac{1}{2} V_1^2 + g*h = \frac{1}{2} V_2^2[/tex]

    Since [tex]Q_1[/tex] must = [tex]Q_2[/tex] then

    [tex]\frac{\pi}{4}D^2V_1^2 = \frac{\pi}{4}d^2V_1^2[/tex]

    That means that

    [tex]V_1 = \left[\frac{d}{D}\right]^2*V_2[/tex]

    Combine that with the original equation and we get

    [tex]V_2 = \sqrt{\frac{2gh}{1-\left(\frac{d}{D}\right)^4}[/tex]

    Anyone familiar with orifice or nozzle flow calculations will see the term [tex]1- \beta^4[/tex] in the final term.
  19. Nov 12, 2008 #18


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    Hi mkkeyan,
    Welcome to the board.

    Fred Garvin has provided a derivation of Bernoulli's equation that indicates a correlation between d and D. Note also, stewartcs has provided essentially the same equation. Both relate V2 to a function of 1/(1-(d/D)4)

    So what does that really look like in practical terms? Let's take a look...

    If d is 38% the diameter of D or smaller, the increase in velocity V2 is only 1% or less. In other words, when d is small relative to D, there is no significant increase in velocity V2. It's only when d grows to be about 64% of D that the velocity can increase by 10%. Generally, when we talk about an 'orifice', we often (not always) neglect the increase in velocity due to the geometry simply because d is generally small in relation to D. For orifice flow meters however, where d is typically large and accuracy is of paramount importance, this factor that Fred is refering to, (d/D)4, is important and is always taken into account.

    Let's neglect the slight increase in velocity due to the increase in valve opening. It's very small, and isn't needed to explain the phenomena you want to understand. As the valve is opened, the flow area through the valve is increased. So although velocity will remain constant, the flow area and thus the volumetric flow increases.

    As flow rate increases, the velocity THROUGH THE PIPE increases. It is this increased velocity in the pipe that is causing the water to come out faster, "striking the ground farther away from the pipe".

    Note also that as velocity in the pipe increases, there will be a higher resistance to flow due to 'frictional' resistance with the pipe walls. This is an irreversible pressure drop which can be calculated using the Darcy Weisbach equation. In fact, the DW equation is typically added into the Bernoulli equation to give overall pressure losses for a piping system. Check my post (#3) on https://www.physicsforums.com/showthread.php?t=271285" for a link to some reference material concerning pipe pressure losses and how they are calculated.
    Last edited by a moderator: Apr 23, 2017
  20. Nov 12, 2008 #19
    i don't agree, this isnt what happens. First you say area increases=>flow increases, then you say flow increases=>velocity increases. That is equivalent to area increases => velocity increases. flow is incompressible & hence violates law of conservation of mass.

    It is the head loss at the valve which reduces, which causes the velocity to increase.
  21. Nov 13, 2008 #20


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    ank_gl. You've taken it out of context. There's a (minimum) flow area through a valve which causes the restriction (ie: "head loss") that's analogous to the orifice. That flow area is typically the curtain area across the valve plug. This minimum flow area is what mkkeyan is asking about when asking why flow velocity through a pipe should increase as this curtain area (and analogous orifice area) increases. So why should increasing the curtain valve flow area (which decreases head loss through the valve) result in increased velocity through the pipe? Answer: because there's more flow! Hope that clears it up for you.
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