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Velocity of efflux

  1. Apr 20, 2016 #1
    1. The problem statement, all variables and given/known data
    A container of height H has a orifice at a distance h from the top, the ratio of the cross section area of container and orifice is. 1, find velocity of efflux

    2. Relevant equations
    Using Bernoullis equation we get v=(2gh)^.5
    Equation of continuity A1V1=A2V2
    3. The attempt at a solution
    Used Bernoullis equation to get v=(2gh)^.5
    I cant think of anything else, thanks in advance :)
     
  2. jcsd
  3. Apr 20, 2016 #2

    haruspex

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    Do you have some reason to doubt that answer?
     
  4. Apr 21, 2016 #3
    That's not the answer, the answer has ratio of cross sections in it. It comes out to be $$\sqrt{2gh} / (1-(a/A)^2) $$ where a and A are cross section of orifice and Vessel respectively
     
    Last edited: Apr 21, 2016
  5. Apr 21, 2016 #4

    haruspex

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    That clearly is not right in general. As a tends to A, the answer must tend to √(2gh) - that's just free fall - and certainly not tend to infinity.
    The question is not entirely clear about state. If this is shortly after the hole was made in the full tank, the velocity would be less than √(2gh) as the flow is still becoming established; if later, the water height is no longer h, so again it would be less than √(2gh). For it even to reach √(2gh), there would need to be a constant head arrangement, with water flowing in to maintain the h height.
     
  6. Apr 21, 2016 #5

    Let me phrase the question directly,
    Water is filled in a container upto 3 m. A small hole of Area a is punched im the wall of a container at a height 52.5 cm from bottom. The cross sectional area of container is A. If a/A=0.1 find velocity of water coming out of hole.
     
  7. Apr 21, 2016 #6

    haruspex

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    As I wrote, the velocity will not be constant. It will rise to a peak, then decline as the water level drops.
    It can never exceed, or even quite reach, √(2gh) as that would violate energy conservation (which is the basis of Bernoulli).
    Here's an analysis of what happens over time:

    A tank of horizontal area A contains liquid density ##\rho## to an initial depth y0.
    At one side near the base, there is a small hole area a.

    Rate of conversion from GPE to KE:
    ##\rho Ay\dot y\ddot y-\frac 12{\dot y}^3\rho A \left(\frac{A^2}{a^2}-1\right)=-\rho A y g \dot y##
    Writing ##k=(\frac Aa)^2-1##:
    ##y\ddot y-\frac 12k{\dot y}^2=-gy##
    Writing ##v=\dot y## and ##v'=\frac{dv}{dy}##, we have ##\ddot y = \frac{d}{dt}v =\frac{dy}{dt}\frac{d}{dy}v= vv'##.
    ##yvv'-\frac 12kv^2=-gy##
    Writing ##w=v^2##
    ##yw'=kw-2gy##
    General solution: ##w=By^{k}##.
    Particular Integral: ##w=cy##
    ##cy=kcy-2gy##
    ##c=\frac{2g}{k-1}##
    Using initial condition:
    ##0=By_0^{k}+y_0\frac{2g}{k-1}##
    ##B=-\frac{2g}{(k-1)y_0^{k-1}}##
    ##w=\frac{2gy}{k-1}(1-\left(\frac{y}{y_0}\right)^{k-1})##
    Peak velocity is at w'=0:
    ##w=\frac{2gy}k##
    ##\frac{k-1}{k}=(1-\left(\frac{y}{y_0}\right)^{k-1})##
    ##ky^{k-1}=y_0^{k-1}##
    ##y=y_0(k)^{-\frac 1{k-1}}##
    ##w_{max}=y_0(k)^{-\frac 1{k-1}}\frac {2g}k##
    ##v_{max}=\sqrt{y_0(k)^{-\frac 1{k-1}}\frac {2g}k}##

    Edit: in the original version of the above I left out a factor of 1/2 in the first equation. This led to 2k appearing subsequently everywhere that should have been just k.
     
    Last edited: Apr 22, 2016
  8. Apr 21, 2016 #7
    If I apply the Bernoulli equation (assuming that, because A>>a, the flow is in quasi-steady-state), I get:

    $$gh+\frac{1}{2}v_1^2=\frac{1}{2}v_2^2$$where ##v_1## is the velocity at the upper fluid surface and ##v_2## is the velocity out the hole. From continuity, ##v_1A=v_2a##. So,
    $$2gh=v_2^2\left(1-\left(\frac{a}{A}\right)^2\right)$$
    This works well for A>>a.
     
  9. Apr 22, 2016 #8

    haruspex

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    But, sadly, it is not valid. It illustrates the pitfalls of applying Bernoulli out of scope.
    Bernoulli's equation assumes steady state. The problem statement asks for the flow when the tank is still full, so this is not steady state unless there is a source of water maintaining the level and, moreover, that water must enter with a downward velocity of v1. It is that velocity which provides the extra energy to exceed the √(2gh) limit at the bottom.
    For a complete-ish analysis of how the flow varies over time as the tank empties (but still ignoring losses) please see my post #6.
     
  10. Apr 22, 2016 #9
    Thanks, this was what i needed :)
     
  11. Apr 22, 2016 #10
    Agreed, but what high school physics demanded was what Chestermiller added. Highly appreciate your effort, I learned something new :)
     
  12. Apr 22, 2016 #11
    Thanks. I'll have to look over what you did more carefully. Certainly, this is an unsteady state situation, and a large fraction of the fluid in the tank is decelerating as time progresses, so the steady state version of Bernoulli is not strictly kosher to use. I had just thought that this approximation would be adequate, but maybe not. I also want to check out the unsteady state version of Bernoulli.

    Chet
     
  13. Apr 22, 2016 #12

    haruspex

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    Perhaps you miss my point. Even in high school physics one is supposed to know when equations apply. The answer given would have been a reasonable approximation if instead it had said that the water had been running out of the hole for some time and asked what the egress velocity is when the head height is h. But in the set-up given, your original answer is more accurate than the 'book' answer.
     
  14. Apr 22, 2016 #13
    So no matter how large the hole is, given the water level is same somehow the velocity will be (2gh)^.5 always?
     
  15. Apr 22, 2016 #14
    Hi @haruspex

    I have a bit of a problem with the analysis in post #6. For a continuum like this, in which the velocity at each location is changing with time, writing $$\frac{dv}{dt}=v\frac{\partial v}{\partial y}$$is not valid. The equation should read:
    $$\frac{dv}{dt}=\left(\frac{\partial v}{\partial t}\right)_y+v\left(\frac{\partial v}{\partial y}\right)_t$$
    I see what you are trying to do here, and it's admirable, but the partial time derivative with respect to t is critical. In addition, the term on the left side of the equation should be an integration over the volume of the tank, since the local kinetic energy per unit mass is varying with position (including horizontal position if the exit hole is vertical) and time.

    I have some ideas on how to modify what you have done, but it's a work in progress.

    Chet
     
  16. Apr 22, 2016 #15

    haruspex

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    No, that's not what I wrote.
    Distinguish carefully these four descriptions:
    1. A hole is made at distance h below the water level in the side of a tank.
    In this arrangement, the water is initially static. As the water flows out of the hole, the water in the tank accelerates downwards. It takes a little time for the flow to become approximately steady, by which time the height of water above the hole will no longer be h. So the speed of the water jet out of the hole was less than √(2gh) while the flow was becoming established, and it will be less than √(2gh) later because the height has fallen.
    Note that √(2gh/(1-(a/A)2)) > √(2gh), so is even less accurate than √(2gh).
    2. At some point after the flow has become reasonably steady, we note that the height of the water level above the hole is x. We can now use Bernoulli's equation to find the speed of the jet, and that gives √(2gx/(1-(a/A)2)). But note x, not h.
    3. As in (1), but a constant head device keeps the tank full. E.g. water flows in from a pipe at the top of the tank, and overfows the tank if it comes in too fast. A while after the hole is made, the flow becomes established, so we can apply Bernoulli. But the water entering the tank at the top has no velocity along the flow line, so does not contribute to the energy. Bernoulli's equation now gives √(2gh).
    4. As in (3), but the water enters the tank from an adjacent reservoir with a level just above the top of the tank. Note that this will not keep the tank completely full. Since the water flows into the tank with a purely horizontal component, it has to fall a short distance to reach the descending surface in the tank. Using h as the distance from hole to top of tank, Bernoulli gives √(2gh). If instead we use x, the distance of the water level in the tank above the hole, Bernoulli gives √(2gx/(1-(a/A)2)).

    Now, I see Chet has some reservations about my analysis in post #6, and he may well be right, but that is largely independent from my much simpler analysis in this post.
     
  17. Apr 26, 2016 #16
    We're back!!! Haruspex and I have been collaborating offline during the past few days to develop a better quantitative understanding of the transient flow behavior of this system. We have developed a time-dependent model of the flow, and have performed calculations using this model. We are here to report on the details of the flow analysis, the results of our calculations, and our conclusions.

    INTRODUCTION
    The usual steady state Bernoulli equation does not correctly describe the effect of the area ratio a/A (where a is the hole area and A is the tank cross sectional area) on the effluent velocity. This is because the Bernoulli equation applies only to steady state flow, and the flow in this system is transient. Because the level of fluid is changing, the fluid velocity at any constant elevation within the tank is varying with time. But the usual Bernoulli equation does not take this part of the fluid acceleration into account. It includes only the advective part of the acceleration. As the ratio of the areas a/A gets higher, the error in the prediction from the usual Bernoulli equation prediction gets worse. For the case where a/A = 1, the Bernoulli equation totally fails to predict the required free fall. To determine the solution to this problem correctly, a time dependent modification to the Bernoulli equation must be used, which properly includes the missing part of the acceleration.

    MODEL DESCRIPTION
    In the present development, we assume that the exit hole is situated in the bottom of the tank. This allows us to assume that the flow velocity in the tank is essentially vertical and 1D, rather than having to contend with a complicated 2D fluid flow approaching the exit hole. This substantially simplifies the determination of the kinetic energy of the fluid in the tank, as well as the rate at which gravitational work is being done on the fluid in the tank.

    Let
    ##v_x(t)## = downward vertical exit velocity from tank
    ##v(t)## = downward vertical velocity of fluid in tank
    ##a##=cross sectional area of exit hole
    ##A##=cross sectional area of tank
    ##h(t)##=location of upper water surface at time t

    From the continuity equation: $$v(t)=\frac{av_x(t)}{A(z)}\tag{1}$$
    From the kinematics, $$\frac{dh}{dt}=-v(t)=-\frac{av_x(t)}{A}\tag{2}$$
    We shall next perform a mechanical energy balance on the system, setting the rate at which gravitational work is being done on the tank contents equal to the rate of change of kinetic energy of the fluid in the tank plus the rate at which kinetic energy is leaving the tank in the exit stream.

    The rate at which gravitational work is being done on the contents of the tank at time t is obtained by multiplying the rate of doing gravitational work per unit volume by the volume of the tank:
    $$\rho g v(t) Ah(t)=\rho g v_x(t)ah(t)\tag{3}$$
    The total kinetic energy of the fluid in the tank at time t is obtained by multiplying kinetic energy per unit volume by the volume of the tank:
    $$\rho \frac{v^2(t)}{2}Ah=\rho \frac{v_x^2(t)}{2}\frac{a^2}{A}h(t)\tag{4}$$
    The rate of change of fluid kinetic energy within the tank is obtained by evaluating the time derivative of the expression in Eqn. 4:
    $$\rho \frac{v_x^2(t)}{2}\frac{a^2}{A}\frac{dh}{dt}+\rho v_x(t)\frac{a^2}{A}h(t)\frac{dv_x(t)}{dt}\tag{5}$$
    Substituting Eqn. 2 into this expression yields:
    $$-\rho \frac{v_x^3(t)}{2}\frac{a^3}{A^2}+\rho v_x(t)\left(\frac{a^2}{A}\right)h(t)\frac{dv_x(t)}{dt}\tag{6}$$
    The rate at which kinetic energy is exiting the tank at time t is given by:
    $$\rho v_x(t)a\frac{v_x^2(t)}{2}\tag{7}$$
    If we now perform a transient mechanical energy balance on the system by setting the rate of doing gravitational work on the fluid in the tank equal to the rate of kenetic energy generation within the tank plus the rate of kinetic energy leaving in the exit stream, we obtain:
    $$\rho g v_x(t)ah(t)=-\rho \frac{v_x^3(t)}{2}\frac{a^3}{A^2}+\rho v_x(t)\left(\frac{a^2}{A}\right)h(t)\frac{dv_x(t)}{dt}+\rho v_x(t)a\frac{v_x^2(t)}{2}\tag{8}$$
    If we divide Eqn. 8 by ##\rho v_x(t)a##, we obtain:
    $$ g h(t)=\frac{v_x^2(t)}{2}\left[1-\left(\frac{a}{A}\right)^2\right]+h(t)\frac{a}{A}\frac{dv_x(t)}{dt}\tag{9}$$
    Eqn. 9 is fully consistent with the standard form of the transient Bernoulli equation presented in the literature.

    The time t can be replaced as the independent variable in this equation by the depth h, by combining Eqn. 9 with Eqn. 2 to yield:
    $$ 2g h=v_x^2\left[1-\left(\frac{a}{A}\right)^2\right]-h\left(\frac{a}{A}\right)^2\frac{dv_x^2}{dh}\tag{10}$$
    Note the comparison between Eqn. 10 and the form of the Bernoulli equation written in previous posts of this thread. The key difference is the second term on the right hand side, involving the derivative of ##v_x^2## with respect to h. This term captures the effects of the portion of the fluid acceleration that is omitted from the steady flow version of the Bernoulli equation.

    I think I'll stop here for now. I'll be back tomorrow with the analytic solution to Eqn. 10, together with some graphs of computational results obtained using the analytic solution, and a discussion of these results.

    Chet
     
    Last edited: Apr 26, 2016
  18. Apr 27, 2016 #17
    MODEL SOLUTION
    Equation Eqn. 10 of the previous post tells us that the velocity of the effluent stream from the tank vx will be a function of the area ratio a/A, the initial depth of the (inviscid) fluid in the tank h0, and the fluid depth at any arbitrary time t, h(t). The general analytic solution to this equation, subject to the initial condition ##v_x=0## at h=h0 is given by:$$v_x=\sqrt{2gh\frac{\left[1-(h/h_0)^{\frac{1-2r}{r}}\right]}{1-2r}}\tag{11}$$where ##r=(a/A)^2##. For the special limiting cases in which ##r=1/\sqrt{2}## and ##r=1##, this solution reduces to:
    $$v_x=\sqrt{-4gh\ln(h/h_0)}\tag{for r=1/√2}$$
    $$v_x=\sqrt{2g(h_0-h)}\tag{for r = 1}$$
    For the case of r =1 (i.e., the case in which the exit hole area is equal to the tank area), the above equation for the efflux velocity vx is, as expected, just that predicted for free fall.

    MODEL RESULTS
    The results calculated from Eqn. 11 for the efflux v_x (normalized by ##\sqrt{2gh_0}## as a function of the (dimensionless) fluid depth ratio h/h0 and the (dimensionless) area ratio a/A are shown in the figure below.
    efflux2.PNG
    In all cases, the efflux velocity is equal to zero initially (i.e., when h = h0), and then rises rapidly as the fluid, both inside the tank and in the efflux, accelerates. However, as the depth of fluid in the tank continues to decrease, the efflux velocity passes through a maximum and then begins to decrease. Eventually, as the fluid depth approaches zero, the efflux velocity, of course, also drops to zero. (In the case of a/A = 1, the maximum velocity is attained just as the tank reaches empty.)

    The results in the above figure are fully consistent with what Haruspex has been saying all along, to wit, the efflux velocity vxis always less than the Torricelli velocity based on the initial depth of fluid in the tank ##\sqrt{2gh_0}##. However, it is also of interest to compare vx with the Torricelli velocity calculated on the basis of the instantaneous depth of fluid in the tank h(t), rather than on the depth at time zero. The figure below shows the efflux velocity ##v_x## normalized in terms of the Torricelli velocity calculated on the basis of the instantaneous depth h(t) plotted as a function of dimensionless depth ##h/h_0## at a selection of values of the area ratio a/A.
    efflux3.PNG
    According to the results in the figure, for (small) values of the area ratio a/A less than 0.5, the dimensionless efflux velocity ##v_x/\sqrt{2gh(t)}## levels off to a constant value as the depth of fluid in the tank decreases. The smaller the value of a/A, the more rapidly the dimensionless velocity levels off. From our analytic solution (Eqn. 11), the value to which the dimensionless velocity levels off is given by:
    $$(v_x)_{level}=\sqrt{\frac{2gh(t)}{[1-2(a/A)^2]}}\tag{12}$$
    For the case of a/A = 0.1 in the OP of this thread, we see from the figure that, once the depth h(t) has decreased to about 90% of the initial depth h0, the velocity has already leveled off.

    For small values of a/A, Eqn. 12 for ##(v_x)_{level}## can be expressed, to linear terms in ##(a/A)^2## by:
    $$(v_x)_{level}~\approxeq \frac{\sqrt{2gh}}{[1-(a/A)^2]}$$
    By an strange coincidence, this is the same relationship for the efflux velocity that our high schooler OP had written in his second post of this thread. He was able to obtain this result by the fortuitous cancellation of two errors: 1. incorrectly assuming that the steady state Bernoulli equation could be applied this transient problem and 2. failing to take the square root of the denominator in obtaining ##v_x## from his equation for ##v_x^2##.

    This pretty much completes what haruspex and I wanted to cover.

    Chet
     
  19. Apr 28, 2016 #18
    Wow, I am a lucky guy :p
     
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