I'm confused about a classical exercise on tank draining, which asks to find the time requested for empty the tank.

I'll put here the derivation I found on textbook and on the web, which I'm not ok with.

Consider a tank with a fluid at height $h$, the surface at the top of the liquid ($_{surface}$ ) has area $\Sigma$ and the hole at the bottom ($_{out}$) has area $s$.

Bernoulli and continuity equations are used between surfaces $\Sigma$ and $s$.

$$s v_{out}(t)=Σ v_{surface}(t)\tag{1}$$

$$\frac{1}{2} \rho v_{out}(t)^2=\frac{1}{2} \rho v_{surface}(t)^2+\rho g h(t)\tag{2}$$

The following is found

$$v_{out}(t)=\sqrt{\frac{2 g h(t)}{1-(s/Σ)^2 }}\tag{3}$$

Then, as we can say that $v_{surface}=-\frac{dh}{dt}$, from $(1)$ follows

$$s\sqrt{\frac{2 g h(t)}{1-(s/Σ)^2 }}=Σ (-dh/dt)\tag{4}$$

Which can be easily integrated.

I'm ok with $(3)$ and $(4)$, the problem is at the beginning.

Bernoulli and continuity equations have, among their conditions of validity, the fact that the flow is stationary, which means $$\frac{\partial v(x,y,z,t)}{\partial t}=0$$

This condition is clearly not satisfied here, as we use the time dependent $v_{out}(t)$, which is the velocity in the same position.

So how can this be correct at all? Is it really possible to use Bernoulli and continuity equations in $(1)$ and $(2)$ here?


3 Answers 3


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 that the second term on the right hand side involves 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.

MODEL SOLUTION Equation Eqn. 10 of the previous post tells us that the velocity of the effluent stream from the tank $v_x$ will be a function of the area ratio a/A, the initial depth of the (inviscid) fluid in the tank $h_0$, 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 $h=h_0$ 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 $v_x$ 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 $h/h_0$ and the (dimensionless) area ratio a/A are shown in the figure below.enter image description here In all cases, the efflux velocity is equal to zero initially (i.e., when $h = h_0$), 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.)

It is also of interest to compare $v_x$ 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. enter image description here

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, for example, we see from the figure that, once the depth h(t) has decreased to about 90% of the initial depth $h_0$, 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: $$(v_x)_{level}~\approxeq \frac{\sqrt{2gh}}{[1-(a/A)^2]}$$


You are completely correct in your statement. Usually we are only 'allowed' to use bernoulli if ${s\over\Sigma}\ll1$ such that ${v_{surface}\over v_{out}}\ll1$. This is known as a quasi-steady state approximation in which the surface velocity is so small compared to the outflow that the surface may be considered approximately static such that: $${\partial v_{surface}\over\partial t}\approx0$$


"Correct or not" is a bit too black and white. The art of physics is often to know when a certain approximation is good enough for practical purposes.

In the case that $\Sigma=s$, i.e., your tank is actually a vertical tube that is open on both sides, the Bernoulli approach is probably not the way to go. You should therefore look at the time scales. The time scale of the tank pressure going down is $$ \tau_1 = \frac{h}{dh/dt}. $$ The other time scale is the amount of time it takes for a volume element of the liquid to gain most of its kinetic energy. That's a bit harder to estimate. If the tank has a conical bottom with a half-angle of 45 deg, the velocity at a distance $z$ above the outlet will be roughly $$ v(z) \approx v(0)\frac{\sqrt s}{z+\sqrt s}.$$ Three quarters of the kinetic energy will be gained in the section of the cone for $0<z<\sqrt s$. The volume of that part of the cone is roughly $V=\frac13 s^{3/2}$ and it takes a time $$ \tau_2 = \frac{V}{s\,v(0)} \approx \frac{\sqrt s}{3 v(0)} $$ for a volume element to pass through.

The Bernoulli condition would be approximately correct if $\tau_2\ll \tau_1$.


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