I wish to explore a slight modification of a well known result found in several physics texts - Torricelli's Law of Efflux.

The most common problem on the above result has the following setup: enter image description here

The container is filled with fluid of a certain density, upto height H and has a hole at a distance h from the water surface. Usually, we assume the density of fluid to be uniform, while applying Bernoulli's Principle to figure out the velocity of efflux. What effect would variable density have on the velocity of efflux?

P.S. Of course, the variation of density with depth $y$ from the fluid surface is known.

P.P.S. Though a qualitative idea would suffice, it is always better to do quantitative analysis of such situations. For the sake of simplicity, let's assume linear and increasing variation of density with depth from the fluid surface.

  • $\begingroup$ Torricelli's law is a specific case of the Bernoulli equation which is only valid for constant density flows. Bernoulli's equation is a statement of conservation of mechanical energy along a streamline in a incompressible, inviscid, stationary flow. $\endgroup$
    – nluigi
    Jun 8, 2018 at 11:51
  • $\begingroup$ @nluigi I beg to differ. Bernoulli's principle should work just fine as long as the fluid is not compressible. $\endgroup$ Jun 8, 2018 at 12:22
  • $\begingroup$ @schrodinger_16 Shouldn't be much of a problem. The Bernoulli's equation gives $$ P_2 + \frac{1}{2}\rho v_2^2 = P_1 + \frac{1}{2} \rho v_1^2$$ ... I'd just change the density in the RHS of the equation. $\endgroup$ Jun 8, 2018 at 12:53

2 Answers 2


I beg to differ with Chester Miller's answer.

You have not stated the cause of density variation. If it is a laboratory scale setup, density variation due to variation in pressure may be neglected (i.e. the fluid may be assumed to be incompressible). The density variation must then be due to some other factor, such as for example due to variation in salinity.

To get the appropriate form of the Bernoulli equation for your case, we must derive it from scratch, beginning with the Navier-Stokes equation. For an incompressible, inviscid, steady flow, the Navier-Stokes equation is: $$\mathbf{u}\cdot\nabla\mathbf{u}=-\frac{1}{\rho}\nabla p+\mathbf{g}$$ in which $\mathbf{u}$ is the velocity of the fluid (bold-face font is used to denote vectors), $p$ is pressure, $\rho$ is fluid density which varies with position in the fluid, $\mathbf{g}$ is the gravitational acceleration vector, and "$\cdot$" is the scalar dot-product.

Now we can write $\mathbf{u}\cdot\nabla\mathbf{u}=\nabla(u^2/2)-\mathbf{u}\times\mathbf{\omega}$, where $u=|\mathbf{u}|$ and $\omega=\nabla\times\mathbf{u}$ is the vorticity vector; this and subsequent results may be easily proved using indicial notation, which I leave to you (or refer Fluid Dynamics by Batchelor, Chapter 3). Further, $\mathbf{g}=\nabla(\mathbf{g}\cdot\mathbf{x})$, in which $\mathbf{x}$ is the position vector from some arbitrarily chosen origin. Substituting these into the previous equation and rearranging we get: $$\nabla\left( \frac{1}{2}u^2-\mathbf{g}\cdot\mathbf{x} \right)+\frac{1}{\rho}\nabla p=\mathbf{u}\times\mathbf{\omega}$$ Since $\rho$ varies with position it cannot simply be pulled inside the $\nabla$ operator.

There are two ways to simplify the equation above. Either you assume that the flow is irrotational, i.e. $\omega=0$ everywhere. Or you integrate above equation along a streamline. The former is the simplest assumption to make, but it is difficult to justify physically; even though it may have begun as irrotational flow because it was initially at rest say, it may not remain irrotational because it is not a barotropic fluid (Kelvin's circulation theorem does not apply). Therefore we do the next best thing, which is to integrate along a streamline.

enter image description here

A streamline is a curve which is tangent to fluid velocity everywhere. Pick a streamline which begins at the free surface and passes through the exit hole. Let $\mathbf{t}$ be a unit tangent vector to that streamline; then by definition of the streamline, $\mathbf{t}\times\mathbf{u}=0$ at every point on the streamline. Therefore $\mathbf{t}\cdot(\mathbf{u}\times\omega)=0$ at every point on the streamline. Forming the scalar product with $\mathbf{t}$ of the previous equation gives: $$\mathbf{t}\cdot\nabla\left( \frac{1}{2}u^2-\mathbf{g}\cdot\mathbf{x} \right)+\frac{1}{\rho}\mathbf{t}\cdot\nabla p=0\\ \frac{d}{ds}\left( \frac{1}{2}u^2-\mathbf{g}\cdot\mathbf{x} \right)+\frac{1}{\rho}\frac{dp}{ds}=0$$ in which $s$ is the parameter of the streamline-curve (for example, $s$ could be distance along the streamline), whose value varies monotonically from $s_1$ at the free surface to $s_2$ at the exit hole.

Next we integrate the equation above from $s_1$ to $s_2$. i.e. we apply the operator $\int_{s_1}^{s_2}ds$. First term is easy: $$\int_{s_1}^{s_2}ds\frac{d}{ds}\left( \frac{1}{2}u^2-\mathbf{g}\cdot\mathbf{x} \right)=\left[ \frac{1}{2}u^2-\mathbf{g}\cdot\mathbf{x} \right]_{s_1}^{s_2}\\ =\frac{1}{2}(u_2^2-u_1^2)+g(z_2-z_1)=\frac{u_2^2}{2}-gh$$ in which I have chosen Z-axis to be vertically upward so that $\mathbf{g}=-g\mathbf{e}_z$, $\mathbf{e}_z$ being the unit vector along Z-axis and $g=|\mathbf{g}|$; I have also assumed that the velocity at the free surface is negligible.

If we write the absolute pressure as $p=p_{atm}+p'$, in which $p_{atm}$ is the atmospheric pressure, then $dp/ds=dp'/ds$. Also $p'=0$ at the free-surface as well as at the exit hole, because pressure is atmospheric there (neglecting surface tension effects). Then the second term may be integrated by parts: $$\int_{s_1}^{s_2}ds\frac{1}{\rho}\frac{dp}{ds}=\int_{s_1}^{s_2}ds\frac{1}{\rho}\frac{dp'}{ds}\\ =\left[ \frac{p'}{\rho}\right]_{s_1}^{s_2}+\int_{s_1}^{s_2}ds\frac{p'}{\rho^2}\frac{d\rho}{ds}\\ =\int_{s_1}^{s_2}ds\frac{p'}{\rho^2}\frac{d\rho}{ds}$$

Thus the Bernoulli equation in your case, applied between free-surface (point 1) and exit-hole (point 2), is: $$\frac{u_2^2}{2}-gh+\int_{s_1}^{s_2}ds\frac{p'}{\rho^2}\frac{d\rho}{ds}=0$$ If density is constant along the streamline then $d\rho/ds=0$, and equation above reduces to the usual form $u_2=\sqrt{2gh}$. Otherwise we have: $$u_2=\sqrt{2}\sqrt{gh+\int_{s_2}^{s_1}ds\frac{p'}{\rho^2}\frac{d\rho}{ds}}$$ where the direction of integration is now from exit-hole to free-surface. We can write $d\rho/ds=(d\rho/dz)(dz/ds)$, in which $d\rho/dz$ is known and $dz/ds$ depends on the streamline chosen. If the fluid was initially at rest, then it must be stably stratified; further if the streamline descends monotonically (as in the figure above), then $d\rho/ds>0$ everywhere along the streamline and value of $d\rho/ds$ increases from exit-hole to free-surface. The gauge pressure $p'$ inside the integral is the troublesome term, because due to presence of flow along the streamline it differs from the hydrostatic value.

Addendum Since the fluid is stratified one may object that the Navier-Stokes equation we began with doesn't account for buoyancy force on the fluid particle. In a stably stratified fluid in equilibrium the constant-density surfaces are horizontal; this means that as long as constant-density surfaces remain horizontal, a fluid particle on that horizontal plane doesn't experience buoyancy force. Buoyancy force comes into play only when constant-density surfaces are tilted or distorted. Distortion will certainly be greatest in the flow region close to the exit-hole, reducing as we move away from the exit-hole. Therefore Navier-Stokes equation we began with is invalid in the region close to the exit-hole. Presumably this region is not big enough, and our equation holds approximately along the rest of the streamline away from the exit-hole. If this is the case, then our final result may yet be a good approximation.

  • $\begingroup$ @schrodinger_16 I have not accounted for buoyancy effects, which is bothering me. I will edit the answer as soon as I am able to do that. Until then you may wish to keep the question open instead of accepting my answer. $\endgroup$
    – Deep
    Jun 9, 2018 at 8:20
  • $\begingroup$ @schrodinger_16 See the addendum. $\endgroup$
    – Deep
    Jun 9, 2018 at 14:18

Let the base of the tank be the datum (z = 0) for zero potential energy. Then the form of the Bernoulli equation that would be valid for this problem would involve an integral of the density variation. Taking the two locations for applying the Bernoulli equation as 1.the upper fluid surface in the tank (assuming it is open to the atmosphere) and 2. the exit hole, we have: $$p_{atm}+\int_0^H{\rho g dz}+0=p_{atm}+\int_0^{(H-h)}{\rho g dz}+\frac{1}{2}\rho_{(H-h)} v^2$$ This reduces to $$\frac{1}{2}\rho_{(H-h)} v^2=\int_{(H-h)}^H{\rho g dz}$$


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