Shape of tank for which water level falls at constant velocity

Question

Greeting, I am working on a problem that involves optimizing the shape of a tank so that the level drops at a constant rate.

Problem and my attempt

Working on the problem, here is what I know. There is an axially symmetrical tank whose wall shape is given by $f(z) = R(\frac{z}{H})^n$ which is shown in the figure. The spigot of the tank has cross-sectional area, a and there is no vena contracta. What I need to do is find $n$ so I can satisfy these conditions.

Here is what I have done so far to start on the problem. I am assuming that the area where the spigot is in the tank will not have any significant effects on the total volume of the tank if I integrate.

Doing a mass balance on the tank I get that

1. $In-out+Generation = Accumulation$ where $Generation; In = 0$

2. $-\rho Q_{out}=\frac{\rho dV}{dT}$ as a consequence of the mass balance. And since density is constant $-Q_{out}=\frac{dV}{dT}$

3. To get $Q_{out}$ use the fact that $u_2a = Q_{out}$

4. Using Bernoulli's Equation to get an expression for $u_2$. Between 1 and 2 I get $\frac{u_1^2}2+gz_1+\frac{P_1}{\rho}=\frac{u_2^2}2+gz_2+\frac{P_2}{\rho}$

5. Using the relation that $P_1=P_2=P_{atm}$ (4) can be simplified to $\frac{u_1^2}2+gz_1+=\frac{u_2^2}2+gz_2$ and saying that $u_1 = \frac{dz}{dt}$ I now have $\frac{\frac{dz}{dt}^2}2+gz_1+=\frac{u_2^2}2+gz_2$

6. Solving (5) for $u_2$ I get that $u_2 = \frac{dz}{dt} + \sqrt{2g(z_1-z_2)}$

My issue

This is where I start to run into issues relating what I know. I know that $\frac{dz}{dt} = constant$ and I know that I need to solve for $n$ in the $f(z)$ function. I am guess that do solve equation (2) for the volume of the tank which is $a_t\frac{dz}{dt}$. I assume since $\frac{dz}{dt}$ is present on both sides it will cancel out. Why does the rate of the flow even matter though? Why is it not valid just to say that the velocity of the water in the tank will be irrelevant in comparison to the velocity coming out of the spigot? Typically that is a valid assumption, or am I in correct in that?

And one small struggle is getting the area of the tank by just knowing $f(z)$. It has been a while since I've done calculus. The first thought in my mind is a solid of revolution. I think that $f(z)$ is a function that relates z and radius so $r = f(z) = R(\frac{z}{H})^n$ and if so, would I be doing $V = \int_a^b \pi r^2 dz$ which is $ V = \int_a^b \pi R(\frac{z}{H})^n dz$? This integral seems very difficult to calculate though and leads me to believe that I have done something wrong. Any push in the right direction would be greatly appreciated!

Answer 1

Thank you for an interesting question, I never considered there is a geometry which would result in a constant decrease in height of the water surface. A couple of points you should consider:

1. Typically for these types of problem we choose a geometry where the area of the water surface $A$ is much larger than the outflow area $a$, i.e. $a/A\ll1$. From a mass balance $u_wA = u_{out}a$, it follows that $u_w/u_{out}=a/A\ll1$ or that the speed at the water surface is much smaller than at the outflow. Now since the Bernoulli equation considers the squares of the velocities this becomes even more pronounced and we can simply assume the water surface is practically stationary in terms of mechanical energy. If you then choose the coordinate system properly, i.e. the outflow is at $z=0$ the Bernoulli equation simplifies to: $$\frac{1}{2}v\left(t\right)^2=gh\left(t\right)$$ where $v$ is the outflow velocity and $h$ is the height of the water surface.
2. The variable area $A\left(t\right)$ of the water surface with height needs to factor in the mass balance: $$\frac{dV}{dt}=-av\rightarrow\frac{d}{dt}\left(Ah\right)=-a\sqrt{2gh}$$ Using the product rule: $$\frac{d}{dt}\left(Ah\right)=A\frac{dh}{dt}+h\frac{dA}{dt}$$ The surface area is related to the radius $r\left(h\right)$ of the tank at a certain height: $$A=\pi r^{2}=\pi R^{2}\left(\frac{h}{H}\right)^{2n}$$ where use was made of your function $r=f\left(h\right)$. Substituting this into the above equation yields: $$\frac{d}{dt}\left(Ah\right)=\left(1+2n\right)\pi R^{2}\left(\frac{h}{H}\right)^{2n}\frac{dh}{dt}$$ and the mass balance is reduced to: $$\left(1+2n\right)\pi R^{2}\left(\frac{h}{H}\right)^{2n}\frac{dh}{dt}=-a\sqrt{2gh}$$ This gives you a differential equation for $h\left(t\right)$ which is solved with the initial condition $h\left(0\right)=H$ using simple integration. It may look complex but it isn't after some rearranging. Pro-tip: I tend to non-dimensionlize these equations to make them easier to handle. To do so rewrite the equation in terms of: $$\theta=\frac{h}{H}\quad\tau=\frac{t}{T}\quad T=\frac{\left(1+2n\right)\pi R^{2}H}{a\sqrt{2gH}}$$ As a bonus this gives you the time-scale $T$ which is an order-of-magnitude estimate of how long it takes to empty the tank
3. While the above integration isn't so difficult, to find the value of $n$ doesn't actually require solving for the profile $h\left(t\right)$. Instead, as you correctly asserted, for the rate to be constant we require: $$\frac{dh}{dt}=K$$ where $K$ is some constant. Unfortunately, we don't know this constant so this relation can't help us in determining $n$. However, if the rate is constant what does that say about the rate of the rate, i.e. $\frac{d^2h}{dt^2}$? Use that to find $n$, which then allows you to find $K$ (which should be independent of $h$; if not then something went wrong) and finally the profile of $h$ from the differential equation (in case you are interested, you should find the profile for $h$ is linear).

Answer 2

I shall assume that the spigot is connected at height $z_0$ where the radius is $r_0$, and that the liquid is at atmospheric pressure at this point. Then from Bernoulli's Equation the speed of descent $\dot z$ of the upper surface of the liquid is given by
$\dot z^2+2gz=\dot z_0^2+2gz_0$.

Assuming the fluid is incompressible then from the continuity condition we have
$\pi r^2 \dot z = \pi r_0^2 \dot z_0$
$\dot z_0 = \dot z (\frac{r}{r_0})^2 = \dot z (\frac{z}{z_0})^{2n}$.
Substitute in Bernoulli's Equation :
$\dot z^2 + 2gz = 2gz_0 + \dot z^2 (\frac{z}{z_0})^{4n}$.

$\dot z$ is constant. If this equation is to hold for all values of $z$ then comparing powers and coefficients of $z^1=z$ and $z^0=1$ on both sides we must have
$n=\frac14$
$\dot z^2 = 2gz_0$.