Funnel: how long does the fluid take to go through?

Posted by Diego Assencio on 2013.12.24 under Physics (Fluid dynamics)

In this post I will compute, under some simple assumptions, how much time a given amount of fluid (e.g. water) takes to go through a funnel. The result presented here is by no means a general formula but will be a good estimate in certain regimes (namely, if the stem radius is much smaller than the mouth radius and if the stem is short compared to the funnel height).

The funnel is shaped like an "upside down" cone whose tip has been cut (see figure 1). It is initially completely filled with fluid. The volume of the neck itself is assumed to be negligible with respect to the total volume of the funnel.

The fluid is assumed to be incompressible and the flow is assumed to be inviscid (meaning we treat the fluid as having zero viscosity).

Fig. 1: A funnel with conical shape. The funnel has mouth radius $b$ and stem radius $a$. The initial fluid height is $h_0$, which is the height of the funnel itself. At time $t$, the height of the fluid is $h(t)$ and the radius of the fluid's surface is $r(t)$.

It will be assumed here that the top surface of the fluid moves down very slowly. This assumption will be good as long as $h(t)$ is not too small, but will break as $h(t) \rightarrow 0$. With this assumption, the problem becomes similar to the "orifice in a tank" problem, meaning we can treat the flow as steady. The advantage this brings is we can use Bernoulli's equation. To be more precise, consider points $A$ and $B$ on figure 1. We can assume both points are on the same flow streamline (the dashed line), meaning: $$ \displaystyle\frac{v_A^2(t)}{2} + \frac{p_A}{\rho} + gh_A(t) = \frac{v_B^2(t)}{2} + \frac{p_B}{\rho} + gh_B(t) \label{post_62f54ab6114d473b6933ad5bf5a5fc88_eq_bern} $$ where $\rho$ is the density of the fluid, $v_A(t)$ is the fluid speed at $A$, $p_A$ is the pressure at $A$ and $h_A(t)$ is the height of point $A$ (all quantities referring to $B$ are equivalently defined). Since both points $A$ and $B$ are in direct contact with the surrounding air, we have $p_A = p_B = p_0$, where $p_0$ is the atmospheric pressure. Also, setting $h_B(t) = 0$ gives us $h_A(t) = h(t)$. Finally, since we assume the surface of the fluid moves down slowly, we can take $v_A(t) \approx 0$. Equation \eqref{post_62f54ab6114d473b6933ad5bf5a5fc88_eq_bern} then becomes: $$ \frac{p_0}{\rho} + gh(t) = \frac{v_B^2(t)}{2} + \frac{p_0}{\rho} $$ and therefore: $$ v_B(t) = \sqrt{2gh(t)} \label{post_62f54ab6114d473b6933ad5bf5a5fc88_eq_vB} $$ so the fluid comes out of the neck's orifice with speed $v_B(t) = v(t) = \sqrt{2gh(t)}$.

Before we proceed, let's obtain a relation between $r(t)$ and $h(t)$. Since we assume $a$ to be small ($a \ll r(t)$ and $a \ll b$), the cross section of the cone shown on figure 1 is close to a triangle, so: $$ \displaystyle\frac{r(t)}{h(t)} = \frac{b}{h_0} \Longrightarrow r(t) = h(t)\frac{b}{h_0} $$

The total volume of fluid contained in the funnel at time $t$ is given by (as a reminder, the volume of the funnel's neck is assumed to be negligible): $$ V(t) = \displaystyle\frac{1}{3}\pi r^2(t) h(t) = \frac{\pi}{3}\left(\frac{b}{h_0}\right)^2h^3(t) \label{post_62f54ab6114d473b6933ad5bf5a5fc88_eq_volume} $$ This approximation will be good as long as $r(t) \gg a$, meaning the volume of the "removed tip" of the cone is negligible with respect to the fluid volume $V(t)$. As $h(t) \rightarrow 0$, $r(t) \rightarrow a$ so this approximation becomes poor, but by that time the small amount of fluid left should not be large enough to break the time estimate.

From equation \eqref{post_62f54ab6114d473b6933ad5bf5a5fc88_eq_vB}, the volumetric rate at which the fluid goes through the orifice of the tunnel is equal to: $$ \Phi(t) = A v(t) = (\pi a^2) v(t) = \pi a^2 \sqrt{2gh(t)} \label{post_62f54ab6114d473b6933ad5bf5a5fc88_eq_flow_rate} $$ where $A = \pi a^2$ is the cross sectional area of the funnel's stem. Since this flow rate is the rate at which the fluid leaves the funnel, we have (below the notation $\dot{q}$ is used to represent $dq/dt$): $$ \dot{V}(t) = -\Phi(t) \label{post_62f54ab6114d473b6933ad5bf5a5fc88_eq_v_phi} $$ But from equation \eqref{post_62f54ab6114d473b6933ad5bf5a5fc88_eq_volume} we have: $$ \dot{V}(t) = \frac{\pi}{3}\left(\frac{b}{h_0}\right)^2 3 h^2(t) \dot{h}(t) = \pi\left(\frac{b}{h_0}\right)^2 h^2(t) \dot{h}(t) $$ Inserting this on equation \eqref{post_62f54ab6114d473b6933ad5bf5a5fc88_eq_v_phi} and using equation \eqref{post_62f54ab6114d473b6933ad5bf5a5fc88_eq_flow_rate}, we get: $$ \pi\left(\frac{b}{h_0}\right)^2 h^2(t) \dot{h}(t) = -\pi a^2 \sqrt{2gh(t)} $$ Cancelling common terms on both sides of the equation above yields: $$ \left(\frac{b}{h_0}\right)^2 h^{3/2}(t) \dot{h}(t) = -a^2 \sqrt{2g} \label{post_62f54ab6114d473b6933ad5bf5a5fc88_eq_hr} $$

Since: $$ h^{3/2}(t) \dot{h}(t) = \displaystyle\frac{2}{5}\frac{d}{dt}h^{5/2}(t) $$ then equation \eqref{post_62f54ab6114d473b6933ad5bf5a5fc88_eq_hr} can be rewritten as: $$ \displaystyle\frac{d}{dt}h^{5/2}(t) = -\frac{5}{2} \left(\frac{h_0}{b}\right)^2 a^2 \sqrt{2g} $$ Integrating both sides with respect to time yields: $$ \begin{eqnarray} \int_{t'=0}^{t'=t} \displaystyle\frac{d}{dt'}h^{5/2}(t')dt' &=& -\int_{t'=0}^{t'=t} \frac{5}{2} \left(\frac{h_0}{b}\right)^2 a^2 \sqrt{2g} dt' \nonumber\\[5pt] h^{5/2}(t)\bigg|_{t'=0}^{t'=t} &=& -\frac{5}{2} \left(\frac{h_0}{b}\right)^2 a^2 \sqrt{2g} t \nonumber\\[5pt] h^{5/2}(t) - h_0^{5/2} &=& -\frac{5}{2} \left(\frac{h_0}{b}\right)^2 a^2 \sqrt{2g} t \end{eqnarray} $$ Therefore: $$ \begin{eqnarray} \displaystyle h(t) &=& \left[ h_0^{5/2} - \frac{5}{2} \left(\frac{h_0}{b}\right)^2 a^2 \sqrt{2g} t \right]^{2/5} \\[5pt] &=& h_0 \left[ 1 - \frac{5}{2} \frac{1}{\sqrt{h_0}b^2} a^2 \sqrt{2g} t \right]^{2/5} \\[5pt] \end{eqnarray} $$ A bit more algebraic manipulation yields: $$ \boxed{ \displaystyle h(t) = h_0\left[ 1 - \frac{5}{2} \left(\frac{a}{b}\right)^2 \sqrt{\frac{2g}{h_0}} t \right]^{2/5} } \label{post_62f54ab6114d473b6933ad5bf5a5fc88_eq_h_vs_t} $$

The time $T$ it takes for the fluid to go through the funnel is such that $h(T) = 0$, which means: $$ 1 - \displaystyle \frac{5}{2} \left(\frac{a}{b}\right)^2 \sqrt{\frac{2g}{h_0}} T = 0 \Longrightarrow \boxed{ \displaystyle T = \frac{2}{5} \left(\frac{b}{a}\right)^2 \sqrt{\frac{h_0}{2g}} } $$ Equation \eqref{post_62f54ab6114d473b6933ad5bf5a5fc88_eq_h_vs_t} can then be written in a simpler form: $$ \displaystyle h(t) = h_0\left( 1 - \frac{t}{T} \right)^{2/5} $$ Figure 2 shows a graph of $h(t)/h_0$ versus $t / T$.

Fig. 2: Fluid height $h(t)/h_0$ vs. $t/T$.

As the graph on figure 2 shows, the fluid's surface moves down slowly until about $t = 0.95T$. After that point, $h(t)$ starts decreasing very quickly, meaning some of the approximations we made will fail to be appropriate. But as initially expected, this should not badly break our computed estimate of $T$.


WDR on Dec 26, 2013:
It will turn to the left or to the right and why?
Cielo on Mar 12, 2016:
Hello. I appreciate your work very much, but could you please elaborate on eq (11)? What did you do exactly to get that expression?
Diego Assencio on Mar 12, 2016:
@Cielo: That equation has not been derived here. It just uses the fact (from Calculus) that for a given function $f(t)$, we have:

\displaystyle \frac{d}{dt}f^\alpha(t) = \alpha f^{\alpha-1}(t) \frac{d f(t)}{dt} = \alpha f^{\alpha-1}(t) \dot{f}(t)

If $\alpha \neq 0$, then we can also write the equation above as:

\displaystyle \frac{1}{\alpha}\frac{d}{dt}f^\alpha(t) = f^{\alpha-1}(t) \dot{f}(t)

This is exactly equation (11) with $f(t) = h(t)$ and $\alpha = 5/2$.

I hope that helps. Let me know if something is still unclear :-)
Amanda on Dec 09, 2017:
Hello. Thanks for your work. I get most of it but don't know how did you get from equation 10 to 11. Can you actually cancel the h(t) since one side is in derivative function and the other one is regular function.

Thank you.
Diego Assencio on Dec 09, 2017:
@Amanda: Equation (11) is not derived from equation (10); it is simply a Calculus fact which I used to convert equation (10) into equation (12). The proof of equation (11) itself is outlined on my previous comment (from March 2016).
Ayaba Floribert Sama on Jan 17, 2018:
Sir, I am so impressed of your work. I would love you post to me more details concerning the forces involve as the fluid (water), is running down the funel shape system. Would love to know if there is a force responsible to cause the water to spin (describing a circular motion) and at what point is the force maximum. Thirdly and lastly, is the force greater is a funel shape system then in a flater shape system with an outlet bellow?
While waiting positively, I remain your sincerely.
Floribert Sama Ayaba.
Mackdaddy on May 25, 2019:
Hi, love your work. How would one go about finding the time to drain a section of a certain cone. For example, from h0-h(t)=0 to h0-h(t)=10 where h0=20.
Diego Assencio on May 25, 2019:
@Mackdaddy: The answer to your question is on equation (16). If $h_0 = 20$ and $h_0 - h(t) = 10$, then $h(t) = 10$. Plug $h_0 = 20$ and $h(t) = 10$ into that equation, then solve for $t$:

\displaystyle 10 = 20\left[ 1 - \frac{5}{2} \left(\frac{a}{b}\right)^2 \sqrt{\frac{2g}{20}} t \right]^{2/5}

I am assuming you are given values for $a$ and $b$ as well. Plug those into the equation above, then raise both sides of the equation to the power of $5/2$. This will give you the time $t$ for the height of the fluid to reach $h(t) = 10$ from the initial height $h_0 = 20$.
J.James on May 26, 2019:
How would you calculate the time taken to drain a cone with the bottom of it cut off.

So for example, from its top at h0 = 0.20 to h(t)= 0.13 where the draining hole is at a height 0.13. So the volume being drained is between the heights 0.2 and 0.13, if the cone was 0.2 units tall, but the hole leaking water is at a height 0.13 (where the cone is cut off).

Hopefully that makes sense.
Diego Assencio on May 26, 2019:
@J.James: In this case, the analysis in the article will lead to $v(t) = \sqrt{2g(h(t) - h_B)}$, where $h_B$ is the height of the orifice. The volumetric rate at which the fluid goes through the orifice then becomes:

\Phi(t) = A v(t) = (\pi a^2) v(t) = \pi a^2 \sqrt{2g(h(t) - h_B)}

You will get the following by the time you reach equation (9):

\pi\left(\frac{b}{h_0}\right)^2 h^2(t) \dot{h}(t) = -\pi a^2 \sqrt{2g(h(t) - h_B)}

This differential equation is significantly harder to solve analytically than what I obtained above, but you can still solve it numerically with an appropriate first-order differential equation solver by rewriting it as:

\dot{h}(t) = - \left(\frac{a}{b}\right)^2 \left(\frac{h_0}{h(t)}\right)^2 \sqrt{2g(h(t) - h_B)}

One very interesting thing to notice is that $\dot{h}(t) \rightarrow 0$ as $h(t) \rightarrow h_B$ when $h_B \gt 0$ (your case), while $\dot{h}(t) \rightarrow \infty$ if $h_B = 0$ (my case). This happens because the assumption that $v_A(t) \approx 0$ breaks as $h(t) \rightarrow 0$, leading to unphysical behavior on my expression for $\dot{h}(t)$.
J.James on May 27, 2019:
Hi Diego, thanks so much for your quick and detailed replies!

Why exactly is this new differential equation much harder to solve?
Diego Assencio on May 27, 2019:
@J.James: In the original problem (from the article), I derived an exact solution for $h(t)$ with little effort, starting at equation (12) and ending at equation (16). I do not see an obvious way to solve the equation obtained for $\dot{h}(t)$ resulting from your modifications to the problem. If there is a trivial way to solve it, it is simply not occurring to me.

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