I am trying to calculate the outflow velocity $v_2(t)$ from the orifice of a funnel for water. I am aware of Torricelli's law derived from the Bernoulli equation assuming $v_1(t)\approx 0$ resulting in $v_2(t)=\sqrt{2gh_1(t)}$. Since $A_2\ll A_1$ does not apply in my case this equation is not precise enough for my calculations, especially for $h_1\to h_2$.

Is there a more exact model describing the outflow velocity $v_2(t)$?



Because this is an unsteady state problem, the usual Bernoulli equation (which applies strictly speaking to steady state flows) is not a good approximation for the flow you are considering. However, I have presented a method in this link

Time taken to reach efflux velocity

(2nd answer) that accurately describes the unsteady state flow of an inviscid fluid draining from a vessel. I suggest you consider applying this method to your draining cone problem.

The basics of this method are presented in Transport Phenomena by Bird, Stewart, and Lightfoot where they present an application of the method to a similar problem.

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  • $\begingroup$ Thank you very much for the answer and the detailed description of the method. $\endgroup$ – theofficialarni Mar 15 '19 at 14:15
  • $\begingroup$ +1 for the considerations on the unsteadiness of the flow. Very impressive. $\endgroup$ – the_candyman Mar 17 '19 at 19:41
  • $\begingroup$ I got two questions concerning the proposed solution: 1) Don't you have $\frac{\mathrm{d}W_G}{\mathrm{d}t}=\frac{\mathrm{d}(h(t) \rho A h(t) g)}{\mathrm{d}t}=2\rho g \frac{\mathrm{d}h}{\mathrm{d}t} A h(t)$ due to the time dependence of the volume of the fluid in Eq. 3? 2) Isn't $\frac{\mathrm{d}h}{\mathrm{d}t}=-v(t)$ in Eq. 3 due to Eq. 2? $\endgroup$ – theofficialarni Mar 18 '19 at 12:58
  • $\begingroup$ No to your first question and yes to your second question. To get the total potential energy of the fluid in the cone at time t, you need to integrate $$\rho g zA(z)dz $$ from z=0 to z=h(t). $\endgroup$ – Chet Miller Mar 18 '19 at 14:36
  • $\begingroup$ The first question was related to your calculations for a cylinder. I was wondering, why Eq. 3 is missing the factor 2. $h(t)\rho Ah(t)g=(h(t))^2\rho Ag$ should be equal to the potential Energy for a cylinder and the time derivative equal to the rate at which gravitational work is being done on the contents of the tank. Don't you get $2h(t)\dot{h(t)}...$ for the derivative using the chain-rule or am I missing something. $\endgroup$ – theofficialarni Mar 18 '19 at 16:10

For the continuity of the flow, we have that:

$$v_1(t)A_1 = v_2(t) A_2.$$

The previous equation reads as follows: the quantity of water entering in the funnel is equal to water exiting from it.

Now, you can use the Bernoulli equation together to the previous one in order to find both $v_1(t)$ and $v_2(t)$.

As a side remark, notice that if $A_1 = A_2$, then the first equation reduces to:

$$v_1(t) = v_2(t)$$ as expected. Moreover, in this very case, you also get that $h_1 = h_2$. Of course, we are assuming that the pressure on the top and on the bottom of the funnel is the same.

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  • $\begingroup$ But this yields $v_2(t)=\sqrt{\frac{2gh_1(t)}{1-\frac{A_2^2}{A_1^2}}}$ which is divergent for $A_1\to A_2$. $\endgroup$ – theofficialarni Mar 14 '19 at 18:19
  • $\begingroup$ you are right. Anyway, $A_1 = A_2$ is a special case. See the remark to my answer. $\endgroup$ – the_candyman Mar 14 '19 at 18:22
  • $\begingroup$ Recall that when you divide somewhere by $A_1-A_2$, you should also say that this mathematical operation is right only for $A_1 \neq A_2$. Therefore, $A_1 = A_2$ is a special case. $\endgroup$ – the_candyman Mar 14 '19 at 18:23
  • $\begingroup$ But does the velocity really grow that much when the water reaches the orifice? Isn't there some equilibrium, which will ultimately be reached and caps the outflow velocity? $\endgroup$ – theofficialarni Mar 14 '19 at 18:30
  • $\begingroup$ I think the last is an additional question that you should ask seperatedly. $\endgroup$ – the_candyman Mar 14 '19 at 18:40

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