# Homework Help: Time needed to half a volume of liquid

1. Jul 6, 2015

### DavideGenoa

1. The problem statement, all variables and given/known data

A barrel, open on the top, full of water, has a radium $R=0.25\text{ m}$ and a height $h=0.75\text{ m}$. At its base there is a hole where a pipe is attached. The pipe's length is $\ell=1.0\text{ m}$ and its radius $r=0.0019\text{ m}$. I would like to find the time necessary for the barrel to lose half its content, if we know that the density of the liquid in the barrel is $\rho=1.00\cdot 10^3\text{ kg/m}^3$ and its viscosity $\eta=2.5\cdot10^{-3}\text{ N sm}^{-2}$.

2. Relevant equations

I know the dependence of the pressure of a static fluid from its height $h$:
$p=p_{\text{atm}}+\rho g h$,​
Bernoulli's equation, which I do not think to be useful in such a context where the liquid is viscous, and the formula for the volume flow rate in a pipe having radius $r$ and length $\ell$ with the difference of pressure $\Delta p$ at its ends
$Q=\frac{\pi r^4\Delta p}{8\eta \ell}$​

3. The attempt at a solution

I think that the pressure at the bottom of the barrel is $p=p_{\text{atm}}+\rho g h$, where $p_{\text{atm}}$ is the external pressure, and the pressure at the exit of the pipe is $p_{\text{atm}}$, therefore $\Delta p=\rho g h$
$\frac{dV}{dt}=:Q=\frac{\pi r^4 \rho g h}{8\eta \ell}=\frac{\pi r^4 \rho g \frac{V}{\pi R^2}}{8\eta \ell}$​
The derivative of the inverse of $t\mapsto V(t)$ is
$\frac{dt}{dV}=\frac{1}{Q}=\frac{8\eta \ell R^2}{ r^4 \rho g V}$​
and the fundamental theorem of calculus would give
$t\big(\frac{V}{2}\big)-t(0)=\int_0^{V/2} \frac{dt(v)}{dv}dv=\int_0^{V/2}\frac{8\eta \ell R^2}{ r^4 \rho g v}dv$​
but this last integral diverges, which means that something went wrong, although I do not see where my error is...
I heartily thank you for any answer!

2. Jul 6, 2015

### Staff: Mentor

dV/dt = - Q (note the negative sign)

The volume is decreasing. Also, the starting volume is not zero, it is V0.

Chet

Last edited: Jul 6, 2015
3. Jul 7, 2015

### DavideGenoa

Thank you, Chestmiller! I understand your remark, but I intended $V$ to be the volume of water leaked out of the barrel.
The flow rate of the volume leaving the barrel is $-Q$ and the flow rate of the the volume filling the exterior is $Q$. I mathematically see that $\int_{0}^{V_0/2}\frac{1}{Q}dV=+\infty\ne-\int_{V_0}^{V_0/2}\frac{1}{Q}dV$ but I think that $-\int_{V_0}^{V_0/2}\frac{1}{Q}dV$ is the time needed for the barrel to "go" from the full volume $V_0$ to its half $V_0/2$ (this integral is finite and I think it to be the correct answer to the exercise) while $\int_{0}^{V_0/2}\frac{1}{Q}dV$ should be the time needed for the exterior to "go" from a dry state to host half the full volume: $V_0/2$. I have some trouble to understand the reason why the two integral are not identical... Thank you so much again!!!

Last edited: Jul 7, 2015
4. Jul 7, 2015

### haruspex

But that's not the way you used it when substituting for $\Delta p$.

5. Jul 7, 2015

### DavideGenoa

I think I have understood now: the variable $V=\pi R^2 h$ in the integral is the volume of the water in the barrel, it is not the volume gone out of the pipe!
Thank you both!!!