- #1

DavideGenoa

- 155

- 5

## Homework Statement

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}##.

## Homework 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}##

## 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!