Question

Kerosene leaks out a hole 0.50 in.in diameterat thebottom of a tank at a rate of 10gal/min. How high is the kerosene in the tank?

Answer 1

Given:

$d=0.50\:\rm in=1.27\: cm$

$\frac{\Delta V}{\Delta t}=10\:\rm gal/min=6.31*10^{-4}\: m^3/s$

The flow rate

$$\frac{\Delta V}{\Delta t}=\frac{Av\Delta t}{\Delta t}=Av=\pi d^2/4*v$$

The speed of a flow

$$v=\sqrt{2gH}$$

Therefore,

$$H=\frac{(4\frac{\Delta V}{\Delta t}/\pi d^2)^2}{2g}$$

$$=\frac{(4*6.31*10^{-4}/3.14*(1.27*10^{-2})^2)^2}{2*9.8}=1.27\:\rm m$$