Fluid Dynamics (In his experiments with balls rolling down inclined planes, Gali

Question

Fluid Dynamics (In his experiments with balls rolling down inclined planes, Galileo timed the descent of the balls by filling a bowl with water flowing out through a pipe at the bottom of a cylindrical tank. He started the flow when the ball passed some point on the incline, and stopped the flow when it had moved past a point further down the track. He took bowls of equal weight to indicate equal intervals of time for different distances traveled down the track. Given that the flow rate through the aperture at the end of the pipe is the time rate of change of volume for fluid in the tank -- Q= dV/dt= v eflux x A aperture -- and given that that the cross-sectional area of the tank is much greater than the cross-sectional area of the pipe, write an expression for the volume of water in the tank as a function of time in terms of the cross-sectional area of the tank, the cross-section of the aperture, the density of water, and the acceleration due to gravity Assume that the tank is open at the top, but otherwise neglect the effect of atmospheric pressure. Hint: You will need to perform an integration

Explanation / Answer

both sides pressure is constant , so pressure term will be dropped from bernoulli equation.

let top side be side 1 and bottom side, at the end of the pipe be side 2

then speed at side 1=0

height at side 1 at any time t=y

speed at side 2=v

height at side 2 =0

then using bernoulli equation:

0.5*pho*v1^2+pho*g*h1=0.5*pho*v2^2+pho*g*h2

where pho=density of water

==>0+pho*g*y=0.5*pho*v^2+0

==>v=sqrt(2*g*y)...(1)

if height decreases by dy,

then volume change=A*dy

where A=area of cross section of the tank

volume flowing through the pipe=a*v

where a=cross sectional area of the pipe

equation both the volumes,

A*dy/dt=a*v=a*sqrt(2*g*y)

==>A*dy/sqrt(y)=a*sqrt(2*g)*dt

integrating both sides,

2*A*sqrt(y)=a*sqrt(2*g)*t

==>y=2*g*(a*t/(2*A))^2

volume=A*y=2*g*A*(a*t/(2*A))^2

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