A tank is in form of a frustum of a cone having top diameter 2m, bottom diameter 0.8 m and height 2m and is full of water. Find time of emptying tank through orifice 100 mm in diameter at bottom. Take `C_d` as 0.625.

SOLUTION:

Here,

Area of orifice,`a=(pi**0.1^2)/4=0.0025pi \ m^2`

 Let us consider  any instant such that the height of water above the centre of orifice is h m. Let us say that during time `dT`, the level of water fall by`dh`. Let r be the radius of the liquid surface.

Now from the principle of continuity, 

`-A*dh=C_d*a*root ()(2gh)*dT`

`-pir^2dh=C_d*a*root ()(2gh)*dT`... (i)

Now from figure,

`0.4/x=1/(2+x)`

or,`0.8+0.4x=x`

`x=4/3`

Also,

SOLUTION:

Here,

Area of orifice,`a=(pi**0.1^2)/4=0.0025pi \ m^2`

 Let us consider  any instant such that the height of water above the centre of orifice is h m. Let us say that during time `dT`, the level of water fall by`dh`. Let r be the radius of the liquid surface.

Now from the principle of continuity, 

`-A*dh=C_d*a*root ()(2gh)*dT`

`-pir^2dh=C_d*a*root ()(2gh)*dT`... (i)

Now from figure,

`0.4/x=1/(2+x)`

or,`0.8+0.4x=x`

`x=4/3`

Also,

`R/r=(2+x)/(x+h)`

`1/r=(10/3)/((4+3h)/3)`

`r=(4+3h)/10`

Now from equation  (i),

`-pi(4+3h)^2/100 dh=C_d*a*root ()(2gh)*dT`

`dT=-pi/(100C_d*a*root ()(2g))*((4+3h)^2)/(root ()(h))dh`

The total time required to empty the given tank can be obtained by integrating this equation from limit h=2 to h=0 since h is assumed to be a  depth of water surface above the orifice i.e,bottom of tank.

`T=-pi/(100C_d*a*root ()(2g))\int_{2}^{0}((4+3h)^2)/(root ()(h))dh `

Solving this using calculator, we get, 

`T=160.211 Sec`