Find time of emptying of cylindrical vessel attached with conical vessel through a orifices of diameter 10 cm at bottom as shown in figure below. There is no inflow into tank. Take `C_d` as 0.6.

SOLUTION:

Here,

Area of orifice,`a=(pi**0.1^2)/4=0.00785\ m^2`

Area of cylindrical portion,`A=pi**2^2=12.57 m^2`

Let,

`T_1` be the time required to empty the cylindrical portion .

`T_2` be the time to empty the hemispherical portion 

Thus the total time to empty the tank is,

`T=T_1+T_2`... (i)

Initially, let us consider an 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`.

Now from the principle of continuity,....Show More

SOLUTION:

Here,

Area of orifice,`a=(pi**0.1^2)/4=0.00785\ m^2`

Area of cylindrical portion,`A=pi**2^2=12.57 m^2`

Let,

`T_1` be the time required to empty the cylindrical portion .

`T_2` be the time to empty the hemispherical portion 

Thus the total time to empty the tank is,

`T=T_1+T_2`... (i)

Initially, let us consider an 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`.

Now from the principle of continuity, 

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

The total time required to empty the cylindrical  tank can be obtained by integrating this equation from limit h=14+2=4.25 to h=4 since h is assumed to be a  depth of water surface above the orifice i.e,bottom of tank.

`T_1=-A/(C_d*a*root ()(2g))\int_{6}^{4}root ()(h)dh`

Solving this using calculator, we get,

`T_1=541.65 sec`

Again, let us assume that the water is at any height h above the bottom orifice at any instant of time t and let the level of water falls by an amount dh in small time interval dT. Let r be the radius of the conical 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,

`1/x=2/(4+x)`

or,`x+4=2x`

`x=4`

Also,

`1/r=4/(x+h)`

`4r=h+4`

`r=(h+4)/4`

Now from equation  (i),

`-pi(h+4)^2/16 dh=C_d*a*root ()(2gh)*dT`

`dT=-pi/(16C_d*a*root ()(2g))*((4+h)^2)/(root ()(h))dh`

The total time required to empty the given tank can be obtained by integrating this equation from limit h=4 to h=0 since h is assumed to be a  depth of water surface above the orifice i.e,bottom of tank.

`T=-pi/(16C_d*a*root ()(2g))\int_{4}^{0}((4+h)^2)/(root ()(h))dh `

`T=-pi/(16C_d*a*root ()(2g))\int_{4}^{0}(h^(3/2)+8h^(1/2)+16h^(-1/2))`

`T=1124.42 Sec`

Thus the total time required to completely emptying the tank is,

`T=T_1+T_1=541.65+1124.42`

`T=1666.07 sec`