A tank has an upper cylindrical portion of 1.25 m radius and 3 m height with hemispherical base. The tank is provided with an orifice of 150 mm diameter at its bottom. Find the time required to empty it if it is initially full of water.Take Cd =0.62 for the orifice.

SOLUTION:

Radius of the tank, R = 1.25 m

Area of cylinder,`A=pi**1.25^2=4.908 m^2`

Area of orifice,`a=pi**0.15^2/4=0.01767 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 .... Show More

SOLUTION:

Radius of the tank, R = 1.25 m

Area of cylinder,`A=pi**1.25^2=4.908 m^2`

Area of orifice,`a=pi**0.15^2/4=0.01767 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=1.25+3=4.25 to h=1.25 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_{4.25}^{1.25}root ()(h)dh`

Solving this using calculator, we get,

`T_1=190.85 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 x be the radius of the liquid surface.

Area of liquid surface is,`A=pix^2`

Now from the principle of continuity, 

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

`dT=-(pix^2 *dh)/(C_d*a*root ()(2gh))`

From figure,

`US=x=root ()(R^2-(R-h)^2)`

or,`x^2=2Rh-h^2`

Now,

The total time required to empty the hemispherical tank can be obtained by integrating this equation from limit h=1.25 to h=0 since h is assumed to be a  depth of water surface above the orifice i.e,bottom of tank.

`T_2=-pi/(C_d*a*root ()(2g))*\int_{1.5}^{0} (2Rh-h^2)/root ()(h)dh`

Solving this using calculator, we get,

`T_2=105.6 sec`

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

`T=T_1+T_1=190.85+105.6`

`T=261.08 sec`