A 600 mm diameter vertical tank contains water to a depth of 2.5m. There are two orifices in the tank. One at the bottom and the other in one of the vertical sides of the tank at a height of 1.5m above the bottom. If the area of each of the orifice is ` 12 cm^2` and`C_d=0.6`, Calculate the time required to empty the tank.

SOLUTION:

Here, the area of the cylindrical tank is,

`A=pi/4**0.6^2=0.2827 m^2`

And, the area of the orifice is,`a=0.0012 m^2`

Let,

`T_1` be the time required to bring water level from AB to CD.

`T_2` be the time to empty the water below the portion CD

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 upper orifice is h ....

SOLUTION:

Here, the area of the cylindrical tank is,

`A=pi/4**0.6^2=0.2827 m^2`

And, the area of the orifice is,`a=0.0012 m^2`

Let,

`T_1` be the time required to bring water level from AB to CD.

`T_2` be the time to empty the water below the portion CD

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 upper 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=(Q_1+Q_2)dT`

`-A*dh=C_d*a*root ()(2g)*(root ()(h)+root ()(h+1.5))`

`dT=-A/(C_d*a*root ()(2g))*(dh)/(root ()(h)+root ()(h+1.5))`

The total time required to drop level of water from AB to CD can be obtained by integrating this equation from limit h=1 to h=0 since h is assumed to be a  depth of water surface above upper orifice.

`T_1=-A/(C_d*a*root ()(2g))int_{1}^{0}(dh)/(root ()(h)+root ()(h+1.5))`

`=-A/(C_d*a*root ()(2g))int_{1}^{0} (root ()(h)-root ()(h+1.5))/(-1.5)dh`

Solving this using calculator,we get,

`T_1=43.96 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.

Now from the principle of continuity, 

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

The total time required to drop level of water from CD to EF (i.e, empty tank) can be obtained by integrating this equation from limit h=1.5 to h=0 since h is assumed to be a  depth of water surface above lower orifice i.e,bottom of tank.

`T_2=-A/(C_d*a*root ()(2g)) int_{1.5}^{0}1/root ()(h)dh`

Solving this using calculator, we get,

`T_2=217.12 sec`

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

`T=T_1+T_1`

`T=261.08 sec`