Question

A rectangular tank is filled to the brim with water. When a hole at its bottom is unplugged, the tank is emptied in time T. How long will it take to empty the tank if it is half filled?

I have to five the answer in terms of T.

Answer 1

Let the actual Height of the tank be `H` and its uniform cross sectional area be `A`.

Now let us consider that at certain moment the height of water level be `h` and the velocity of water emerging through the orifice of cross sectional area `a` at the bottom of the tank be `v`.As the surface of water and the orifice are in open atmosphere, then by Bernoulli's theorem we have

`v=sqrt(2gh)`

Let `dh` represents decrease in water level during infinitesimally small time interval `dt` when the water level is at height `h`. So the rate of decrease in volume of water will be `-A(dh)/(dt)`.

Again the rate of flow of water at this moment through the orifice is given by `vxxa=asqrt(2gh)`.

These two rate must be same by the principle of cotinuity.

Hence `asqrt(2gh)=-A(dh)/(dt)`

`=>dt=-A/(asqrt(2g))*h^(-1/2)dh`

If the tank is filled to the brim then height of water level will be `H` and time required to empty the tank T can be obtained by integrating the above relation.

`T=int_0^Tdt=-A/(asqrt(2g))*int_H^0h^(-1/2)dh`

`T=-A/(asqrt(2g))[h^(1/2)/(1/2)]_H^0`

`T=(A/a*sqrt(2/g))H^(1/2)`

If the tank be half filled with water and the time to empty it be `T'` then

`T'=(A/a*sqrt(2/g))(H/2)^(1/2)`

So `(T')/T=1/sqrt2`

`T'=T/sqrt2`

Answer 2

Bernoulli's theorem states that the total mechanical energy of the flowing fluid, comprising the energy associated with fluid pressure due to the gravitational potential energy associated with height `h`, and the kinetic energy of fluid motion associated with velocity `v`, remains constant.

In the instant case we need to apply Torricelli's law, which is a special case of Bernoulli's theorem.

For a water particle of mass `m` at height `h`, when it reaches bottom of the tank all the potential energy has been converted to its kinetic energy. We have
`mgh=1/2mv^2`
`=>v=sqrt(2gh)` ....(1)

Assuming that when tank is completely empty the velocity of water coming out of hole is `0`.
The average velocity of the emerging water for tank full up to height `h` is
`v_"full"=sqrt(2gh)/2` .....(2)
Similarly, for the tank when half full, the average velocity is given by
`v_"half"=sqrt(2gh/2)/2` ....(3)

Let `T_"full" and T_"half"` be the times taken to completely empty the tank in respective cases. Also let `a` is area the orifice.

For case of full tank we have
Volume discharged`=T_"full"xxv_"full"xxa` .......(4)

Similarly in case of half full tank we have
Volume discharged`=T_"half"xxv_"half"xxa` ........(5)

We know that Volume discharged in case of equation (4) is twice the volume discharged in case of equation (5)

Hence we have the equality from (4) and (5)
`T_"full"xxv_"full"xxa=2(T_"half"xxv_"half"xxa)`

Using (2) and (3) and solving for `T_"half"`, we get
`T_"half"=1/2xxT_"full"xxv_"full"/v_"half"`
`=>T_"half"=1/2xxT_"full"xx(sqrt(2gh)/2)/(sqrt(2gh/2)/2)`
`=>T_"half"=1/2xxT_"full"xx(sqrth)/(sqrt(h/2))`
`=>T_"half"=1/sqrt2T_"full"`