Hi,
I have a question regarding How much energy is required to push the water back to the tank, II have a tank size 10x10 (in meters), I have a tap on the bottom of the tank, The TAP size is 1 meter in diameter, I want to push back 785 liters (water present in cylinder Clinder size 1 Meter diameter, one meter depth),
Consider Tap is present in side bottom of the Tank.
Is there any formulae to find the answer ?

Thanks in Advance,
Regards,
Karthik

I don't understand the scenario you are trying to describe, but perhaps this will help:

Energy is calculated by the force required to do the task times the distance the force acts through. I gather you want to know about the energy required to inject water into the bottom of a tank of water from a cylinder through a tap, right? Well, the force required to overcome the pressure of the water in the tank above the tap is given by F= \rho g h A where \rho is the denisty of water (nominally 10^3 kg/m^3), h is the depth of the water above the tap (head) in meters and A is the cross-sectional area of the tap where the water is coming in. The work that has to be done is this force times the length of that the injected water has to move, which I assume is the length of the cylinder you mentioned:

W = \rho g h A L


Note that the units work out to be kg/m^3 * m/s^2 * m * m^2 * m = kg-m^2/s^2, or joules.

Thanks Ebaines, Same thing I am looking for.

Is this calculation done below is OK,

Density of water=1000
Gravity=9.8
Height of the Tank = 10
Crosssection Area of the Pipe= Pi *r * r
= 3.14*0.5 *0.5
= 0.785
Hence Pressure = 1000*9.8*10*0.785=
= 76930

to calculate the Work,
= Pressure * height of the cylinder
=76930 *1=76930 Joules

How do I convert to watts, is one juole=1 watt

then it means in horse power = 76930/745=103.26 HP
Is this calculation is Ok.

Thanks & Regards,
Karthik Raman

joules =Watts * seconds

Thus in order to go to Watts, you need to know how fast you have to have the water.