I have a tank of capacity 1300 m3 with a 150mm diameter exiting pipe. This pipe is located 5m from the top of the tank and I'm wondering what is the flowrate in m3/sec through this pipe?

It depends on the shape of the tank and how full is the tank. Also, is that exiting pipe at the bottom of the tank so that if water is not entering, the tank will eventually completely be emptied?

The tank is a proposed reservoir with the reservoir wall being 15m wide at top and a perpenpdicular drop of 5m.tank will always be full and exiting pipe is at the bottom of reservoir wall.

It's too difficult for me... I worked out until the average force the water would exert at the position of the pipe, but I need more knowledge about fluid physics to solve this. (force I obtained was 853.8 N)

EDIT: seems that you have to use the formula here http://www.engineeringtoolbox.com/mannings-formula-gravity-flow-d_800.html

You can estimate the flow rat out of the tank using:


v = \sqrt {2 g h}


where h = height of the head of water above the drain. Notice that the width of the tank is immaterial - all that matters is the depth of the exit pipe below the surface. In your case you have h = 5 m, so:


v = \sqrt { 2 * 9.8 \frac m {s^2} * 5 m } \\
\ \ = 9.9 \frac m s


This formula ignores flow friction and the viscosity of the water, so it's basically an upper limit to the correct answer.

Oh, I see that formula is derivated from kinetic energy and gravitational potential energy. But then, doesn't that give the speed of water falling a height of 5 m? :confused:

Oh, I see that formula is derivated from kinetic energy and gravitational potential energy. But then, doesn't that give the speed of water falling a height of 5 m? :confused:

Exactly right! If you ignore friction, then the energy gained as a molecule of water descends 5 meters - no matter how slowly - is converted to KE. Another way to think of it is this - the water at the bottom of the tank is under pressure from the weight of water above, and this pressure is a form of energy that gets converted to KE when the molecule of water is released out the drain pipe to a low pressure area.

Bernoulli's law states that for flow of an ideal liquid the total energy per unit density remains constant:


P + \frac 1 2 \rho v^2 + \rho g h = Constant.


where P = hydrostatic pressure and \rho = density of the fluid (modelled as a constant for water).

Ah thanks ebaines, you helped me into understanding how to solve the problem! :) I saw Bernouilli's equation in my past papers physics book in the fluid physics section, which is optional for next year. I feel like already having some 'advanced knowledge'! :p