Calculating Water Flow Rate Under Compression

[Idea04]

I would like to know how to calculate the flow rate of water under compression by a hydraulic weight. How do you calculate using the value of the weight on top of the water and the height of the water in a straight column?

 

[Andrew Mason]

I would like to know how to calculate the flow rate of water under compression by a hydraulic weight. How do you calculate using the value of the weight on top of the water and the height of the water in a straight column?

Do you want to know the flow rate of the water out of a hole in the side fo the cylinder at a certain depth?

The flow rate is determined by the pressure in the water and the area of the hole and the pressure outside the hole using the rate of change in potential energy of the water (as the level drops, the change in potential energy of all the water in the tank is converted into the kinetic energy of the water leaving the tank.

AM

 

[Idea04]

well the problem I have is that the equation I used was .6 x s x((2 x p)^2)
where s was the area of the hole in square meters and p was the pressure in bars. I think this could be the wrong equation because from a 7 foot height column .5" diameter hole I was gettting around .01644308 liters a second. And from another site I saw a chart where over 7 feet head of water was getting 22 feet per second velocity rate.

And another thing atmosphere pressure has to play a role. so for example if you take a bottle of water and turn it upside down, the water flows kind of in clumps because air is rushing into equalize the pressure. But if you have a hole on top where air can enter the water flows more smoothly and faster. Because the water rushing out the bottom doesn't have to fight the force of air rushing in through the bottom. So my question if how do you add atmospheric pressure to water pressure? will the pressure increase in the example of the open top bottle of water.

 

[Andrew Mason]

well the problem I have is that the equation I used was .6 x s x((2 x p)^2)
where s was the area of the hole in square meters and p was the pressure in bars. I think this could be the wrong equation because from a 7 foot height column .5" diameter hole I was gettting around .01644308 liters a second. And from another site I saw a chart where over 7 feet head of water was getting 22 feet per second velocity rate.

But velocity does not just depend on the height of water. Velocity depends on the area of the hole, and the extra weight that is added.

And another thing atmosphere pressure has to play a role. so for example if you take a bottle of water and turn it upside down, the water flows kind of in clumps because air is rushing into equalize the pressure. But if you have a hole on top where air can enter the water flows more smoothly and faster. Because the water rushing out the bottom doesn't have to fight the force of air rushing in through the bottom. So my question if how do you add atmospheric pressure to water pressure? will the pressure increase in the example of the open top bottle of water.

The difference in pressure between the inside and outside of the tank multiplied by the area of the hole gives you the force pushing the water out. The higher the outside the pressure, the less force pushing the water out (hence less velocity).

AM

 

[Idea04]

Thanks for the help. But the actual equation that I am looking for uses the values of depth of the water, fluid density, spout exit diameter and discharge coefficient. It is to give you the value of velocity in meters per second. Can someone give me that equation? Thanks.

 

[Andrew Mason]

Thanks for the help. But the actual equation that I am looking for uses the values of depth of the water, fluid density, spout exit diameter and discharge coefficient. It is to give you the value of velocity in meters per second. Can someone give me that equation? Thanks.

I was hoping you would work it out from what I gave you.

Use Bernoulli's equation:

P_{tank} - P_{atm} = \frac{1}{2}\rho v^2

You can work it out from first principles using energy. Recall that P\Delta V is the work done on the water, which must be equal to the kinetic energy of the water leaving the tank (assume that the tank is large enough so that the kinetic energy of the water inside the tank is very small). So the rate that energy is transferred to the water is:

P_{net}\Delta V/\Delta t = dW/dt = \frac{d}{dt}(KE)

P_{net}Av = \frac{d}{dt}\right{(}\frac{1}{2}mv^2\left) = \frac{1}{2}(\rho Av)*v^2

so:

P_{net} = \frac{1}{2}\rho v^2

AM