How High Can A Vacuum Lift Water And Velocity at Given Height (Bernoulli Equation Viewer Request) скачать в хорошем качестве

How High Can A Vacuum Lift Water And Velocity at Given Height (Bernoulli Equation Viewer Request)

I got a viewer request! The problem is if you were to dig down till you hit the water table and connect this reservoir to another reservoir that is a perfect vacuum. In an ideal situation when will the water stop flowing from the first reservoir to the second reservoir. Well the short answer is as long as the head pressure is less then the atmospheric pressure the water will continue to flow albeit slower as you approach the head pressure that is equal to atmospheric pressure. Lets dive into how I proved this…. We will once again use the Bernoulli equation which looks at the energy in an ideal system… so the pressure energy plus the potential energy plus the kinetic energy is the bernouilli equation Because energy is conserved we can set the Bernoulli equation at 2 different points equal to one another… I decided to make point 1 at the water line of the first resvoir and the second point the inside of the pipe at the second resvoir. So now we need to see what cancels out. First we have that the potential energy at point one is zero because we will be defining the height as zero. Next we have that the kinetic energy is zero because the water line on the first resvoir is always constant and not moving and we are interested in the point at which the water will stop flowing at point 2 And finally we have a vacuum at point 2 so the pressure energy is equal to zero at this point. So we have that the pressure at point 1 which is atmospheric pressure is equal to the head pressure or potential energy point at point 2 the water stops flowing. So as long as the atmospheric pressure is greater than the head pressure water will continue to flow. This is an ideal example that does not include friction loss, or the difficulty of getting a perfect vacuum but can be used as a ball park estimate. So now lets solve for velocity at point 2 given a height of 3 meters The potential energy and velocity at point 1 is zero and the pressure at point 2 is zero so all of these values can be removed Next we can rearrange the formula to get velocity 2 by it self We are left with the square root of 2 times the pressure at point 1 minus the density times gravity time height difference at point 2 all divided by the density is equal to velocity 2. Plugging in the atmospheric pressure in pascals the density in kilogram per meters cubed acceleration due to gravity and the height difference in this example which we will use 3 meters we get.. A velocity at point 2 of 11.99 meters per second. Disclaimer These videos are intended for educational purposes only (students trying to pass a class) If you design or build something based off of these videos you do so at your own risk. I am not a professional engineer and this should not be considered engineering advice. Consult an engineer if you feel you may put someone at risk.