Calculating Water Speed & Volume from Pressure

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SUMMARY

This discussion focuses on calculating the speed and volume of water from pressure data provided by three wells in a town. Using the principles of energy conservation, the kinetic energy of the water stream is equated to its potential energy at the maximum height of 12.6 meters. The derived formula, v = √(2gh), leads to a calculated water speed of 15.7 m/s. This method effectively demonstrates the relationship between pressure, height, and velocity in fluid dynamics.

PREREQUISITES
  • Understanding of basic physics concepts, specifically energy conservation
  • Familiarity with fluid dynamics principles
  • Knowledge of the formula for kinetic and potential energy
  • Basic algebra for manipulating equations
NEXT STEPS
  • Study the derivation of Bernoulli's equation in fluid dynamics
  • Explore the application of the continuity equation in fluid flow
  • Learn about pressure measurement techniques in hydraulic systems
  • Investigate the impact of friction and air resistance on fluid motion
USEFUL FOR

Engineers, physicists, and professionals in the water industry who are involved in fluid dynamics calculations and water resource management will benefit from this discussion.

Physics is Phun
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Maybe this should be in the math section, I'm not sure. I have this friend that works in the water industry that tells me my town runs off of 3 wells that each have the pressure to shoot a stream of water 8" diameter 12.6m up. I was wondering from this information how I could figure out how fast that water must be moving and in turn the volume. So how could I go about doing this?
 
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anyone?
 
You can use energy conservation to solve that.
Take the potential energy as 0 at the output of the well.

The energy of the stream at the output of the well is

[tex]\frac{Mv^2}{2}[/tex]​

(no potential energy, only kinetic energy, speed is v, M is the mass of a piece of water)

The energy of the stream at the highest reach, when the velocity drops to zero is:

[tex]Mg[/tex]​

(no kinetic energy, only potential energy)

Since energy is conserved during the motion (without air friction), we have:

[tex]\frac{Mv^2}{2}=Mgh[/tex]​
(h is the highest reach)

We get easily:

[tex]\frac{v^2}{2}=gh[/tex]​
[tex]v={(2gh)}^{1/2}[/tex]​

Finally, this gives v = 15.7 m/s .
 

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