Why stream of water sent in upward direction spread like fountain?

[vkash]

why stream of water blown in upward direction spread like a fountain. I failed to explain it can you please help.

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[boneh3ad]

As it goes up, it loses energy and momentum, so eventually it has to stop. When that happens, it has to go somewhere since gravity is still acting. It can't go up because it has no more kinetic energy. It can't go down because the rising column of water is in the way. It can't go in because it would run into itself. Therefore, the only way it can go is out.

 

[vkash]

As it goes up, it loses energy and momentum, so eventually it has to stop. When that happens, it has to go somewhere since gravity is still acting. It can't go up because it has no more kinetic energy. It can't go down because the rising column of water is in the way. It can't go in because it would run into itself. Therefore, the only way it can go is out.

So according to your explanation water will spread after becoming it's kinetic energy zero.

 

[Khashishi]

It's not important that the kinetic energy is near zero. You can point a stream of water at an upward angle, so the water maintains some horizontal motion, but it still spreads out. This is because water is pretty much incompressible. If the stream slows down, it needs to get wider to conserve volume.

 

[Andy Resnick]

why stream of water blown in upward direction spread like a fountain. I failed to explain it can you please help.

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I thought this was a silly question, but there are some nice simple results, based on jets that point downward.

For the simple case, neglect viscosity and surface tension, and assume the exit velocity of the jet is uniform across the cross-section. Then you use Bernoulli's law (conservation of energy), written as \frac{1}{2} \rho U^{2}_{0}=\frac{1}{2} \rho U^{2}_{\infty}-\rho gz and conservation of mass U_{0} R^{2}_{0} = U_{∞} R^{2}_{∞} to get:

R(z) = R_{0}(1-\frac{2gz}{U^{2}_{0}})^{-1/4}

Which is sort of symmetric to the drawing down of a dropping jet. Adding a parabolic velocity profile at the exit (Poiseuille flow), viscosity effects, and surface tension complicate the solution considerably. The solution above can be written in terms of the Reynolds number and Froude number as well.

It is claimed that an exact analytic solution of jet draw-down cannot be obtained (Middleman, "modeling axisymmetric flows"), and I suspect the upward-directed jet has the same problems.

 

[boneh3ad]

It's not important that the kinetic energy is near zero. You can point a stream of water at an upward angle, so the water maintains some horizontal motion, but it still spreads out. This is because water is pretty much incompressible. If the stream slows down, it needs to get wider to conserve volume.

It doesn't matter that the kinetic energy is beig lost while moving up? Why do you suppose the water slows down an therefore widens then?

It is losing energy, aka slowing down. They are one and the same.

 

[DrewD]

It doesn't matter that the kinetic energy is beig lost while moving up? Why do you suppose the water slows down an therefore widens then?

It is losing energy, aka slowing down. They are one and the same.

Kashishi was agreeing with you, but pointing out that the kinetic energy could still be greater than zero. The water will spread out as it loses KE and slows, but it may still have a lot of KE.