- #1
dom_quixote
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Are these hydrodynamic effects the same?
B Hydrodynamics Effects: Are They the Same?
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The discussion clarifies that the Coanda effect and ram pressure are distinct hydrodynamic phenomena. The Coanda effect involves fluid flow parallel to a surface, remaining attached due to viscosity and boundary layer velocity variations. In contrast, ram pressure arises from perpendicular flow creating pressure due to unbalanced kinetic energy. Observations of a hose nozzle being "sucked in" during an experiment relate to the Bernoulli effect, where fluid velocity beneath the nozzle exceeds that above it. Understanding these principles is crucial for accurately explaining observed fluid dynamics.
- #2
Baluncore
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No, they are different effects.
Coanda effect; https://en.wikipedia.org/wiki/Coandă_effect
Ram pressure; https://en.wikipedia.org/wiki/Ram_pressure
In one, the flow is parallel to the surface and remains attached due to fluid viscosity and the velocity variation in the boundary layer.
In the other, the flow is perpendicular and provides a pressure due to unbalanced kinetic energy; KE = ½·m·v² .
Maybe you are seeing some other phenomenon. You need to explain the process you are referring to, or observing, in each case.
Coanda effect; https://en.wikipedia.org/wiki/Coandă_effect
Ram pressure; https://en.wikipedia.org/wiki/Ram_pressure
In one, the flow is parallel to the surface and remains attached due to fluid viscosity and the velocity variation in the boundary layer.
In the other, the flow is perpendicular and provides a pressure due to unbalanced kinetic energy; KE = ½·m·v² .
Maybe you are seeing some other phenomenon. You need to explain the process you are referring to, or observing, in each case.
- #3
dom_quixote
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Thanks Baluncore!
I did a variation of the second experiment with a bucket. I did not film the experiment due to the difficulty of observing the phenomenon. However, the hose nozzle is also "sucked in" even when the water level rises. Would it be the same effect "RAM Pressure"?
I did a variation of the second experiment with a bucket. I did not film the experiment due to the difficulty of observing the phenomenon. However, the hose nozzle is also "sucked in" even when the water level rises. Would it be the same effect "RAM Pressure"?
- #4
Lnewqban
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Then, for the second part of the video you were referring to the sucking effect between hose end and flat lid.dom_quixote said:
This happens because the Bernoulli effect, as the fluid velocity under the hose nozzle is greater than the one above.
Please, see:
https://en.wikipedia.org/wiki/Bernoulli's_principle
Imagine a charged sphere at the origin connected through an open switch to a vertical grounded wire.
We wish to find an expression for the horizontal component of the electric field at a distance ##\mathbf{r}## from the sphere as it discharges.
By using the Lorenz gauge condition:
$$\nabla \cdot \mathbf{A} + \frac{1}{c^2}\frac{\partial \phi}{\partial t}=0\tag{1}$$
we find the following retarded solutions to the Maxwell equations
If we assume that...
Maxwell’s equations imply the following wave equation for the electric field
$$\nabla^2\mathbf{E}-\frac{1}{c^2}\frac{\partial^2\mathbf{E}}{\partial t^2}
= \frac{1}{\varepsilon_0}\nabla\rho+\mu_0\frac{\partial\mathbf J}{\partial t}.\tag{1}$$
I wonder if eqn.##(1)## can be split into the following transverse part
$$\nabla^2\mathbf{E}_T-\frac{1}{c^2}\frac{\partial^2\mathbf{E}_T}{\partial t^2}
= \mu_0\frac{\partial\mathbf{J}_T}{\partial t}\tag{2}$$
and longitudinal part...
Is it true that in any mechanical set-up, it is possible to predict the nature of Normal Reaction ( magnitude, direction, etc. ) without solving through the dynamical equations of motion and constraints for the set-up as Normal Reaction is completely unknown? I mean is it true that we can explain NR intuitively beforehand?
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