8
The average flow rate must stay the same all the way down, unless water is piling up somewhere. A possible exception is if the fall is so far that some of the water evaporates on the way down. But all the water has to go somewhere.
If the falling water column has cross-sectional area $A$ and speed $v$, the volume flow rate is $Av$. If the water is in free ...
7
Let's just use the ideal gas law. Let $P_{ab}$ = absolute pressure, $P_g$ = gauge pressure and $P_a$ = atmospheric pressure. $$P_{ab} V = NkT$$ From this is it obvious if I double Pab and keep the volume the same, then N must double. Now write Pab as $$(P_g + P_a)V = NkT$$ If I only double $P_g$, then N does not double.
Your question about pressure being ...
3
The reason that pressure acts normal to all surfaces is that, in reality, it is not actually a scalar. At static equilibrium, the (2nd order) stress tensor is isotropic, and the parameter we call pressure is the magnitude of this stress tensor: $$\boldsymbol{\sigma}=-p\mathbf{I}$$where I is the so-called (isotropic) identity tensor or metric tensor. The ...
1
To make the situation a little simpler:
1.let us neglect the viscosity of water for a while.
2.density of water remains the same at every point.
Now, using Bernoulli's equation:
Let us compare two points in the stream at different heights but near the edge of the stream. So, the pressure at both these points will be almost equal to the atmospheric pressure. ...
1
If you hold a bowl under the stream at every height, you do not expect the rate at which the bowl is being filled to change; it should be exactly the same rate as water is getting out of the tap, otherwise either extra water is being created in order to compensate or it is being vanished, which is in contrary to the conservation of water! Therefore the flow ...
1
You will find tables of this sort of data within the materials science literature under key words wettability, spreading, capillary rise, contact angle, and surface energy.
You will also find surface energy test fluid sets consisting of ranked series of specially-prepared chemical solutions with distinct and well-controlled surface energies. placing droplets ...
1
If the surface tensions are such that $\gamma_{\rm water,air}> \gamma_{\rm oil,air}+\gamma_{\rm oil,water}$ then here is no solution to the (simplified for a flat water surface) contact angle equation
$$\gamma_{\rm water,air}= \gamma_{\rm oil,water}+ \gamma_{\rm oil,air} \cos\theta_{\rm contact}
$$
In that case the oil will spread until it is a layer ...
1
If the plate and cylinder are rigid, there is not enough information to answer. The container below the plate is basically a pressure vessel. The water within that vessel could be at virtually any pressure. It is disconnected from the upper container by the strength of the iron plate. This strength means the upper and lower pressures don't have a fixed ...
1
To a good approximation yes. Pascal's law won't hold exactly for an emulsion because the interface between the two fluids will have a non-zero interfacial tension, and there will be small excess pressure inside the emulsion droplets given by the well known formula:
$$ \Delta P = \frac{2S}{r} $$
where $S$ is the interfacial tension and $r$ is the radius of ...
1
Lattice-Boltzmann algorithms have several applications ranging from the traditional fluid flow to other topics such as scalar transport (such as species or energy equation) or radiation. Depending on the application you will find a lot of different discretisations:
with/without rest node
with a single speed for the non-rest node velocities or with multiple ...
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