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1

As you correctly point out there is a pressure gradient behind the object that causes the flow to return to its original path. As to why that pressure gradient exists, image what would happen if the flow continued straight after bending round the object. Behind the cylinder there would be no fluid as it is blocked by the cylinder, essentially there would ...


2

First, Navier-Stokes governs the fluid in your setup. So, anything apart from the fluid will be an external force in N-S equation. Body-force means an external force that applies in the bulk of the fluid, like gravity or a magnetic force. Interaction with a "body", as a wing, which is external to the fluid domain, is done through boundary conditions : the ...


0

Maybe the force $f_i$ can be thought of as being defined by the navier stokes equations: $f_i = \rho(\frac{\partial v_i}{\partial t} + v_j\frac{\partial v_i}{\partial x_j}) +\frac{\partial p}{\partial x_i} - \mu\frac{\partial^2 u_i}{\partial x_j \partial x_j} $ for $i =1,2,3$ In this sense a body force will be zero if momentum is balanced by pressure ...


2

A gravity-driven free surface flow on an incline is unstable at moderate Reynolds number, as was shown by Yih, 1963, http://deepblue.lib.umich.edu/handle/2027.42/69956 : surface waves form. These waves are, I believe, also unstable in the transverse direction, leading to the criss-cross pattern you mention. See Liu et al 1995, ...


1

Before the kink it had momentum density $\frac{\vec{p}}{V} = \frac{\rho V \vec{v} }{V} = \rho \vec{v_1}$; afterward $\rho \vec{v_2}$. The change in momentum density is $\rho (\vec{v_2} - \vec{v_1})$. Multiplying by $A v t$ (the volume that moves past the kink in time $t$), we get the change in momentum that must be supplied to maintain the kink (i.e. the ...


0

For a given pressure drop (pressure upstream minus pressure downstream), the flow rate is proportional to the 4th power of radius, according to the Hagen–Poiseuille equation. As you close the valve, flow across the valve will decrease, and pressure between the fan and the valve will increase. Because the pressure on the downstream side of the fan ...


3

Let's assume a one litre $1000$ W electric kettle, filled with $0.5$ kilograms of water at $20^o$ C: It takes 4.2 joules to warm one gram of water one degree Celsius. So, to warm the $500$ grams of water $80$ degrees from $20$ to $100$ takes $168,000$ joules. The kettle will supply $1000$ joules per second, so it'll take $168$ seconds for the kettle to ...


0

Before the water boils, the rate that the water becomes water vapor is slow compared to after it is boiling. The pressure inside the kettle does not increase significantly until after the water is boiling. Only after the water is boiling, is there enough pressure inside the kettle to send a stream of air through the whistle.


0

Flows in which the maximum Mach number is less than 0.3 are usually assumed to be incompressible. Consider gas at rest with density $\rho_0$ that is then accelerated isentropically to Mach number $M$. The density of the gas will change in this new state and is given by: $$\frac{\rho_0}{\rho} = \left(1 + \frac{\gamma-1}{2}M^2\right)^{1/(\gamma-1)}$$ As it ...


0

You say, ignore currents and I assume, other extraneous factors. If that is the case, then, considering a hydrostatic balance in the water column where z is the vertical coordinate. Then, the motion of a water parcel with density, $\rho$, displaced upwards by a distance, $\Delta z$, in a fluid with a reference density, $\rho_0$, is governed by $\rho_0 ...


1

OK so in the end I found an answer: $\textbf{Derivation}$ Mass is conserved, so $$\frac{dm}{dt} = 0$$ But $$m = \iiint \rho(\mathbf{r}, t)d^3\mathbf{r} $$ so that $$ \frac{dm}{dt} = \iiint \frac{\partial \rho(\mathbf{r}, t)}{\partial t}d^3\mathbf{r} = 0$$. From the continuity equation $$ \frac{\partial \rho(\mathbf{r}, t)}{\partial t} = -\nabla ...


2

The relevant equation is the kinematics with linear drag. In this case, there is a resistant force that acts opposite gravity (i.e., upwards) and is linear to the velocity at which it travels: $$ \mathbf F_D=-b\mathbf v $$ where $b$ is some fluid- and object-dependent constant. Using Newton's 2nd law, $$ m\ddot{\mathbf x}=m\mathbf g - b\dot{\mathbf x} $$ If ...


0

The form of the viscosity term can be derived on physical grounds. You can most likely do no better than to consult Landau and Lifshitz, volume 6, Chapter II. The Euler equation is the conservation law for momentum, it can be written in the form $$\frac{\partial}{\partial t} (\rho v_i) = -\partial_i \Pi_{ij}$$ that is the change of momentum density is the ...


1

There are three distinct things: 1) The rate-of-strain tensor is $ 2\underline{\underline{D}} = \nabla u + \nabla u^T $. Thus viscous stresses of homogeneous viscosity will be $\eta\nabla\cdot D$, which is equal to $\eta \nabla^2 u$ only if $\nabla\cdot u =0$ (incompressible and constant density). 2) The viscosity may be heterogeneous in space, in case ...


0

The problem in your post is that you don't consider the force resisting the water pressure, which is exerted by the walls of your container. To do this, imagine a two-chamber container, with outer walls infinitely strong, but with a piston in between them which is maintained at its central position by some force that you exert. Chamber A is filled with air ...


0

We can evaluate it from the observer's rest frame. Then $v^\mu = (1, 0, 0, 0)$ and $u_\mu = (\gamma , -\gamma\mathbf v)$ where $\gamma = \frac{1}{\sqrt{1-v^2}}$ and $\mathbf v$ is the relative velocity. (We put $c = 1$.) Then $$T_{\mu\nu}v^\mu v^\nu = \rho_0 \gamma^2.$$ However, $\rho_0$ is the density in the rest frame of the fluid. In the observer's rest ...


0

Pressure is synonymous with energy - its a kind of kinetic potential energy, not unlike heat. Kinetic theory relates these two quantities, together with volume. Let's make some assumptions. When we bottle something we expect its volume to stay about the same - the glass shouldn't flex. Also we pretend the bottom of the ocean is a comfy room temperature, so ...


4

I henceforth assume $c=1$. Consider a 3-volume $\Delta \Sigma_0$ at rest with the dust. For the rest observer (this is its definition) $u^\mu$ has only (unit) temporal component. The energy (i.e the mass) associated with that portion of system is $\Delta \Sigma_0 \rho_0$. The 4-momentum of that portion is therefore $\Delta p^\mu := \Delta \Sigma_0 \rho_0 ...


0

Power is defined as $P = Fv$ where $F$ is the driving force and $v$ is the velocity of the moving object. In this case, determine the values of both $F$ and $v$, and use this to calculate the power. If you need additional help, feel free to ask in the comments.


1

I'll assume that you have a well-defined interfacial energy $\gamma$ between water and "vacuum". If there's only mechanics involved, Navier-Stokes equations with a boundary condition that the total stress along the normal $\vec n$ of the boundary (left hand side below) is normal and equal to the product of mean curvature $H$ and interfacial energy, along ...


1

In the case of a pebble falling from some height into water, I believe surface tension will be altogether negligible. You should calculate the Weber number to check this, http://en.wikipedia.org/wiki/Weber_number The dominant effect will be the pressure generated by the displacement of water by the pebble entering. Again, I am not sure that the viscous drag ...


5

This term looks like Faraday's law that is used in Ideal magnetohydrodynamics (MHD). So yes, it is true. In order to see the mathematical advection, you'll need to apply some vector calculus: $$ \nabla\times\mathbf a\times\mathbf b = \mathbf a\left(\nabla\cdot\mathbf b\right) - \mathbf b\left(\nabla\cdot\mathbf a\right) + \left(\mathbf ...


0

see Fig. 4 of this reference: http://www.cibsejournal.com/cpd/2011-10/ Pressure is low just upstream of the fan and high just downstream of the fan.


0

You seem to be looking for boundary layer separation, which occurs for sufficiently large Reynolds number. See Batchelor's Introduction to fluid dynamics, sections 5.8 to 5.10. See also: http://en.wikipedia.org/wiki/Flow_separation http://en.wikipedia.org/wiki/Boundary_layer


1

At low Renyolds number, the vortex shedding described in the question does not occur, based upon the following simulation videos: http://www.youtube.com/watch?v=ElmTA0t3bEc http://www.youtube.com/watch?v=8o-JC3R9YBY http://www.youtube.com/watch?v=sN9LP5dNWhc However, for higher Renyolds number see Nakamura et al. "Experiments on vortex shedding from flat ...


1

It all depends on the length scale. At human scale, surface tension is sure not to contribute. At the one of an insect, it will. The Weber number allows one to compare surface tension to inertial effects. (http://en.wikipedia.org/wiki/Weber_number)


0

Surface tension of a liquid may change when something is mixed with it. For e.g. the substances like soaps, detergents, etc decrease the surface tension of water. See http://en.wikipedia.org/wiki/Surfactant for more information.


1

Dyson fans are NOT efficient (despite claiming otherwise). They emit about 1 gram of high-speed air for every 10 grams of air accelerated. Sounds efficient, right? No! The Dyson fan is analogous to throwing a 1kg dart at 10 m/s toward at a 9kg target sitting on ice. The resulting dart+target has a velocity of 1 m/s (momentum conservation). However, the ...


0

It will wick down, if the wicking action is strong enough. To demonstrate this with a stronger wicking action, get a one-inch wide strip of paper towel. Fold it over so one end goes to the bottom of the cup or glass, and the other end goes down to the desktop. Now pour in water (or tea) to fill the cup, and come back in an hour to see the result.


5

As best I have been able to tell, vortex air intakes are mostly a scam designed to sell useless car modifications to people, as discussed on this HowStuffWorks article. In case of the inevitable future link rot, I'll paste the article below: The internal combustion engines in cars and trucks are essentially large air pumps: The action of the pistons ...


5

Unless I have made a conceptual mistake (which is very possible), surface tension plays essentially no role in the damping of the impact of a fast-moving object with a liquid surface. To see this, a simple way to model it is to pretend that the water isn't there, but only its surface is, and see what happens when an object deforms this surface. Let there ...


1

I recommend Turbulence: The Legacy of A. N. Kolmogorov by Uriel Frisch. It explains how turbulence is a top-down behavior, large scale turbulence in turn causes turbulence at smaller and smaller scales. This continues until a small enough scale is reached where the energy from the turbulence serves to create more heat on the molecular level. Because of this ...


1

Your method seems fine to me. Essentially you're measuring volume by filling it with a fluid, and this is a method commonly used for determining the densities of powders or other materials where it's hard to measure the volume. The method is known as pycnometry if you feel the urge to Google it. To improve the accuracy I would: weigh the tube fill it with ...


0

Changes are huge, I would recommend to re-derive the pipe flow rate with a (linear) temperature dependent formula for viscosity and density. You'll get $Q(T)$, from this can get the heat flux and thus will have a nonlinear differential equation for $T(x)$, which you can integrate numerically. Then find the intercept of $T(x)$ with desired outlet temperature. ...


2

We are asking how the rod moves through the viscous medium if we apply a force to it. Since the Reynolds number is so low, inertial forces must be small, and the externally applied force must be balanced by a viscous force. Also since we are in the low Reynold's number limit, the viscous force is linear in the velocity of the object. Thus there must be a ...


0

What is your objective in calculating a characteristic Reynolds number of the droplet formation? If you believe you can rule out inertia from this (which may be, droplet pinching happens in many regimes, including visous forces/capillary forces balance), then pick the characteristic velocity, density, etc., so that the result will be largest: if this upper ...


0

The porosity is needed if you are interested in the flow velocity within the porous medium. If you look at $Q/A$ this is also a velocity but with reference to the void or empty channel, as $A$ represents void and matrix. To get the velocity within the medium (in straight flow direction) one needs to correct $A$ by the porosity or void fraction $n$: ...


1

As long as we are far from compressible effects (low Mach) and the 0 pressure under which cavitation may appear, yes, pressure is defined up to a constant if boundary conditions are given in terms of velocity. If you provide boundary condition in terms of normal stress (e.g. at an inlet), then you provide a pressure value with this boundary condition. But ...


0

One way would be to use a radioactive tracer gas. Krypton-85 has been used to study atmospheric flow: http://onlinelibrary.wiley.com/doi/10.1029/JC075i015p02985/abstract Otherwise, spectroscopic techniques such as absorbance or emission spectra could be used to measure the concentration of various gases at different points in space and time.


2

The motion of the fluid parcel only depends on the gradient of the pressure, not the pressure itself. Think about it this way: we know that pressure increases the deeper you go into water. But if I place a cylinder in the water, the flow will always look like (the colors represent pressure with red being high pressure and blue low pressure; the arrows are ...


0

Inside the solar system the solar wind consists of sub-atomic particles that form a plasma, one of the four states (phases) of matter along with solid liquid and gas. Plasma is basically a gas of charged particles--normal gas laws don't apply as moving charges create magnetic fields and magnetic fields influence moving charges, thermodynamic gas laws assume ...


2

Helicopter rotor is a rotating wing. It produces lift the same way aircraft wing does, but instead of relying on forward motion of the aircraft it has it's own motion. The lift generated depends on coefficient of lift, air density and forward speed. Formally $L = \frac{1}{2}\rho v^2\alpha C_L$ where $L$ is lift force, $\rho$ is air density, $v$ is forward ...


1

If you were to draw out the velocity field, you would see that it is going outward. In fact if you were really astute, you would notice that your velocity has the same form as the electric field from a wire. This should help you visualize the field. Because of the symmetry, it makes sense to work in cylindrical coordinates. The position along the axis will ...


0

Usually the properties are taken at mean temperature (and pressure) between in- and outlet, often by iteration. If your problem is that sensitive to changes in thermal properties I would calculate the problem sectionwise to account for the non-linearity. When calculating Nusselt Numbers, often the wall temperature needs to be taken into account as well. ...


2

The Torricelli Formula you used is certainly a good way to start if the level of the water source is stationary. In addition there will be losses which depend on geometry of your flow restriction and on Reynolds Number. Loss coefficients $\zeta_{loss}$ for free jet discharge, valves, etc. can be found in literatute, for example here: ...


3

I think the simplest answer to this question would be that the stream of water has a number of forces acting on it (gravity, air drag) from many directions. Some torque is bound to be produced as the stream falls through the air. If you throw a ball or any small object from a height, it rotates, no matter how you drop it. Same logic applies here. As far as ...


1

When an object moves in a fluid, the fluid molecules just around it get entrained along with it due to friction. In turn, this layer of fluid entrains the next layer at a lower velocity, and so on. This is due to the fact that individual molecules switch layer and thus "diffuse" the momentum (a molecule coming from a fast layer to a slower transfers some ...


7

This is not altogether correct. In particular, it does not matter what the pressure of the room is (unless you reach extreme values), because actually what is holding the water column is not some "high" value of the pressure at the bottom of it, it is the fact that the pressure at the top of the water column (at the water-finger interface, or in the air ...


0

Weight of water is too more to get hold from bottom just by h$\rho$g pressure but actually when you put thumb you cutoff above pressure and then the pressure at bottom then dominates and hold water in straw.You can find the force now using formula above and it will be right to say that the force just due to pressure on bottom surface is $1.01325*10^5*\pi ...


1

Yes, what you have formulated is fine. The pressure acting on the water from the bottom of the straw will be equal to the weight of the water times the cross section area. So $101325\pi r^2$ is the force acting from below also. That is precisely why there is an equilibrium and the water is not falling.



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