# Tag Info

0

Perhaps this can be done by the readings collected by doing simple archimedes displacement experiment. The volume of liquid overflow into the eureka can be considered as volume of object submerged since the water is initially full.

0

Assume the contrary, the volume of liquid displaced is greater than, or less than, the volume of the object. Then there is either a volume containing neither water nor object, or there is a volume occupied by both water and object.

3

Any kind of "funnel" - an area where tall buildings create an obstacle to the free flow of air - acts as an amplifier to wind. That is, even a little bit of air moving from point A to point B will notice the "obstacle" that is a pair of buildings; it will build up pressure in front of the buildings and result in a faster flow of air through the passage. An ...

1

When the sugar dissolves, it is broken up into individual particles; each molecule is surrounded by water molecules. Essentially, you are trying to decide if a single molecule is a solid or liquid when dissolve. The best fit is a liquid since the molecules are free to move independently and will take on the shape of the container.

1

One of my favorite quotes, and I think this complements Ján Lalinský's answer: "Does the engineer ever predict the acceleration of a given body from a knowledge of its mass and of the forces acting upon it? Of course. Does the chemist ever measure the mass of an atom by measuring its acceleration in a given field of force? Yes. Does the physicist ...

1

What makes you believe that there us an unbalanced net force in a Venturi tube? The force required to change the momentum of the liquid comes from the walls of the tube. This can be easily observed with an open nozzle (a garden hose will do!) and is used in rocket motors. In closed form a U-shaped tube is used as one of the most precise flow sensors, which ...

3

Is there any causality implied by Newton's 2nd Law? It depends on what you mean by "causality" and "Newton's 2nd law". I would rather not go into various meanings. Newton's original formulation of the 1st law seems causal: Every body perseveres in its state of rest, or of uniform motion in a right line, unless it is compelled to change that state by ...

0

I would say that it depends what your requirements are. If you just need approximately a constant flow rate your design is not bad. You can separate the functions of maintaining the level water level and supplying air. You could even totally separate reservoir from the basin and connect it by tubes. If you need a simple and cheap design, give it some more ...

0

"what causes the unbalanced net force in the first place?" - potential. The energy of molecular motion upstream of the flow is greater than downstream. And we see the energy as pressure or potential (energy). But regarding the main question - Is there causality wrapped up in Newton's 2nd law? Maybe. Consider F = d/dt(momentum). The fact that I had to ...

0

The highest wind speed ever measured outside of a tornado was 113m/s, approx. one third the speed of sound. Needless to say, no wind turbine will be operational at that wind speed, let alone operate with its wing tips close to supersonic mode. Smells like the pseudoscience it is.

1

The statement in the first paragraph "In fact, today’s standard turbines are based on the same physical principles as 18th century windmills." is marketing hooey. They are hanging their hat on the fact that the windmills were unducted props and most of today's turbines are the same, which is true. The airfoils used today are not 18th century designs. They ...

1

As @user3823992 points, this does not work: the (excess) pressure of water at the level of the hole is equal to $\rho g h$, where $h$ is the altitude difference between the hole and the free surface in contact with atmosphere. As water flows out, $h$ decreases and thus the pressure and flow rate too. In the tube now, the pressure in the water column is less ...

0

Great problem! Simple to state, incredibly complex to solve. I have three suggestions: I'm not sure that it's a good idea to consider an infinite cylinder. It will have an infinite mass an might just sit there. Your question is not formulated very precisely. To study the time evolution of a dynamical system you need to fully specify it's initial ...

0

This is assuming no stirring, since that is a hassle. There is a third possibility of pouring both in at the same time. That would ensure the fastest convection. Cold then warm will be slowest since warm water rises. Warm then cold will be slightly faster since it promotes convection. This all depends on your personal preferences. The fastest route to ...

0

There doesn't appear to be speed at which you can fall on water, and have it be the same as falling on a solid. The Water vs. Pavement episode of Mythbusters did tests involving falling from a variety of heights, and even at a height great enough for the human analog to achieve terminal velocity, the effect of the impact was notably greater for landing on ...

1

Looks like it is possible to some extent: http://iusti.univ-provence.fr/Local/iusti/dir/user-4263/documents_CPI/pdf/2003_CR.pdf (C. R. Mecanique 331 (2003) 61–67 ): "In 1908,Worthington [1] observed with a high-speed photography technique an odd but frequent phenomenon: milk drop impacts on shallow milk surfaces. His observations show complex dynamics of ...

0

I can only make the following hypothesis. It explains why the drop explodes and why the explosion is symmetric. It does not explain why it happens always in the same plane though. As the water falls along the building, the airflow around it is not uniform because of the presence of the building on one side. Then it is natural that the water bubble starts ...

0

What physical effect are you studying? If you are investigating the properties of the steady state, I would try to start as close to it as possible. I would recommend something that only depends on the transverse coordinate: linear velocity change with the potential. If you are investigating turbulence you should break all the symmetries and try many ...

4

As I had a MSc thesis that had to regard this problem within it, did look into this extensivelly. For Beer formation it would be good for you to read http://edepot.wur.nl/202245, tho it's an older PhD thesis, it was checked and backed by experimental data. Alot about foam formation and breakdown was explained. Found also that many newer works also used much ...

1

There's static pressure (P) and dynamic pressure ($\frac{1}{2} \rho v^2$). A pitot tube placed in the center section of your drawing will measure the stagnation pressure, which is the sum of the two terms. Physically, the dynamic pressure will be larger at the narrow section, and the static pressure will be smaller there. The sum will be constant ...

1

I disagree with the most voted answer, by CAGT. He says "This area is completely different to the one above", but this means nothing. The equation $p = {F \over A}$ mentioned by the author does hold, and there is no contradiction or paradox in it. In fact, the equation $p = {F \over A}$ holds not only here but anywhere else in physics. You may write it in ...

3

A vortex ring has a finite core radius $a$. The circular line vortex is a vortex ring with an infinitesimal core radius $a\to 0$. It is unphysical and only occurs in an inviscid fluid.

0

As far as I can tell this condition is not apparent at this order via these BC considerations. But consider the conservation of energy (to the order you've considered) in a time independent system, ie $$\frac{d(c_g E)}{dx} = 0,$$ here $E\sim a^2$ and $c_g \sim \sqrt{h}$ so that $a\sim h^{\frac{-1}{4}}$, which is a relationship between the two scales ...

1

Why can't we measure the pressure at 1000 m for different temperatures? Meteorologists certainly can do that, and in fact do do that, all the time. They use weather balloons, sounding rockets, and all other kinds of instrumentation to measure conditions in the atmosphere. The resulting picture is rather complex. Conditions vary with place, the seasons, ...

0

A fan changes the average velocity of air molecules. This can be seen as a molecular scattering process for very thin gases, whereby every single molecule hits the surface of the angled rotating fan blade in such a way, that an axial velocity component is imparted on the molecule. Since a single rotating fan blade can not do this without also imparting a ...

1

For aviation purposes, standard atmosphere is considered to be dry air at mean sea level, at 15 degrees C (59F). It is true that pressure decreases with increasing altitude, and temperature usually does, but not always. It is not a simple relationship, because it depends on humidity, heat transfer from above and below, vertical circulation, horizontal ...

2

Think about the air around the fan at any given time: The amount of air flowing into the fan must equate to the air flowing out of the fan. The amount of air that passes through an area in a given time is related to the velocity of the air i.e. the faster the air is moving, the more air that can flow through a fixed area/hole/slot. The fan blades apply a ...

6

Here's a standard fan with some (hard to see) arrows indicating air flow. The fan works by pulling air in and then making it move faster. The air flow behind the fan is slow moving and wide (you can see the arrows behind the fan coming from above and below the fan blades) whereas the air flow in front of the fan is fast moving and narrow (which follows ...

24

There is a YouTube video that visualizes the air flow around a propeller for various configurations. I caught a screen shot of a moment that more or less shows what is going on: As you can see, this happens at 2:07 into the clip - this happens to be for a dual rotor configuration (two counter rotating blades) but the principle is the same. Behind the ...

3

$T^{ij}$ is nothing more or less than the flow of $i$-momentum across surfaces of constant $j$.1 As a result, the force exerted across a surface $S$ with unit normal one-form components $n_j$ has components $$F_{(S)}^i = \int\limits_S T^{ij} n_j \,\mathrm{d}^{d-1}x$$ in $d$ dimensions. The argument for symmetry is not that the cube is static. The argument ...

1

What you're feeling there is the wake behind truck. As the truck passes through the air, it imparts momentum and generates quite a lot of turbulence (allowing it to impart even more momentum). It takes a little bit of time for that momentum to spread outward and reach you. As that air is 'pulled' behind the truck more will be drawn in from the sides. These ...

0

I'm not an expert in fluid mechanics, but let me try to answer your question. Let us write the stress tensor as $\tau_{ij}$, with $i,j=x,y,z$. For definiteness, we look at $\tau_{xz}$. The meaning of this component is as follows: consider a directional face element of area $dA$ with normal pointing along $z$-direction, $\hat{z}dA$. Then, $\tau_{xz} dA$ gives ...

0

No. If the fluid is compressible and the flow is not stationary we can choose initial conditions with an inhomogeneous density. Then the flow will equilibrate and the density will become become stationary but during this time anything can happen. One can build a counter example simply by choosing appropriate initial conditions. If your are asking about a ...

0

I don't think that you should think of particle conservation as a conservation law in the context of classical physics. As Qmechanic says in a classical system, the particle number is the number of degrees of freedom and is fixed as a definition of the problem. Once we have decided how many particles will be there we can write a Lagrangian, investigate its ...

0

Not sure about this - best guess is that you have observed the formation of an irrotational vortex. When you stir, the liquid tends to rotate as a rigid body. When you stop stirring, then In the absence of external forces, a vortex usually evolves fairly quickly toward the irrotational flow pattern, where the flow velocity u is inversely proportional ...

3

I can not answer you question mathematically, but my experience with Burgers equation tells me that there is no such transformation. If you think of Euler equation as Navier-Stokes in the limit where the viscosity vanishes $\nu \to 0$, then time reversal symmetry is simply spontaneously broken. As long as you have a finite viscosity the system is ...

1

However, If I orient the x axis along initial direction of velocity, I should get: $$m\ddot{x}= - (\mu_1 \dot{x} + \mu_2\dot{x}^2)$$ That's not quite true. You implicitly assumed that $\sqrt{\dot x^2} = \dot x$. That is only true if $\dot x\ge 0$. Otherwise you need to use the more general $\sqrt{\dot x^2} = |\dot x|$. You don't want to go there. You ...

2

I've found in this website an interesting plot. In particular figure 12 shows some polar curves vs weight*: It is possible to note that bigger weights requires more engine power to maintain the altitude for every given fixed speed. Plus we must take into account the extra fuel spent to take an extra bag from ground to 900 km/h at 12 km altitude (that ...

0

Some numerical values: Althought he didn't explain his calculations, according to Tony Webber, former Qantas Group chief economist the costs of 2 extra kilograms are: These increases represent weight gains of around 0.23 per cent and 0.20 per year for woman and men, respectively. Since 2000, the extra loading that an average adult passenger carries is ...

1

Any tensor $A_{ij}$ can be decomposed into symmetric and antisymmetric parts, regardless of whether or not it is diagonalizable. \begin{align} S_{ij} & = \frac{1}{2}\left(A_{ij} + A_{ji}\right) \\ \Omega_{ij} & = \frac{1}{2}\left(A_{ij} - A_{ji}\right) \end{align} so that $$A_{ij} = S_{ij} + \Omega_{ij}$$

0

The milk is a thick liquid (more viscous than water). When it boils the bubbles formed don't break as quickly as they do in water. This means more bubble form which take up room in the pot and this causes the milk to boil over. Water rises, too, just not as much since the bubble break more quickly.

0

The direction of the Resultant force $R$ is always dependent on the direction of the $V\infty$ But however, the direction/orientation of the Normal force $N$ is dependent on the orientation of the body itself ($N$ is perpendicular to the body and axial force $A$ is parallel to the body.) In the above case, since there is not much of surface interaction, ...

1

Let there be given an $n$-dimensional manifold $(M,\nabla)$ endowed with a connection $\nabla$. [In particular, we do not assume that the manifold $M$ is equipped with a metric tensor.] Let there be given a curve $\gamma:\mathbb{R}\to M$. Here the reader should think of $\mathbb{R}$ and $M$ as time and space, respectively. If $f: M\times \mathbb{R}\to ... 2 You correctly identify the residual velocity of the water after bouncing off the bucket as a critical parameter in the calculation. Where you go wrong is in assuming that you can assign any value you want to it. If your bucket's bottom was shaped in such a way as to "turn around" the water jet hitting it, then you would have the maximum possible momentum ... 3 You've pretty much answered your own question; you've motivated it and derived the expression, just with a different notation. To rephrase what you've said: the material derivative is the derivative along the path defined by the integral curves of$\vec{u}$. This is useful because this is the path taken by a small element of the fluid. As a motivating ... -1 You have to think according to the situation. Consider a tank with high pressure, if we open the tap means water flows with high velocity and eventually after sometime the velocity will start to reduce because of pressure is decreasing accordingly. Velocity and pressure is directly proportional. 0 I'm not quite sure if this correct, but I think you can solve this with an energy balance, the equation of state and the mass flow: Assuming$\dot{v}$is zero, no losses in the pipe: $$\dot{E}_{kin}=\frac{1}{2} \dot{m} v^2$$ $$\dot{m}=\rho A V$$ $$P = \rho R T$$ Substitute them, and evaluate them for both the incoming and outgoing location(s) On second ... 0 You are right, difference in the bounds of the integrals represents the difference between the Eulerian and Lagrangian perspectives. The Lagrangian frame of reference follows a volume of fluid, so Newton's laws can be directly applied. However, the bounding surfaces of the volume can change over time. This makes it difficult to actually apply the ... 1 Short answer: I think the notation is the main problem here. In your second equation, the LHS$\rho\mathbf{u}$is a function of$\mathbf{x}_0$and$t$, while your RHS$\rho\mathbf{u}$is a function of$\mathbf{x}$and$t$. The subtle difference is that$\mathbf{x}_0$should be treated as a particle label, not an actual position. As you suspected, the ... 3 If$V = \frac{m}{\rho}$,$dV = \frac{m}{d\rho}$This is the source of your error. You can re-write the above as$\rho V = m$, and this yields$\rho dV + Vd\rho = 0$, or$V \frac d {dV} = -\rho\frac d {d\rho}\,$as a differential operator. This leads directly to the alternate form for the bulk modulus$B = \rho \frac{dP}{d\rho}\$. Being a bit more ...

Top 50 recent answers are included