New answers tagged

0

Your roof won't rise in this case. Your roof will simply crush. If tornado wants to lift your roof, its outer air would exert an equal downward force on ground. If its of exact diameter that force would be exerted on your walls that support roof and crush down your roof.


0

I find Pulliam's notes for the Euler equations to be a pretty good introduction to this topic using the equations of fluid motion. The idea is that you start with a conservation law: $$ \frac{\partial \vec{Q}}{\partial t} + \frac{\partial \vec{F}\left(\vec{Q}\right)}{\partial x} = 0$$ where $Q$ is your variable vector and $F$ is your flux function. You can ...


1

Pouring anything into the cup adds mass but doesn't add angular momentum. Conservation (plus friction between fluid and cinnamon) of angular momentum requires the new total mass to rotate slower than the original contents.


1

On a vortex mixer the liquid is swirling inside a test tube. In a centrifuge, the test tubes (containing the liquid) are held in a fixed position inside a rotor, and the rotor is spinning inside the centrifuge.


2

A vortex is a local disturbance caused by turbulent flow. It carries fluid and particles inward toward a central axis of rotation. The central axis is not fixed. It moves within a larger body of fluid. The speed of the vortex is greater near the axis of rotation and decreases toward the edge of the vortex. Unlike a vortex, a centrifuge creates ...


4

There's no theoretical method that will reliably predict how the cryogenic check valves will respond to flow/ pressure given the possibility of multiphase flow and that the check valve is a complex, active restriction. You really have to instrument your setup and take data. True a manometer will not work, but you can use cryogenic pressure transducers like ...


2

Draft and cavitation. Ships cannot afford to have big propeller diameters, they have to make do with the smallest diameter available in order to stay within the draft of the ships' hull. They operate in a medium which is 800 times denser than air, and one important concern is to avoid cavitation. This again means to limit suction peaks and leads to very ...


0

The issue with massive waves on a 1 meter deep ocean is that the waves cannot propagate fast enough on a planetary size object. We get fast moving shallow tsunami waves in the open ocean over a thousand meters deep. The tsunami piles up when the wave slows down due to contact with a shallow shoreline. Hundred meter high waves could never propagate fast ...


3

First of all, don't think of multipole moments as separate things that have their own individual meaning. Instead, think of them as parts of one thing. Once we have all the parts written down, we can start naming and organizing each one to determine its contribution to the whole. Now, for your question Is there a physical interpretation for multipole ...


3

Yes, you are right. The conservation of mass (for incompressible fluids, as water is) states that $$ A_1 V_1 = A_2 V_2 $$ where $A$ is the area of the stream and $V$ the velocity of the fluid. Hence, as water accelerates due to gravity, $V_2>V_1$, so $$ A_2 = \frac{V_1}{V_2}A_1 \rightarrow A_2<A_1 $$ and the stream becomes thinner as the water flows ...


3

Is it because the velocity of water increases as it flows down which results in a decrease in cross sectional area? I believe you're right. We can also add that in the usual case of flow conservation, that water flow is constrained to "stick in one piece" by the tube inside of which it is flowing. In the case of the tap, the water flow binds together ...


2

The key concept needed here is that the Hemholtz decomposition is not necessarily unique. Non-uniqueness can occur because there exist nontrivial vector fields which are both irrotational and divergence-free. For example, the constant 2D velocity field $\vec v = (1,0)$ can be expressed as either $\vec v=-\nabla \phi$ with $\phi(x,y)=-x$, or as $\vec ...


1

Full derivation of volumetric flow for a siphon: Bernoulli’s equation in the absence of friction: $p_1+\frac12 \rho v_1^2+\rho gz_1=p_2+\frac12 \rho v_2^2+\rho gz_2$ Head loss in the case of friction (Darcy – Weisbach): $\Delta p=f \frac{L}{D} \frac12 \rho v_2^2$, where $f$ is the Moody friction coefficient. So: $p_1+\frac12 \rho v_1^2+\rho ...


2

Well, let's look at the equations for a flow initially at rest: $$ \frac{\partial \rho u}{\partial t} = -\frac{\partial p}{\partial x}$$ So, if we assume density is constant in the flow, you get: $$ \frac{\partial u}{\partial t} = -\frac{1}{\rho}\frac{\partial p}{\partial x}$$ and so you can see that in both your cases, the gradient in pressure is the ...


3

As I mentioned in my comments, there are many approaches you can take here, and full-on Navier-Stokes is not something you want to attempt. Either incompressible or compressible. Fully solving the Navier-Stokes is very, very far from real-time capable. Even for the most basic problems I can think of. You said you have incompressible figured out. That's ...


1

The initial velocity of the water through the hose doesn’t affect the velocity of the water that the siphon quickly settles down to. After only a brief moment, if water were a perfect fluid, the steady-state velocity of the water in a siphon would be $$v=\sqrt{2 g h} ,$$ where $g$ is the acceleration due to gravity at the Earth's surface, and $h$ is the ...


2

i.e. everything else being equal, will I get a faster siphon if I start the process with a powerful pump than with a weaker pump? Or does that become irrelevant once I disconnect the pump and the siphon starts flowing? Simply put, only three factors significantly affect flow rate: Difference in height between inlet and outlet of your siphon. Roughly, ...


2

Your setup of the problem isn't correct. Let V be the volume of water in the container. Then the rate of accumulation of heat in the container is equal to the heat in minus the heat out. The correct equation for this is: $$V\rho C_p\frac{dT}{dt}=q\rho C_p(T_{NEW}-T)-kV\rho C_p(T-T_a)$$This assumes that the tank is well-mixed so that the exit temperature ...


-2

The spot that you injected the red ink will turn red first of course.


2

This is not limited to hydrofoil hulls. You can sail across the wind ("reaching") on other low-drag shapes such as windsurfers at speeds greater than the windspeed. The fastest such craft, so far as I've heard, are iceboats -- which are sort of an ultimate "hydrofoil," in that they skate along the frozen lake surface with almost no friction at all. ...


2

Just look at it as a wedge. To a 0th approximation, a sail or wing through fluid is just like a knife through jello. It can only move along a line. A sailboat has two of those. It has a sail cutting through the air, and it has a keel (centerboard, leeboard, hull) cutting through the water. Neither one of those can move sideways in its fluid, so if the air ...


0

The Reynolds number gives a ratio between forces of inertial origin and those of viscous origin. For a given geometry of the problem, increasing the Reynolds number will lead to turbulent flow from a certain threshold. However, this threshold is strongly dependent on the geometry: this is actually common knowledge, a better design (more "aerodynamic" we say ...


4

Use the Reynolds number equation: $Re={vL \over \nu}$ where $\nu\approx 1.5\times10^{-5}m^2/s$ is the kinematic viscosity for air. If you enter this into the equation, you end up with $Re\approx 67000{v\over{m/s}}{L\over m}$ i.e. for a race car traveling at 40m/s and with a length of 4m it comes out to be around 10 million, which is certainly ...


0

Can i also use this equation to calculate the drag force on a car going through air (which is a fluid)? No. Stokes really only applies for perfect spheres and low speeds (laminar flow). For higher speeds and non-spherical objects you need to use this approach.


1

The characteristic length is the dimension that defines the length scale of a physical system. This implies that for any system, which my contain several length scales, there may only be one characteristic length scale. This is also generalizable to the other characteristic scales such as time, speed, etc. As you may have read in my other answer, generally ...


1

In my view, the objective of knowing the friction factor, is for one to be able to calculate what is the pressure drop needed to push a given flow $Q$ through a given pipe diameter. This kind of relations exists for several models of non Newtonian fluid, take for example the power law model: $\tau=K\gamma^n$ In this case the solution gives: ...


0

Let me start off with something that should be obvious but I feel the huge need to say it -- I am not trained in the biomedical field and what I put in my answer should not be construed as definitive and ready-for-use in actual patients. I will do my best and present what I think is correct, but I am not a doctor and don't even pretend to play one on TV. ...


3

Hitting the air twice as fast takes four times as much energy ($E = {1\over 2} m v^2$). So there's an extra factor of two. To expand a little; you're fighting drag ($F_d$) in a rowing machine and drag increases as the square of velocity. ($F_d \propto v^2)$. Now, $Energy = force \times distance$ and $Power = {energy \over time}$ so $Power = force \times ...


1

Let me write $\Pi=\nabla\cdot u$. The shear strain is given by $$ S_{ij}=\nabla_iu_j+\nabla_ju_i-\frac{2}{3}\delta_{ij}\Pi $$ and the vorticity tensor is $$ \Omega_{ij}=\nabla_iu_j-\nabla_ju_i . $$ The most general stress tensor linear in gradients of $u$ is $$ \tau_{ij} = \eta S_{ij} + \zeta \delta_{ij} \Pi + \eta_R \Omega_{ij}. $$ We want to show that ...


0

The behavior of a material must be independent of the motion of any observer. So, if an observer were rotating at the same angular velocity as the local vorticity, he would observe the same local rate of deformation tensor, but with the vorticity removed. This would have to result in the same stress tensor, because the motion of the observer can not affect ...


2

For an incompressible viscous fluid, the equation of continuity (mass conservation) for an axisymmetric deformation is given by$$\frac{1}{r}\frac{\partial (ur)}{\partial r}+\frac{\partial w}{\partial z}=0$$ where u is the radial velocity, w is the axial velocity, r is the radial coordinate, and z is the axial coordinate. The key components of the stress ...


2

For the sake of the argument, assume that the stream of water is 10 cm tall. The water drops at the bottom of the stream have been falling for a longer period of time than the water drops 1 cm above them, and this is true for all the water drops in the falling stream. Because of this, the water drops at the bottom of the stream are moving at a higher ...


1

It is due to a matter of fact that same mass and hence same volume of water has to flow through a given cross section.This is termed as law of continuity in fluid mechanics. It is a known fact that velocity of a falling object increases with height, and hence to satisfy the law of continuity the cross section has to decrease.(to keep the volume constant).And ...


5

Gravity acts as a source term in the equations, and it is a source term on the energy and momentum equations. The mass conservation equation is not modified by gravity. So, looking at the momentum equation with gravity, we have: $$ \frac{\partial \rho u_i}{\partial t} + \frac{\partial \rho u_i u_j}{\partial x_j} = -\frac{\partial p}{\partial x_i} + \mu ...


2

There are three points to be noticed: If you just blow without closing the lips, you would change the boundary condition. The trumpet waveguide is not "nicely predictible", the approximation of an open tube does not work cause the bore variations $S(x)$. You need to solve this kind of beasts for reasonable 1D propagating pressure approximation: $$ ...


1

I agree with don_Gunner94's answer. If the fluid come out from the constricted passage to atmosphere,it will experience atmospheric pressure,which is same as the pressure acting at top of the container. Even according to Bernoulli's principle, Static pressure + Dynamic Pressure = Constant Therefore, the pressure acting on the fluid when it is inside the ...


0

I suppose that the best way to answer this question is by using an analogy. Take a glass of water(at room temperature) and place it in a refrigerator. What happens to the water? It cools down. Now you take this glass of cold water and keep it back outside. What happens now? The temperature of the water comes back to the room temperature. Why? Because the ...


3

Assuming the equation you found is correct, it's just the 1-dimension heat equation: $$ D^2 \frac{\partial^2 v}{\partial r^2} = \frac{\partial v}{\partial t}, $$ where $D^2 \equiv \mu/\rho$. We want to solve it on the domain $r \in [0, R]$, $t \in [0, \infty)$, subject to the boundary conditions $v(r,0) = 0$, $v(R, t) = R \Omega$, and $v(0,t) = 0$. (This ...


1

Edit: This use of non-dimensionalization is wrong because it is incompatible with the boundary conditions. More information here. We're looking for a solution to the fluid velocity $\vec u$ that looks like $\vec u = u(r,t)\hat\theta$. Given the symmetries of the problem, we have to solve Navier Stokes: $$ \partial_t u=\nu(\nabla^2 u - \frac u{r^2}) $$ The ...


1

First of all, I think the equation of motion is not correct. I believe the equation of motion is $$ \dot{v}_\theta = \nu \left( v^{\prime\prime}_\theta + \frac{v^\prime_\theta}{r} - \frac{v_\theta}{r^2} \right) , $$ where $\nu=\mu/\rho$. This is easiest to derive using the result for the Laplacian of a vector field in cylindrical coordinates. In order to ...


0

Question; I am not sure how to proceed to derive $vθ(r,t)$ It may be easier to make a clean simple start rather than trying to seek a mistake. If the fluid is rotating slowly, and it's viscous. You basically have a laminar flow conditions, which makes the whole really simple as the velocity distribution is an exponential curve. Ie parabel. At your ...


-1

At 50cm 'depth' the ball was not actually submerged, because the water did not have sufficient time to refill the 'crater' created by the ball entering the water. The ball can be said to be 'floating' on the base of the crater. As the water returns at very high speed to refill the crater the bottom of the crater with the ball 'floating' on it is effectively ...


0

I didn't forget about this one! I cannot find fault with your derivation and the resulting PDE. I may have missed something also though, so who knows. But what I can do is come up with a solution for the PDE you do have, which should help if it turns out the original isn't right for one reason or another. Starting from (where $\nu = \mu/\rho$): $$ ...


2

Not really. For laminar flows, the solution will be the Blasius solution, but the solution is still numerical. There are analytical functions that can approximate it fairly well. For turbulent boundary layers, there is even less hope for an analytical solution. The flow is non-linear and time dependent, but mean equations can be found. These will still need ...


0

As heat is essentially a measure of the kinetic energy of the particles in a substance, technically a thermometer with excellent precision would measure a difference. The particles of air colliding with the thermometer (in Situation 2) will have a smaller relative speed to the thermometer (compared with situation 1), thus their collisions will have lower ...


1

If the velocity is constant then ratio is 1:1


0

I believe these attractors you are referring to are generally referred to as eddies in the ocean. These features are similar to the hurricanes, and low and high pressure systems in the atmosphere, and just like in the atmosphere they move around. With monthly mean data (monthly climatology) you can advect particles around in a variety of different ways. ...


1

A rule of thumb exists if coriollis force is the dominant force balancing the pressure gradient. This is known as the geostrophic balance : $$ \overrightarrow{V_g} = {\hat{k} \over f} \times \nabla_p \Phi $$ However if only a pressure gradient is being maintained by some source then the velocity will keep increasing as the pressure gradient results in ...


0

It seems like you are complicating the process a bit too much. As noted in the earlier response: boiling occurs when the vapor pressure of the liquid reaches the surrounding pressure (e.g. of atmosphere or atmosphere + pressure of overlying liquid (heating usually done from bottom)). Evaporation, in the context of initiation of boiling, serves only to ...


1

You say: Horizontal movement dictates that hydrofoil has to be at zero angle (minus angle of attack necessary for hydrofoil overcome the weight). You could make the same argument about a surfboard; if it weren't horizontal, its movement across/through the water would make it go up or down. In both cases, you'd be missing the point. At the point of ...



Top 50 recent answers are included