New answers tagged

2

What I fail to see is how moving too quickly could also impair cooling performance as stated in a lot of online forums. One argument I clearly remember from reading about this a while back was: and: You shouldn't crank the pump speed too fast or the water won't have time to pick up the heat from the waterblock as well. The latter statement you quoted ...


4

What comes first? Pressure or flow? Without potential, a pressure differential, there is no flow. It's pressure that needs to be there first. So rather than ask what pressure drop you get, it's better to ask what flow you get from applied pressure. We can never measure pressure drop across an infinite pipe because we can never reach the infinite point. ...


1

You can wait three or four minutes and let the pressure drop or..... If you feel lucky you can try tapping the can a dozen times, as is shown on this video https://youtu.be/NQYO3Dp8lCA


2

Consider lubrication of bearings involving rotating shafts. The gap between the shaft and the outer static surface of the bearing is small, so that the curvature can be neglected and locally, the flow can be considered as occurring between parallel plates. In bearings, the shaft and static member are not concentric, so that it is not exactly Couette flow, ...


1

My apologies for a short answer, but it really depends on many factors I don't know about, such as virtually anything that may have disturbed the airflow along the fuselage. These extracts are based a pdf file from the University of Leipzig but unfortunately I can't get my tablet to copy the url. My basic point is that, despite the direction of the ...


1

Extrusion seems to be one application... Couette flow is a classical problem of primary importance in the history of fluid mechanics (1-4), which is a typical example of exact solutions for Navier-Stokes equation. Couette flow is perhaps the simplest of all viscous flows, while at the same time retaining much of the same physical characteristics of ...


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A simple way to understand this is as follows. Every system can be linearized and decomposed into linear eigenmodes. If the system was truly linear, the amplitude and phase of each eigenmode would remain constant, and each eigenmode would evolve dynamically by its corresponding eigenfrequency. In the weak turbulence limit, we allow non-linearities, but ...


0

You provide some constant power input to motor. This power is proportional to the product of torque exerted by the motor and its r.p.m. Neglecting losses, power input to the motor goes into two parts: one, raising liquid from sump level to outlet level (outlet is where flow comes out of the pipe into atmosphere), and two, kinetic energy of the flow at the ...


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Another alternative: make sure that you are using a positive displacement pump. When such a pump shuts off, water cannot keep flowing through the pump.


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This looks like a positive displacement pump. Assuming that it is, the speed of the liquid at the inlet is directly related to the rotational speed of the gears, which are directly related to the speed of the liquid at the outlet. Without more information regarding piping diameters on the inlet and outlet, and the trapped volume around the gear teeth, the ...


1

Textbook derivation is for a fluid element, and you need to find net force on it. If you consider the fluid element to be cubical in shape, then considering only z-direction, viscous forces act on both top and bottom face of the cubical element. These are given respectively by $\eta\frac{\partial u}{\partial z}\bigr|_z$ and $\eta\frac{\partial u}{\partial z}\...


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Why? The expression given by: $$ \int_{S} \ dS \ \hat{\mathbf{n}} \cdot \left( \nabla \times \mathbf{A} \right) = \oint_{C} \ \mathbf{A} \cdot d\mathbf{l} $$ is just a vector calculus rule called Stokes' theorem and is valid regardless of the type of flow. Because the $S$ is not simply connected? The $S$ in the expression represents the closed ...


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Exactly because the region is not simply connected. The stokes (or Green in 2d) theorem no longer holds. Consider for example the two dimensional vector field $$\vec F(x,y)=\frac{-y\hat i+x\hat j}{x^2+y^2},$$ which has vanishing curl and circulation $2\pi$ around a unit circle centerd at the origin. If this vector field is meant to be a flow velocity field ...


2

The point is there's several $W$'s. $$\rho_i=\sum_j W(\mathbf{p}_i-\mathbf{p}_j,h)=W(\mathbf{p}_i-\mathbf{p}_1,h)+W(\mathbf{p}_i-\mathbf{p}_2,h)+\ldots$$ dropping the mass as they do in the paper. The thing is they're treating the particles as indistinguishable e.g. they all have equal mass, so the form of the functions/constraints applied to them are the ...


2

Compute the gradient of this velocity field: $$\operatorname{grad}{\bf u}=\nabla_i u_j= \begin{bmatrix} \partial_x v_{x}& \partial_x v_{y}&\partial_x v_{z}\\ \partial_y v_{x}& \partial_y v_{y}&\partial_y v_{z}\\ \partial_z v_{x}& \partial_z v_{y}&\partial_z v_{z} \end{bmatrix}$$ This matrix can be represented as a sum of an ...


0

$\frac{D}{Dt}(\cdot)$ operator gives the change in property of a $\textit{specific}$ fluid particle. In particular, $\frac{D\rho}{Dt}$ evaluated at $(x,y,z,t)$ gives change in density of the fluid particle located at position $(x,y,z)$ at time $t$. Therefore $\frac{D\rho}{Dt}=0$ everywhere means that at any spatial point and at any time, density of fluid ...


1

https://en.wikipedia.org/wiki/Incompressible_flow Yes a flow can be incompressible (rather isochoric) and unsteady. However, the unsteady term in Conservation of Mass equation is cancelled by advection term regardless of whether the flow is incompressible or compressible CASE 1: Any fluid material that undergoes an incompressible flow ASSUME: -No ...


2

In "Adventures in Friedmann cosmology: A detailed expansion of the cosmological Friedmann equations" by Robert J. Nemiroff and Bijunath Patla in the American Journal of Physics volume 76, on page 265 (2008); http://dx.doi.org/10.1119/1.2830536 the authors call them "cosmic strings" But this is in the context of cosmology, so its for a universe that on very ...


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There are three possible contributors to the leaking: 1) a reduction in ambient pressure as you drive to a higher altitude; 2) a possible expansion of the liquid if it heats up during transport, which makes the liquid expand and decreases the vapor space (hence increasing the air pressure inside the container); 3) If the liquid has a moderate vapor pressure,...


1

Consider Fick's second law of diffusion, in one dimension, where $u$ is the concentration of the diffusing gas (in $\mathrm{mol/m^3}$) and $D$ the diffusion coefficient: $$\frac{\partial u}{\partial t}=D\frac{\partial u}{\partial x}$$ If we assume the concentration of gas outside the container to be much smaller than inside (a reasonable assumption), then ...


1

The pressure at any point open to the atmosphere is the atmospheric pressure. 1) You are using concepts of fluid statics and applying it to fluid dynamics. When an entire fluid is in complete equilibrium, pressure must increase proportionally with depth, since the mass of the fluid above a certain depth must be completely balanced by the pressure forces ...


0

(Subscripts $w,i,m$ correspond to water, insecticide, and mixture respectively.) First, $Q_m=Q_w+Q_i$ only if $\rho_w=\rho_i=\rho_m$. Otherwise you must equate mass flow rates, $\dot{M}_m=\dot{M}_w+\dot{M}_i$, to find $v_m$, assuming that mixture density is uniform over the cross-section at point 3. Second, the form of Bernoulli equation you have written ...


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Part of the answer is, at low Reynolds number, in streamline flow, there's little if any drag. At higher speeds, though, the wake of a moving object creates vortices (and does so in random fashion, a spontaneous symmetry breaking occurs). Those random vortices cause a velocity-squared retarding force, which is energy-losing whether your motion is ...


0

In principle you can't apply Bernoulli to what is in effect a (simple) network of pipes but in some cases approximations will do. Let's apply Bernoulli's equation to the left and middle sections of the pipe: $$P_1+\frac12 \rho v_1^2=P_2+\frac12 \rho v_2^2$$ As liquids are incompressible ($A$ is the cross-section of the pipe): $$A_1v_1=A_2v_2$$ So with ...


1

I think the sound makes the particles move more because there's more interaction when the sound particles are included, but I'm probably wrong because I don't know much about physics...


0

You can either measure $F$ and $v$ then calculate $C_D$ from the eqn, or you can look up $C_D$ in a table for the corresponding Reynolds number [1], and use that to calculate $F$. You cannot calculate both quantities using one equation. [1] http://www.thermopedia.com/content/707/


2

Total energy of a closed system is always conserved (time translation, Noether theorem). However, this energy might go from macroscopic motion (fluid flow) into random vibrations of the molecules. I am not 100% sure of the thermodynamics, but I will try. The total energy $U$ is still there but it has changed to heat. The velocity $v$ of the molecules also ...


2

@ChesterMiller: answer is good, but I would just simplify it. What you say is that the total momentum is the momentum of the contents of the control volume plus the sum of that which traverses the boundaries. If you take the change in momentum per unit time, you have momentum flux, which equals force.


3

The forces acting on the fluid are changing its momentum. There are forces acting on the stationary portions of the control volume boundary, and there are also forces being applied to force fluid into, and acting to prevent it from flowing out of the control volume. The total rate of change of momentum in the control volume at any instant of time is equal ...


2

Pressure is momentum transfer due to molecular collisions once you have subtracted out their average motion. So decrease in pressure due to increase in average speed may be construed as transfer of kinetic energy from random molecular motion to mean motion. This means that random molecular motion (by which I only mean molecular motion with average subtracted ...


1

Because water "squishes out", allowing the sole of your show (which has treads for that reason) to contact the ground directly, and experience friction. Shoes with smooth flat soles are quite slippery on water, if the floor is also smooth and flat. Oil, because it is more viscous, does not "squish out".


0

The following is academic knowledge. I'm no pipe flow expert. For laminar flow through a rectangular pipe, the volumetric flow rate $\dot V$ calculates to $$ \dot V=\frac{K \cdot h^3 \cdot b}{12 \, \eta \, l} \cdot \Delta p $$ with $b$ and $h < b$ being the pipe dimensions, $l$ pipe length, $\eta$ viscosity, $\Delta p$ pressure drop along the pipe and $K$...


2

My theory is that since water, and indeed all liquids, are incompressible, they both form at the same time. As the pebble hits the surface of the water and pushes it down, the surrounding liquid is pushed up (relative to the original surface level of the liquid), as you stated. Any displacement of a liquid must be accounted for.


0

If you set parabolic profile for axial velocity at inlet then flow will be axial throughout pipe's length. This is because, apart from hydrostatic contribution, there is no pressure variation in radial direction. If you set parabolic profile for pressure at inlet, then you will have radial motion too (at least at inlet), and how the flow shall develop ...


0

The percentage by which the pressure lowers on the leeward side of your finger is minuscule, so it barely affects the rate of evaporation.


0

What will happen if we set inlet having variable pressure profile (e.x parabolic profile)? Can someone explain the flow behaviour? varying the inlet pressure profile will vary the flow rate, possibly cause boundary layer separation, eddies, vorticity, etc. There are many possibilities. Also, will there be difference between compressible and ...


1

Turbulence is well defined in the bulk of the flow, hence in 3D in the examples you quote, while surface tension acts, well, at the surface! I believe a good reformulation of your question would be "will a flow be less chaotic with larger surface tension?" Surface tension is highly dependent on geometry. It stabilizes some flows (e.g. a not-too-shallow ...


2

First, give an illustration of a linear system. You know you have to eat healthy, so you are prepared to pay 50 cents for an apple. But the next day, if apple prices double, you will forget the health benefits. Now a non linear system, because it's based on people's reaction to supply and demand. An everyday example based on the Uber taxis system. Uber ...


0

No, you don't have to write flow equations for solid part of the billet (sounds like a trivial statement), but you will have to take into account vertical velocity $u_z$ of solid billet while solving energy equation. Under certain conditions you may neglect this as well and treat the problem as one of pure conduction to a good approximation. This happens ...


3

First, the reason why the finger becomes more wind-sensitive with some saliva isn't that the saliva evaporates but because the saliva, or water, is a good thermal conductor. The finger has to be warmer than the air so the heat flows from the finger to the air and a good thermal conductor such as saliva helps this flux to take place. Second, because it's the ...


1

Your question is short on details, so to answer your question I am assuming a particular configuration. Hope it helps with whatever your actual configuration is. A closed porous container consists of a gas at partial pressure $p_1$ which is less than partial pressure $p_2$ of that gas in the ambient. We shall assume that $p_2$ is constant and to further ...


2

The equation you wrote is for steady state flow through a porous wall of thickness d. Q is the volumetric flow rate. Q/A is the so-called superficial velocity. The equation inherently assumes that the pressures on both sides of the wall are constant, and not varying with time. The only way that you would get an increase in pressure difference with time ...


1

The weight functions should be continuous and vary from a maximum value at $r=0$ to zero at the cut-off. The linear function is the simplest and computationally least expensive function that satisfies this requirement. That's why it's used. Note, however, that non-linear functions are occasionally used, e.g. Yaghoubi, S., et al. "New modified weight ...


3

Sugar has a solubility in water of 909 g/1 L at 25 °C (wikipedia). Even if coffee is not water, let's take this as an estimate. In other words, if you do not heavily overload coffee with sugar, the sugar will get dissolved. The kinetics of this process are not known to me, but in general it seems fairly quick. The reason you stir is to distribute the water ...


2

Querido beto, sugar not only get mixed by convection, or fluid transport, as you do when you revolve the spoon, if that were the case you would be right. But sugar also transport to the liquid through diffusion (more precise mathematical description here). Even if you do not revolve the coffee with a spoon, diffusion will act fast enough to, say, edulcorate ...


1

The vortices are formed near the hole. There may be multiple suvh vortices, but they soon coalesce into one vortex due to perturbations.


5

As always, read John Denker's wonderful ebook. To quickly answer the question, the equal-time argument is a wrong application of Bernoulli. Bernoulli is right. Bernoulli plus the Kutta condition (air can't flow up over the trailing edge) is what makes flight possible. If one accepted the equal-time argument it would not explain how airplanes can fly ...


2

Equal transit time would violate Newton's laws of motion in the sense that the computer simulations that are based on Newton's laws of motion show that equal transit time is false. I don't think you can show that using a simple heuristic argument or a simple equation, but the computer simulations do count as a valid proof, albeit perhaps not one you find ...


0

Let's remove this from the list of unanswered questions. The derivation in the book is a bit odd. I favor the derivation in Schlichting's book "Boundary-Layer Theory", because it's cleaner. Usually the derivatives of $\sigma_{xx}$, $\sigma_{yy}$, $\tau_{xy}$ and $\tau_{yx}$ in equations (12-5) and (12-6) are treated in combination. For (12-6) the right ...


0

You could treat this as a case of effusion of the mixture of two gases through a small hole in the chamber. See Graham's Law (https://en.wikipedia.org/wiki/Graham%27s_law). The pumping speed is equivalent to a rate of effusion (number n of moles per unit time t) for each gas : $\frac{dn}{dt}= -kn $ where k is a constant depending on type of gas. This has ...



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