Tag Info

New answers tagged

2

I would say pressure is better defined by $$ \vec{F} = P \vec{A}. $$ Yes, we are defining a quantity without having it all alone on the left-hand side. And yes, area is a vector. And as you guessed trying to divide one vector by another leads to trouble, so we won't do it. Let me explain where this comes from and what it is shorthand for. In continuum ...


0

There are different mathematical way to define pressure (all equivalent), but perhaps the most common one is using the component of the force normal to the surface. That is why in the definition you are using you are actually dividing scalars. For this way to define pressure, you see http://en.wikipedia.org/wiki/Pressure. You can also consider area as a ...


1

Yes, area is a vector, which is the normal to the surface. ($\vec{A}=A\vec{n}$) $\vec{F} = -P\vec{A}$ In this case P is simply the proportionality constant to the vectors F and A, which also mean that F and A has to be in the same direction (F is the normal force and not shear forces). The negative sign accounts for the fact that the force and normal ...


3

Pressure is a scalar and does not have a direction. This is discussed in some detail in the answers to Define Pressure at A point. Why is it a Scalar?, though this might be a bit technical. When you measure a pressure you are actually measuring the force applied to a surface. For some small bit of surface $\delta {\bf A}$, the force produced on that surface ...


0

Pressure at a point in a static fluid is independent of direction. http://www.southampton.ac.uk/~jps7/Aircraft%20Design%20Resources/Sydney%20aerodynamics%20for%20students/fprops/statics/node4.html The force exerted by the walls on the liquid will be pointing inwards. Imagine if there is a hole in the container and water is liquid out, it is easy to see ...


0

A very pragmatic solution would be to introduce a 3rd camera looking from a 3rd angle to make the problem well-defined. This camera doesn't have to be as good or fast a camera as the other ones (assuming that those are highspeed cameras), because you basically only need 1 frame for which you know for sure which of the 2 possibilities it is. This does mean ...


0

No, it is not equal. Without sucking and without friction losses, the kinetic energy would be equal if the inlet and outlet pressures were equal (at equal cross sections). But with sucking, a part of kinetic energy is lost in the inelasic process of mixing of two masses. It is like in a usual classical mechanics: if one body hits another one and sticks to ...


1

B will move faster, the reason is that the acceleration, $a$, of A is smaller for two reasons (remember that $F_{applied}-F_{drag}=ma$) : 1) the same force forward is applied so the contribution to the acceleration on the smaller ball will be larger 2)the drag force on the larger ball will be larger (see Rennie's comment) on A because the cross sectional ...


0

I'll just throw some ideas out. Initial angle of tilt. Higher the angle the higher the back pressure has to get before air will hold the fluid from flowing until it equalizes (glugs). It will have longer time between "glugs" but this will also tend to slow the flow down. At a very low angle air can enter while fluid exists at the same time even ...


2

I can't answer the question as posed - but I can point you to something related. If you have an (infinite) cylinder perpendicular to a flow, you can calculate the flow around it (see for example this interesting page where these images and equations come from). The velocity profile would look like this: and the pressure profile is similar: They give ...


0

Decreasing diameter indeed leads to increasing flow resistance, and that is the dominant factor in lung airway resistance. But branching itself also contributes to airway resistance. When flow is forced to change direction there are energy losses which also cause pressure drop and this adds to the resistance. The trachea is more or less a straight pipe with ...


0

I believe the question can be rephrased as asking why is a single big pipe better than two smaller pipes with the same area Since that's really what you are asking. The bronchial tree splits into smaller branches, but tries to carry the same amount of air. Now the answer should be obvious: for the same area A = $2\pi r^2$, where $r$ is the radius of ...


1

If you pump gas along a pipe then pressure drop per unit length of the pipe depends on the diameter of the pipe. The smaller the pipe the harder it is to pump the gas through it. The pressure drop is given by the Darcy-Weisbach equation: $$ \Delta P = f_D \frac{\rho v^2}{2} \frac{\ell}{d} $$ though with the complication that the density of the gas depends ...


0

I once had the same problem understanding. But one day I realized pressure is just a measure of energy and Bernoulli's law is just another way of expressing the conservation of energy. The analogy: Total Pressure = Dynamic Pressure + Static Pressure >> Total Energy = Kinetic Energy + Potential Energy. In either case we assume no losses, or otherwise ...


1

Since you know the force on the bottle (roughly 200N), you would have to get an approximate area over which this force is distributed. You could try by spreading some ink on either the bottle or the weight to estimate the contact area. While this is not completely correct (the wall of the bottle does redistribute the pressure on the outside to a larger area ...


2

Yes, it does protect against G forces because it spreads the pressure on the support surfaces of the body evenly. For example, and interesting article here


1

The ideal shape for the water to just 'hang' is actually the cube that you've proposed. In this case all of the forces on the water are uniform across the interface. For one region to slip down, another needs to move up (as you've indicated in your diagram). But since every location is experiencing the same forces, there's no reason any spot to start ...


1

The definition that you are using is not the most general. If you insist on applying the way you do, it only applies for a uniform flow in between two counter-mooving walls. Then, the velocity profile is indeed linear. A Newtonian fluid is defined by the approximation that local stress (or drag) is proportional to local strain. I would write your equation ...


0

I'm assuming incompressible fluid here and rigid pipes (no compliance). Before the actual physical split is there any appreciable resistance compared to RB and RA? If not then PA = PB = 19" wg. If there is a significant resistance you need to include it in the model as Rin, and then PA = PB = 19" *(RA || RB)/(Rin + RA || RB) In any event PA = PB


2

As an another attempt, I calculated the coefficients of the cubic regressions that describe the NIST data. I first calculated the cubic regression coefficients as a function of pressure for viscosity as a function of temperature. In other words, I calculated the coefficients for each isobar. In equation form that is, $$\mu ...


1

Steady would mean that flow at a point defined in that coordinate system does not change in time. If you mark a spot on the ground and look at the air flow above it you'll see it change over time. It will start out being still, it will move as the cyclist passes and then become still again. If you look at a spot say 1 $m$ ahead of the cyclist (moving with ...


4

The nonlinear term, $\left( \mathbf{V} \cdot \nabla \right) \mathbf{V}$, determines the steepening of a wave. This can be balanced/offset by loss terms like dispersion, diffusion, viscosity, resistivity, friction, etc. If the loss term dominates over the nonlinear term, then the wave cannot steepen as there is too much damping. If the loss term balances ...


2

It depends on the fluid. Consider, for example, an ideal gas at fixed temperature near the surface of the Earth. Does the density vary in such a column? Yes. Let's investigate as follows. Imagine that the column is in the $z$-direction and has cross-sectional area $A$. Let $z=0$ at the ground. Consider a small, vertical "piece" of the column between ...


0

There are many resources on the web which help to answer this question, but not necessarily very recent ones. Take NACA TN 2674 which shows tufted delta wings, or NACA Research Memorandum L57A30. Generally, a delta wing shows separating flow at the leading edge beginning at moderate angles of attack. This separation leads to the formation of a vortex which ...


4

This technique is called supercavitation. So far it has only been applied to objects no larger than, say, a torpedo, because it not easy to produce a stable bubble the size of a ship when moving at high velocities.


0

I suspect the little bubbles are actually CO2. CO2 is water soluble, and you could find it in most forms of tap water. "hard" water tend to have high concentrates of CO2 while softer water have less. If indeed this is the case, the bubbles form because it is energetically efficient to them to adhere to your hand, which is a lower energy state for them. I ...


1

I think that it helps to define appropriate control volumes. See the image below where I define surfaces A and B. Here, we can say that the pressure at A is given by $\rho g h_A$ and the pressure at B is given by $\rho g h_B$, recognizing that $h_a$ and $h_b$ are functions of time. If the tank is open to atmosphere the $P_A$ and $P_B$ terms will be equal ...


0

I would tell her the bubbles contain water, and that water is sticky. I would remind her that even after she lets water run off her hands by gravity, she still needs to dry them off with a towel (unless you use an electric hand dryer), because some of the water sticks. It's easier to see the foam than it is to see the water, because the foam is puffed up ...


0

a liquid is not perfectly incompressible - see cavitation in fluids. still, i think we can assume that here . we can calculate the maximum possible height reachable. If i assume that the wall is perfectly rigid and it cannot move when the wave hits it, and if i also assume that the wave system is perfectly conservative , i.e: no energy is lost as heat or ...


0

I would recommend that you derive an empirical answer by running a few experiments. Use different colored water, and measure the resulting splatters. Use the results to derive the formula.


2

The parameter $h$ is the maximum distance that two smoothed particles, $a$ and $b$, can be before the distance between them is negligible for SPH purposes. If the distance, $\vert r_a-r_b\vert>h$ then the weight is zero. For any kernel, the integral over the particular region, e.g. $r\in(-h,\,h)$, is necessarily 1. Since $h$ is an unknown parameter, then ...


1

How do slight changes in these properties result in a large change in pressure, microscopically? Slight change of volume is not so easy to accomplish for solids - it takes a great force to achieve it. Considerable external force applied by different body (wall) needs to be maintained. The pressure is a measure of this force per unit area and since the ...


6

There's no magic behind it. It was done by non-dimensionalizing the momentum equation in the Navier-Stokes equations. Starting with: $$\frac{\partial u_i}{\partial t} + u_j\frac{\partial u_i}{\partial x_i} = -\frac{1}{\rho}\frac{\partial P}{\partial x_i} + \nu \frac{\partial^2 u_i}{\partial x_i x_j}$$ which is the momentum equation for an incompressible ...


2

The way it was explained to me: you start by thinking of all the possible factors that could play in drag (size, velocity, density, viscosity, ...); then you do dimensional analysis and find dimensionless combinations - these tend to be "special" since they remain constant over different scales of time and space. Reynolds number is one such combination. The ...


0

Alright, so here's how I'm thinking about the scenario now. I'd appreciate any comments/suggestions on my logic here. If the bubbles are close, density gradients in each bubble will cause some of the gas to diffuse out into the liquid, and, since the small bubble has a larger pressure, the concentration outside of the smaller bubble will be higher than ...


5

The intuitive way to think about this is to consider a gas inside a glass container (that cannot expand). If the gas expands, then what must happen as a result? The gas leaks out of the container. Similarly, if try I put more gas into the container, then the gas compresses. The vector field $\mathbf F$ is what we use to describe the flow of a fluid. The ...


0

Beautiful article (great graphics - not sure if they are photos or GCI) at http://math.berkeley.edu/~hutching/pub/bubbles.html - not sure if this addresses "your" kind of bubbles, but worth a look anyway. In general surface tension prevents bubbles from becoming a single larger bubble as there is an intermediate phase when the surface would have to be ...


1

Here we assume that OP is mostly interested in the Eulerian fluid picture (as opposed to the Lagrangian fluid picture). Both fluid pictures are discussed in great detail in Ref. 1. Note however that in the methods of Ref. 1, the mass density $\rho$ is a dynamical variable. The variation of $\rho$ is important in order to obtain a full set of eoms. But OP ...


4

The key is the Reynolds number, $$ Re=\frac{\rho LV}{\mu}=\frac{LV}{\nu}\tag{1} $$ where $L$ and $V$ are characteristic lengths and velocities of the particular problem and $\mu$ & $\nu$ are the dynamic & kinematic viscosities, respectively. If you multiply (1) by $\rho LV/\rho LV$, you get $$ Re=\frac{\rho L^2V^2}{\mu LV} $$ The numerator is the ...


2

Sorry for the enormous delay, I was caught with more work than I thought, then, here it is. @DanielSank, relativity is not necessary, it would help with what you said, since it would pinpoint exactly you are calling momentum in your system. My answer would be an extension of DanielSank comment. When there is the conservation of a continuous quantity, the ...


0

The term in equation is: $$\frac{\partial u_i}{\partial x_j}\frac{\partial u_j}{\partial x_i}$$ So let's take a step back and think about what kinds of terms can appear in conservation equations. There can be a production term, a transport term, and a dissipation term. The transport term is the $\vec{u}\cdot\nabla q$ term that you noted. When you look at ...


2

For the sake of the explanation I will assume you mean a gas bubble in a liquid*. David Hammen names a few conditions for a bubble to be spherical, in fact you could summarize these all as: for a bubble to be spherical the surface tension has to dominate over other forces (per unit length). If surface tension is indeed dominant than the pressure in the ...


4

Rising bubbles of air in a liquid oftentimes are anything but spherical. These bubbles have haphazard shapes because they are rising and because they are interacting with other nearby bubbles. The combination of drag, turbulence, and mutual interactions prevents those bubbles from taking on a nice, simple spherical shape. Here's a rather non-spherical ...


1

The thing you'll notice about a sphere is that it's symmetrical. very symmetrical. No matter how you rotate it, it looks the same. the surface tension pulls the surface of the bubble into a shape that has even surface tension over the entire bubble. The shape with even surface tension is a sphere. a sphere has the smallest possible surface area for an ...


0

It is a case of flow through an orifice. It depends on the shape and area of the orifice, and on the viscocity of the fluid. At a low ratio of pressure to viscocity, flow rate is proportional to pressure. At a high ratio of pressure to viscocity, flow rate is proportional to square root of pressure. You're going to have to write a differential equation, and ...


0

The subscript $j$ represents all particles, including $i$, from 1 to $n$. This should be obvious in the example below Equation (9), To give an example, let us consider the distance constraing function $C(\mathbf p_1,\,\mathbf p_2)=\vert\mathbf p_1-\mathbf p_2\vert-d$. The derivative with respect to the points are $\nabla_{\mathbf p_1}C(\mathbf ...


1

You can get nothing out of equilibrium thermodynamic considerations for the rate at which pressure will equalize. What will matter is the speed of sound in the gas, as that is the rate at which density fluctuations travel in a fluid and assuming an equation of state, say $p(\rho)=\rho^{\gamma}$, the pressure is then enslaved to the density. So the sound ...


1

The answer is due to the area-Mach number relation for hydrodynamic shocks. G.B. Whitham has a great book (check out Chapter 8) on all sorts of various waves and has a good discussion of this topic. The idea is that one can define the Mach number as a function of the cross-sectional area of a ray tube. The simple form is: $$ \frac{ 1 }{ A } \frac{ d A }{ ...


1

Unless the two containers are separate, i.e. have a wall sealing them off completely, the right set of tools for this question is fluid-dynamics rather than thermodynamics. for the sealed off problem, assuming ideal gasses, the end state for the coupled baths will be that of equal temperature. in that case it is essential you have the right number of ...


2

The rotation is part of the key to the storm itself. Primarily the pressure and temperature differences are what causes these systems to take the shape and forms that they do. Once a tropical depression starts to form you can already see rotation in the moisture around the low pressure zone, even through it typically looks nothing like a hurricane. Not ...



Top 50 recent answers are included