New answers tagged

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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 ...


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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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The basic point is that the picture in wikipedia shows one very particular flow, a sheared laminar flow, but viscosity also acts in more complicated situations. Viscosity arises from diffusion of momentum, which acts to equalize the fluid velocity in adjacent fluid elements, no matter what the geometry of the flow. Purely based on rotational symmetry and ...


4

For plastic water bottles of the market, Msha gave a video link ,in a deleted, unfortunately, answer because it just gave the link.For this case, plastic bottles, it should be the answer chosen. Crush the bottle fast, beginning from the bottom. The same will be true for any elastic type bottle. It is only glass bottles that have an unsqueezable limit. I ...


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The fastest way to get that liquid out of a bottle is to insert a lot of air in while the liquid comes out. How you insert a lot of air only depends on your imagination.


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I have two answers for this and one response to your strategies. 1 If the flow is infinity and you're trying to find pour rate averaged over infinite time. 2 response to what is mentioned. 3. Are strategies if the water flow is not infinite, hypothetical and is timed. 1.a It splatters or fluctuates in pouring because when water pours it builds a less ...


1

Even if you call the term dynamic pressure, and has units of pressure, it is not a pressure at all. It is a necessary contribution to the total static pressure, though, And it is the static pressure that matters when you move a piston in a gas. The static pressure is also related thermodynamically to the temperature, in the first case as the average ...


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go into a vacuum chamber open the bottle turn it upside down gravity will do it in a second or two.


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Hold the bottle upside down, your hand grasping it near the base. Extend your arm downwards. Make a circular, conical-shaped motion with your hand. Keep it up until you have a nice vortex going. Spinning too fast is counter productive; centrifugal force keeps too much water hugging the sides. You need a speed just a bit more than what's necessary to ...


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I assume you are not worried about the few drops that are always left in the bottle after pouring out the water. The reason I make this assumption is that without "evaporating lasers" being allowed every method suggested would need to wait hours for the drops to naturally evaporate. Even then, there would still technically be a tiny bit of water vapor in the ...


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Put an straw thru the open cap of the bottle. Bend the straw so that you can blow into it while it is upside down. Turn the bottle upside down and blow as hard as you can.


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This is really three separate questions. Static and dynamic pressure: It is the static pressure that really matters in practical situations. The dynamic pressure is related to the kinetic energy of the fluid which, when it changes, causes a corresponding change in the static pressure. Condenser/evaporator application: The basic Bernoulli equation ...


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You could look at viscosity and its effects for yes, and at some never happened story of a guy dropping something from a tower and concluding something for no. We're not allowed to give straight answers, so I hope this is vague enough.


2

The convection-advection equation resembles the mathematical structure of Burgers equation. Despite differences (convection-advection has an "extra" variable $c(x,t)$) you can think that Burgers describes compressible material fluid transportation by means of advection (the $ \vec{v} \cdot \nabla $ term) and diffusion (the $D \nabla^2$ term) without ...


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Actually. the velocity can be determined as follows. V = sq. root of 2gh where: V = velocity in ft./sec. g = acceleration const. 32.2 ft per sec per sec. And h = head in feet of liquid. h = P * 2.31/SG where: P = pressure in psi. SG = liquid specific gravity. With the above information you can now calculate F.


2

I doubt if the person who made that animated gif knew how the pump works. It's not at all clear why the valves open and close, and the regular movement of the water is nothing like an actual ram pump. It's more likely to confuse than enlighten people imo. This one (found on http://www.meribah-ram-pump.com) does a better job showing the strong waterhammer ...


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I hear this regularly. I think its a combination of the effects above. The first effect is a front of compressed air being forced ahead of each of the trains. But then, as the fronts of the trains meet and pass, the Bernoulli effect leads to lower pressure between the trains. But this isn't uniform, each carriage has its own mini air front, and the carriages ...


1

I am telling this from my practical experience. In my home we have very poor water supply so we purchase 15 ltr water dispenser for drinking water. The guy needs to take the bottle back. So he shakes the bottle such that water make a kind of whirlpool while still keeping it straight and then quickly turns it upside down at an angle. The water flows quickly ...


1

You need to get $p_4-p_3$. Taking the datum of elevation z as that of points 3 and 4, we have $$p_{atm}+(10)\rho g=p_2+\frac{1}{2}\rho v^2$$ $$p_3+\frac{1}{2}\rho v^2=p_2+\frac{1}{2}\rho v^2$$ $$p_{atm}+(120)\rho g=p_5+\frac{1}{2}\rho v^2+(120-h)\rho g$$where h is the depth of point 5 below the surface of the tank on the right. $$p_4+\frac{1}{2}\rho v^2=...


2

You spin the bottle so that the water comes out like a tornado. This lets the air come in faster through a tube of space in the middle. The air pushes up forcing the water out leaving a space in the middle for air to come in. The rate of water flow is exponential. This is because the water lessens letting air come in faster, pushing the water out faster with ...


1

Using Bernoulli's equation and the momentum conservation equation, we can show that water flowing out of a pipe with cross-section $A$ at speed $v$ exerts a force $F$ on a wall (at 90 degrees), acc.: $$F=\rho Av^2$$ With $\rho$ the density of the water. But your specification of "8" pipe with 500psi stream of water exiting it and hitting a wall at 90 ...


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After some searches I found this little paper. I will give only a short answer to the question, for details you can read the paper. So, according to the results obtained for the model in the paper: the flame length: increases as the gravity level increases from $0g_e$ to $3g_e$ decreases from $3g_e$ to $60g_e$ and blows off at higher gravity levels ...


0

i understand that the check valve catches the pressure from the Water hammer that is correct what i do not understand is why does the waste valve open (and close) the water hammer opens and closes the waste valve (see bellow) (and how does it sync with the check valve?) it does not sync, (the waste valve opens and closes without relying on the ...


1

what i do not understand is why does the waste valve open (and close) The waste valve is normally open. In your figures, the plug is weighted and falls away from the valve. The water flow through the waste creates drag which pulls the valve closed. and how does it sync with the check valve? When the waste valve closes, it causes the pressure in ...


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Your premise "but higher the velocity, greater is the temperature and pressure must be high" is wrong. To see why, you must recognize that the velocity field $\mathbf v $ that you calculate in fluid dynamics refers to a "fluid particle", which is composed of maybe some $10^{10\pm 5}$ molecules. On the other hand, the (local) temperature, pressure, density ...


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In my judgment, this problem was not quite interpreted correctly by the OP, and by the member who provided the answer. This seems to me to be mainly an "oscillating manometer" problem. The fluid in tank A oscillates up and down, out of phase with the fluid in tank B, which also oscillates up and down. Fluid flows periodically back and forth between tanks ...


0

If you just look at the values given, you know that the water that exits from between the disks started with a potential energy of $\rho g H$ per unit volume. Ignoring everything you might know about fluid dynamics, you should know enough physics to know that the water coming out cannot have more energy $\frac12 \rho v^2$ than the potential energy it started ...


2

The question is: what is the optimal way to pour the water so that it [the bottle] completely empties fastest? I conclude the aim is to have the empty bottle, not the water in another container. Solution: Create a centrifuge-like setup, bottle opening to the outside. The setup will generate artificial gravity for the water in non-inertial frame of reference ...


4

Option 3: Squeezing the bottle. Of course it depends how much pressure you are able to do, the strongest you are the fastest, and if you compress the bottle in an industrial press, the water will leave in a split second without breaking the bottle (well, it gets deformed, but that was an option).


1

This is a correct way to solve exercises involving viscosity (within certain constraints, e.g., constant viscosity). Eqn. 1 is a version of the Bernoulli equation, modified to include a frictional head loss, and is definitely valid, provided the velocities used are the average velocities. Eqn. 1 without the $h_L$ is valid along a streamline, even for a ...


2

Cut the bottle base and squeeze the bottle The air pressure has a major role to play in this situation. If you keep the bottle vertical, there wouldn't be any room for the air to move in as the water falls through which is the reason why you see turbulence and interruptions. There are various ways to tackle the issue. The best method would be to punch a ...


2

What I have observed is that if we turn the filled bottle (open) upside down and plug in a straw then the water starts to flow out faster. This happens because by plugging in a straw we make a way for air to come inside the bottle and fill empty space. Another observation that I've made is that when water is flowing out of a bottle (upside down) just shake ...


5

I should start with the disclaimer that I don't know the answer to this, however I have seen very similar patterns in flocculating systems and I would guess that the same principles are involved. The patterns are produced by adding a drop of pigment to a layer of slip. Both of these are colloidal suspensions. Slip is a suspension of aluminosilicate ...


2

UPDATED ANSWER Sorry, I interpreted your question too narrowly. Couette flow occurs without a pressure gradient, due to viscous drag from a boundary surface, and is laminar. If the drag force is increased the flow can become turbulent. If a transient inertial flow begins laminar I think it must remain laminar as it dies out, because the speed of flow ...


3

For a candle flame, following processes occur. 1. heat transfer (from flame to surrounding and to candle) 2. material transfer (wax vapor diffused outwards and oxygen diffused inwards) 3. heat generation (chemical reaction at stoichiometric mixture location) with gravity, the above will shift due to free convection flow. This will accelerate heat transfer, ...


2

A bit of 1, a bit of 3... The technical name is flow velocity, as correctly stated in the Wikipedia article about NS equations. But one could ask what "flow velocity" means. From the Wikipedia article: flow velocity [...] is a vector field which is used to mathematically describe the motion of a continuum. Although correct, this definition is ...


2

You are correct, it is the velocity of a small volume of fluid centered at the point, that is a macroscopic motion, but it is also the result of the average velocity of the particles in that volume.


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The usual picture you see in wkipedia and other sources is indeed over-simplified. Viscosity resists more general velocity gradients, not just pure shear flows $\nabla_x v_y\neq 0$. For example, if you have a compressible fluid undergoing non-isotropic scaling expansion $$ \vec{v} = (\alpha x,\beta y,\gamma z) $$ then shear viscosity will try to equalize ...


0

Your assumption is correct. At point (a), assuming the liquid is static, the pressure is: P = (H x SG)/2.31 Where: P = pressure in PSI: SG = Liquid specific gravity However, at point (B), there is a dynamic component and some of the pressure energy P is converted to velocity energy (hv). hv = (V^2/2g) where: hv = Velocity head in feet of liquid. ...


1

Your guess is correct. I think you are expected to assume that $a$ is sufficiently far from the hole that $v_a=0$. If the question does not ask you to make this assumption you should state it explicitly yourself. It does not follow that there is a discontinuity in velocity at the hole just because there is a discontinuous change in cross-section. ...


2

In your approximation there is a velocity discontinuity where the pipe joins the container. Assuming that is a good approximation then you are correct and $P_a=P_b+\rho v^2/2$. In real life though, the pressure will change smoothly because the closer you get to the pipe, the larger the effect of the motion of the water inside the tank.


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My understanding is as follows. Fluid passing the leading edge nearest to the surface is slowed by viscous forces between the surface of the body and the free stream. Adjacent layers of fluid in the stream consequently move relative to this reduced flow, and are hence affected by viscous forces themselves. The result is that the adjacent layers of fluid ...


1

The question is also: is there a pressure difference between point 11 in the center of the tube and a point on the same section but very close to the tube boundary? (Does the weight of the fluid between 11 and the boundary create a pressure difference between these two points?) The answer to this is Yes. Even with a horizontal tube, the pressure ...


1

In a sense, you can consider chromatography to be a kind of 'filter', and it can certainly separate out coffee components, including water. There's no lack of other ways, though, to extract water. Freeze drying would carry the water vapor away, and you can sell the residue as instant coffee...


0

Huh, people? :-) He says the main tank is closed! That means it can be pressurised, and the pressure will ofc push out water, depending on pump efficiency. Nothing else needed than a valve that hinders the water from flowing back to the open reservoir when the pump is off. In the pic below a self-adjusting system. - If there is too little water in the ...


2

Pascal's law: Pascal's law or the principle of transmission of fluid-pressure (also Pascal's Principle) is a principle in fluid mechanics that states that pressure exerted anywhere in a confined incompressible fluid is transmitted equally in all directions throughout the fluid such that the pressure variations (initial differences) remain the same. Due ...


5

Your analysis and intuition are correct. The force needed in the second setup is larger, even though the weight of the water is the same. To understand why, consider the horizontal part of the container, $0.5\ \text{m}$ off the ground. This wall is above the water, so the water's pressure pushes up on it. Then in reaction, the wall pushes down on the water, ...


-2

The quick answer is that a long as the surface area of the piston, and the volume/mass of the remain unchanged, the pressure stayed the same. So the erreanous step is to assume that $ P_2 = \rho g h $. This invalid as the cross section area changes with height.


0

In Bernoulli equation, the only term which corresponds to what one usually calls "pressure" is $p$, the other are still pressures dimensionally but their meaning is linked to kinetic and potential energy per unit volume. Nevertheless I get confused especially for the term $ρgh$. That is corect. The term $\rho gh$, better written as $\rho g(z_2-z_1)$, is ...


2

Coffee is a homogeneous solution hence it can not be separated by usual methods. I have few suggestions may not be very accurate but good for brainstorming. Distillation : Although you have mentioned not to mention it but I would like to add that coffee has several aromatic organic compounds that make the smell of coffee hence you can not get rid of ...



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