I'm working on a project. We have two balloons, connected by a hose with a valve.We know when we open the valve, by according to internal pressure of each balloon, air goes form the higher pressure balloon to lower pressure  balloon. I want to know the relation between internal pressure of the balloons and the flow rate between these two balloons. Pay attention, I don't want the air transfer direction, I want the FLOW RATE.  Thanks.
93318
Last day , when i was working on two interconnected balloons , a question was kicking my brains !!! This is the explanation of the question:  First , suppose a system that composed of two spherical membranes filled with air (two balloons have different initial volumes {means that the pressure inside which balloons are different} and the air pressure is 1 a.t.m) . We connect them with hollow tube and a valve. When we open the valve , one of them shrinks and the other one expands (It depends on their pressure) .  So how can we get the velocity of the flowing air between two balloons? (Consider everything but if you have reasons for not considering one of them -for example the ratio of friction in the tube- Don't consider them and just tell me the reason)  Thanks
98409
Since the cross sectional area is smaller in the middle the water flows faster and experiences a smaller hydrostatic pressure. Additional mass is sucked in from the middle tube and gets accelerated so that the speed of the fluid at the exit should be less than at the entry.  Is the kinetic energy of the fluid at the exit equal to the kinetic energy of the water flow at the entry?
25835
If I built a tube from Earth's surface to the exosphere, would all the air be sucked out to space?  If this pipe reached to a big planet, like Jupiter, would its gravity through the pipe suck our atmosphere?  If one end of the pipe was at the Earth's core, and other in the exosphere, would the magma go there, like in giant volcano?
124049
A thought experiment:     A U-shaped tube with semi-permeable membrane at the base. The tube is completely thermally isolated from its surroundings. The liquid (solvent) is at some temperature $T$. When solute is added to one side of the tube the fluid level on that side rises until the increased pressure due to gravity ($\rho gh$) equals the osmotic pressure.   If I understand correctly, osmotic pressure could be used to do work (i.e. lift a weight as the solvent level in one side of U-shaped tube increases). Where does this energy come from? I guess it must be from the internal energy of the fluid.  At this point has the temperature of the fluid decreased?? If so, has it decreased by exactly the work done against gravity divided by the specific heat of the fluid? What does the quantitative treatment of this problem look like?
61704
I have a hard time understanding whether or not a wing placed in a potential flow, assuming there is no viscosity and no friction with the wing, will produce a lift. I've seen several contradictory (to me) versions of what happens in this case, in particular:   In Landau - Lifshitz, Vol 6, §11 it is shown that the force vector on any object in a potential flow is zero (i.e. there is neither lift nor drag). The mathematics behind this derivation are above my understanding, thus I can't really follow how this is shown and what assumptions are made along the way. The Wikipedia article on the  states that a body in a potential flow does not experience drag. On the other hand the article does not say anything about lift forces and the German version explicitly states that it does not apply for lift forces. A common explanation for the lift produced by wings seems to be that its asymmetric shape causes a higher velocity of the flow above the wing and a lower below. Due to Bernoulli's equation that would result in a pressure-difference and thus a force. Honestly I don't quite get this explanation as I don't see where the different velocities are supposed to come from (without friction and thus without the circulation resulting from turbulences behind the wing).   So now I'm wondering, do or do not wings have lift in irrotational, inviscid, incompressible fluids.
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I'm trying to model the ascent of a helium filled weather balloon from 0km to 25km altitude. The plan is to eventually use a python script to calculate the time taken to reach 25km. However, I don't really know where to start.  I have worked out an expression for acceleration in terms of the balloons volume and the density of surrounding air. I now need to find an way of calculating the volume at a given altitude such that I can model the acceleration throughout the ascent.  So if anyone could help me with this I would greatly appreciate it.
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The explanation of shear rate in laminar flow is straightforward: We imagine small layers of fluid that glide on each other. Now, in turbulent flow, this does not work as there are no layers. I'm not even sure that shear rate is a meaningful concept in turbulent flow.  If I want to know the apparent viscosity of a shear thinning (or other non-Newtonian) liquid, I need to know the shear rate. How do I know the shear rate in turbulent flow?
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Pressure can be explained at microscopic level for a gas with kinetic theory of gases. From that the pressure $p$ is linked to the velocity of molecules (and it is caused by the high amount of collisions in the gas).  $$p=\frac{m N_a}{V} \frac{\bar{v}^2}{3}$$  Where $m$ is the mass of a molecule, $N_a$ Avogadro's number, $V$ volume, $\bar{v}^2$ the quadratic average velocity of molecules.   Nevertheless I did not find a similar microscopic interpretation in the case of liquids. In that case molecules are not as free as in a gas, so it looks like pressure is not linked to the higher or lower velocity of molecules. So what is responsible for liquid pressure, at a microscopic level?  Is there a quite simple microscopic description for pressure in liquids, as there is in the kinetic theory of gases?
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The following diagram pictures a simple container filled with water. At the bottom of the container is a chamber with bottom B. Also directly above the chamber is an aperture that can completely open and close, sealing off the lower chamber or allowing unrestricted water flow.  My question is this: as pictured, is the pressure at B the same as if the aperture were completely open? If not, what is it?    I may be mistaken, but I think the pressure on B is proportional to the width of the aperture opening relative to the width of the chamber itself.
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I understand the common explanation of lift, which describes the airflow over the top of the wing as moving faster than the air below the wing.  However, I don't quite understand why the air moves faster.  I've read some explanations citing a circulation effect among others, but haven't found a good, clear explanation for the exact cause.  Can someone help me out?
84920
2 diffrent liquids in a u shaped tube can have height difference at equilibrium but when I try to analyse the pressure at the same level of c,on the surface of more denser liquid there is only atmospheric pressure but on the other hand the less denser liquid have both the atmospheric pressure and the pressure due to the column of liquid above it,so they are not same but they must be.where am I getting wrong?
109285
I was reading the altitude records for hot air balloons on Wikipedia, and noted that the . It didn't really say if there was a reason why. I know that closed envelope balloons frequently climb much higher.   I was wondering... If you have a pressurized cabin (for keeping the operator conscious), and you mix oxygen into the fuel to heat up the air, is there really any reason that a balloon cannot ascend higher? Does the thinning of the air prevent it from holding heat as well? Do other atmospheric events come to play in not allowing a hot air balloon to ascend higher? I don't understand hot air balloons that well and could be missing a major component =)
59234
Suppose I have a wooden block that floats in water and I put it in an empty  glass beaker. Then I poured water in the beaker so carefully that no water goes under the block. In that case, would the block float or it will remain under the water?  As there is no water under the block there is no upward thrust on the base. Therefore it seems that the block would not float.  On contrary, there is tangential upward thrust on the sides of the block. Because of this thrust the block will float.   Which is the correct answer?  EDIT  I think the right answer is : the block would not float. Thanks to sammygerbil, anna v, Kieran Moynihan.  This reminds me another question.  A wooden block is floating on water in a air tight container. Is it going to float more or float less in the following cases?   All the air has been pulled out from the container. We compress the air above the water.   My Thought  Case 1:  When all the air has been pulled out from the container there is no buoyant force due to air, so total downward force acting on the block would increase and the block would float less.  OR,  As there is no air below the block no buyout force due to air. When all the air has been pulled out from the container there is no air pressure, so the block would float more.   Case 2:  When we compress the air, the density of air would increase. So the weight of the displaced air as well as the buoyant force due to air would also increase and the block would float more.  OR,  As there is no air below the block no buyout force due to air. When we compress the air, so the block would float less as air puts more downward pressure on the block.   Now which is correct or incorrect and why?
1711
When I was in high school, my teacher did an experiment to show the power of atmospheric pressure.  Experiment: Prepare a glass bottle, fill with water, put a glass plate on the bottle, make sure there is no air in the bottle, just water. Hold the glass bottle and plate, and invert them. Slowly, release the hand which hold the plate. Because the atmospheric pressure, the glass plate will not fall down.    Question 1: As we know, if we didn't put water into the glass bottle, just air in the glass bottle the glass plate will fall down. So, If we use other liquid instead of water, this liquid has smaller density than water, like alcohol, the plate will not fall down. But, if we continue to chose the smaller density liquid until the liquid has density like air the plate will fall down. So, if there is a threshold density of liquid make the plate between falling  down and not falling down?  Question 2: If we put the bottle and plate into water slowly, when the plate and part of bottle are in the water but part of the bottle still above the water, will the plate fall down?  Question 3: Continuing question 2, when both bottle and plate are in the water I know the plate will fall down. So, how does the pressure of water and atmosphere change?  Is there a good description of this phenomena?  Why does the trick continue to work with lower density fluids and why does placing the plate in the water cause it to fall?
94119
So I was thinking about this while driving home the other day.  I've never been quite clear on why when you drive with the windows down air rushes into your car. I thought this might be explained by Bernoulli's equation for incompressible flow, but I ran into what seems to be a contradiction. If we consider the problem from the reference of the car, the air in the car is stationary and the air outside the car has a certain velocity. Then, Bernoulli's equation implies the pressure outside the car is lower than that inside the car. However, if we take the reference frame of the road, the air in the car is moving and then the pressure in  side the car is lower. Intuitively, this second situation seems to be correct since air apparently flows into the car (from high pressure to low pressure). However there seems to be a contradiction, as the pressure gradient depends on reference frame. So my question is what has gone wrong here? Is this a situation in which bernoulli's principle simply isn't applicable? Did I make some sort of mistake in my application of the principle?
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Let us say I have a beaker floating in a water tank. Suddenly I put some water in the beaker also. The figure here shows the exact situation   Now I want to draw the FBD of the beaker. I am somehow confused with the role of the atmospheric pressure in drawing the FBD. I thought of two ways of drawing this and I am just unable to figure out which one is right.    Here since the beaker is floating an $F_{buoyant}$ is there in the upward direction which the net mass (beaker+water in beaker) would put its weight downward.  But while solving one question I came to know that total force that a liquid exerts on the bottom of the container is $P_0+m_{liquid}g$ ($P_0$ is the atmospheric pressure). So should the downward force on the beaker be $(M_{beaker}+M_{water in beaker})g+P_0 S$ (S being the base area of the beaker) as in the second diagram down here.    Here the $P_0S$ due to atmospher on the tube gets cancelled by the $P_0S$ exerted by the water in tank on the tube. But what about the $P_0S$ from the water present in the tube? Do the FBD's apply like the one in the new image?
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$K.E=\frac{1}{2}mv^2$ , $P=mv$ thus there is a relation between them. I really can't understand why there is no decrease in momentum when kinetic energy is decreased in an inelastic collision. I am only a high school passed student(ready for college). Should I leave this question until I learn &quot;Lagrangian&quot; and &quot;Noether's Theorem&quot;? Because I literally read most of the answers about these kinds of questions and spend almost half a day on it, BUT I do not get it. Can anyone please answer this question? please help me.