# Tag Info

4

This is actually something not easily answered because it is part of the definition of pressure. I will instead point you to other answers which hopefully make sense. The following is an excerpt from Lumen Physics. *The force exerted on the end of the tank is perpendicular to its inside surface. This direction is because the force is exerted by a static or ...

1

From the inside of the flak the pressure is $P_0+ hg\rho$. If you are looking for the force that secrets force on the base then that force is $(P_0+ hg\rho- P_0 )A$.

0

Because of the fear of being accused of witchcraft, of all things. The mercury barometer is the oldest type of barometer, invented by the Italian physicist Evangelista Torricelli in 1643. Torricelli conducted his first barometric experiments using a tube of water. Water is relatively light in weight, so a very tall tube with a large amount of water had to ...

3

Since the density for the both the cases are different. The density of mercury will remain the same. The density of the air will vary with pressure and that's exactly what you want to measure. The air pressure pushes the mercury up the tube and, since the top of the tube is a vacuum, there is no counter-pressure. Mercury has a density of 13.6 g/cm3. 1 ...

1

Most labs have a mercury barometer in them to give the atmospheric pressure. Usually taken at the start and end of a test - especially when dealing with engines, also ambient temperature and humidity are taken as well. The "lab standard" mercury barometers also have a temperature correction chart to correct the indicated reading due to the ...

1

I'll give you a hint, use the conversation of linear momentum equation. For a fixed control volume $CV$: $$\sum \vec{F} = \frac{d}{dt} \int_{CV} \rho \vec{V} dV + \int_{CS} \rho \vec{V}(\vec{V}_r . \vec{n}) dA$$ wher $\vec{V}_r = \vec{V} - \vec{V}_{CS}$ is the relative flow velocity exiting the control volume relative to the control surface. Assuming a ...

0

Shouldn't it be that the diameter should approach a finite threshold value [no matter how small it is] rather than zero for continuum hypothesis to hold true? Undeniably, yes. If there are not enough particles in the control volume to achieve a sufficient Hamiltonian to represent the system (continuum check), then the measurement may not be ergodic even in ...

-1

The pressure at P1 is P2 plus the friction head between the two locations plus the bend head loss.

0

As you have stated, a partial vacuum in the grey water space could be causing the water level in the pressure sensing tube to rise. It’s drawing water up the tube. The actuation level is probably set for atmospheric pressure in the grey chamber. Level Probe from Wikipedia

1

An interesting question :) There is a theoretical limit on how far a water jet can pass through air. In this instance, we neglect hydrodynamic effects, gravity, water jet breakup, etc. and just consider momentum. Newton solved this some time ago, and derived his impact depth equation based on momentum considerations alone, i.e. the jet transfers momentum to ...

1

The vaporization and dissolution of gases out of and into, respectively, of water is a dynamic equilibrium, meaning that gases are continually coming out of solution, and re-entering. We may assume that water that has been sitting for some time has reached its dynamic equilibrium point, of a certain quantity of dissolved gases for a fixed mass of water, ...

1

"negative pressure downstream, dissolved gases coming out" - this is correct, and that's a great video showing it. It's called cavitation in some contexts. The question of at what pressure this happens has three parts, as far as I understand it. (1) The local temperature and pressure (i.e. resulting from valve action) need to be such that you are ...

0

Depending on your types of boundary conditions, you have a few options to increase the pressure drop: Lower the exit static pressure Increase the mass flow rate Increase the inlet total pressure All of these will increase your static pressure drop and hence your velocities. Since the thickness is inversely proportional to square root of Reynolds number (...

0

As you're trying to measure small volumes would a volume marked supply vessel feeding your supply valve in conjunction with a timer (watch,phone or stopwatch) give you the answer you require? The only cost would be your time setting valve to required flow, which you'd have to do anyway with an expensive meter.

0

When I look at your scheme, there should be no influence between the two systems. (no connection) You state that there is an influence. So there must be a connection. If the air pressure in the machine space room falls, a deformation off the tank may occur. This could possibly be the cause.

0

If your sensing tube has barometric compensation, the air pressure would have no effect. If not, then a drop in air pressure would be interpreted as a drop in liquid level. It's just working from an assumed air pressure. So it kind of makes perfect sense what you are saying. Can you estimate the air pressure change? If so, compare it to the change in head ...

1

If I understand your question correctly your qualm is about $\tau$ being negative. I will use the same image as karman because it is a very good one: you will notice that shear stress is defined as: $$\tau = \mu\frac{d u}{dy}$$ Now notice in your image: that for increasing y, the velocity increases up to the middle of the pipe. (That is because the fluid ...

-1

If the cylinders are ejected flying parallel to the relative wind, basically descending in an appropriate arc, yes the drag is almost the same if we ignore small additional skin friction due to added length. If they are ejected horizontally or any other way that creates an angle of attack WRT the relative wind then the drag will be significantly higher.

1

If I get your point correctly, It is only a sign convention. say in a pipe with the positive flow gradient we consider a plane C parallel to the pipe axis. The shear stress above the Yc plane is considered positive and the shear stress below that same plane is considered negative. both of these stresses are due to $\tau \$ which is positive. and the sum of ...

0

A cheap way to realize low pressure differences (and a method that is nearly unbeatable for smoothness) is gravity. Fill everything with liquid, run input and outlet tubes to two beakers/buckets/etc, with the one on the input side be higher than the output. Approximately 1 mbar per centimeter height difference for water at room temp.

0

Same answer as above, "Yes BUT..." This is from a practical point of view: Other than HPLC, nearly all "microfluidics" applications are far into the laminar region. Poiseuille's law works very well most of the time. When going below 100$\mu$m dimensions, the problems come. lack of stiffness -> violates assumption that fluid into a ...

0

The setup you are referring to is called "fans in series" and unless the 2nd fan is much weaker than the first laser fan it improves the performance of the system moderately. But it is only better than parallel systems in high resistance ducting, like narrow windy ducting, not your case. Fans in series add to the outlet pressure (not its volume) by ...

0

In the problem as you have stated it, equilibrium is reached when the forces, and not the pressures, at the top and bottom of the stop cock are equal. Thus, P1A1 + mg =ρghA2 where: P1 is net of atmospheric pressure i.e. the gauge pressure. If the absolute pressure is used, then there's P2 = 1 atm on the RHS * A2 m is the weight of the cock h is the height of ...

1

Does this mean that if I keep the manifold open and pump I air at 3.5 liters/minute it will flow liquid out of the outlet at the same rate? Short answer: No. The 3.5 L/min rate is a flowrate of a compressible fluid. The keyword here is compressible. The true flow rate of the water-like liquid being driven by the air pressure (from your 3.5 L/min of airflow) ...

Top 50 recent answers are included