Fluid mechanics : Barometer - slight confusion.

[ludic]

What would happen to the mercury in the column if a barometer is accelerated upwards?


In static condition, acceleration is g.

When the system moves upward with an acceleration a, effective acceleration = g + a.

The air above the mercury in the trough will press it with higher force (i.e the normal reaction force at this interface increases), so the pressure at this interface will be higher than atm. P . To make the pressure at the base of the column equal to this increased pressure, mercury rises in the column.

Am I not right in thinking that on being accelerated upward the mercury would rise in the column?
Please do point out where I am amiss if I am wrong.

 

[russ_watters]

Your answer assumes that the reservoir is exposed to an airstream while accelerating upwards. What about when it first starts accelerating? What if it is in an elevator? The air above the barometer need not apply additional pressure.

 

[ludic]

But why would we assume it otherwise ?

And my answer and explanation are sound if the air in the surrounding is not confined as in the elevator, right?

 

[K^2]

Acceleration will cause the mercury to get effectively heavier. That will make the column go down, not up.

Ram pressure will depend on velocity, not acceleration. That's currently an undefined quantity, because you only stated that there is an acceleration. At what moment in time should we consider the situation? What initial velocity was?

 

[ludic]

Please be so kind and patient to clarify this doubt:

I understand the part where mercury in column becomes heavier and goes down because of increased effective acceleration

What I don't understand is why - if we call the pressure of atmosphere it's weight upon unit surface area being considered - would the effective weight of atmosphere, and hence its pressure not increase on the barometer trough due to this same increased effective aceleration, and force the mercury in the column back up... effectively keeping the height of mercury in the column same as when the barometer was static.

Sorry I have never studied about Ram pressure... please tell me what is the flaw in the above reasoning.....

I am thinking that maybe what I reason considers the atmosphere above the trough static, but in actuality, it won't have any increased pressure because the air will shift sideways as the barometer ascends.

 

[russ_watters]

But why would we assume it otherwise ?

Because maybe the barometer is not exposed to a moving airstream - such as in the barometer in an elevator example I gave...

And my answer and explanation are sound if the air in the surrounding is not confined as in the elevator, right?

No.

If there is an airstream, the airstream starts off zero speed, while the acceleration will be non-zero. Velocity pressure is a function of velocity (squared), not acceleration, so when the airstream is moving slowly, the pressure will be very small.

The pressure and the acceleration are not proportionally related.

What I don't understand is why - if we call the pressure of atmosphere it's weight upon unit surface area being considered - would the effective weight of atmosphere, and hence its pressure not increase on the barometer trough due to this same increased effective aceleration... [/emphasis added]

The velocity pressure is a function of the velocity. So if the acceleration of the barometer is 10 m/s/s, the instant that acceleration begins, the velocity is still zero and the velocity pressure is therefore also still zero.

 

[K^2]

What I don't understand is why - if we call the pressure of atmosphere it's weight upon unit surface area being considered - would the effective weight of atmosphere, and hence its pressure not increase on the barometer trough due to this same increased effective aceleration, and force the mercury in the column back up... effectively keeping the height of mercury in the column same as when the barometer was static.

Ah, good question.

The reason is that the entire column of air above is not being accelerated when you accelerate the barometer upwards.

First of all, air can flow around. So you are only accelerating a fraction of air around barometer. That situation gets worse higher up, as air just parts to let barometer through, rather than accelerate with it.

Secondly, even if you did force all of the air to accelerate with barometer, it will take a rather long time for the disturbance to propagate upwards, and make the entire height accelerate.

If you were to put barometer at the bottom of an extremely tall cylinder, one that's tall enough to hold the entire height of the atmosphere (or just enough so that the everything above doesn't contribute significantly) then accelerate the whole structure, and let things settle, then when the entire column of air is accelerating at that constant rate, the barometer will again equalize.

But this is very far removed from situation where you just take barometer and accelerate it upwards.

 

[ludic]

I thank you both profoundly for your patient and analytical explanations.