Deriving Barometric Equation: Force Balance Confusion

[WCL]

I'm having trouble deriving the barometric equation using the force balance.

Say, for the simplest case with a tube of mercury inverted in the middle of a beaker full of mercury. For the sake of simplicity, both are cylinders, with the beaker having radius R, tube having radius r and height of the mercury h.

So, doing a force balance on the stationary system would give:

$$\begin{gather*} \sum F = P_\textrm{atm} \pi (R^2-r^2) - \rho g h \pi r^2 = 0 \\ P_\textrm{atm} = \frac{\rho g h r^2}{ R^2-r^2} \end{gather*}$$

I do understand the force balance on the top and bottom areas of a control volume of fluid, which leads to the areas canceling out, but that doesn't happen in this case. Every book just says the pressures at the surface of the liquid must be equal and spits out a formula. I suppose that I could do a pressure balance and use the formula $P = \rho g h$ for a column of fluid, but the force balance should still make sense.

In the end, I don't see how the radii cancel. I don't think the barometric equation is an approximation, so I must be missing something.


Sorry for any sloppy LaTeX...I'm improvising the best I can.

Thanks.

 

[Filip Larsen]

It does not make sense to equate the force on the free surface with the force on the base of the tube because a static fluid cannot transfer directional forces (it has zero shear stress everywhere). A model of the weight of a small static fluid column will give you the normal hydro-static equation relating pressure with height of fluid and surface pressure.

It can be instructive to calculate the total pressure on the bottom of your cylindrical beaker in two cases and explain why they are equal:

1. the beaker is filled with fluid (but no tube) and the position of the surface level is marked.
2. the tube is inserted into the fluid so it fills up and then placed and held in position by another device. More fluid is inserted into the beaker until the previous surface mark is reached.

If you can explain this, you can also explain "where" the weight of the additional fluid in the tube go when its not going into increasing fluid pressure and why that weight indeed does depend on r.

 

[WCL]

For the first case, it should be a simple $P = P_\textrm{atm} + \rho g H \pi R^2$, right?

For the second case, I would just basically add the weight of the column of fluid and distribute the force over the bottom of the beaker, right? Similar to how a hydraulic press works?
So I'd have $P = P_\textrm{atm} + \rho g H \pi R^2 + \frac{\rho g h r^2}{R^2}$
where H is height of the fluid in the beaker and h is the height of the fluid in the (I'm assuming inverted) tube.


But you're saying the fluid pressure would not increase in that case? Then I must be missing something.

 

[Filip Larsen]

For the first case, it should be a simple $P = P_\textrm{atm} + \rho g H \pi R^2$, right?

No, you are mixing pressure (unit N/m²) and weight (unit N). You can rectify this by dividing the last term by area.

For the second case, I would just basically add the weight of the column of fluid and distribute the force over the bottom of the beaker, right? Similar to how a hydraulic press works?

Not quite. It is a bit difficult to explain, since this "equal pressure rule" is very fundamental for a fluid.

For each fluid column you can calculate pressure at the bottom of the column as the sum of the pressure at the surface and the weight of the column. For this there are two "types" of columns: those with a free surface outside the tube and those with a free surface inside the tube. For the columns outside the tube the pressure is straight-forward calculated like mentioned above (using the corrected version of your first equation in post 3). For the columns inside the tube the surface pressure is zero (assuming the tube is lifted so high that a free surface with vacuum is created above the fluid inside the tube) and the weight of such a column is a bit more than the columns outside the tube since there is more fluid.

You can now either do an experiment measuring the height of the additional fluid inside the tube and the pressure on the bottom, or you can invoke the theoretical principle of no shear stresses in a fluid, to get to the conclusion that the bottom pressure ends up being the same for both types of columns, or more generally, the fluid will have equal pressure at the same height in the fluid.

You should notice, that the vertical component of the atmospheric pressure on the tube is not carried by the fluid but by some other means, for instance by a person holding the tube up. This "missing" atmospheric pressure on the fluid inside the tube is why the fluid is pressed up by the fluid outside the tube such that the pressure at the mouth of the tube is the same as the pressure in the immediate surrounding fluid.

If you still have trouble grasping it, you may perhaps just consider it a strange fact of fluids that you can get the same static pressure from a column of fluid no matter if is absolutely tiny or wide as an ocean.

 

[WCL]

Ah...I should have looked at what I wrote and figured that the first part made no sense. It should just be $P = P_\textrm{atm} + \rho g H$

For the second case, I think I understand. The pressure fluid in the tube above the level of the fluid in the beaker has to equal atmospheric pressure in order for there to be static equilibrium, right? So the total pressure on the surface of the fluid in the beaker would still be atmospheric.

 

[Filip Larsen]

Ah...I should have looked at what I wrote and figured that the first part made no sense. It should just be $P = P_\textrm{atm} + \rho g H$

For the second case, I think I understand. The pressure fluid in the tube above the level of the fluid in the beaker has to equal atmospheric pressure in order for there to be static equilibrium, right? So the total pressure on the surface of the fluid in the beaker would still be atmospheric.

That is correct.

 

[WCL]

Thank you very much! :D