Measuring changes in barometric pressure without a barometer

[Jehannum]

Could you use a water u-gauge (manometer), zeroed and with one side of the u-gauge plugged and the other open to air, to show an increase or decrease in barometric pressure after a period of time (e.g. 2 hours)?

The idea is that you trap some air at the starting atmospheric pressure on one side of the gauge. Then, if the current atmospheric pressure changes significantly it will cause the water column to rise or fall. At the end of the required period the pressure change can then simply be read off the u-gauge.

 

[russ_watters]

Yes, and such a device exists, but unfortunately it is called a "barometer".

Also, it is generally done under a vacuum and with a heavier liquid like mercury, but it doesn't need to be.

 

[Jehannum]

I realize the question reduces to "can a u-gauge be used as a barometer", of course. I just needed confirmation that there isn't some important detail I've omitted or something that would cause inaccuracy, etc.

 

[Carrock]

One possibly substantial cause of inaccuracy would be trapped air dissolving in the water or more air etc coming out of solution.
Temperature change could also also cause significant inaccuracy.
This is nevertheless an interesting idea.

 

[Jehannum]

I was wondering about temperature change myself.

I will be measuring ambient temperature at the start (T₀) and end (T₁) of the process anyway. Can I apply a temperature correction as follows?

u-gauge reading = P

additive correction for u-gauge reading = (P + 1000) * (T₁ - T₀) / (T₀ + 273)

 

[rootone]

One possibly substantial cause of inaccuracy would be trapped air dissolving in the water or more air etc coming out of solution.
Temperature change could also also cause significant inaccuracy.

Both of which potential innaccuracies are addressed by using Mercury instead of water,
though Mercury will still expand with temperature, the effect is much less than water.

 

[Cutter Ketch]

The weather station toy I received as a child some ... ahem ahem ... 40 years ago came with just such a water U and a scale to act as the barometer. A quick look on the internet shows some weather kits still do.

 

[Jehannum]

Unfortunately, mercury wouldn't be usable (UK legislation) because of its harmful vapour.

Can anyone confirm my temperature correction formula is OK a couple of posts ago?

 

[Nidum]

Balloon barometers are quite fun things to make .

Two basic varieties :

(1) Just the balloon .

(2) Balloon skin over a jam jar .

Use a long straw as a motion amplifier .

 

[Jehannum]

Upon thinking about this sealed-one-side u-gauge idea a little more I've come up against a few problems:

1. The trapped air on the sealed side will compress or rarefy if atmospheric pressure changes, so the pressure scale reading won't be accurate (the u-gauge will balance atmospheric pressure against water height pressure PLUS trapped air pressure).

Workaround: make the volume of air in the sealed side bigger, using a long tube attached to the gauge, with the seal at the end of the tube. This means changes in water level will have less effect on the pressure of the trapped air, reducing the 'piston' effect.

2. Changes in temperature will cause changes in volume of the trapped air, which will cause pressure readings to be incorrect.

Workaround: make the volume of air in the sealed side smaller so that changes in ambient temperature don't cause so much expansion or contraction of the compressed air.

Unfortunately, the workarounds oppose each other.

Do these problems arise when using a purpose-made barometer?

 

[Cutter Ketch]

Upon thinking about this sealed-one-side u-gauge idea a little more I've come up against a few problems:

1. The trapped air on the sealed side will compress or rarefy if atmospheric pressure changes, so the pressure scale reading won't be accurate (the u-gauge will balance atmospheric pressure against water height pressure PLUS trapped air pressure).

Workaround: make the volume of air in the sealed side bigger, using a long tube attached to the gauge, with the seal at the end of the tube. This means changes in water level will have less effect on the pressure of the trapped air, reducing the 'piston' effect.

2. Changes in temperature will cause changes in volume of the trapped air, which will cause pressure readings to be incorrect.

Workaround: make the volume of air in the sealed side smaller so that changes in ambient temperature don't cause so much expansion or contraction of the compressed air.

Unfortunately, the workarounds oppose each other.

Do these problems arise when using a purpose-made barometer?

Problem number one isn't a problem. It is how the thing works. The ideal gas law expansion and contraction of that trapped volume is what the tool is measuring. The size of the trapped volume determines the sensitivity and dynamic range of the barometer. A larger volume is more sensitive, but with a smaller dynamic range.

Problem two is a problem. You can't fix it by reducing the volume. Not only does that reduce your sensitivity, but since the change in that volume is what you are measuring the change due to temperature is the same fraction of the dynamic range regardless of the volume. The way to fix the temperature problem is to measure the temperature and calibrate it out.

 

[Jehannum]

Could you help me understand your response to 'problem' 1?

Let's say atmospheric air pressure increases by 10 mbar. The water column is pushed down on the open side and the water level rises on the sealed side.

The air trapped in the sealed side is now at a higher pressure.

Now, if the atmospheric pressure increases by another 10 mbar, it won't move the water column as much as before because the trapped pressure is exerting more pressure against it.

On a normal u-gauge, the scale for, say, millibars, is marked at even distances (5.1 mm per mbar, I believe).

So for this to work, surely the distances between millibars marked on the scale will have to decrease the higher the atmospheric pressure it's measuring.

So a normal u-gauge can't be used this way unless you change its printed scale.

 

[Nugatory]

So for this to work, surely the distances between millibars marked on the scale will have to decrease the higher the atmospheric pressure it's measuring.

By how much? Don't guess, calculate it.

Start with a reasonable assumption about the maximum and minimum pressure the device will have to measure and about the volume of the trapped gas.

 

[Cutter Ketch]

Could you help me understand your response to 'problem' 1?

Let's say atmospheric air pressure increases by 10 mbar. The water column is pushed down on the open side and the water level rises on the sealed side.

The air trapped in the sealed side is now at a higher pressure.

Now, if the atmospheric pressure increases by another 10 mbar, it won't move the water column as much as before because the trapped pressure is exerting more pressure against it.

On a normal u-gauge, the scale for, say, millibars, is marked at even distances (5.1 mm per mbar, I believe).

So for this to work, surely the distances between millibars marked on the scale will have to decrease the higher the atmospheric pressure it's measuring.

So a normal u-gauge can't be used this way unless you change its printed scale.

Let's say you draw a vacuum and the measure of the atmospheric pressure is just the weight of the lifted column of working fluid. This is how a Mercury barometer works. The reason they use Mercury is that it is insanely dense and heavy. Even so it takes most of a meter of Mercury to equal an atmosphere. If you want to use water, you need a column almost 10 m high! This is impractical. Instead for a water U you leave a trapped volume of gas. The height of the water is part of the answer, but mostly the compression and expansion of the trapped gas by the ideal gas law tells the story.

You note that P and V are inversely related. That does make a precise absolute scale more complicated, but still easily calculated. What's more the atmospheric pressure doesn't change by a large amount. The highest pressure ever recorded to the lowest pressure is a change of less than 20%. Normal day to day changes at one location are much smaller. Over that small range the curvature of the inverse relation is barely noticable and the graduations are nearly equally spaced.

 

[Jehannum]

Thank you.

I'll now try to apply this to the actual equipment I would typically want to use:

I have a water u-gauge that measures to 60 mbar. The u-tube carries on beyond the scale and has a full-bore connector, so the effective length is 70 mbar.

I zero the gauge at atmospheric pressure and then carefully plug one side of the gauge to seal air at that pressure. I leave the other side open to atmosphere.

After a few hours I look at the gauge reading. Say it has gone down by 2 mbar on the open side (meaning atmospheric pressure has increased).

The volume of the sealed side has decreased by an amount proportional to a change from 70 mbar to (70 - 2) = 68 mbar. The air pressure will have therefore increased by a factor of 70/68 = 1.029, i.e. 2.9%.

Let's say the original atmospheric pressure was 1000 mbar. This means trapped air is now at 1029.4 mbar. So, the increase in air pressure, which looked like 2 mbar from water movement, is actually 29.4 + 2 = 31.4 mbar.

This suggests that this won't be accurate enough unless I include the trapped air pressure increase in the calculations, or I increase the trapped volume so that changes in atmospheric pressure don't significantly change the trapped pressure.

Are my above calculations and conclusions correct?

* Note: I'm assuming constant temperature. Let's get the pressure stuff right first!

 

[Cutter Ketch]

Thank you.

I'll now try to apply this to the actual equipment I would typically want to use:

I have a water u-gauge that measures to 60 mbar. The u-tube carries on beyond the scale and has a full-bore connector, so the effective length is 70 mbar.

I zero the gauge at atmospheric pressure and then carefully plug one side of the gauge to seal air at that pressure. I leave the other side open to atmosphere.

After a few hours I look at the gauge reading. Say it has gone down by 2 mbar on the open side (meaning atmospheric pressure has increased).

The volume of the sealed side has decreased by an amount proportional to a change from 70 mbar to (70 - 2) = 68 mbar. The air pressure will have therefore increased by a factor of 70/68 = 1.029, i.e. 2.9%.

Let's say the original atmospheric pressure was 1000 mbar. This means trapped air is now at 1029.4 mbar. So, the increase in air pressure, which looked like 2 mbar from water movement, is actually 29.4 + 2 = 31.4 mbar.

This suggests that this won't be accurate enough unless I include the trapped air pressure increase in the calculations, or I increase the trapped volume so that changes in atmospheric pressure don't significantly change the trapped pressure.

Are my above calculations and conclusions correct?

* Note: I'm assuming constant temperature. Let's get the pressure stuff right first!

I'm confused. Why are you giving lengths in mbar? Is this a piece of equipment you found that already has a pressure scale? If so, don't you think the scale is already marked to read correctly for some designed trapped volume?

 

[Jehannum]

I'm giving lengths in mbar purely for convenience. It's a water u-gauge / manometer consisting of a tube of constant cross section. So volume is proportional to length and to the pressure scale markings on the manometer.

I'm only using volume ratios, not absolute volumes, so it seems convenient to just use the scale that's on the manometer rather than perform separate measurements.

The equipment is a manometer designed to be open to atmosphere, not sealed; millibars are marked at even intervals.

A barometer is occasionally required when tightness testing (leak testing) large pipe systems because the long test times required mean atmospheric pressure may change between the start and end of the test.

The guys who do this testing are extremely unlikely to have a barometer but they'll definitely have a manometer. So I just wanted a simple way of using a manometer as a moderately accurate (+/- 0.5 mbar) barometer, but it looks like the compression factor from sealing one end is too significant for this idea to work.

 

[Cutter Ketch]

This suggests that this won't be accurate enough unless I include the trapped air pressure increase in the calculations, or I increase the trapped volume so that changes in atmospheric pressure don't significantly change the trapped pressure.

Yes, as I explained in order to measure significant fractions of an atmosphere the weight of the water column is insufficient and you will want the compression of the trapped volume to dominate the calculation. But why do you say it won't work? As you have seen the pressure change is readily calculable.

 

[Jehannum]

I can calculate the pressure change in the trapped air due to compression or expansion as a factor F_P. It's the ratio of trapped air volume at the start to that at the end, which is equal to the ratio of the original length of air and the final length of air in the tube:

F_P = scale length / (scale length - water movement)
​

At the end, the atmosphere is balancing the water weight (shown directly in mbar) PLUS F_P x P_A0, where:

P_A0 = atmospheric pressure at the start​

But if I don't know what P_A0 was, I'm left with it as an unknown.

 

[Cutter Ketch]

But if I don't know what P_A0 was, I'm left with it as an unknown.

I suppose if you had a barometer to calibrate against you wouldn't need the water U. However, (he says uselessly) that is all you need: a single point calibration.

 

[lychette]

do you realize that the pressure in the sealed tube of the U tube arrangement is the saturated pressure of the fluid (water or mercury) at the appropriate temperature?
Mercury is used because its saturated vapor pressure is so low.

 

[Cutter Ketch]

do you realize that the pressure in the sealed tube of the U tube arrangement is the saturated pressure of the fluid (water or mercury) at the appropriate temperature?
Mercury is used because its saturated vapor pressure is so low.

Only if you draw a vacuum. If you seal the volume with an atm of air then that dominates. over reasonable temperatures the vapor pressure of water varies from a fraction of 1% of an atm to 7 or 8 %. The trapped air will be at 100% relative humidity all the time so that is another complication with temperature.