Vertical Well Manometers

[rwdford]

Hello, I will be building a vertical well manometer to suit Dwyer red oil .826 SG and would like to size the Well ID to Tube ID so that the scale reads directly in inches

To make a formula first some manometer principles

Whatever fluid volume travels up the tube must be taken from the well below the zero line

Water has a specific gravity of 1 (at 15.6*C) which is 1cc of water = 1 gram

Dwyer fluid has a SG of 0.826 which is lighter than water

With the same vacuum Dwyer fluid will move 1/.826 = 1.2107 times the distance, so 28" water vacuum = 33.898" total distance

All basic info so far

Now the question is if I want to use 0.826 SG fluid and an acrylic tube with a 6mm ID then what is the ideal Well ID so that under 28" vacuum the fluid travels exactly 28" up from the zero line?

Can anyone enlighten me on how this is calculated?

I have what looks like a working theory about but would like to hear how other people are working this out first

Some of the calculations: Tube Area: 0.2827cm2

Vertical travel: 71.12cm

Volume above line: 20.1087cc (also equals volume below line removed from well)

I will for sure be upgrading to PTS DM later on but want to start with red oil manometers first and it will be useful to have both as a reference to confirm accuracy

[86rocco]

A long time ago, I made the spreadsheet found HERE you might find if useful, download it, go to the tab labelled "Inclined Scale Compensation" for vertical manometers, just make the scale height equal to the scale length, the column labeled Δs is how far the fluid has moved along the scale in inches and the column labeled Δp is pressure applied in inches of water.


Anyways, the direct answer is this:

If T= diameter of the tube, R= diameter of the reservoir and g=specific gravity of the fluid

and if we want the fluid rise on the tube inches to equal the pressure applied in inches of water then

R =√(T²/(1/g-1))




In your particular example, it works out to 13.073 mm

[rwdford]

Thank you Rocco

That is Exactly! the answer I had worked out

I found a basic rule with manometers that the total vertical travel always equals the same, so as the well gets larger the tube travel is increased

With water if wanting to get a full 28" rise the well would need to be very large so that the well fluid level drops only a tiny amount giving all of the travel to the tube end of the scales

Due to size limitations using red oil is much more practical

For inclined manometers I am thinking that by tilting the tube over the area is increased and well will react to the tube as if it were a larger diameter but with the same vertical travel, if the tube vertical rise was 10" and the scales were 20" long it would act the same as a straight vertical tube 10" long that has double the area, looking at it that way calculating the ideal well ID is the same as with the vertical manometer calculations, with the well being larger due to the extra volume of the length of the scale

[86rocco]

You're welcome. I added a diagram to my first post and to anyone that might be interested, the formula for R in that post is derived from the equations in the diagram by setting Δp = Δs then solving for R

[rwdford]

That is a very clear image and formula

Done a bit more work on my spreadsheet today, my formula is slightly different but hopefully it matches up well with your calc's

Let me know what you think

[86rocco]

Looks like you've got a good handle on the math, when I run your numbers, I get the same results.

[rwdford]

Thank you Rocco, good to know it is working well, I have done some hands on testing and it is working out very well so far with water manometers, next up is to get some Dwyer red oil