Inclined Manometer Again (sorry)

[hollywood63]

Ok let me try this once more :p My incline is 25 in long and has a raise of 8.75 in. When I set up for a test I have a test plate that flows a calibrated 274cfm @ 28 in. (after it was made it was flowed @10in). When I run the test plate the incline goes up to the 24-24.25 .in Now can I
1: say that I am flowing 274 cfm at this time?
2: if so when I am reading at say 12 inches is this 50% of 274cfm?
3: or according to spreadsheets that are around I am flowing 69% of 274
4: I am have the darnedest time trying to comprehend a incline.
5: I plan on going electronic but not til I have a half assed handle on the mechanical side.
6: I have searched the forum but I just can not find a explanation that I can understand.

Thanks Art

[86rocco]

It's a square root type relationship, so half the distance along the scale is sqrt(1/2) the flow.

The way you do that calculations is you take the distance the column has moved along the scale, divide by the total length of the scale then take the square root of that number and what you end is the percentage of full scale.

Let's use your numbers as an example, your scale is 25" long, so if you're flowing 274 cfm at 24" along the scale, 24" corresponds to sqrt(24/25)=.9798% of full scale so, full scale(100%) on your scale would be 274/sqrt(24/25)= 279.65 cfm

Now 12" along the scale is sqrt(12/25)=69.28% of 279.65 or 193.75 cfm

I hope that helps.

P.S. I'll say this to save someone else from pointing out my "error", this method makes certain assumptions about the way your inclined manometer is constructed specifically, the diameter of the tube and the surface area of the well must remain consistant through out the full range of fluid movement but that's the way almost everyone makes them so chances are this'll work fine for you.

[Socket]

[color=#000000]I understand the calulation with the formula that 86rocco posted. What I

[86rocco]

[B. Elliott]

I've also had quite a bit of confusion with the manometers. I thought I understood them pretty well... till I actually started testing.

86rocco, on the spreadsheet you made it states that the pressure difference is the vertical rise on the incline. The part that confuses me is, how is it calculating flow if there is no input for the pressure drop across the orifice?

Isn't the inclined mano measuring just the velocity? Where's the reference to atmosphere (test pressure)?

[Thomas Vaught]

Hollywood63,

86rocco gave you a calculation for an example of a 12" vertical rise in the inclined manometer and what the cfm would be.

Your original question asked for a rise of 8.75 inches.

The rise is directly related to the delta P across the measurement orifice. If your measurement orifice is calibrated for 100% on the inclined scale at 6" delta P and if the orifice hole size was calculated to flow 200 cfm at that same at 6 inches delta p then at 100% on the scale you would be flowing 200 cfm.

Same applies to your deal:

If your measurement orifice is calibrated for 100% on the inclined scale at 8.75" delta P and if the orifice hole size was calculated to flow 200 cfm at that same at 8.75 inches delta p then at 100% on the scale you would be flowing 200 cfm.

Ok!

Tom V.

[B. Elliott]

Ahh... Gotcha.

Something somewhere finally clicked i guess. I feel a little stupid for asking now!

[Thomas Vaught]

Nothing stupid about it, It took 86rocco a while to get me squared away on his spread sheet with the well and tube calculations so that I understood dwyers calculations.

Keep asking the questions.

Tom V.