Is the difference in water levels in the manometer (tank test) linearly related to the difference in pressure between inside and outside the tank? I think it must be, but my engineering training was a long time ago (and wasn't Mechanical eng).
If so, we have the following formula:
P = AD+B
Where P is the pressure difference, D is the water height difference and A and B are constants. From two of Vans test-kit instructions, we have the two data points (P=2.4 PSI, D=66") and (P=1 PSI, D=27"). Solving for A and B, we get the formula
P=0.036D+0.031
where P is in PSI and D is in inches. Does this seem right to the engineers out there?
The reason I want this formula is I'm doing an extended-time pressure test, and want to be able to compensate for changes in atmospheric pressure. If I keep a barometer in the shop, I should be able to compute updates of what the difference in pressure between inside and outside the tank should be (assuming the tank is airtight and making the simplifying assumption that the tank skin is inflexible), and then use the above formula to compute how the water levels in the manometer should be changing. Of course, I should probably factor in the temperature, which would affect the pressure in the tank to some degree.
Overnight I had the water separation D decrease by about 2.5", and I'm not sure if it's reasonable to expect this is just due to the atmospheric pressure change. The atmos press increased by about 0.2kPa(=0.029PSI) overnight (according to weather info), which by my formula should only mean a -1.7 inch change in D. Except I haven't accounted for temperature, and I'm not sure how to.
If so, we have the following formula:
P = AD+B
Where P is the pressure difference, D is the water height difference and A and B are constants. From two of Vans test-kit instructions, we have the two data points (P=2.4 PSI, D=66") and (P=1 PSI, D=27"). Solving for A and B, we get the formula
P=0.036D+0.031
where P is in PSI and D is in inches. Does this seem right to the engineers out there?
The reason I want this formula is I'm doing an extended-time pressure test, and want to be able to compensate for changes in atmospheric pressure. If I keep a barometer in the shop, I should be able to compute updates of what the difference in pressure between inside and outside the tank should be (assuming the tank is airtight and making the simplifying assumption that the tank skin is inflexible), and then use the above formula to compute how the water levels in the manometer should be changing. Of course, I should probably factor in the temperature, which would affect the pressure in the tank to some degree.
Overnight I had the water separation D decrease by about 2.5", and I'm not sure if it's reasonable to expect this is just due to the atmospheric pressure change. The atmos press increased by about 0.2kPa(=0.029PSI) overnight (according to weather info), which by my formula should only mean a -1.7 inch change in D. Except I haven't accounted for temperature, and I'm not sure how to.
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