h = p / (rho g)
= 101670 Pa / [(999.209 kg/m^3)(9.806 m/s^2)]
= 10.4 m
Since "g" varies somewhat with latitude, the 3rd digit of this result is as far as you can go.
Are you measuring air pressure or water pressure?
That air pressure sounds like a sea level air pressure reading (14.746 psi) , which comes from the weight of the atmosphere from above pushing down on the surface of the earth.
Water pressure is the exact opposite and increases linearly with every cubic liter in depth you go. So since 1 liter has a mass of 1 kg, then if you went 1 meter deep, there would be 10 cubic liters in that depth so the pressure at the bottom of the meter would be 10 kg/ 10 cm^2 = .1 kg/ cm^2
1 Meter deep water column normally exerts a water pressure at the bottom in pascals of 9800 pascals or about 10 Kpa
About 10.4 metres. If by "corrected" you mean that the air pressure at sea level was subtracted from the measured pressure.
pressure = density x depth x gravitational acceleration
If you want to get the depth to 6 places of decimal, first you have to measure the local gravitational field to 6 places of decimal. It varies by about 1%, just on this planet.
The question is contradictory.
If 101.67kPa is corrected for barometric variation
(atmospheric pressure), then this is surface pressure
(no depth, since STP is 101.3kPa).
If density (999.209kg/m3) is corrected for temperature (14.05°C),
then pressure will be 101.3kPa × 999.209/1,000
for every 10.33 meters of water + 101.67 atmospheric
Depth (meters) = (Absolute pressure - 101.67kPa)/9.80kPa × .999209
Pressure of 101.670 kPa (14.746 PSI)
Water Density of 999.209 kg/m3 (Temperature 14.05 degrees C)
I'm getting a mad result and I was wondering if I, or the figures were wrong.
Cheers
