Absolute and Gauge Pressure
Consider a cylinder of water bounded on the top by y_{1} and the bottom by y_{2} in Fig 1.
— The forces on the cylinder of water are:
— F_{1} on the top surface.
— F_{2} on the bottom surface.
— mg is the weight of the cylinder (caused by gravity).
— The pressures on the cylinder of water are:
— P_{1} on the top surface.
— P_{2} on the bottom surface.
— The dimensions of the cylinder:
— y_{1} - y_{2} is the height
— A is the area of the top or bottom surface.
Note: The water is at rest (no translation or rotation) so it is in static equilibrum.
ΣF = 0
F_{2} + (-F_{1}) + (-mg) = 0
F_{2} = F_{1} + mg --- Eq1
By definition F_{2} = P_{1 }A and F_{2} = P_{2 }A
So Eq 1 becomes P_{2}A = P_{1}A + mg --- Eq2
Mass of water in cylinder = ρV where V is the volume of the cylinder
Mass of water in cylinder = ρ [A (y_{1} - y_{2})]
So Eq2 becomes P_{2}A = P_{1}A + ρg [A (y_{1} - y_{2})]
P_{2} = P_{1} + ρg (y_{1} - y_{2})
This becomes the famous equation P= ρgh where h is the depth of the liquid if y_{1} becomes 0 . This is called gauge pressure as it is the pressure difference between the suface and the point as measured by a gauge.
If the pressure at the surface is p_{0} then the absolute pressure is p_{0} + ρgh
Gauge Pressure = ρgh
Absolute Pressure = p_{0} + ρgh
Divers experience p_{0} + ρgh
Implication: Pressure at a point in a ﬂuid in static equilibrium depends on depth at that point and is the same on a particular horizontal level. |