Fluid Statics Example
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Example 1 of Mercury Barometer Consider
the mercury barometer shown in the figure on the left below. Find
the atmospheric pressure in terms of the vapor pressure of the gas above the mercury
column, the specific weight of mercury, γ,
and the height, h, of the column of
mercury. Let the
radius of the tube be R. Let
the cross-hatched liquid be mercury.
Note the specific gravity of mercury is 13.6. The
column of mercury in the circular tube of radius, R, has a height, h. The pressure at points
2 and 3 in the liquid are at the same since they are at the same level. The pressure at
point 1 is the vapor pressure of the gas above the mercury. The
figure on the right is a (FBD) free body diagram of the column of liquid in
the circular
tube of radius, R and height h. For
equilibrium the net force in the vertical direction
must sum to zero. Σ Fz =
0 Patm
πR2 – γ πR2
h – Pvapor πR2 = 0 Therefore Patm = Pvapor + γh (result) Click
here to go on to another application of an inclined manometer with three |
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