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Unformatted text preview: n Equations of Hydrostatics
The integral formulation for the equations of hydrostatics is
Ρg V 0
m pn A t m (15) t We can use the divergence theorem to convert the area integral in (16) to a volume integral
Ρg V pV
m t m (16) t Since our volume m t is arbitrary, the only way (17) can hold is if the integrands are equal. Thus we
have the point equations for hydrostatics
p Ρg (17) 4 Hydrostatics.nb This is a vector equation and in terms of components we have
p p p Ρgx , Ρgy , x Ρgz y (18) z Let us consider a fluid body such that gravity acts in the z direction. Then (19) becomes
p p p Ρg, 0, x y 0 (19) z Integrating the first equation in (20) gives
Ρg z C x, y p x, y, z where C x , y is a function of integration. If we now differentiate this equation with separately respect to
x and y we get
p C x x p C y y 0 C y, 0 C constant Thus the hydrostatic pressure field in our fluid is
pz Ρgz C (20) Let us suppose that the pressure at z=L is atmospheric patm . Then the pressure in the fluid is
pz patm Ρg L z (21) Often it is convenient to work with the pressure relative to atmospheric p
pressure.
pgage pz patm patm . This is called the gage Ρg L z (22) Applications
Example 1
Atmospheric pressure is measured by means of a barometer, a schematic of one is shown below Let us define the origin of our coordinate system at the mercury level in the bath. The equations of
hydrostatics become
p p
Ρhg g, x p
0, y 0
z (23) Hydrostatics.nb 5 Integrating we get
Ρhg g z C pz
Now at z (24) h, the pressure is pvap which for mercury is about 3 10 6 atm. Thus we set pvap 0 and solve for the constant C in (27). the result is
Ρhg g h z pz (25) Now at z=0,
p0 patm (26) Substituting this result into (28) gives
Ρhg g h patm (27) Because of a barometer readings, atmospheric pressure is often reported in terms of inches of mercury. Example 2
A manometer like a barometer is used to measure differences in pressure. It is usually in the form of a
Utube and may consist of several liquids. In the example shown in figure 4 we have a manometer
consisting of two fluids A and B We will assume that gravity acts in the vertical direction, and the origin of out coordinate system is
located somewhere below the Utube. The location is material for the analysis. Again the hydrostatic
pressure field is given by
p p
Ρg, p
0, x 0 y (28) z Now if the pressure in the bulb is p1 then the pressure at the fluidA/b interface is
p2 ΡA g H1 p1 (29) At the fluid B/air interface it is
p3 patm p2 ΡB g H2 (30) Thus the difference in pressure read by the manometer is
patm p1 ΡA g H1 ΡB g H2 (31) so that
p1 patm g ΡA H1 ΡB H2 Thus a manometer opened at one end to atmospheric pressure reads gage pressure (32) 6 Hydrostatics.nb References
The following texts were helpful in preparing these notes 1. Anthony Esposito, Fluid Mechanics with Applications, Prentice Hall,1998
2. Stephen Whitaker, Introduction to Fluid Mechanics, Kreiger Publishing Company, 1968...
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 Spring '14

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