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ECH141- Hydrostatics

# Fluid at rest is given by t pn n 14 note that the

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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 U-tube 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 U-tube. 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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