{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

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

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

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...
View Full Document

{[ snackBarMessage ]}

Ask a homework question - tutors are online