Fluid statics Study of fluid at rest i e. not in motion or not flowing On any plane, shear force is zero(no velocity gradient or shear deformation hence only pressure forces exist Sx ss Pasca’sLaw For fluid at rest, pressure at a point is the same in all direction
Fluid Statics • Study of fluid at rest, i.e. not in motion or not flowing • On any plane, shear force is zero (no velocity gradient or shear deformation) hence only pressure forces exist Pascal’s Law • For fluid at rest, pressure at a point is the same in all direction P = Px = Py = Pz
Proving pascal’sLaw P,8 δxδz 6 As element is in equilibrium ∑F,=0 P、δxδz=P6 x SS Sin0 Ss sin e= sz Hence Similarly,forΣF=0
Proving Pascal’s Law As element is in equilibrium : Fy = 0 Py x z = Ps x s Sin s Sin = z Hence : Py = Ps Similarly, for Fz = 0 Pz = Ps
Basic hydrostatic equation for Pressure field Rectangular free body P+↓↑ of top area a W P+ dP Body of fluid in equilibrium ∑F,=0 PA+W=(P+ dP).A .A+pgadh=(P+ dP)a dP/dh=pg=y dP/dy =-pg-Y
Basic Hydrostatic Equation for Pressure Field Body of fluid in equilibrium : Fy = 0 P.A + W = (P + dP) . A P.A + g A dh = (P + dP) A dP/dh = g = dP/dy = - g = - h dh P P + dP W Rectangular free body of top area A y
Incompressible fluid Pressure difference between two points in a body of fluid dP/dh=pg dP=pg dh Integrating from P, to P2 dp=p gdh h If fluid is incompressible and homogeneous p=constant 2-PI=pg(h2-h1=pg(Ah)
Incompressible Fluid Pressure Difference between two points in a body of fluid dP/dh = g dP = g dh Integrating from P1 to P2 : If fluid is incompressible and homogeneous = constant P2 – P1 = g (h2 – h1 ) = g ( h) P1 P2 h1 h2 h = 2 1 2 1 h h P P dP gdh
Incompressible fluid If Pi is on free water surface, i.e. PI=Pat P=P 2-Pi=pg h P P-P atm p g P2.abs =pgh+P atm For gauge pressure measurement, Patm=0 gauge pg h + P ,ga auge Gauge pressure at a point h below free surface =pg h
Incompressible Fluid If P1 is on free water surface, i.e. P1 = Patm : P2 - P1 = g h P2 - Patm = g h P2,abs = g h + Patm For gauge pressure measurement, Patm = 0 P2,gauge = g h P2,abs = P2,gauge + Patm Gauge pressure at a point h below free surface = g h h P2 P1=Patm