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1、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 Pascals Law For fluid at rest, pressure at a point is the same in all direction P = Px = Py = Pz,Proving Pascals Law,
2、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 Equationfor 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 = - ,Incompressible Fluid,Pres
3、sure 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),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 + Pa
4、tm 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,Definition of Pressure Head,Pressure head at point = P/ g (m) Piezometric head = P/ g + y (m) Piezometric pressure = P + g y In fluid statics, gauge p
5、ressure head at depth h below free surface = P/ g = g h / g = h,Example,P1 = F/A = 1000 / ( 0.0252) = 509.3 kPa P2 = P1 - g (2.5) = 484.77 kPa F = P2 A2 = 484.77 x 0.12 = 15.23 kN,2.5m,1 kN,F ?,0.2 m diam,0.05 m diam,P1,P2,Pressure at liquid interface,P1 = 1gh1 P2 = 1gh1 + 2gh2 P3 = 1gh1 + 2gh2 + 3g
6、h3,Liquid 1 1,Liquid 2 2,Liquid 3 3,h1,h2,h3,P1,P2,P3,Pressure Measurement,Absolute Pressure : Measured relative to perfect vacuum Perfect vacuum : 0 absolute pressure Gauge Pressure : Measured relative to local atmospheric pressure If no specified, pressure reading is usually assumed to be gauge Ga
7、uge pressure is positive if higher than atmospheric pressure, and negative if lower than atmospheric pressure Negative gauge pressure is also known as suction or vacuum pressure,Atmospheric Pressure,Atmospheric pressure is usually measured using a mercury barometer Patm = Hg g h 101,300 N/m2 abs = 0
8、 N/m2 gauge 1.013 bar abs 760 mm Hg abs 10.3 m water abs,Manometers,Devices for pressure measurement : Piezometer tube (stand pipe) Simple U-tube Differential U-tube manometer Inclined tube manometer Other types : Micro-manometer Barometer,Manometer Calculation,Within a continuous fluid of constant
9、P1 = P3 P2 = P1 - g h1 = P1- h1 P4 = P1 + g h2 = P1 + h2 Start computation at one end, and compute pressure from meniscus to meniscus till the other end Alternatively, select two points along same liquid column at same level, equate the pressures at the two points,Piezometer Tube,PA.gauge = g h1 Sim
10、ple U-Manometer P2 = P3 PA + 1gh1 = 2 g h2 PA,gauge = 2 g h2 - 1gh1,Differential U-tube,PA + 1 g h1 - 2gh2 - 3gh3 = PB PA PB = 2gh2 + 3gh3 - g h1 Inclined Manometer Pipes A & B contain gas PA - 2 g l2 Sin = PB PA PB = 2 g l2 Sin ,Pressure and force on a small plane area,Pressure at small plane : P =
11、 g h Force on small plane dF = P x dA = g h dA P depends on vertical depth h below surface Force dF = P x plane area P and dF perpendicular to plane,Pressure on plane P,h,Small plane area dA,Hydrostatic thrust on a plane (flat) surface,Force on dA : dF = g h dA = g y Sin dA Resultant thrust F on pla
12、ne of area A given by integration of dF over A,Hydrostatic Thrust on a Plane (flat) Surface,dF = g h dA = g y Sin dA Resultant thrust F on plane of area A given by integration of dF over A Thrust F = Resultant of pressure acting on every point of the plane surface F acts perpendicular to plane Magni
13、tude of F = g A hc F does not act through centroid of plane, C, in general,Line of Action of Thrust F,Moment of dF about O : dM = dF . y = g y2 Sin dA,Line of Action of Thrust F,Parallel axes theorem : Io = Ic + Ayc2 yR = (Ic + Ayc2) / (Ayc) = yc + Ic / (Ayc) y = yR - yc = Ic / (Ayc),Summary :Thrust
14、 on Plane Surface,Magnitude of thrust F : F = g A hc Acting at a point P y below the centroid of plane C : y = Ic / (A yc),Properties of common shapes,Concept of Pressure Prism,Average pressure Pav = g h/2 Resulting force FR = Pav x A = g h/2 x b x h = g A h/2 = Volume of pressure prism and passes t
15、hrough the centroid of the pressure prism,Pressure Prism,Resulting force FR = Volume of pressure prism = ABDE + BCD = F1 + F2 Location of FR can be determined by taking about some axis, like A : FR yA = F1y1 + F2y2,Effect of Atmospheric Pressure,Atmospheric pressure acts on both sides of wall, the n
16、et effects on the wall cancel out Resultant fluid force on a surface in contact with fluid is caused by gauge pressure,Pressure on Surfaces,P1 = g h1 P2 = g h2 P3 = g h3 P4 = g h4,h1,h2,h3,h4,P1,P2,P3,P4,Thin weightless plate submerged in water,Hydrostatic thrust F acts on both sides of plate and ca
17、ncel each other,F,F,Thrust on surface in contact with liquid,F,F1,F2,F,P,P,Rectangular Vertical Surface,F,h,h/3,P,Inclined Rectangular Surface,F,h,h/3,F,h, h/3,P,P,Horizontal Plane Surface,F,C=P,Summary :Thrust on Plane Surface,Magnitude of thrust F : F = g A hc Acting at a point y below the centroi
18、d of plane C : y = Ic / (A yc),Example : Circular Surface,Magnitude : F = g A hc y = Ic / Ayc For circle : Ic = D4 / 64 y = ( D4 / 64) / ( D2 /4) (D/2) = D/8 or (3/8) D from base,Example : Circular Gate,Show that moment about hinge is independent of h F = g A hc = g ( D2 /4) (h + D/2) y = Ic / Ayc = ( D4 /64) / ( D2 /4) (h + D/2) Moment about hinge = F y = g ( D4 /64) = independent of h,Hydrostatic Thrust on Curved Surface,Compute horizontal component Fh and vertical component Fv of thrus
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