CEE 3310 – Hydrostatics, Sept. 10, 2010 412.7 Review• Compressible fluids ⇒ Isothermal, adiabatic• The U.S. standard atmosphere is based on the concept of adiabatic conditions(γ = γ(θ)) for the troposphere (0 < z < 11.0 km) and isothermal conditions(γ = γ0) for the stratosphere (11.0 < z < 20.1).• If the surface S is horizontal, P = a constant ⇒ FR= P A2.8 Pressure PrismConsider the following example:42We can solve this directly as just shown, however, for many situations (or just for manypeople who prefer to think in a different manner!) a decomposition of the pressures intoa series of pressure prisms is often easier. Consider the decomposition such thatFyR= F1y1+ F2y2= (γbh2h1)h1+h22+γbh222h12h23Therefore yR=h1h1+h22+h22h1+2h23h1+h22= h1+h22h1+2h23h1+h22= 4 + 34 + 44 + 3= 7.43′CEE 3310 – Hydrostatics, Sept. 10, 2010 432.9 Review• Hydrostatic force on an inclined plane surfaceFR= PCAxR=Ixycγ sin θPCAyR=Ixcγ sin θPCA• Pressure prism → break the force up into a part arising from the constantpressure due t o everything that happens to the point just above the surface andthe depth var ying part over the surface.⇒ Most easily applied to shapes with no lateral variation and constant density overthe surface.2.10 Hydrostatic Force on a Curved SurfaceThere are two approaches:1. dFR= P dA ⇒ non- planar hence the integration is hard!2. Apply a static control volumeXFx= 0 = FAC− FH⇒ FAC= FHXFz= 0 = −FAB− W + FV⇒ FV= FAB+ Q44Where is the center of pressure?FHmust be collinear (no shear) with FACFVmustbe collinear with the resultant of FAB+ WExample – Oil TankerFH= FAC= PCA = γhCA = γ(d −r2)rbNow assume we are int erested in solving for a unit width (i.e., b = 1 m). ThereforeFH= 10kNm3(24m −1.5m2)(1.5m)(1m) = 349kN2.10.1 Line of ActionHow do we determine the line of action fo r the resultant force on a curved surface? Wecould use our moment of inertia based method reviewed above but let’s use pressureprisms here.F1= γ(d − r)rb = 10 · 22.5 · 1.5 · 1 = 3 37.5kNy1= d −r2= 23.25mCEE 3310 – Hydrostatics, Sept. 10, 2010 45F2= γr2rb = 10 ·1.52· 1.5 · 1 = 1 1 .2 5kNy2= d −r3= 23.5mThereforeyACFAC= F1y1+ F2y2⇒ yAC=337.5kN · 23.25m + 11.25kN · 23.5m348.75kN= 23.26mSimilarlyFV= 355.3kN and xBC= −0.756m (x = 0 at the tanker’s vertical surface)2.11 BuoyancyFB=Zbody(P1− P1) d AH= −γZ(z2− z1) d AH= γ · (body volume)ThereforeFB= FV2− FV1= Fluid weight above S2− fluid weight above S1= weight of fluid occupying the volume of the body2.11.1 Archimedes’ PrincipalFB= Weight of fluid displaced by a body or a floating body displaces its own weight ofthe fluid on which it floats.462.11.2 Stability of Floating BodiesThe stability of a floating body depends on the location of the buoyancy force and theweight of the body. They each exert a moment – one is a rig hting moment (the tendencyto rotate the object to an upright position) while the other is an overturning moment(the tendency to flip the body
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