Fluid Mechanics-61341

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An-Najah National University College of Engineering Fluid Mechanics-61341 Chapter [2] Fluid Statics 1 Fluid Mechanics-2nd Semester 2010- [2] Fluid Statics

Fluid Statics Problems Fluid statics refers to the study of fluids at rest or moving in such a manner that no shearing stresses exist in the fluid These are relatively simple problems since no velocity gradients exist. Thus, viscosity does not play a role Applications include the hydraulic pressure, manometry, dams, and fluid containment (tanks) 2 Fluid Mechanics-2nd Semester 2010- [2] Fluid Statics

Pressure (P) is the force per unit area, where the force is perpendicular to the area F (N) P (N/m 2 or Pa) = A (m2 ) Pressure 1 kn/m 2 = 1 kpa 1 kpa = 0.145 psi Pressure in a fluid acts equally in all directions Pressure in a static liquid increases linearly with depth pressure increase p= γ h increase in depth (m) 3 Fluid Mechanics-2nd Semester 2010- [2] Fluid Statics

Pressure at a Point Pressure is a scalar quantity that is defined at every point within a fluid Force balance in the x-direction: 4 Fluid Mechanics-2nd Semester 2010- [2] Fluid Statics

Pressure at a Point Force balance in the z-direction: Vertical force on DA Vertical force on lower boundary Total weight of wedge element = specific weight 5 Fluid Mechanics-2nd Semester 2010- [2] Fluid Statics

From last slide: Pressure at a Point Divide through by to get Now shrink the element to a point: This can be done for any orientation, so 6 Fluid Mechanics-2nd Semester 2010- [2] Fluid Statics

Pressure at a Point The result shows that pressure at any point in a fluid at rest has a single value, independent of direction as long as there are no shearing stresses (or velocity gradients) present in the fluid For fluids in motion with shearing stresses, this result is not exactly true, but is still a very good approximation for most flows 7 Fluid Mechanics-2nd Semester 2010- [2] Fluid Statics

Pressure Transmission In a closed system, pressure changes from one point are transmitted throughout the entire system (Pascal s Law). F out A A out in F in Hydraulic Lift 8 Fluid Mechanics-2nd Semester 2010- [2] Fluid Statics

Pressure Variation with Elevation Static fluid: All forces must balance as there are no accelerations Look at force balance in direction of D l 9 Fluid Mechanics-2nd Semester 2010- [2] Fluid Statics

Pressure Variation with Elevation From the previous figure, note that Shrink cylinder to zero length: From the previous slide: or 10 Fluid Mechanics-2nd Semester 2010- [2] Fluid Statics

Pressure Variation with Elevation The pressure-elevation relation derived on the previous slide, is perfectly general (applies also to variable g) But if g is constant, the above equation is easy to integrate: The quantity and is known as the piezometric pressure is called the piezometric head 11 Fluid Mechanics-2nd Semester 2010- [2] Fluid Statics

Pressure Variation with Elevation For an incompressible fluid (g is constant), pressure and elevation at one point can thus be related to pressure and elevation at another point as: p p 1 2 z 1 z2 or z 2 or z 1 h p 1 Constant p p p p z z ) 1 2 2 ( 2 1 h 12 Fluid Mechanics-2nd Semester 2010- [2] Fluid Statics

Absolute and Gage Pressure 13 Fluid Mechanics-2nd Semester 2010- [2] Fluid Statics

Absolute and Gage Pressure 14 Fluid Mechanics-2nd Semester 2010- [2] Fluid Statics

What is the pressure at the faucet? What do you do if you want more pressure at the faucet? 15 Fluid Mechanics-2nd Semester 2010- [2] Fluid Statics

Blood Pressure 26.8 k Pa 16 Fluid Mechanics-2nd Semester 2010- [2] Fluid Statics

Example 1 Solution: 17 Fluid Mechanics-2nd Semester 2010- [2] Fluid Statics

Example 2 What is the water pressure at a depth of 35 ft? Solution: With the information given, all we can calculate is the pressure difference between points 1 and 2 18 Fluid Mechanics-2nd Semester 2010- [2] Fluid Statics

Example 3 What is the gage pressure at point 3? Solution: Two step solution: 1) Calculate 2) Calculate s.g.=0.8 (relative to atmospheric pressure at point 1) 19 Fluid Mechanics-2nd Semester 2010- [2] Fluid Statics

Pressure Measurement Mechanical Pressure Gages The Bourdon pressure and Aneroid barometer are typical mechanical devices for measuring gage and absolute pressures, respectively 20 Fluid Mechanics-2nd Semester 2010- [2] Fluid Statics

Pressure Measurement Liquid Pressure Gages Manometer: gravimetric device based upon liquid level deflection in a tube Mercury barometer: evacuated glass tube with open end submerged in mercury 21 Fluid Mechanics-2nd Semester 2010- [2] Fluid Statics

The Manometer Simple, accurate device for measuring small to moderate pressure differences Rules of Manometry: pressure change across a fluid column of height h is gh pressure increases in the direction of gravity, decreases in the direction opposing gravity two points at the same elevation in a continuous static fluid have the same pressure 22 Fluid Mechanics-2nd Semester 2010- [2] Fluid Statics

The Manometer p p 1 l p 2 0 h x 1 23 Fluid Mechanics-2nd Semester 2010- [2] Fluid Statics

The Manometer p l p 4 p p x 1 1 y 2 2 3 5 l h 24 Fluid Mechanics-2nd Semester 2010- [2] Fluid Statics

) 0( 1 gage p p atm The Manometer l p p p l h p p p h p m m 3 4 4 1 3 2 1 Fluid Mechanics-2nd Semester 2010- [2] Fluid Statics 25

Example 4 Find the location of the surface in the manometer Solution: The distance Dh is the height of the liquid in the manometer above the heavier liquid in the tank A B D C p C h m p B h 10 10 m w w 10 1 3 3.33cm 26 Fluid Mechanics-2nd Semester 2010- [2] Fluid Statics

Example 5 Find the gage pressure at the pipe center A C Solution: Manometer equation from the pipe center to the open end of the manometer p p A A 0( gage) p C 2.5(1)(62.4) 0 1(2)(62.4) 0.5(1)(62.4) p C 27 Fluid Mechanics-2nd Semester 2010- [2] Fluid Statics

Example 6 28 Fluid Mechanics-2nd Semester 2010- [2] Fluid Statics

Example 6 (Solution) 29 Fluid Mechanics-2nd Semester 2010- [2] Fluid Statics

Find the specific weight of the fluid which filled part CD of the tube Solution: D V 2 d h 4 (0.5) 4 2 h 2cm Manometer Equation p A liq ( h 0.05) h Example 7 3 liq ( h 0.05) h (0.10186 0.05) 0.10186 h 10.186cm p D (9810) liq p A 4995 N A B p D / m 3 h C 2 cm 3 o (gage) 30 Fluid Mechanics-2nd Semester 2010- [2] Fluid Statics

Differential Manometer Used for measuring pressure differences between points along a pipe g w l p 1 wl mh w l h) ( p 2 g m p p p ( ) h 2 1 w m 31 Fluid Mechanics-2nd Semester 2010- [2] Fluid Statics

Find the change in piezometric pressure and in piezometric head between points 1 and 2. Solution: Manometer equation from point 2 to point 1 In general Example 8 (piezometric pressure) (piezometric head) 32 Fluid Mechanics-2nd Semester 2010- [2] Fluid Statics

Example 9 33 Fluid Mechanics-2nd Semester 2010- [2] Fluid Statics

Example 9 (Solution) 34 Fluid Mechanics-2nd Semester 2010- [2] Fluid Statics

Example 9 (Solution) 35 Fluid Mechanics-2nd Semester 2010- [2] Fluid Statics

Pressure Forces on Plane Surfaces Surfaces exposed to fluids experience a force due to the pressure distribution in the fluid The resultant force on vertical, rectangular surfaces can be found using a graphical interpretation known as the pressure prism 36 Fluid Mechanics-2nd Semester 2010- [2] Fluid Statics

The differential force is df Pressure Forces on Plane Surfaces pdah dal sin da Integrating to get the total force on the gate yields: F sin lda From basic mechanics, we recall for first moment of area, So, lda F γl c l c A Asin α Recognizing that l c sin α h c F h c A 37 Fluid Mechanics-2nd Semester 2010- [2] Fluid Statics

In general, the resultant force on an area equals the pressure at the centroid of the area (gh c ) times the area (A) To complete the analysis, we must compute the location of the center of pressure where the resultant force F can be assumed to act Pressure Forces on Plane Surfaces da l da l l ldf F l p sin ) sin ( 2 38 A l I l l l A l I F l A l I A l F l A l A l I A l I F l A l I I da l c c c p c c c c c c c p c c c c c p c c o sin / by ) multiplying ( sin Recognizing that 2 2 2 Fluid Mechanics-2nd Semester 2010- [2] Fluid Statics

Example 10 39 Fluid Mechanics-2nd Semester 2010- [2] Fluid Statics

Example 10 (Solution) 40 Fluid Mechanics-2nd Semester 2010- [2] Fluid Statics

Example 10 (Solution) 41 Fluid Mechanics-2nd Semester 2010- [2] Fluid Statics

Example 10 (Solution) 42 Fluid Mechanics-2nd Semester 2010- [2] Fluid Statics

Example 10 (Solution) 43 Fluid Mechanics-2nd Semester 2010- [2] Fluid Statics

Example 10 (Solution) 44 Fluid Mechanics-2nd Semester 2010- [2] Fluid Statics

Example 11 45 Fluid Mechanics-2nd Semester 2010- [2] Fluid Statics

Example 11 (Solution) 46 Fluid Mechanics-2nd Semester 2010- [2] Fluid Statics

Example 11 (Solution) 47 Fluid Mechanics-2nd Semester 2010- [2] Fluid Statics

Example 11 (Solution) 48 Fluid Mechanics-2nd Semester 2010- [2] Fluid Statics

Example 12 Find the normal force required to open the rectangular gate if it is hinged at the top. The gate is 5 m wide and θ =30 o Solution: First find the total hydrostatic force acting on the plate F h c A 9810N / m 3 10m5m5m 2.45MN 3 Ic o 55 /12 l p lc ((8/ cos30 ) 2.5) 11. 92 m l A 11.74 25 M c hinge o o F 5m 2.45MN ((11.92 (8/ cos30 )) F 1.31MN 49 Fluid Mechanics-2nd Semester 2010- [2] Fluid Statics

Example 13 Given: Gate AB is 4 ft wide, hinged at A and Gage G reads -2.17 psi. Find the horizontal force at B to hold gate. Air G 5ft 18ft Water A gate Oil SG=0.75 6ft B 50 Fluid Mechanics-2nd Semester 2010- [2] Fluid Statics

Example 13 (Solution) First convert negative pressure in tank to ft of water p 2.17144 h 5 ft 62.4 Total hydrostatic force acting on the gate from oil F F oil h c A o. 7562.43 4 6 3370Ib water h c A 62.4(18 5 3) 46 14976Ib l p( oil) 4 Total hydrostatic force acting on the gate from water ft 3 Ic 46 /12 l p( water) lc 10 10. 3 l A 10 24 c ft 51 Fluid Mechanics-2nd Semester 2010- [2] Fluid Statics

Example 13 (Solution) A 3.3ft 4ft F w F oil M A 0 F F water B B 3.3 5990lb F oil F B 4 F 149763.3 3370 4 B 6 F B 6 52 Fluid Mechanics-2nd Semester 2010- [2] Fluid Statics

Pressure Forces on Curved Surfaces Resultant pressure forces on curved surfaces are more difficult to deal with because the incremental pressure forces, which are normal to the surface, vary continually in direction There are two ways to approach the problem. One is to use direct integration and the second method is to utilize the basic mechanics concepts of a free body and the equilibrium equations Fig. 2.11 53 Fluid Mechanics-2nd Semester 2010- [2] Fluid Statics

Pressure Forces on Curved Surfaces 54 Fluid Mechanics-2nd Semester 2010- [2] Fluid Statics

Pressure Forces on Curved Surfaces 55 Fluid Mechanics-2nd Semester 2010- [2] Fluid Statics

Example 14 56 Fluid Mechanics-2nd Semester 2010- [2] Fluid Statics

Example 14 (Solution) 57 Fluid Mechanics-2nd Semester 2010- [2] Fluid Statics

Example 14 (Solution) 58 Fluid Mechanics-2nd Semester 2010- [2] Fluid Statics

Example 14 (Solution) 59 Fluid Mechanics-2nd Semester 2010- [2] Fluid Statics

Example 14 (Solution) 60 Fluid Mechanics-2nd Semester 2010- [2] Fluid Statics

Example 14 (Solution) 61 Fluid Mechanics-2nd Semester 2010- [2] Fluid Statics

Example 14 (Solution) 62 Fluid Mechanics-2nd Semester 2010- [2] Fluid Statics

Example 14 (Solution) 63 Fluid Mechanics-2nd Semester 2010- [2] Fluid Statics

Example 14 (Solution) 64 Fluid Mechanics-2nd Semester 2010- [2] Fluid Statics

Example 14 (Solution) 65 Fluid Mechanics-2nd Semester 2010- [2] Fluid Statics

Example 15 66 Fluid Mechanics-2nd Semester 2010- [2] Fluid Statics

Example 15 (Solution) 67 Fluid Mechanics-2nd Semester 2010- [2] Fluid Statics

Example 15 (Solution) 68 Fluid Mechanics-2nd Semester 2010- [2] Fluid Statics

Buoyancy and the Stability of Floating Bodies The familiar laws of bouncy (Archamedes principle) and flotation are usually stated (1) a body immersed in a fluid is buoyed up by a force equal to the weight of fluid displaced; and (2) a floating body displaces its own weight of the liquid in which it floats 69 Fluid Mechanics-2nd Semester 2010- [2] Fluid Statics

Buoyancy and the Stability of Floating Bodies The resultant buoyant force on a submerged or partially submerged object in a static fluid is given by Archimedes principle as: F B ( volumeof submerged object) or F B ( volumeof liquid displaced) The buoyant force is equal to the weight of the fluid displaced by the object and is in a direction opposite the gravitational force 70 Fluid Mechanics-2nd Semester 2010- [2] Fluid Statics

Buoyancy and the Stability of Floating Bodies The line of action of the buoyant force passes through the centroid of the displaced volume, often called the center of buoyancy (COB) The stability of submerged objects (balloon and sub-marine) is determined by the center of gravity (G): Stable: G is below COB Unstable: G is above COB 71 Fluid Mechanics-2nd Semester 2010- [2] Fluid Statics

Buoyancy and the Stability of Floating Bodies For floating objects, stability is complicated by the fact that the COB changes with rotation The stability of floating objects (Ship) is determined by the metacentric (M) (the point of intersection of the vertical line through B with the centerline of the ship): Stable: G is below M Unstable: G is above M 72 Fluid Mechanics-2nd Semester 2010- [2] Fluid Statics

Example 16 Solution: 73 Fluid Mechanics-2nd Semester 2010- [2] Fluid Statics

Example 17 74 Fluid Mechanics-2nd Semester 2010- [2] Fluid Statics

Example 17 (Solution) 75 Fluid Mechanics-2nd Semester 2010- [2] Fluid Statics

Example 17 (Solution) 76 Fluid Mechanics-2nd Semester 2010- [2] Fluid Statics

Example 18 77 Fluid Mechanics-2nd Semester 2010- [2] Fluid Statics

Example 18 (Solution) 78 Fluid Mechanics-2nd Semester 2010- [2] Fluid Statics