FCCIV HIDRCANA: Channel Hydraulics Flow Mechanics Review Fluid Statics

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1 FCCIV HIDRCANA: Channel Hydraulics Flow Mechanics Review Fluid Statics Civil Engineering Program, San Ignacio de Loyola University
2 Objective Calculate the forces exerted by a fluid at rest on plane or curved submerged surfaces Analyze the stability of floating and submerged bodies
3 Hydrostatic Forces on Submerged Plane Surfaces On a plane surface, the hydrostatic forces form a system of parallel forces, and we often need to determine the magnitude of the force and its point of application, which is called the center of pressure. Hoover Dam.
4 Hydrostatic Forces on Submerged Plane Surfaces P Po gh Po gysin The resultant hydrostatic force: F R PdA ( P0 gysin ) da P0 A g sin A A A But the first moment of area y c A yda yda A F yda is related to the ycoordinate of the centroid: R P gy sin ) A ( P gh ) A ( 0 C 0 C P C A P avg A
5 Hydrostatic Forces on Submerged Plane Surfaces The pressure at the centroid of a surface is equivalent to the average pressure on the surface.
6 Hydrostatic Forces on Submerged Plane Surfaces The resultant force acting on a plane surface is equal to the product of the pressure at the centroid of the surface and the surface area, and its line of action passes through the center of pressure. The moment of result force = moment of distributed pressure force about the xaxis y P F R ypda y( P0 gysin ) da P0 yd A g sin A A A A y PFR P0 yc A g sini xx, O I I xx, O xx, O y A I 2 xx, C da y 2 C A y for P y P P y o y C C 0 I y y y 2 xx, C C C da P A I xx, C o /( g sin ) A
7 The centroid and the centroidal moments of inertia for some common geometries.
8 Special Case: Submerged Rectangular Plate Tilted rectangular plate: F R P C A P O g( s b/ 2) sin ab Tilted rectangular plate (s=0): FR PO g( bsin ) / 2ab
9 Special Case: Submerged Rectangular Plate vertical rectangular plate: FR PO g( s b / 2) ab Vertical rectangular plate (s=0): FR PO gb / 2ab
10 Special Case: Submerged Rectangular Plate Horizontal rectangular plate: FR ( PO gh) ab Hydrostatic force acting on the top surface of a submerged horizontal rectangular plate.
11 Hydrostatic Forces on Submerged Curved Surfaces Horizontal force component on curved surface Vertical force component on curved surface F F H V FX 2 2 FR FH FV F W V tan F F V H
12 Hydrostatic Forces on Submerged Curved Surfaces 1. The horizontal component of the hydrostatic force acting on a curved surface is equal (in both magnitude and the line of action) to the hydrostatic force acting on the vertical projection of the curved surface. 2. The vertical component of the hydrostatic force acting on a curved surface is equal to the hydrostatic force acting on the horizontal projection of the curved surface, plus the weight of the fluid block.
13 Hydrostatic Forces on Submerged Curved Surfaces When a curved surface is above the liquid, the weight of the liquid and the vertical component of the hydrostatic force act in the opposite directions.
14 Hydrostatic Forces on Submerged Curved Surfaces The hydrostatic force acting on a circular surface always passes through the center of the circle since the pressure forces are normal to the surface and they all pass through the center.
15 Hydrostatic Forces on Submerged Curved Surfaces The hydrostatic force on a surface submerged in a multilayered fluid can be determined by considering parts of the surface in different fluids as different surfaces. Plane surface in a multilayered fluid F R FR, i PC, i Ai where: P C, i PO i ghc, i
16 Example 39 Schematic for Example 3 9 and the freebody diagram of the liquid underneath the cylinder.
17 Example 39
18 Example 39
19 Example 39
20 Problem 372 The water side of the wall of a 100mlong dam is a quarter circle with a radius of 10 m. Determine the hydrostatic force on the dam and its line action when the dam is filled to the rim.
21 Buoyancy Buoyant force is the upward force of a fluid exerts on a body immersed in it. The buoyant force acting on the plate is equal to the weight of the liquid displaced by the plate.
22 Archimedes Principle The buoyant force acting on a body immersed in a fluid is equal to the weight of the fluid displaced by the body, and it acts upward through the centroid of the displaced volume.
23 Buoyancy The buoyant forces acting on a solid body submerged in a fluid and on a fluid body of the same shape at the same depth are identical. The buoyant force F B acts upward through the centroid C of the displaced volume and is equal in magnitude to the weight W of the displaced fluid, but is opposite in direction. For a solid of uniform density, its weight W s also acts through the centroid, but its magnitude is not necessarily equal to that of the fluid it displaces. (Here W s > W and thus W s > F B ; this solid body would sink.)
24 Buoyancy For floating bodies, the weight of the entire body must be equal to the buoyant force, which is the weight of the fluid whose volume is equal to the volume of the submerged portion of the floating body: Why A solid body dropped into a fluid will sink, float, or remain at rest at any point in the fluid?
25 Buoyancy The altitude of a hot air balloon is controlled by the temperature difference between the air inside and outside the balloon, since warm air is less dense than cold air. When the balloon is neither rising nor falling, the upward buoyant force exactly balances the downward weight.
26 Example 311
27 Example Seawater (g=10.1 kn/m 3 ) Buoy Spherical buoy has a diameter of 1.5 m, weighs 8.50 kn, and is anchored to the sea floor with a cable as shown. The buoy normally floats on the surface, at other times the water depth increases so that the buoy is completely immersed as shown. What is the tension in the cable? F B W T Cable F 0 y y F B W T F B g g d (10,100 N / m ) (1.5 m) 17, 850 N 6 6 T F W B 17,850 8,500 N 9,350 N Source:
28 Hydrometer Buoyant force F B = weight of the hydrometer must remain constant Hydrometer floats deeper or shallower depending on the specific weight of the fluid Source:
29 Example h A hydrometer weighs N and has a stem at the upper end that is cylindrical and 2.8 mm in diameter. 1 2 How much deeper will it float in oil of S=0.78 than in alcohol of S=0.821? S = S = For position 1: W hydrometer *9810* V V 1 For position 2: W hydrometer W displaced water 2.68x10 W 6 m *9810*( V 0.780*9810*[2.68x10 h m 23.2 mm 3 1 displaced water 1 Ah) 6 (0.0028) 4 h] 2
30 Stability For floating bodies such as ships, stability is an important consideration for safety.
31 Stability An immersed neutrally buoyant body is (a) stable if the center of gravity G is directly below the center of buoyancy B of the body, (b) neutrally stable if G and B are coincident, and (c) unstable if G is directly above B.
32 Variation of the Pressure Differential fluid element
33 Variation of the Pressure Pressure gradient P The only body force acting on the fluid element is the weight of the element acting in the negative zdirection, which is expressed as vector: The total force acting on the element becomes:
34 Variation of the Pressure Substituting into Newton s second law of motion And canceling dx dy dz, the general equation of motion for a fluid that act as rigid body (no shear stress) is given by: F m. a dxdydz. a In scalar form in he three orthogonal direction as:
35 Problem 375E The flow of water from a reservoir is controlled by a 5ftwide Lshaped gate hinged at point A, as shown below. If it is desired that the gate open when the water height is 12 ft, determine the mass of the required weight W.
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