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Introduction to Fluid Mechanics

CEVE 101. Introduction to Fluid Mechanics . Dr. Bedient Civil and Environmental Engineering. Fluids: Statics vs Dynamics. Atmospheric Pressure. Pressure = Force per Unit Area

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Introduction to Fluid Mechanics

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  1. CEVE 101 Introduction to Fluid Mechanics Dr. Bedient Civil and Environmental Engineering

  2. Fluids:Statics vs Dynamics

  3. Atmospheric Pressure Pressure = Force per Unit Area Atmospheric Pressure is the weight of the column of air above a unit area. For example, the atmospheric pressure felt by a man is the weight of the column of air above his body divided by the area the air is resting on P = (Weight of column)/(Area of base) Standard Atmospheric Pressure: 1 atmosphere (atm) 14.7 lbs/in2 (psi) 760 Torr (mm Hg) 1013.25 millibars = 101.3 kPascals 1kPa = 1Nt/m2

  4. Fluid Statics Basic Principles: • Fluid is at rest : no shear forces • Pressure is the only force acting • What are the forces acting on the block? • Air pressure on the surface - neglect • Weight of the water above the block • Pressure only a function of depth

  5. Units SI - International System Length Meter Time Sec Mass Kg Temp 0K = 0C + 273.15 Force Newton = Nt = 1 kg m / s2 Gravity 9.81 m/s2 Work = Fxd Joule = Nt-m Power = F/t Watt = Joule/sec

  6. Units English Length in Ft Time in Sec lbm (slug) - 1 slug = 32.2 lbm Force - lb Gravity - 32.2 ft/sec2 Work = slug-ft/s2

  7. Properties of Fluids • Density = r(decreases with rise in T) • mass per unit volume ( lbs/ft3 or kg/m3 ) for water density = 1.94 slugs/ft3 or 1000 kg/m3 Specific Weight = g (Heaviness of fluid) • weight per unit volume g = rg for water spec wt = 62.4 lbs/ft3 or 9.81 kN/m3 Specific Gravity = SG • Ratio of the density of a fluid to the density of water SG = rf / rw SG of Hg = 13.55

  8. Ideal Gas Law relates pressure to Temp for a gas P = rRT T in 0K units R = 287 Joule / Kg-0K Pressure Force per unit area: lbs/in2 (psi), N/m2, mm Hg, mbar or atm 1 Nt/m2 = Pascal = Pa Std Atm P = 14.7 psi = 101.33 kPa = 1013 mb Viscosityfluid deforms when acted on by shear stress m = 1.12 x 10-3 N-s/m2 Surface tension - forces between 2 liquids or gas and liquid - droplets on a windshield.

  9. Section 1: Pressure Pressure at any point in a static fluid not fcn of x,y,or z Pressure in vertical only depends on g of the fluid P = gh + Po Gage pressure: relative to atmospheric pressure: P = gh Thus for h = 10 ft, P = 10(62.4) = 624 psf This becomes 624/144 = 4.33 psi P = 14.7 psi corresponds to 34 ft 10 ft

  10. Pressure in a Tank Filled with Gasoline and Water What is the pressure at point A? At point B? gG = 42.43 lbs/ft3 SG = 0.68 gW = 62.4 lbs/ft3 At point A: PA = gG x hG + PO = 42.43 x 10 + PO 424.3 lbs/ft2 gage At point B: PB = PA + gW x hW = 424.3 + 62.4 x 3 611.5 lbs/ft2 gage Converting PB to psi: (611.5 lbs / ft2) / (144 in2/ft2) = 4.25 psi

  11. Measurement of Pressure Barometer (Hg) - Toricelli 1644 Piezometer Tube U-Tube Manometer - between two points Aneroid barometer - based on spring deformation Pressure transducer - most advanced

  12. Manometers - measure DP • Rules of thumb: • When evaluating, start from the known    pressure end and work towards the    unknown end • At equal elevations, pressure is    constant in the SAME fluid • When moving down a monometer,    pressure increases • When moving up a monometer,    pressure decreases • Only include atmospheric pressure on    open ends

  13. Manometers Simple Example: P = gx h + PO Find the pressure at point A in this open u-tube monometer with an atmospheric pressure Po PD = gWx hE-D + Po Pc = PD PB = PC - gHgx hC-B PA = PB

  14. Section 2: Hydrostatics And the Hoover Dam For a fluid at rest, pressure increases linearly with depth. As a consequence, large forces can develop on plane and curved surfaces. The water behind the Hoover dam, on the Colorado river, is approximately 715 feet deep and at this depth the pressure is 310 psi. To withstand the large pressure forces on the face of the dam, its thickness varies from 45 feet at the top to 660 feet at the base.

  15. Hydrostatic Force on a Plane Surface Basic Concepts and Naming Pressure = g h g = spec gravity of water h = depth of water C = Center of Mass of Gate CP = Center of Pressure on Gate Fr = Resultant Force acts at CP γh

  16. Hydrostatic Force on a Plane Surface Basic Concepts and Naming C = Centroid or Center of Mass CP = Center of Pressure Fr = Resultant Force I = Moment of Inertia For a Rectangular Gate: Ixc = 1/12 bh3 Ixyc = 0 For a circle: Ixc = p r4 / 4 Ixyc = 0 γh

  17. Hydrostatic Force on a Plane Surface The Center of Pressure YR lies below the centroid - since pressure increases with depth FR = g A YC sinq or FR = g A Hc YR = (Ixc / YcA) + Yc XR = (Ixyc / YcA) + Xc but for a rectangle or circle:XR = Xc For 90 degree walls: FR = g A Hc

  18. Hydrostatics Example Problem # 1 What is the Magnitude and Location of the Resultant force of water on the door? gW = 62.4 lbs/ft3 Water Depth = 6 feet Door Height = 4 feet Door Width = 3 feet

  19. Hydrostatics Example Problem #1 Important variables: HC and Yc = 4’ Xc = 1.5’ A = 4’ x 3’ = 12’ Ixc = (1/12)bh3 = (1/12)x3x43 =16 ft4 Magnitude of Resultant Force: FR = gW A HC FR = 62.4 x 12 x 4 = 2995.2 lbs Location of Force: YR = (Ixc / YcA) + Yc YR = (16 / 4x12) + 4 = 4.333 ft down XR = Xc (symmetry) = 1.5 ft from the corner of the door

  20. Section 3: Buoyancy Archimedes Principle: Will it Float? The upward vertical force felt by a submerged, or partially submerged, body is known as the buoyancy force. It is equal to the weight of the fluid displaced by the submerged portion of the body. The buoyancy force acts through the centroid of the displaced volume, known as the   center of buoyancy. A body will sink until the buoyancy force is equal to the weight of the body. FB = g x Vdisplaced FB = gW x Vdisp = Vdisp FB W =FB FB

  21. Buoyancy Example Problem # 1 A 500 lb buoy, with a 2 ft radius is tethered to the bed of a lake. What is the tensile force T in the cable? FB gW = 62.4 lbs/ft3

  22. Buoyancy Example Problem # 1 Displaced Volume of Water: Vdisp-W = 4/3 x p x R3 Vdisp-W = 33.51 ft3 Buoyancy Force: FB = gW x Vdisp-w FB = 62.4 x 33.51 FB = 2091.024 lbs up Sum of the Forces: SFy = 0 = 500 - 2091.024 + T T = 1591.024 lbs down

  23. Will It Float? Ship Specifications: Weight = 300 million pounds Dimensions = 100’ wide by 150’ tall by 800’ long Given Information: gW = 62.4 lbs/ft3

  24. Assume Full Submersion: FB = Vol x gW FB = (100’ x 150’ x 800’) x 62.4 lbs/ft3 FB = 748,800,000 lbs Weight of Boat = 300,000,000 lbs The Force of Buoyancy is greater than the Weight of the Boat meaning the Boat will float! How much of the boat will be submerged? Assume weight = Displaced Volume WB = FB 300,000,000 = (100’ x H’ x 800’) x 62.4 lbs/ft3 H = Submersion depth = 60.1 feet

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