1. Fluids

2. Introduction The 3 most common states of matter are: –Solid: fixed shape and size (fixed volume) –Liquid: takes the shape of the container and has a fixed volume (incompressible) –Gas: expands to fill its container and volume is not fixed (compressible) Fixed ShapeFixed Volume Solid  LiquidNo  GasNo

3. Introduction (cont.) A fluid is a substance that flows –Liquids and gases are fluids because they do not have a fixed shape Fluids exert pressure on the walls of their containers –Pressure is force per area –Examples: party balloons, garden hose Gas

4. Introduction (cont.) Fluid mechanics is the branch of physics that studies fluids and the forces on them –Fluid statics: study of fluids at rest Hydrostatic pressure Hydraulics Buoyancy –Fluid dynamics: study of fluids in motion Continuity equation Bernoulli equation

5. Introduction (cont.) Fluid mechanics assumes that matter is a continuum (“the continuum hypothesis”) –Fluids are continuous and continuously dividable –The fact that matter is composted of atoms and molecules is ignored –Properties such as density, pressure, temperature and velocity are taken to be well- defined at infinitely small points

6. Introduction (cont.) Fluid mechanics assumes every fluid obeys –conservation of mass –conservation of energy –conservation of momentum –Newton’s laws of motion Fluids have a property called viscosity which is a measure of resistance to flow –Viscosity is related to friction –Syrup has high viscosity, water has low viscosity –An inviscid fluid has zero viscosity

7. Density The density of a solid, liquid, or gas is defined as mass per unit volume: density =  = m/V –  is the Greek letter “rho” –SI unit for density: kg/m 3 –m and V are mass and volume, respectively Another common unit for density is g/cm 3 –1 g/cm 3 = 1000 kg/m 3 The density of liquid water is 1.00 g/cm 3

8. Specific Gravity The specific gravity of a substance is defined as the ratio of the density of that substance to the density of liquid water: SG =  –Example: The density of gold is 19300 kg/m 3 SG of gold is  gold /  H2O = 19300/1000 = 19.3 –Note: SG is a dimensionless quantity –Note: SG of any substance is equal numerically to its density in g/cm 3  substance  liquid water

9. Pressure Pressure is defined as force per unit area: pressure = P = F/A –F is the force acting perpendicular to the surface area A –SI unit for pressure is the pascal (Pa) 1 Pa = 1 N/m 2 –The American unit for pressure is “pounds per square inch” or “psi” 1 psi = 1 lbf/in 2 = 6,900 Pa

10. Pressure (cont.) Example: What is the pressure exerted on the ground by a 60-kg person whose two feet cover an area of 500 cm 2 ? P = F/A = mg/A = (60 kg)(9.8 m/s 2 )/(0.050m 2 ) = 12,000 N/m 2 = 12 kPa If the person stands on one foot, what will the pressure will be? 24 kPa!

11. Atmospheric Pressure Atmospheric pressure (P 0 ) is the pressure of the air around us –P 0 = 101.3 kPa (14.7 psi) at sea level –Atmospheric pressure decreases with increasing altitude (height above sea level) –The cells of living organisms maintain an internal pressure that counter balances atmospheric pressure –A common unit of pressure is the atmosphere (atm): 1.000 atm = 101.3 kPa

12. Gauge Pressure Gauge pressure (P G ) is the pressure over and above atmospheric –Tire gauges read “zero” when open to the air because they read gauge pressure The total or absolute pressure is abs. pressure = P 0 + P G = 1.0 atm + P G For many problems in fluids, only gauge pressure is needed –Decide for each problem whether “P” is gauge or absolute

13. Compressibility of Fluids A gas will readily change volume and density when it experiences a change in pressure –We say gases are compressible fluids Most liquids do not readily change volume and density, even under a wide range of pressures –Such liquids are called incompressible fluids –Incompressible   = constant

14. Fluid Statics (Hydrostatics) Hydrostatics is the study of fluids at rest Most hydrostatic issues take place on Earth  The fluid is in a gravitational field  Usually we are interested in gauge pressure (over and above Earth atmospheric pressure)  Usually we assume the fluid is incompressible (such as a liquid) Because the fluid is at rest, we know the pressure is always perpendicular to container walls

15. Hydrostatic Pressure Earth’s gravitational field causes the pressure inside a fluid to increase with depth –At the surface: P = 0 (gauge) –At a depth of h below the surface we have hydrostatic pressure: P =  gh –This formula assumes  = constant (incompressibility) g h P 

16. Pascal’s Principle If a fluid in a confined space is squeezed, the added pressure shows up everywhere in the fluid This is called Pascal’s principle: Pressure applied to a confined fluid increases the pressure throughout by the same amount Pascal’s principle is the basis of hydraulics F

17. Hydraulic Systems Hydraulic systems use an incompressible fluid under pressure to amplify force Pascal’s principle says P in = P out so: F in /A in = F out /A out  F out /F in = A out /A in F in F out A in A out P in P out

18. Buoyancy Any object partially or totally submerged in a fluid in a gravitational field experiences a buoyancy force upward Archimedes principle states that the buoyancy force on an object is equal to the weight of the displaced fluid: F B =  F gV –  F is the density of the fluid –V is the volume of the object

19. Apparent Weight A dense object submerged in a fluid appears to weigh less –This is called apparent weight, F g –Apparent weight is equal to the weight (F g ) minus the buoyant force (F B ) F g = F g  F B An object is said to be neutrally buoyant if the weight and the buoyant force are equal –In other words, apparent weight is zero If the weight is less than the buoyant force, the object floats FgFg m

20. Fluid Dynamics Fluid dynamics is the study of fluids in motion under the influence of forces Simplifying assumptions: –Fluid is incompressible (  = constant) –Fluid is inviscid (viscosity = 0) –Laminar, steady-state flow We will assume incompressibility for some problems involving air even though air is a gas and thus compressible

21. Continuity Equation We can apply conservation of mass to a fluid flowing through a pipe of changing size: Mass flowing into the left side (per time) equals mass flowing out of the right side:  V 1 /  t =  V 2 /  t  A 1 v 1 =  A 2 v 2 A 1 v 1 = A 2 v 2 (incompressible) A1A1 v1v1 A2A2 v2v2

22. Continuity (Example) Water flows at a velocity 14 cm/s out of a circular pipe of diameter 1.5 cm into a rectangular trough 3.3 cm wide. If the velocity of the water in the trough is 6.2 cm/s, what is the water level in the trough? A 1 =  d 2 /4 =  (1.5cm) 2 /4 = 1.76 cm 2 A 1 v 1 = A 2 v 2 = whv 2 h = A 1 v 1 /wv 2 = {(1.76 cm 2 )(14 cm/s)} / {(3.3 cm)(6.2 cm/s)} = 1.2 cm h = ? v2v2 v1v1 d

23. Bernoulli Equation Bernoulli’s principle states that where the velocity of a fluid is high, the pressure is low, and where the velocity is low, the pressure is high Consider an inviscid, incompressible fluid in steady-state, laminar flow: –Pressure (P) may change from point to point –Velocity (v) may change from point to point –Elevation (y) above a reference level may change from point to point

24. Bernoulli Equation (cont.) Bernoulli’s equation is a statement of conservation of energy between two points: P 1 + ½(  v 1 2 ) +  gy 1 = P 2 + ½(  v 2 2 ) +  gy 2 –P  elastic potential energy (per volume) –½(  v 2 )  kinetic energy (per volume) –  gy  gravitational potential energy (per volume) P1P1 P2P2 v1v1 v2v2 y1y1 y2y2

25. Bernoulli Eqn. (Example) Water flows from a large water main (20.0 cm diameter) under the street into a house and up 6.50 m of elevation to a 0.750-cm diameter faucet on the second floor. The velocity of the water is 1.10 m/s as it comes out of the faucet. What is the pressure in the main? Main: d 1 = 20.0 cm, v 1  0 (continuity says its 0.0015 m/s) y 1 = 0.00 m (ref. level), P 1 = ? Faucet: d 2 = 0.750 cm, v 2 = 1.10 m/s, y 2 = 6.50 m, P 2 = 0 (gauge pressure is zero in atmospheric air) P 1 + ½(  v 1 2 ) +  gy 1 = P 2 + ½(  v 2 2 ) +  gy 2 P 1 = ½(  v 2 2 ) +  gy 2 =  [½  v 2 2 + gy 2 ] =(1000 kg/m 3 )[½(1.10 m/s) 2 + (9.80 m/s 2 )(6.50m)] = 6.43  10 4 N/m 2 = 64.3 kPa