PHYSICS OF FLUIDS

1. PHYSICS OF FLUIDS

2. Fluids • Includes liquids and gases • Liquid has no fixed shape but nearly fixed volume • Gas has neither fixed shape or volume • Both can flow

3. Density Mass per unit volume r = m/V (Greek letter “rho”) m = rV Example: Density of Mercury is 13.6 x 103 kg/m3 What is the mass of one liter? M = rV = 13.6 x 103 kg/m3 x 10-3 m3 = 13.6 kg Specific Gravity is ratio of its density to that of water (1.00 x 103 kg/m3 =1.00 g/cm3)

4. Density of water • 1000 Kg per m3 • 1 Kg/liter • 1 gram /cubic centimeter (cc) • 1 cubic meter = 1000 liters • 1 cubic centimeter = 1 milliliter • 1 cubic meter = 1,000,000 cc

5. Table of Densities

6. Pressure in Fluids • Force per unit area • Pressure = P = F/A • Unit N/m2= Pascal(Pa) • Exerted in all directions • Force due to pressure is perpendicular to surface in a fluid at rest

7. rgh Pressure Varies with Depth as • Let depth be h. Assume incompressible. • Force acting on area is mg = rVg = rAhg • P = F/A = rgh • Pressure at equal depths is the same • If external pressure is also present it must be added DP = rg Dh h A

8. Example: Pressure at Bottom of a Lake • What is thepressure at the bottom of a 20.0 meter deep lake? P = rgh = 1.0 x 103 kg/m3 x 9.8 m/s2 x 20m = 1.96 x 105 N/m2 due to the water What about the atmosphere pressing down on the lake?

9. Atmospheric, Gauge and Absolute Pressure • Average sea level pressure is 1 atm = 1.013 x 105 N/m2 (14.7 lbs/sq inch) • Pressure gauges read pressure above atmospheric • Absolute (total) pressure is gauge + atmospheric P = PA + PG What is total pressure at bottom of lake? Add 1.01 x 105 N/m3 to 1.96 x 105

10. Example: Water in a Straw • Finger holds water in straw • How does pressure above water compare with atmospheric? (hint: atmospheric pushes up from below) Pressure less because it plus weight of water must balance atmospheric

11. What is the tallest column of water that could be trapped like this? • rgH = 101,300 Pa; H = 101,300/(9.8 N/kg x 1000 Kg/m3) =10.3 m

12. Pascal’s Principle • Probably not tested • Pressure applied to a confined fluid increases pressure throughout by the same amount • Pout = Pin • Fout/Aout = Fin/Ain • Fout/Fin = Aout/Ain • Multiplies force by ratio of areas • Principle of hydraulic jack and lift Diagram courtesy Caduceus MCAT Review

13. Pascal’s Principle in Action

14. Measuring Pressure • Open tube manometer simulation The pressure difference is rgh The (greater) pressure P2 = P1 + rgh How would this look if P1 was greater than P2 ? Diagram courtesy Sensorsmag.com

15. Buoyancy • Submerged or partly submerged object experiences an upward force called buoyancy • Pressure in fluid increases with depth • Studied by Archimedes over 2000 years ago.

16. Buoyant Force on Cylinder h = h2 – h1 • FB = F2 – F1 = P2A – P1A = rFgA(h2 – h1) = rFgAh = rFgV = mFg • Buoyant Force equals weight of fluid displaced h1 F1 A h2 h F2

17. Archimedes Principle • Buoyant force on a body immersed (or partly immersed) in fluid equals weight of fluid displaced. • Argument in general: consider immersed body in equilibrium of any shape with same density as fluid. FBup must equal weight down. Replacing body by one with different density does not alter configuration of fluid so conclusion would not change.

18. Weighing Submerged object Sfy =0 T +B - W = 0 T = W – B T = W – rFVg T is apparent weight W’ Diagram courtesy Caduceus MCAT Review

19. Example: King’s Crown • Given crown mass 14.7 kg but weighed under water only 13.4 kg. Is it gold? • W’ = W – FB W = rogV • W-W’= FB = rF gV • W/(W-W’) = rogV / rF gV = ro / rF ro / rH2O= W/(W-W’) = 14.7/(14.7 – 13.4) = 14.7/1.3 = 11.3 LEAD Another way to solve: isolatero = W/gV and then get V from FB /rfg

20. Floating Objects Objects float if density less than that of fluid. FB = W at equilibrium rFVdisp g = ro Vo g Vdisp / Vo = ro /rF Vo Vdisp Example: If an object’s density is 80% of the density of the surrounding fluid, 80% of it will be submerged

21. Example: Floating Log • 15 % of a log floats above the surface of the ocean. What is the density of the wood? Vdisp / Vo = ro /rF ro =Vdisp / Vo x rF = 0.85 x 1.025 x 103 kg/m3 = 0.87125=0.87 x 103 kg/m3

22. Example: Lifted by Balloon • What volume of helium is needed to lift a 60 Kg student? FB= (mHe + 60 kg)g rair Vg = (rHe V + 60 kg) g V = 60 kg/(rair – rHe) = 60 kg/(1.29 – 0.18kg/m3) = 54 m3

23. Fluid Flow • Equation of Continuity • Volume rate of flow is constant for incompressible fluids (not turbulent) • A1v1 = A2 v2 • v is velocity

24. Laminar vs. Turbulent Flow Erratic, contains eddies Fluid follows smooth path Courtesy MIT Media Laboratory

25. Example: Narrows in a River • A river narrows from 1000m wide to 100m wide with the depth staying constant. The river flows at 1.0 m/s when wide. How fast must it flow when narrow? 10 m/s

26. Example: Heating Duct • What must be the cross sectional area of a heating duct carrying air at 3.0 m/s to change the air in a 300 m3 room every 15 minutes? • A1v1 = A2v2 = A2l2/t = V2/t • A1 = V2/ v1t = 300m3 /(3.0 m/s x 900s) = 0.11m2 A2 A1 l2

27. Bernoulli’s Equation • Where velocity of fluid is high, pressure is low; where velocity is low, pressure is high • Consequence of energy conservation • P + ½ rv2 + rgy = constant for all points in the flow of a fluid • P + ½ rv2 = constantifall on same level

28. Question If A1 is six times A2 how will the pressure in the narrow section compare with than in the wide section? Hint: P + ½ rv2 = constant

29. Speed of Water Flowing Through Hole in Bucket P1 = P2 + ½ rv2 + rgz P1 = P2 since both open to air ½ rv22 + rgz = 0 v2 =(2gz)1/2 Torricelli’s Theorem

30. Speed and Pressure In Hot Water Heating System If water pumped at 0.50 m/s through 4.0 cm diameter pipe in basement under 3.0 atm pressure, what will be flow speed and pressure in 2.6 cm diameter pipe 5.0m above? 1) find flow speed using continuity A1v1 = A2 v2 v2 = v1A1/A2 = v1pr12/pr22 = 1.2 m/s

31. 2) Use Bernoulli’s Eq. to find pressure P1 + ½ rv12 + rgy1 = P2 + ½ rv22 + rgy2 P2 = P1 +rg(y1 – y2) + 1/2r(v12 –v22) =(3.0 x 105 N/m2) + (1.0 x 103 kg/m3)x (9.8 m/s2)(-5.0m) + ½ (1.0 x 103 kg/m3)[(0.50 m/s)2 – (1.18m/s)2] = = 2.5 x 105 N/m2

32. No Change in Height • P1 + ½ rv12 = P2 + ½ rv22 • Where speed is high, pressure is low • Where speed is low, pressure is high

33. Why Curveballs Curve Courtesy Boston University Physics Dept. web site

34. How an Airfoil Provides Lift Where is the pressure greater, less? Courtesy The Aviation Group

35. Crowding of streamlines indicates air speed is greater above wing than below Courtesy http://www.monmouth.com/~jsd/how/htm/airfoils.html

36. Lift Illustrated Courtesy NASA and TRW, Inc.

37. Sailing Against the Wind Sails are airfoils Low pressure between sails helps drive boat forward Courtesy Dave Culp Speed Sailing

38. Venturi Tube Courtesy http://www.abdn.ac.uk/physics/streamb/fin13www/sld001.htm

39. Bernoulli’s Principle also • Helps explain why smoke rises up a chimney (air moving across top) • Explains how air flows in underground burrows (speed of air flow across entrances is slightly different) • Explains how perfume atomizer works • Explains how carburetor works

40. Giancoli Website