1. 5.1 Hydrostatics 5.2 Fluid flow 5.3 Pascal Law for pressure 5.4 Archimedean Law 5.5 Continuity equation 5.6 Bernoulli equation. 5.7 Diffusion and endosmosis 5.8 Temperature and thermal expansion 5.9 Calorimetry and heat transfer Chapter 5 FLUIDS AND THERMAL

2. Getting point from the Photo ? Motion of fluids

3. Part 1 Hydrostatics

5. By definition, the fluid is at rest. Or, no there is no relative motion between adjacent particles. No shearing forces is placed on the fluid. There are only pressure forces and no shear. Results in relatively “simple” analysis Note for the pressure variation in the fluid The force per unit area across any surface is normal to the surface and is the same for all orientations of the surface. Fluid Statics

6. Learning Check You are sitting on the area of 1m 2, If you change your seat to area of 0,7m 2, How does the pressure you make to the seat change ?? Increase 1/0.7=1.42

7. Part 2 Fluid flow

8. What do we mean by “fluids”? –Fluids are “substances that flow”…. “substances that take the shape of the container” –Atoms and molecules are free to move.. No long range correlation between positions. What parameters do we use to describe fluids? –Density è LIQUID: incompressible (density almost constant) è GAS: compressible (density depends a lot on pressure)

12. Part 3 Pascal Law for pressure Blaise Pascal (1623-1662)

13. Review Pressure

14. Pressure Pressure is defined as a normal force exerted by a fluid per unit area. Units of pressure are N/m 2, which is called a pascal (Pa). Pa is too small, in practice, kilopascal (1 kPa = 10 3 Pa) and megapascal (1 MPa = 10 6 Pa) are commonly used. Other units include bar, atm, kgf/cm 2, mm Hg.

15. Units Standard atmosphere is defined as the pressure produced by a column of mercury 760 mm in height at 0°C (29.92 in Hg or of water about 10.3 m ) * (rHg = 13,595 kg/m3) under standard gravitational acceleration (g = 9.807 m/s 2 ). 1 atm = 760 torr and 1 torr (mmHg) = 133.3 Pa

16. Pressure Units Other units include bar, atm, kgf/cm 2, mm Hg. 1 bar = 10 5 Pa 1 atm = 101325 Pa = 1.01325 bars 1 kgf/cm 2 = 9.807 N/cm 2 = 9.807  10 4 N/m 2 = 9.807  10 4 Pa = 0.9807 bar = 0.9679 atm 1 atm = 14.696 psi. 1 kgf/cm 2 = 14.223 psi. Mm Hg = 9,8.13,6 =133 Pa

17. Pressure at a Point: Pascal ’ s Law normal force per unit area * Pressure is the normal force per unit area at a given point acting on a given plane within a fluid mass of interest. Pascal’s Law: Pascal’s Law: the pressure at a point in a fluid at rest, or in motion, is independent of the direction as long as there are no shearing stresses present.

18. Measurement of Pressure: Barometers Evangelista Torricelli (1608-1647) The first mercury barometer was constructed in 1643-1644 by Torricelli. He showed that the height of mercury in a column was 1/14 that of a water barometer, due to the fact that mercury is 14 times more dense that water. He also noticed that level of mercury varied from day to day due to weather changes, and that at the top of the column there is a vacuum. Animation of Experiment: Schematic: Note, often p vapor is very small, 0.0000231 psia at 68° F, and p atm is 14.7 psi, thus:

19. Learning Check Torricelli formula The Pressure in a homogenous, incompressible fluid at rest depends on the depth of the fluid relative to some reference and is not influenced by the shape of the container. p = p o p = p 1 p = p 2 Lines of constant Pressure For p 2 =  gh + p o h1h1 For p 1 =  gh 1 + p o Compute: P 1 and P 2 ?? Give P 0 =760 mmHg, h 1 =3cm, h=5cm,  =1500kg/m 3

20. Learning Check Compute: F 2 = ?? Give A 1 / A 2 =3, F 1 =45mmHg

21. Learning test What happens with two different fluids?? Consider a U tube containing liquids of density  1 and  2 as shown: –Compare the densities of the liquids: A)  1 <  2 B)  1 =  2 C)  1 >  2 If we use the same liquids in a U tube of twice the cross-sectional area as the first, compare the distances between the levels in the two cases (depth of liquid 2 same in both cases). A) d I < d II B) d I = d II C) d I > d II I 11 22 dIdI II 11 22 d II

22. Hint C)  1 >  2 At the depth of the interface, the pressures in each side must be equal. Since there’s more liquid above this depth on the left side, that liquid must be less dense! The pressure depends ONLY on the depth and the density of the fluid. e.g. consider case I : B) d I = d II I 11 22 dIdI p p II 11 22 d II

23. Part 4 Archimedean Law

25. Learning Check Calculate D =? If L=10m, H=3m, W=500000 N, g=10 m/s 2  water = 1000 kg/m 3 H

26. Learning Check

27. Example: Submarine Buoyancy and Ballast Submarines use both static and dynamic depth control. Static control uses ballast tanks between the pressure hull and the outer hull. Dynamic control uses the bow and stern planes to generate trim forces.

28. Submarine Buoyancy and Ballast Normal surface trim SSN 711 nose down after accident which damaged fore ballast tanks

29. Submarine Buoyancy and Ballast Damage to SSN 711 (USS San Francisco) after running aground on 8 January 2005.

30. Submarine Buoyancy and Ballast Ballast Control Panel: Important station for controlling depth of submarine

31. Test A lead weight is fastened to a large styrofoam block and the combination floats on water with the water level with the top of the styrofoam block as shown. –If you turn the styrofoam+Pb upside down, what happens? styrofoam Pb A) It sinks C) B)

32. Hint styrofoam Pb C) If the object floats right-side up, then it also must float upside-down. However, when it is upside-down, the Pb displaces some water. Therefore the styrofoam must displace less water than it did when it was right-side up (when the Pb displaced no water).

33. Part 5 Continuity equation

35. Learning Check Write an equation for steady conditions (see ficture below) ???

36. Part 6 Bernoulli equation

38. Static P Dynamic P Hydrostatic P = const Static Pressure, Dynamic Pressure and Hydrostatic Pressure have the same unit

40. Learning Check Hint: Use the dynamic pressure P=F/A=  v 2 /2

41. Learning Check Explain this experiment ??