Fluids

1. Fluids AP Physics B Chapter 10 Notes

2. Back to Basics—Phases of Matter • Three basic states of matter • Solid—maintains a fixed shape and size • Liquid—takes on shape of container, not really compressible • Gas—has neither a fixed shape nor volume Note: Plasma is a fourth state and is ionized atoms

3. Back to Basics—Phases of Matter • Fluids consist of liquids and gasses—both have the ability to flow

4. Fluids Can Be Really Useful • Liquids are (nearly) incompressible so pressure is transmitted through out the fluid—pushing one end is like pushing directly on an object at the other end

5. Density and Specific Gravity • Density, ρ, of a substance is defined as its mass per unit volume: • Units are kg/m3 • Specific to substance (see table on p. 256) • Specific gravity is the ratio of a substance’s density to the density of water at 4.0 ˚C (1000 kg/m3)

6. Pressure in Fluids • Pressure is force per unit area: P = F/A • SI unit is Pa (Pascal) = N/m2 • Fluids exert equal pressure in all directions at a given depth • It always acts perpendicular to the surface it contacts—there is no pressure parallel to the surface

7. Pressure in Fluids • Pressure at a depth h is due to the weight of the fluid above it • Force of water is: F=mg=(ρV)g =ρAhg • Since P=F/A: P= ρAhg /A P = ρgh • Good only for incompressible fluids (liquids not gasses) • And ∆P = ρg∆h

8. Example Pressure Problem Example 10-3 pg. 258: A water tank is 30 m above the faucet in a house. Find the difference in water pressure between the tank and the faucet.

9. Atmospheric and Gauge Pressure • At sea level air pressure is 1.013 x 105 N/m2 • This is 1 atm = 101.3 kPa = 14.7 psi • Pressure gauges measure pressure above atmospheric pressure so: P = PA + PG (PG = ρgh) • If car tire pressure gauge says 32 psi, total pressure is 46.7 psi

10. Pressure in Fluids—Pascal’s Principle • Pascal determined that if an external pressure is applied to a confined fluid, the pressure at every point within the fluid increases by that amount • So Pin = Pout or (F/A) in = (F/A) out • This leads to mechanical advantage: F out /Fin = Aout/A in • Applications

11. Pressure Measurement • Manometer • Tire gauge • Table 10-2 on pg. 261 has conversion factors

12. Barometric Pressure Measurement • Open tube • Modern

13. Buoyancy • Submerged objects appear to weigh less due to buoyancy • Hydrostatic pressure depends upon depth • Note length of force vectors—pressure is greater on the bottom than top of objects

14. Buoyancy • Buoyant force is: FB = F2 – F1 But F = PA and P = ρFgh so… FB = ρFgA∆h FB = ρFVg FB = mFg Eureka!

15. Buoyancy—Archimedes’ Principle • The buoyant force on an object immersed in a fluid is equal to the weight of the fluid displaced by the object • Fluid displaced is volume (VF = VO) • General proof using Newton

16. Buoyancy—Archimedes’ Principle • What happens when an object is submerged • The displaced fluid is equal in weight to the FB • New scale reading is: FTꞌ= mg - FB FT FT’ FB mg mg W=FT – FTꞌ

17. Buoyancy—Example Problem • P. 24 pg. 282: A crane lifts the 18,000 kg steel hull of a ship out of the water. Determine tension in the crane’s cable when a) the hull is submerged and b) when the hull is completely out of the water.

18. Buoyancy—The Case of Floating • What happens when an object floats • Archimedes’ Principle still holds • Objects float when their density is < water FB = ρFVg

19. Buoyancy—The Case of Floating • For a floating object, the fraction that is submerged is given by the ratio of the object’s density to that of the fluid. ρO/ρF = VDispl/VO

20. Buoyancy—Another Example • P 77 pg. 285: A Cu weight is placed on top of a 0.5kg block of wood (dens=0.60x103 kg/m3). What is the mass of Cu if the top of the wood block is exactly at the water’s surface?

21. Fluid Dynamics--Continuity • Ideal Fluid • Non-viscous (no internal friction) • Incompressible (constant ρ) • Flows smoothly (non-turbulent--laminar)

22. Fluid Dynamics--Continuity • Consider laminar flow in a pipe Mass flow rate = ∆m/∆t ∆m1/∆t = ρ1∆V1 /∆t = ρ1A1∆l1 /∆t = ρ1A1v1 Flow through A1 and A2 must be the same ∆m1/∆t = ∆m2/∆t and ρ1A1v1 = ρ2A2v2

23. Fluid Dynamics--Continuity • For most cases, density doesn’t change so A1v1 = A2v2 • The product Av is the volume flow rate since ∆V/∆t = A∆l/∆t = Av Small A, big v and big A small v

24. Continuity—Example Problem • Ex. 10-12 pg. 269 What area must a heating duct have if air moving 3 m/s can replenish the air every 15 min. in a room of 300 m3?

25. Continuity—More Complex • P. 83 pg. 285 Four lawn sprinklers are fed by a 1.9 cm diameter pipe. Water comes out at 9.134 m/s. a) If the output diameter of each head is 3 mm, how many liters of water do the four heads deliver per second? b) How fast is water flowing in the pipe?

26. Bernoulli’s Principle • The Swiss physicist discovered that when the velocity of a fluid is high, the pressure is low (2) and when velocity is low, pressure is high (1) 2 1 2 1

27. Bernoulli’s Equation • Consider the work done on a volume of fluid flowing a distance ∆l1 at point 1 W1 = F1∆l1 = P1A1∆l1 • At point 2 the fluid does work back on the fluid moving forward: W2 = -F2∆l2 = -P2A2∆l2

28. Bernoulli’s Equation • At the same time gravity is doing work as a mass of fluid moves from point 1 to 2 (y1 to y2) W3 = -mg(y2 -y1) Total Work: W = W1 + W2 + W3 W = P1A1∆l1 - P2A2∆l2 - mg(y2 -y1)

29. Bernoulli’s Equation • By work energy principle W = ∆KE so 1/2mv12 + 1/2mv22 = P1A1∆l1 - P2A2∆l2 - mg(y2 -y1) • Substitute m = ρA∆land divide each side by A∆l P1 + 1/2 ρ v12 + ρgy1=P2 + 1/2 ρ v22 + ρgy2 • Generally: P + 1/2 ρ v2 + ρgy= constant

30. Bernoulli’s Equation--Applications A person with constricted arteries will find that they may experience a temporary lack of blood to the brain (TIA) as blood speeds up to get past the constriction, thereby reducing the pressure. Air flow across the top helps smoke go up a chimney, and air flow over multiple openings can provide the needed circulation in underground burrows.

31. Bernoulli’s Equation--Applications For fluid flowing in a reservoir with a large surface area so that water level doesn’t really change, the Bernoulli’s equation becomes: 1/2 ρ v12 + ρgy1= ρgy2 v1 = √ 2g(y2 – y1) This is known as Torricelli’s theorem

32. Example Problem • Prob. 46 pg. 283. Water at gauge pressure 3.8 atm flows into a building at 0.60 m/s through a pipe 5 cm diameter. The pipe tapers down to a 2.6 cm pipe 18 m above, where the faucet has been left open. Find the flow velocity and gauge pressure on the top floor.