Chapter 10: Fluids

1. Chapter 10: Fluids • Fluids: substances which flow • Liquids: take the shape of their container but have a definite volume • Gases: take the shape and volume of their container • Pressure in a fluid: force per area p = F/A • Force = normal force, pressure exerts a force perpendicular to the surface. • Units for pressure: • 1 N/m2 = 1 Pa • 1 Bar = 105 Pa ~ atmospheric pressure (14.7 psi) • atmospheric pressure varies from .970 bar to 1.040 bar

2. Example: A flat roof of a house is 10.0 m by 8.0 m, and has a mass of 7500kg. Just before a severe storm the windows of the house are shut so tightly that the air pressure inside remained at 1.013 bars even when the out side pressure fell to 980 mb. Compare the net force on the house due to the difference in pressure to the weight of the roof.

3. Most pressure gages detect pressure differences between the measured pressure and a reference pressure. • absolute pressure: the actual pressure exerted by the fluid. • gauge pressure: the difference between the pressure being measured and atmospheric pressure. • p = pgauge + patm

4. Some important aspects of pressure in a fluid • The forces a fluid at rest exerts on the walls of its container (and visa versa) always perpendicular to the walls. • An external pressure exerted on a fluid is transmitted uniformly throughout the volume of the fluid. • The pressure on a small surface in a fluid is the same regardless of the orientation if the surface.

5. An application of pressure in a fluid: the hydraulic press F2 = pA2 F1 = pA1 p

6. pexternal • Pressure and Depth • A fluid supports itself against its weight with pressure. • The fluid also must support itself against external pressure • p = F/A = pexternal + w • w = mg = rVg V = Ah • p = pexternal + rgh • h • A • p

7. Example: Find the pressure on a scuba diver when she is 15.0 m below the surface. How many atmospheres of pressure is this?

8. Buoyant force: pressure balances gravity for a fluid to support itself. Fnet= w = rVg Fnet= rfluidVg Buoyant force = weight of fluid displaced Fbuoyant = Vrg

9. Example: An iron anchor weighs 200 N in air. How much force is necessary to lift it if it is submerged in water? The densities of iron and water are 7800 Kg/m3 and 1000 Kg/m3 respectively. Question: Since a floating object needs only displace enough water to offset its weight, the displaced volume is less than the total volume of the floating object. Consider a man with a rock in a rowboat . If the man takes the rock and throws it overboard into the pond, will the water level of the pond go up, go down or stay the same?

10. Fluid Flow • with approximations: • incompressible fluid • no viscosity (friction) • laminar flow (a.k.a. streamline flow) • in contrast with turbulent flow • Rate of flow: volume per time V = vtA A A vt

11. If no fluid is added/lost, flow rate must be the same throughout Rin = v1A1 Rout = v2A2 Rin = Rout v1A1 = v2A2 Example 10.8: Water flows at 2.5 m/s through a garden hose whose inside radius is 6.0 mm. (a) What should the nozzle radius be for water to leave it at 10.0 ms? (b) What is the rate of flow of water through the hose? (c) How many cubic meters of water will flow in an hour?

12. p2, v2 A2 p1, v1 h2 A1 h1 • Bernoulli’s Equation: flow with changing heights and pressure • Work-Energy Theorem + incompressible fluid

13. 1.5m Water flows through the pipe as shown at a rate of .015 m3/s. If water enters the lower end of the pipe at 3.0 m/s, what is the pressure difference between the two ends? A2 = 20 cm2

14. h v • Applications of Bernoulli’s Equation • Liquid at rest: • p2-p1 = r g(h1 - h2 ) • No pressure difference, one part “at rest”: • Torricelli’s theorem • typically atmospheric pressure for both

15. A boat strikes an underwater rock and opens a pencil-sized crack 7mm wide and 150mm long in its hull 65 cm below the waterline. The crew takes 5.00 minutes to locate the crack and plug it up. How much water entered the boat during those 5 minutes?