1. Day 63, Wednesday, 2 December, 2015 Fluid Mechanics Problems Quick review Pressure, Pascal’s Principle, Archimedes's principle, Pressure at depth Problems Car tire Tornado vs. door Hydraulic piston Tension on wire from submerged object Pressure at depth 1 Pressure at depth 2

2. Pressure P = F/A Perpendicular to surface In the metric (SI or mks) system the unit of pressure is the Pascal abbreviated Pa 1 Pa = 1N/m 2 1 atm = 101,325 Pa = 101.325 kPa 1 atm = 14.7 lb/in 2 1 atm = 1 bar 1 atm = 1,000 millibar

3. Pressure of a car on the ground A car tire makes contact with the ground on a rectangular area of 12 cm by 18 cm. The car’s mass is 925 kg. What pressure does the car exert on the ground? F = mg, A = 4(L x W) P = F/A = [(925)(9.81)]/[(4)(.12)(.18)] P = 1.05 x 10 5 N/m 2 = 1.05 x 10 5 Pa

4. Tornado pulls on a door In a tornado the pressure can be 15% below normal atmospheric pressure. Sometimes the tornado can move so quickly this pressure drop can occur in 1 second. Suppose a tornado occurred outside your front door, which is 182 cm high and 91 cm wide. What net force would be exerted on the front door? In what direction would the force be exerted?

5. Tornado on door F net = F outside – F inside P = F/A ⇒ F = PA ⇒ F net =  PA F net = (P inside – P outside ) A F net = (1.00 x 10 5 –.85 x 10 5 ) (1.82)(.91) F net = 2.5 x 10 4 N (towards the outside) (equivalent to 2 ½ tons of mass on the door)

6. Pascal's Principle Pressure is transmitted equally to every point in the fluid and to the walls of the container.

7. Hydraulic Press

8. Force on a piston to lift an Engine In a machine shop a hydraulic lift is used to raise heavy equipment for repairs. The system has a small piston with a cross sectional area of 7.0 x 10 -2 m 2 and a large piston with a cross sectional area of 2.1 x 10 -1 m 2. An engine weighing 2.7 x 10 3 N rests on the large piston. What force must be applied to the small cylinder in order to lift the engine?

9. Lifting an engine P 1 = P 2 F 1 / A 1 = F 2 / A 2 F 1 = F 2 A 1 / A 2 F 1 = (2.7 x 10 3 N)(7.0 X 10 -2 m 2 ) / 2.1 X 10 -1 m 2 F 1 = 9.0 x 10 2 N

10. Archimedes's principle F b = ρ v g Density of a fluid is rho or ρ The magnitude of the buoyant force is equal to the weight of the displaced fluid

15. (below the waterline)

18. Tension on wire supporting submerged object What is the tension in a wire supporting a 1250 N camera submerged in water? The volume of the camera is 8.3 x 10 -2 m 3. F tension + F buoyant = F gravity where F gravity is the air weight of the camera F T = F g – F b = F g – ρ v g F T = 1250N – (1000)(.083)(9.81) F T = 4.36 x 10 2 N

19. Pressure at depth P = F/A = mg/A P = ρ g h

20. An aluminium scuba tank is rated to 3,000 psi. How deep can you dive with this tank? P = ρ g h  h = P/ ρ g P = 3,000 psi lb/in 2 (kg/lb)(in/m)(in/m) = kg/m 2 Kg/m 2 (m/s 2 ) = N/m 2 = Pa 3,000 (1/2.2)(39.4/1)(39.4/1) = 2.11 x 10 6 2.11 x 10 6 (9.81) = 2.077 x 10 7 Pa ρ = 1,000 kg/m 3 g = 9.8 m/s 2

21. h = 2.077x 10 7 /[(1,000)(9.8)] H = 2117 m = 2000 m Checks At 10 m/atm that is 200 atm 3,000 psi/(15 psi/atm) = 200 atm

22. What pressure must a submarine withstand? 300 m submarine operating depth P = ρ g h = (1,000)(9.8)(300) = 2.9 x 10 6 Pa 2.9 x 10 6 / 1.01 x 10 5 = 29 atm (atm)(psi/atm) = psi 29 (15) = 435 psi

23. Credits Hydraulic Press picture hyperphysics.phy- astr.gsu.edu/.../hpress.gif Archimedes's graphics www.aeic.alaska.edu/.../index.html Archimedes picture www.daviddarling.info/images/Archimedes_princ...

24. Credits II Physics problems Physics: Principles and Problems published by Glencoe, 2002 ISBN 0-07-825934-7 Barn door photo http://www.wrh.noaa.gov/images/otx/phot o_gallery/spokane_tornado/P5220353.JPG