1. CHAPTER 9 Solids and Fluids Fluid Mechanics (only) pages 261-277

2. Fluids and Fluid Mechanics Density: Density:mass of substance divided by its volume  = M V M = kg V = m 3  = kg/m 3 Specific Gravity: (sp.gr.) The ratio of a substances density to the density of water at 4°C (which is 1.0x10 3 kg/m 3 ) The specific gravity of a substance is: 1. Dimensionless 2. Magnitude differs from the substances density by a factor of 10 3 Pressure: Pressure:A force applied to an area P = F/A F = N A = m 2 P = N/m 2 = Pascal (Pa)

3. Example Problem Example Problem (Density/Sp.gr./Pressure) A lead cube with a side of.250m rests on the floor. Calculate the mass of the cube, the force exerted on the floor, and the pressure on the floor beneath the cube (  lead = 11.3x10 3 kg/m 3 )  = M/V M =  v M = (11.3 x 10 3 kg/m 3 )(.250m x.250m x.250m) M = 177 kg F = mg F = (177 kg)(9.8m/s 2 ) F = 1730 N P = F/A P = (1730 N) / (.250m x.250m) P = 27,700 N/m 2 P = 27,700 pascals NOTE: 1 atmosphere of pressure = 101,300 pascals

4. Pressure In/Under Fluids P o = atmospheric pressure above the water P = pressure in the water at a point under the shaded block of water If the shaded block of water is not falling or rising, then F Net,y = 0 Mg + P o A = PA M =  V  Vg + P o A = PA  Vg + P o = P A  hg + P o = P P = P o +  gh Static head pressure

5. Example Problem Example Problem (Head Pressure) A water tower sits on top of a 50.m platform in the middle of the city (which is on level ground). The tower’s water tank is 10.m high. Assuming the tank is full of water, what is the pressure of the bottom of the tank and what is the pressure in the water pipe at grade level below the tower. Sketch P A = P o +  gh P A = 101.3kpa + (1.0x10 3 kg/m 3 )(9.8m/s 2 )(10.m) P A = 101.3kpa + 98,000N/m 2 1 N/m 2 = 1 pa P A = 101.3kpa + 98.0kpa P A = 199.3kpa P B = P A + (1.0x10 3 kg/m 3 )(9.8m/s 2 )(50.m) P B = 689.32kpa NOTE: 101.3kpa = 1.00atm = 14.7psi = 760mmHg 10.m B A o 50.m To water users

6. Example Problem Example Problem (Barometer) How high will liquid Mercury rise in a barometer tube when the air pressure is 1.00atm? [note:  Hg = 13.6x10 3 kg/m 3 ] Analysis Pressure at bottom of rising mercury column is equal to P o = 1.00atm. P o is forcing the liquid up. When in equilibrium, the static head pressure of the column plus the pressure above the column (0.00atm) equals P o. P o =  gh +P 1.00atm = 101,300N/m 2 101,300N/m 2 = (13.6x10 3 kg/m 3 )(9.80m/s 2 )h h =.760mHg h = 760mmHg 760mmHg 1 cm 1 inch 10mm 2.54cm h = 29.9 inches Hg As in “the barometric pressure is at 29.2inches and rising.”

7. Buoyant Forces Archimedes’s Principle: Any body completely or partially submerged in a fluid is buoyed up by a force whose magnitude is equal to the weight of the displaced fluid. (What else did Archimedes say 2200 years ago that is still repeated today?) P 1 A = F 1 (Downward Force) P 2 A = F 2 (Upward Force) F Net,up = F Buoyant = F 2 – F 1 = P 2 A – P 1 A = A (P 2 – P 1 ) = A (  gh 2 –  gh 1 ) = A  g (h 2 – h 1 ) = A  g  h = V  g = Mg F Buoyant = Mg M = mass of displaced fluid Fluid 1 2 A hh

8. Example Problem Example Problem (Floating Boat) A particular boat could displace 4.50m 3 of water before water would pour over the sides of the boat. The boat itself weighs 35,000N when filled with gasoline and other provisions. If each fisherman wishing to go out in the boat has a mass of 100.kg, how many fisherman can the boat hold? Analysis At maximum load the F Buoyant = W Boat + W Men F Buoyant =  Vg = (1.0x10 3 kg/m 3 )(4.50m 3 )(9.80m/s 2 ) = 44,100N 44,100N= 35,000N + N(100.kg)(9.80m/s 2 ) N = 9.29 Fisherman N = 9 Fisherman

9. Example Problem Example Problem (Sunken Treasure) A box containing.250m 3 of gold bars has been located at the bottom of the sea. How much force will it require to lift the box upward through the water? (Neglect the mass of the box.) FgFg  gold = 19.3x10 3 kg/m 3 V =.250m 3 FTFT F Buoyant  water = 1.0x10 3 kg/m 3 Analysis If lifting at constant velocity, F up = F down F T + F Buoyant = F g F T + (  water )(V gold )(g) = (  gold )(V gold )(g) F T = 44,800 N Notice: F T = (  gold –  water ) Vg F T =  Vg

10. Continuity Equations for Flowing Fluids Analysis Mass passing point 1 each second must equal mass passing point 2 each second (at steady-state).  v 1 A 1 =  v 2 A 2 v 1 A 1 = v 2 A 2 Continuity Equation v1v1 A1A1 A2A2 v2v2 1 2

11. Bernoulli’s Equation for Flowing Fluids Bernoulli’s Equation is really a conservation of energy equation and pressure x volume has units of energy (N/m 2 · m 3 = N · m = Joule). Kinetic and Potential Energy are also conserved. P 1 V + ½ Mv 1 2 + Mgy 1 = P 2 V + ½ Mv 2 2 + Mgy 2 V V P 1 + ½ (M/V) v 1 2 + (M/V) gy 1 = P 2 + ½ (M/V) v 2 2 + (M/V) gy 2 P 1 + ½  v 1 2 +  gy 1 = P 2 + ½  v 2 2 +  gy 2 P + ½  v 2 +  gy = Constant Energy in a flowing fluid can be transferred pressure velocity elevation

12. Bernoulli’s Equation w/Closed Systems v2v2 v3v3 A1A1 A2A2 1 2 3 A1A1 v1v1 At constant elevation: P 1 + ½  v 1 2 = P 2 + ½  v 2 2 = P 3 + ½  v 3 2 = constant v 2 > v 1 v 3 < v 1 P 2 < P 1 P 3 > P 1

13. Flight v 1 > v 2 P 2 > P 1  p = ½  v 1 2 – ½  v 2 2  p · Area (wing) =  F  F = F Net F Net is in upward direction vv1v1 v2v2