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Outline. Review Pressure and Density. Begin Buoyant Forces. Continuity Equation. Bernoulli’s Principle. CAPA and Examples. Pressure and Density. Density = Mass/Volume A property of the material. Pressure = Force/Area Depends on the height of the fluid. Same in all directions.

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**Outline**• Review Pressure and Density. • Begin • Buoyant Forces. • Continuity Equation. • Bernoulli’s Principle. • CAPA and Examples.**Pressure and Density**• Density = Mass/Volume • A property of the material. • Pressure = Force/Area • Depends on the height of the fluid. • Same in all directions. • Units are: • Force/Area = N/m2. • Pascals 1 Pa = 1 N/m2. • Atmosphere 1 atm = 1.013 X 105 N/m2.**Pressure**• Pressure = Force/Area. • P = F/A. • Depends on the height of the fluid. • Same in all directions. • Units are: • Force/Area = N/m2. • Pascals 1 Pa = 1 N/m2. • Atmosphere 1 atm = 1.013 X 105 N/m2.**Pressure**• Pressure = Force/Area. • P = F/A. • Depends on the height of the fluid. F = Mg = (V)g = ( (A*h) )g F/A = ( (A*h) )g / A = gh P = gh**Buoyancy**• Fluid or gas exerts a force on materials it touches • For an immersed object, the sum of forces from surrounding material is the buoyancy • If the net force on an object is upward, then it “rises”, if it is downward then it “sinks”, if zero then it “floats”**Buoyant Forces**• Force exeted by a displaced liquid. Ft-Fb = B gAht - gAhb = B B = gA(ht – hb) = Wt - Wb = B B = A(ht – hb) * g**Ice and Water**• Why does ice float? • Solids are generally denser than liquids • Liquids are generally denser than liquids • Ice should sink • In fact, “really” cold water is denser than ice (I.e. water at 34o F is denser than ice at 32o F) • If this wasn’t true, oceans and lakes would freeze from the bottom up**Equation of Continuity**• The flow of a liquid is constant throughout a system (if no liquid is added or subtracted) • The amount of water going into a pipe is the same as coming out • Amount of liquid flow is Flow = Av**Equation of Continuity**Flow1 = Flow2 1A1v1 = 2A2v2 assuming 1 = 2 (same liquid) A1v1 = A2v2 A1 so v2 = x v1 A2**Bernoulli’s Principle**where the velocity of a fluid is high the pressure is low and where the velocity is low, the pressure is high • This is why baseballs spin, aspirators work, a roof can blow off a house in high winds and sailboats can sail into the wind Let’s see it work, take out a piece of paper…**Bernoulli’s Equation**Energy in a moving fluid: Pressure: W = Fd F = PA (pressure does work) Gravity: fluid flowing up or down changes its potential energy Kinetic: fluid speeding up or slowing down changes its kinetic energy Since energy is conserved, we can relate the pressure, velocity and height between two points in a flowing system**1**1 _ _ 2 2 Bernoulli’s Equation P = Pressure v = velocity = density of fluid y = height g = acceleration due to gravity P1 + v12 + gy1 = P2 + v22 + gy2 2 1**Example 10-12**Water heater in the basement, faucet on the 2nd floor What is the flow speed and pressure on the 2nd floor? First, find the flow rate (continuity): v1A1 = v2A2 v2 = A1/A2 x v1 v2 = r12/r22 x v1 v2 = 2.02 /1.32 x 0.50 v2 = 1.2 m/s Basement: dpipe = 4.0cm v = 0.50m/s P = 3 atm 2nd floor: dpipe = 2.6cm height = 5m**1**1 1 _ _ _ P2 = P1 + (v12 - v22) + g(y1 - y2) P1 + v12 + gy1 = P2 + v22 + gy2 2 2 2 1 1 _ _ P2 = P1 + v12 - v22 + gy1 - gy2 2 2 Example 10-12(cont) Basement: dpipe = 4.0cm v = 0.50m/s P = 3 atm What is the flow speed and pressure on the 2nd floor? Next, find the pressure (Bernoulli): P2 = (3.0 N/m2) + (1.0x103 kg/m3)(9.8 m/s2)(-5.0 m) . + 0.5(1.0x103 kg/m3)[(0.5 m/s)2 – (1.2 m/s)2] P2 = 2.5 x 105 N/m2 2nd floor: dpipe = 2.6cm height = 5m v2 = 1.2 m/s**CAPA #1**What is the absolute pressure on the bottom of a swimming pool 20.0 m by 11.60 m whose uniform depth is 1.92 m? Pw = gh = (1.0x103 kg/m3)(9.8 m/s2)(1.92m) = 1.89x104 N/m2 But, we need absolute pressure… P = Pw + Patm P = 1.89x104 N/m2 + 1.013x105 N/m2 = 1.20x105 N/m2**CAPA #2-3**2. What is the total force on the bottom of that swimming pool? Area = 20.0 m x 11.60 m F = P x A = (1.20x105 N/m2)(20.0 m)(11.60 m) = 2.79x107 N 3. What will be the pressure against the side of the pool near the bottom? The pressure near the bottom is the same as on the bottom P = 1.20x105 N/m2**CAPA #4**4. A dentist’s chair of mass 236.0 kg is supported by a hydraulic lift having a large piston of cross-sectional area 1434.0 cm2. The dentist has a foot pedal attached to a small piston of cross-sectional area 76.0 cm2. What force must be applied to the small piston to raise the chair? Fchair F? Fchair = mg = (236.0 kg)(9.80 m/s2) = 2312.8 N**Fchair**F? CAPA #4(cont) 4. A dentist’s chair of mass 236.0 kg is supported by a hydraulic lift having a large piston of cross-sectional area 1434.0 cm2. The dentist has a foot pedal attached to a small piston of cross-sectional area 76.0 cm2. What force must be applied to the small piston to raise the chair? P1 = P2 P = F/A F1 F2 A1 A2 F2 = (A2/A1)F1 = ((76.0 cm2)/(1434.0 cm2)) x (2312.8 N) F2 = 123 N ___ = ___**Next Time**• Larry Dennis will be back Monday. • Continue Chapter 10. • CAPA Problems. • Prepare for Quiz 7. • Please see me with any questions or comments. See you Monday.

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