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Outline

Review Pressure and Density.

Begin Buoyant Forces. Continuity Equation. Bernoulli’s Principle.

CAPA and Examples.

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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.

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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.

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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

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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”

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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

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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

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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 isFlow = Av

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Equation of Continuity

Flow1 = Flow2

assuming 1 = 2 (same liquid)

1A1v1 = 2A2v2

A1v1 = A2v2

so v2 = x v1

A1

A2

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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…

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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 energyKinetic: 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

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Bernoulli’s Equation

P = Pressurev = velocity = density of fluidy = heightg = acceleration due to gravity

1_2

1_2P1 + v1

2 + gy1 = P2 + v22 + gy2

1

2

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Example 10-12

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/r2

2 x v1

v2 = 2.02 /1.32 x 0.50

v2 = 1.2 m/s

Water heater in the basement,faucet on the 2nd floor

2nd floor: dpipe = 2.6cm height = 5mBasement: dpipe = 4.0cm v = 0.50m/s P = 3 atm

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Example 10-12(cont)

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

Basement: dpipe = 4.0cm v = 0.50m/s P = 3 atm

v2 = 1.2 m/s

1_2

1_2P2 = P1 + v1

2 - v22 + gy1 - gy2

1_2P2 = P1 + (v1

2 - v22) + g(y1 - y2)

1_2

1_2P1 + v1

2 + gy1 = P2 + v22 + gy2

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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

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CAPA #2-3

2. What is the total force on the bottom of that swimming pool?

Area = 20.0 m x 11.60 mF = 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

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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

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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

Fchair F?

___ = ___

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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.