Physics 101 lecture 18 fluids ii
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Exam III. Physics 101: Lecture 18 Fluids II. Textbook Sections 9.6 – 9.8. Review Static Fluids. Pressure is force exerted by molecules “bouncing” off container P = F/A Gravity/weight effects pressure P = P 0 + r gd Buoyant force is “weight” of displaced fluid. F = r g V

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Physics 101: Lecture 18 Fluids II

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Physics 101 lecture 18 fluids ii

Exam III

Physics 101: Lecture 18 Fluids II

Textbook Sections 9.6 – 9.8


Review static fluids

Review Static Fluids

  • Pressure is force exerted by molecules “bouncing” off container P = F/A

  • Gravity/weight effects pressure

    • P = P0 + rgd

  • Buoyant force is “weight” of displaced fluid.

    • F = r g V

      Today include moving fluids!

      A1v1 = A2 v2

      P1+rgy1 + ½ rv12 = P2+rgy2 + ½rv22

08


Archimedes principle

Archimedes’ Principle

  • Buoyant Force (FB)

    • weight of fluid displaced

    • FB = fluidVoldisplaced g

    • Fg = mg = object Volobject g

    • object sinks if object > fluid

    • object floats if object < fluid

  • If object floats…

    • FB = Fg

    • Therefore: fluid gVoldispl. = object gVolobject

    • Therefore: Voldispl./Volobject = object / fluid

10


Preflight 1

CORRECT

Preflight 1

Suppose you float a large ice-cube in a glass of water, and that after you place the ice in the glass the level of the water is at the very brim. When the ice melts, the level of the water in the glass will:

1. Go up, causing the water to spill out of the glass.

2. Go down.

3. Stay the same.

12


Preflight 2

CORRECT

Tub of water + ship

Tub of water

Overflowed water

Preflight 2

Which weighs more:

1. A large bathtub filled to the brim with water.

2. A large bathtub filled to the brim with water with a battle-ship floating in it.

3. They will weigh the same.

Weight of ship = Buoyant force =

Weight of displaced water

15


Continuity of fluid flow

Continuity of Fluid Flow

  • Watch “plug” of fluid moving through the narrow part of the tube (A1)

    • Time for “plug” to pass point Dt = x1 / v1

    • Mass of fluid in “plug” m1 = r Vol1 =r A1 x1 or m1 = rA1v1Dt

  • Watch “plug” of fluid moving through the wide part of the tube (A2)

    • Time for “plug” to pass point Dt = x2 / v2

    • Mass of fluid in “plug” m2 = r Vol2 =r A2 x2 or m2 = rA2v2Dt

  • Continuity Equation says m1 = m2fluid isn’t building up or disappearing

    • A1 v1 = A2 v2

22


Faucet preflight

A1

V1

A2

V2

Faucet Preflight

A stream of water gets narrower as it falls from a faucet (try it & see).

Explain this phenomenon using the equation of continuity

As the the water falls, its velocity is increasing. Since the continuity equations states that if density doesn't change...Area1*velocity1=Area2*velocity2. From this equation, we can say that as the velocity of the water increases, its area is going to decrease

I tried to do this experiment and my roommate yelled at me for raising the water bill.

Adhesion and cohesion…

23


Fluid flow concepts

Fluid Flow Concepts

r

A1P1

v1

A2 P2

v2

  • Mass flow rate: Av (kg/s)

  • Volume flow rate: Av (m3/s)

  • Continuity: A1 v1 = A2 v2

    • i.e., mass flow rate the same everywhere

    • e.g., flow of river

24


Pressure flow and work

Pressure, Flow and Work

Recall:

W=F d

= PA d

= P Vol

  • Continuity Equation says fluid speeds up going to smaller opening, slows down going to larger opening

  • Acceleration due to change in pressure. P1 > P2

    • Smaller tube has faster water and LOWER pressure

  • Change in pressure does work!

    • W = P1A1Dx1 - P2A2Dx2 = (P1 – P2)Volume

Demo

28


Pressure act

Pressure ACT

  • What will happen when I “blow” air between the two plates?

    A) Move Apart B) Come Together C) Nothing

There is air pushing on both sides of plates. If we get rid of the air in the middle, then just have air on the outside pushing them together.

31


Bernoulli s eqs and work

Bernoulli’s Eqs. And Work

Note:

W=F d

= PA d

= P V

  • Consider tube where both Area, height change.

    • W = DK + DU

      (P1-P2) V = ½ m (v22 – v12) + mg(y2-y1)

      (P1-P2) V = ½ rV (v22 – v12) + rVg(y2-y1)

      P1+rgy1 + ½ rv12 = P2+rgy2 + ½rv22

34


Bernoulli act

Bernoulli ACT

  • Through which hole will the water come out fastest?

P1+rgy1 + ½ rv12 = P2+rgy2 + ½rv22

A

Note: All three holes have same pressure P=1 Atmosphere

B

rgy1 + ½ rv12 = rgy2 + ½rv22

C

gy1 + ½ v12 = gy2 + ½v22

Smaller y gives larger v. Hole C is fastest

37


Physics 101 lecture 18 fluids ii

CORRECT

Act

A large bucket full of water has two drains. One is a hole in the side of the bucket at the bottom, and the other is a pipe coming out of the bucket near the top, which bent is downward such that the bottom of this pipe even with the other hole, like in the picture below:

Though which drain is the water spraying out with the highest speed?

1. The hole

2. The pipe

3. Same

Note, the correct height, is where the water reaches the atmosphere, so both are exiting at the same height!

40


Lift a house

Lift a House

Calculate the net lift on a 15 m x 15 m house when a 30 m/s wind (1.29 kg/m3) blows over the top.

P1+rgy1 + ½ rv12 = P2+rgy2 + ½rv22

P1 – P2 = ½ r (v22 – v12)

= ½ r (v22 – v12)

= ½ (1.29) (302) N / m2

= 581 N/ m2

F = P A

= 581 N/ m2 (15 m)(15 m) = 131,000 N

= 29,000 pounds! (note roof weighs 15,000 lbs)

48


Fluid flow summary

r

A1P1

v1

A2 P2

v2

Fluid Flow Summary

  • Mass flow rate: Av (kg/s)

  • Volume flow rate: Av (m3/s)

  • Continuity: A1 v1 = A2 v2

  • Bernoulli: P1 + 1/2v12 + gh1 = P2 + 1/2v22 + gh2

50


Practice problems

Practice Problems

Chapt. 9, problems 1, 5, 9, 13, 15, 17, 21, 29, 33, 35, 41, 45,47.


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