Fluid Dynamics

Two Parts

1. Fluid Flow 2. Bernoulli’s Equation and Applications

Assumptions for Fluid Flow: 1.

2.

3.

4.

Non-viscous. (isn’t “sticky”) Incompressible (constant ρ) All particles in cross section travel at the same speed (flow rate) Flow is laminar (no turbulence) Streamline flow Turbulent flow

Laminar Flow Laminar flow, type of fluid (gas or liquid) flow in which the fluid travels smoothly or in regular paths Laminar flow over a horizontal surface may be thought of as consisting of thin layers, or laminae, all parallel to each other.

• Video: Laminar Flow

Flow Rate Flow Rate (ƒ): Volume of fluid that passes a particular point in a given time Units used to measure Flow Rate = m³/sec Equation for: Flow Rate ƒ = Aν = (m 2 )(m/s) (A = cross sectional area) (ν = velocity of fluid)

R



Avt



vA t

Rate of Flow

V A



Avt vt

Volume = A(vt)

Rate of flow = velocity x area

Since A 1 > A 2 … For an incompressible, frictionless fluid, the velocity increases when the cross-section decreases:

R



v A

1 1 

v A

2 2

v

1 <

v

2

Continuity Equation Flow rates are the same at all points along a closed pipe

Continuity Equation: ƒ₁ = ƒ₂

A₁ν₁ = A₂ν₂ Reminder: the equation for Area of a circle: A = πr²

• Fluid Flow PHet

Question: Water travels through a 9.6 cm diameter fire hose with a speed of 1.3 m/s. At the end of the hose, the water flows out through a nozzle whose diameter is 2.5 cm. What is the speed of the water coming out of the nozzle?

Venturi Meter The higher the velocity in the constriction at Region-2, the lower the pressure... Wait what?

Venturi Effect

Venturi Effect

Airplane Wings

Airplane Wings

How do Plane’s Fly Video

The Physics of Sailing Video http://science.kqed.org/quest/video/the physics-of-sailing/

Question A small ranger vehicle has a soft, ragtop roof. When the car is at rest the roof is flat. When the car is cruising at highway speeds with its windows rolled up, does the roof a. bow upward b. remain flat c. bow downward?

Question A small ranger vehicle has a soft, ragtop roof. When the car is at rest the roof is flat. When the car is cruising at highway speeds with its windows rolled up, does the roof a. bow upward b. remain flat c. bow downward?

Fluid Flow Questions 1. MC - 4,14,21,42,47 2. Homework: Watch Bernoulli Video 3. MOST IMPORTANTLY: Paper Airplane Competition next class Go to: http://www.funpaperairplanes.com/index.html

a. Pick a plane and build it for the start of class b. Make TWO of the same design c. Planes will be thrown in players hall

d. Winner will be determined by displacement from initial throw

Sports Science • Record Paper Airplane

Conservation of Energy of Fluids within a Pipe Bernoulli's Principle

PRESSURE

plus

ENERGY

is CONSTANT!

1. P + E = P + E 2. P + U + K = P + U + K 3. P + ρgh + ½ρ ν ² = P + ρgh + ½ρ ν ²

This hold at ANY point!

P 1 + ρgh 1 + ½ρ ν 1 ² = P 2 + ρgh 2 + ½ρ ν 2 ²

Bernuolli Effect

1. High Velocity: _____ Pressure 2. Low Velocity: _____ Pressure

Bernuolli Effect

1. High Velocity: LOW Pressure 2. Low Velocity: HIGH Pressure

Special Case #1 – Horizontal Pipe

P

1  

gh

1  ½ 

v

1 2 

P

2  

gh

2  ½ 

v

2 2 Horizontal Pipe (h 1 = h 2 ) Horizontal Pipe 

gh

 ½ 

v

2 2  ½ 

v

1 2

Question Suppose the pressure in the fire hose is 350 kPa. What is the pressure in the nozzle? ν 1 = 1.3 m/s ν 2 = 19.17 m/s

Special Case #2 – Constant Velocity

P

1  

gh

1  ½ 

v

1 2 

P

2  

gh

2  ½ 

v

2 2 Constant velocity (ν 1 = ν 2 ) Notice how a difficult problem becomes easier when we remove constants!

Question Water flows with constant speed through a garden hose that goes up a step 20.0 cm high. If the water pressure is 143 kPa at the bottom of the step, what is its pressure at the top of the step?

ν 1 = ν 2

Special Case #3 – Fluids at Rest

P

1  

gh

1  ½ 

v

1 2 

P

2  

gh

2  ½ 

v

2 2

P 1 - P 2 =

 gh 2  gh 1  P =  g(h 2 - h 1 ) We have already seen this!

Special Case #4 – No Change in Pressure Know as Torricelli’s Theorem

P

1  

gh

1  ½ 

v

1 2 

P

2  

gh

2  ½ 

v

2 2

h 2 h h 1

v 2  0

v

 2

gh

Torricelli ’ s theorem:

v

 2

gh

Question: A dam springs a leak at a point 20.0 m below the surface. What is the emergent velocity?

h v

 2

gh

v = 19.8 m/s 2

Summary of Hydrodynamics Streamline Fluid Flow in Pipe :

R



v A

1 1 

v A

2 2 Fluid at Rest:

P A - P B =



gh

Horizontal Pipe (h 1

P

1 

P

2  ½ 

v

2 2 = h 2 )  ½ 

v

1 2

P

1  Bernoulli ’ s Theorem: 

gh

1  ½ 

v

1 2 

Constant

Torricelli ’ s theorem:

v

 2

gh

Bernoulli’s Principal 1. MC: 5,13,22,25,27,28,33,36,37,44 2. Homework: Review Free Response Questions Posted on Website 3. Next Class: Hydrodynamics Quiz