08 Fluids In Motion.pptx

The lower falls at Yellowstone

National Park: the water at the top of the falls passes

through a narrow slot, causing the

velocity to increase at that point. In this chapter, we will

study the physics of fluids in motion.

Fluid Motion

Objectives: After completing this

module, you should be able to:

Define the rate of flow for a fluid and solve problems using velocity and cross-section. Write and apply Bernoulli’s equation for the

Fluids in Motion

All fluids are assumed in this treatment to exhibit streamline flow.

• _{Streamline flow is the motion of a fluid in} which every particle in the fluid follows the same path past a particular point as that

followed by previous particles.

• _{Streamline flow is the motion of a fluid in}

which every particle in the fluid follows the same path past a particular point as that

Assumptions for Fluid Flow:

Streamline flow Turbulent flow

• _{All fluids move with streamline flow.}

• _{The fluids are incompressible.}

• _{There is no internal friction.}

• _{All fluids move with streamline flow.} • _{The fluids are incompressible.}

Rate of Flow

R =

R =

velocity

·

Area

(m

/s)

(m/s



m

)

Volume

Flow = Flow

z

Constant Rate of Flow

For an incompressible, frictionless fluid, the velocity increases when the cross-section decreases:

A

1

A

2

R = A₁v₁ = A₂v₂

v₁

v₂

v

·A

= v

·A

v

·π·r

= v

2

·π·r

v

·π·(½·d

)

= v

2

·π·(½·d

)

v

·π·¼·d

= v

2

·π·¼·d

v

·d

= v

2

·d

Example 1: Water flows through a rubber hose 2 cm in

diameter at a velocity of 4 m/s. What must be the diameter of the nozzle in order that the water emerge at 16 m/s?

The area is proportional to the square of diameter, so:

16 m/s dd22 = = 1 cm 1 cm

2 2

1 1 2 2

v d



v d

2 2

2 _{1 1}

2 ₂

2

(4 m/s)(2 cm)

(20 cm)

v d

d

v

Example 1 (Cont.): Water flows through a rubber hose 2 cm in diameter at a velocity of 4 m/s. What is the rate of flow in m3_/min?

R₁ = 0.00126 m3/s

R₁ = 0.0754 m3/min

R₁ = 0.0754 m3/min

s 60 s min 2 2 1 1 1

(4 m/s) (0.02 m)

4 4

d

R  v





1 1 2 2

R v A





v A

2 1 1 1; 1

4

d R v A A 



3

1

m 1 min 0.00126

min 60 s

R  _ _

Problem Strategy for Rate of Flow:

Read, draw, and label given information. The rate of flow R is volume per unit time. When cross-section changes, R is constant. Read, draw, and label given information. The rate of flow R is volume per unit time. When cross-section changes, R is constant.

•

Be sure to use consistent units for area

and velocity.

•

Be sure to use consistent units for area

and velocity.

1 1 2 2

Problem Strategy (Continued):

Since the area A of a pipe is proportional to its diameter d, a more useful equation is:

Since the area A of a pipe is proportional to its diameter d,

a more useful equation is:

• _{The units of area, velocity, or diameter chosen} for one section of pipe must be consistent with those used for any other section of pipe.

• _{The units of area, velocity, or diameter chosen}

for one section of pipe must be consistent with those used for any other section of pipe.

2 2

1 1 2 2

http://www.google.com/search?

q=venturi+tube&safe=on&source=lnms&tbm=isch&sa=X&ei=GQmmUqjmL4zxrAGlzICoDA&sqi=2&ved=0CAcQ_AUoAQ&biw=1024&bi h=676&surl=1#facrc=_&imgdii=_&imgrc=kITrHykgRdsWYM%3A%3BpmXtoSSwkJ7-6M%3Bhttp%253A%252F

%252Fwww.physics.umd.edu%252Flecdem%252Fservices%252Fdemos%252Fdemosf5%252Ff5-24hires.jpg%3Bhttp%253A%252F %252Fwww.physics.umd.edu%252Flecdem%252Fservices%252Fdemos%252Fdemosf5%252Ff5-24.htm%3B1600%3B1200

The Venturi Meter

The higher velocity in the constriction B causes a difference of pressure between points A and B.

P_A - P_B = rgh

P_A - P_B = rgh

Demonstrations of the Venturi Principle

The increase in air velocity produces a difference of pressure that exerts the forces shown.

Work

– Work

= ΔK + ΔU

Wo

rk

tion

Work

ΔU

ΔK

transfers of energy through work done.

Work in Moving a

Volume of Fluid

P₁ A₁ P₁ A₁ P₂ A₂ A₂ P₂ h Volume V Note differences in pressure DP and area DA

Fluid is raised to a height h. F₁

, F₂

2

2 2 2 2

2

;

F

P

F

P A

A





1

1 1 1 1

1

;

F

P

F

P A

A

Δ Pressure

Work on a Fluid (Cont.)

Net work done on fluid is sum of work done by input force F₁ less the work done by resisting force F₂, as shown in figure.

Net work done on fluid is sum of work done by input force

F₁ less the work done by resisting force F₂, as shown in figure.

Net Work

= P

V - P

V = (P

- P

) V

Net Work

= F

S

- F

S

Conservation of Energy

Kinetic Energy K:

Potential Energy U:

Net Work = DK + DU also Net Work = (P₁ - P₂)V

F₁ = P₁A₁

F₂ = P₂A₂

v₁

v₂

A₁

A₂

h₂ h₁ s₁

s₂

1

2 2

2 1

½

½

K

mv

mv

D 



2 1

U

mgh

mgh

D 



2 2

1 2 2 1 2 2

Conservation of Energy

Divide by V: (P₁ – P₂)V = (½ ρV·v₂2 - ½ ρV·v

12) + (ρVgh2 – ρVgh1)

Recall that density ρ  m/V, m = ρV then simplify:

then simplify: (P₁ – P₂) = (½ ρ·v₂2 - ½ ρ·v

12) + (ρgh2 – ρgh1)

1

(P₁ – P₂)V = (½ m·v₂2 - ½ m·v

12) + (mgh2 – mgh1)

2 2

1 1

½

½

2 2

1 1

½

½

Bernoulli’s Theorem (Horizontal Pipe):

h₁ = h₂

r

v_{1 v}

2

Horizontal Pipe (h₁ = h₂)

h

Now, since the difference in pressure DP = rgh,

Horizontal Pipe

2 2

1 1

½

½

P



r

gh



r

v



P



r

gh



r

v

2 2

1 2

½

½

P P





r

v



r

v

2 2

2 1

½

½

P

r

gh

r

v

r

v

Example 3: Water flowing at 4 m/s passes through a Venturi tube as shown. If h = 12 cm, what is the velocity of the water in the constriction?

r

v₁ = 4 m/s

v₂

h

h = 12 cm

Bernoulli’s Equation (h₁ = h₂)

2gh = v₂2 - v 12 Cancel r, then clear fractions:

v₂ = 4.28 m/s

v₂ = 4.28 m/s _{Note that density is not a factor.}

2 2

2 1

½

½

P

r

gh

r

v

r

v

D 





2 2 2

2 2 1 2(9.8 m/s )(0.12 m) (4 m/s)

Bernoulli’s Theorem for Fluids at Rest.

For many situations, the fluid remains at rest so that v₁

and v₂ are zero. In such cases we have:

P₁ - P₂ = rgh₂ - rgh₁ DP = _{DP =} r_rg(h_{g(h2 2} - h_{- h11}) ₎

h r = 1000 kg/m3

This is the same relation seen earlier for finding the pressure P at a given depth h = (h₂ - h₁) in a fluid.

2 2

1 1

½

½

Torricelli’s Theorem

h₁ h₂ h

When there is no change of pressure, P₁ = P_2.

Consider right figure. If surface v₂  0 and P₁= P₂ and v₁ = v we have:

Torricelli’s theorem:

v₂  0

ρgh₁ + ½ ρ v₁2 = ρgh

2 + ½ ρ v22

gh₁ + ½ v2 = gh

2

½ v2 = gh

2 - gh1

U_top = K_hole mgh = ½mv2

2gh = v2

2

v  gh

2 2

1 1

½

½

P



r

gh



r

v



P



r

gh



r

v

2

Interesting Example of

Torricelli’s Theorem:

v v

v

Torricelli’s theorem:

• _{Discharge velocity}

increases with depth.

• _{Holes equidistant above and below midpoint}

will have same horizontal range.

• _{Maximum range is in the middle.}

2

Example 4: A dam

springs a leak at a point

200 m below the

surface. What is the emergent velocity?

Torricelli’s theorem:

Given: h = 200 m

g = 9.8 m/s2

v = 62.6 m/s

v = 62.6 m/s

v = 140 mi/hr

v = 140 mi/hr

2

Strategies for Bernoulli’s Equation:

• _{Read, draw, and label a rough sketch with givens.}

• _{The height h of a fluid is from a common reference} point to the center of mass of the fluid.

• _{In Bernoulli’s equation, the density} r _{is mass} density and the appropriate units are kg/m3.

• _{Write Bernoulli’s equation for the problem and} simplify by eliminating those factors that do not change.

• _{Read, draw, and label a rough sketch with givens.} • _{The height h of a fluid is from a common reference}

point to the center of mass of the fluid.

• _{In Bernoulli’s equation, the density} r _{is mass}

density and the appropriate units are kg/m3.

• _{Write Bernoulli’s equation for the problem and}

simplify by eliminating those factors that do not change.

2 2

1 1

½

½

Strategies (Continued)

• For a stationary fluid, v₁ = v₂ and we have:

DP = rg(h₂ - h₁) DP = rg(h₂ - h₁)

• For a horizontal pipe, h₁ = h₂ and we obtain: h r = 1000 _kg/m3

2 2

1 1

½

½

P



r

gh



r

v



P



r

gh



r

v

2 2

1 2

½

½

• For no change in pressure, P₁ = P₂ and we have:

Strategies (Continued)

Torricelli’s Theorem

2 2

1 1

½

½

P



r

gh



r

v



P



r

gh



r

v

2

General Example: Water flows through the pipe at the rate of 30 L/s. The absolute pressure at point A is 200 kPa, and the point B is 8 m

higher than point A. The lower section of pipe has a diameter of 16 cm

and the upper section narrows to a diameter of 10 cm. Find the velocities of the stream at points A and B.

8 m

A

B R=30 L/s

A_A = (0.08 m)2 _{= 0.0201 m}3

A_B = (0.05 m)2 = 0.00785 m3

v_A = 1.49 m/s v_B = 3.82 m/s

R = 30 L/s = 0.030 m3/s

2_;

2

D A   R R 

3 3

2

2 2

2

0.030 m /s 0.030 m /s

1.49 m/s; 3.82 m/s

0.0201 m 0.00785 m

General Example (Cont.): Next find the absolute pressure at Point B.

8 m

A

B R=30 L/s

Consider the height h_A = 0 for reference purposes. Given: v_A = 1.49 m/s

v_B = 3.82 m/s

P_A = 200 kPa

h_B - h_A = 8 m

P_A + rgh_A +½rv_A2 _{= P}

B + rghB + ½rvB2

0

P_B = P_A + ½rv_A2 - rgh

B - ½rvB2

P_B = 200,000 Pa + ½(1000 kg/m3_{)(1.49 m/s)}2

– (1000 kg/m3)(9.8 m/s2)(8 m) - ½(1000 kg/m3)(3.82 m/s)2

P_B = 200,000 Pa + 1113 Pa –78,400 Pa – 7296 Pa

P_B = 115 kPa

Summary

Bernoulli’s Theorem:

Streamline Fluid Flow in Pipe:

P_A - P_B = rgh

Horizontal Pipe (h₁ = h₂) Fluid at Rest:

Torricelli’s theorem:

2 1 1 ½ 1

P  r gh  rv  Constant

1 1 2 2

R v A





v A

v d



v d

2 2

1 2

½

½

P P





r

v



r

v

2

Summary: Bernoulli’s Theorem

• _{Read, draw, and label a rough sketch with givens.}

• _{The height} h _{of a fluid is from a common reference} point to the center of mass of the fluid.

• _{In Bernoulli’s equation, the density r is mass} density and the appropriate units are kg/m3.

• _{Write Bernoulli’s equation for the problem and} simplify by eliminating those factors that do not change.

• _{Read, draw, and label a rough sketch with givens.} • _{The height} h _{of a fluid is from a common reference}

point to the center of mass of the fluid.

• _{In Bernoulli’s equation, the density r is mass}

density and the appropriate units are kg/m3.

• _{Write Bernoulli’s equation for the problem and}

simplify by eliminating those factors that do not change.

2 2

1 1

½

½