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 followed by previous particles.

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.

• There is no internal friction.

Rate of FlowThe The rate of flow Rrate of flow R is defined as the volume is defined as the volume VV of a fluid of a fluid that passes a certain cross-section that passes a certain cross-section AA per unit of time t. per unit of time t.

The volume The volume V V of fluid is given by of fluid is given by the product of area A and the product of area A and vtvt:: V AvtV Avt

AvtR vA

t Rate of flow = velocity x area

vt

Volume = A(vt)

A

Constant Rate of FlowFor an incompressible, frictionless fluid, the For an incompressible, frictionless fluid, the velocity increases when the cross-section velocity increases when the cross-section decreases:decreases:

1 1 2 2R v A v A

A1

A2

R = A1v1 = A2v2

v1

v2

v2

2 21 1 2 2v d v d2 2

1 1 2 2v d v 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:

2 21 1 2 2v d v d

2 22 1 12 2

2

(4 m/s)(2 cm)

(20 cm)

v dd

v d2 = 0.894 cmd2 = 0.894 cm

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?

2 21

1 1

(4 m/s) (0.02 m)

4 4

dR v

R1 = 0.00126 m3/s

1 1 2 2R v A v A

21

1 1 1; 4

dR v A A

3

1

m 1 min0.00126

min 60 sR

R1 = 0.0754 m3/minR1 = 0.0754 m3/min

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.

1 1 2 2R v A v A

• Be sure to use consistent units for area and velocity.

• Be sure to use consistent units for area and velocity.

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 21 1 2 2v d v d

The Venturi Meter

The higher velocity in the constriction B The higher velocity in the constriction B causes a difference of pressure between causes a difference of pressure between

points A and B.points A and B.

PA - PB = ghPA - PB = gh

h

AB

C

Demonstrations of the Venturi Principle

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

Examples of the Venturi Effect

Work in Moving a Volume of

Fluid

P1

A1

P1

A1

P2

A2

A2

P2

h

Volume V

Note differences in pressure P and area

A

Fluid is raised to a height h.

22 2 2 2

2

; F

P F P AA

11 1 1 1

1

; F

P F P AA

F1

, F2

Work on a Fluid (Cont.)

Net work done on fluid is sum of work done by input force Fi 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 Fi less the work done by resisting force F2,

as shown in figure.

Net Work = P1V - P2V = (P1 - P2) V

F1 = P1A1

F2 = P2A2

v1

v2

A1

A2

h2

h1 s1

s2

Conservation of EnergyKinetic Energy

K:2 22 1½ ½K mv mv

Potential Energy U:

2 1U mgh mgh

Net Work = K + U

2 21 2 2 1 2 2( ) (½ ½ ) ( )P P V mv mv mgh mgh

also Net Work = (P1 - P2)V

F1 = P1A1

F2 = P2A2

v1

v2

A1

A2

h2

h1 s1

s2

Conservation of Energy2 2

1 2 2 1 2 2( ) (½ ½ ) ( )P P V mv mv mgh mgh

Divide by V, recall that density m/V, then simplify:

2 21 1 1 2 2 2½ ½P gh v P gh v

Bernoulli’s Theorem:2

1 1 1½P gh v Const

v1

v2

h1

h2

Bernoulli’s Theorem (Horizontal Pipe):2 2

1 1 1 2 2 2½ ½P gh v P gh v

h1 = h2

v1 v2

Horizontal Pipe (h1 = h2)

2 21 2 2 1½ ½P P v v

h

Now, since the difference in pressure P = gh,

2 22 1½ ½P gh v v Horizontal

Pipe

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?

v1 = 4 m/s

v2h

h = 12 cm

2 22 1½ ½P gh v v

Bernoulli’s Equation (h1 = h2)

2gh = v22 - v1

2Cancel , then clear fractions:

2 2 22 12 2(9.8 m/s )(0.12 m) (4 m/s)v gh v

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

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:

2 21 1 1 2 2 2½ ½P gh v P gh v

P1 - P2 = gh2 - gh1P = g(h2 -

h1)

P = g(h2 - h1)

h = 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.

Torricelli’s Theorem

2v gh

h1

h2h

When there is no change of pressure, P1 = P2.

2 21 1 1 2 2 2½ ½P gh v P gh v

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

Torricelli’s theorem:

2v gh

v2

Interesting Example of Torricelli’s Theorem:

v

vv

Torricelli’s theorem:

2v gh

• Discharge velocity increases with depth.

• Holes equidistant above and below midpoint will have same horizontal range.

• Maximum range is in the middle.

Example 4: A dam springs a leak at a point 20 m below the surface. What

is the emergent velocity?

2v ghhTorricelli’s theorem:

2v gh

Given: h = 20 m g = 9.8 m/s2

22(9.8 m/s )(20 m)v

v = 19.8 m/s2v = 19.8 m/s2

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 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 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 21 1 1 2 2 2½ ½P gh v P gh v

Strategies (Continued)

2 21 1 1 2 2 2½ ½P gh v P gh v

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

P = g(h2 - h1)

P = g(h2 - h1)

• For a horizontal pipe, h1 = h2 and we obtain:

h = 1000

kg/m3

2 21 2 2 1½ ½P P v v

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

Strategies (Continued)

2 21 1 1 2 2 2½ ½P gh v P gh v

2v gh

Torricelli’s Theorem

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

BR=30 L/s

AA = (0.08 m)2 = 0.0201 m3

AB = (0.05 m)2 = 0.00785 m3

2; 2

DA R R

3 3

22 22

0.030 m /s 0.030 m /s1.49 m/s; 3.82 m/s

0.0201 m 0.00785 mAA

R Rv v

A A

vA = 1.49 m/s vB = 3.82 m/s

R = 30 L/s = 0.030 m3/s

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

8 m

A

BR=30 L/s

Consider the height hA = 0 for reference purposes.

Given: vA = 1.49 m/s vB = 3.82 m/s PA = 200 kPa hB - hA = 8 m

PA + ghA +½vA2 = PB + ghB + ½vB

2 0

PB = PA + ½vA2 - ghB - ½vB

2

PB = 200,000 Pa + ½1000 kg/m3)(1.49 m/s)2

– (1000 kg/m3)(9.8 m/s2)(8 m) - ½1000 kg/m3)(3.82 m/s)2

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

PB = 115 kPa PB = 115 kPa

Summary

Bernoulli’s Theorem:2

1 1 1½P gh v Constant

1 1 2 2R v A v A 2 21 1 2 2v d v d

Streamline Fluid Flow in Pipe:

PA - PB = ghHorizontal Pipe (h1 = h2)

2 21 2 2 1½ ½P P v v

Fluid at Rest:

Torricelli’s theorem:

2v gh

Summary: Bernoulli’s Theorem

2 21 1 1 2 2 2½ ½P gh v P gh v

• 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.