Chapter 11 - Fluids in Motion

Sections 7 - 9

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

Paul E. Tippens

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 general case and apply for (a) a fluid at rest, (b) a fluid at constant pressure, and (c) flow through a horizontal pipe.

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.

Assumptions for Fluid Flow:

Streamline flow Turbulent flow

• All fluids move with streamline flow.

• The fluids are incompressible.

• There is no internal friction.

Rate of Flow The rate of flow is defined as the volume V of a fluid that passes a certain cross-section A per unit of time t.

The volume V of fluid is given by the product of area A and Δl, but Δl = vΔt, so V = A Δl, or

Δl = the distance traveled

Volume = A(vt)

A

tAvV

mV tAv

m

Av

t

m

Which is the equation for the MASS FLOW RATE.

Constant Rate of Flow The MASS that flows into a region = The MASS that flows out of a region. For an incompressible, frictionless fluid, the velocity increases when the cross-section decreases:

A1

A2

R = A1v1 = A2v2

v1

v2

v2

tvAtvA 222111

Continuity of Equation The,vAvA 2211

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?

2

2

21

2

1

221

vrvr

vAvA v

cm.cm.r;cm.r

)s/m(

)s/m()cm(

v

vrr

50250250

16

41

22

2

22

2

2

2

1

2

12

2

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?

R = 0.00126 m3/s

R1 = 0.0754 m3/min

1

2

11 vrvAt

mR

)s/m()m.(vrR 4010 2

1

2

11vAR

min 1

s60x

s

m 0.00126 R

3

1

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.

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

2211 vAvA

Problem Strategy (Continued):

• 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.

2 2

1 1 2 2v d v d

The Venturi Meter

The higher velocity in the constriction B causes a difference of

pressure between points A and B.

PA - PB = gh

h

A B

C

11.9 Bernoulli's Principle

• The Swiss Physicist Daniel Bernoulli, was interested in how the velocity changes as the fluid moves through a pipe of different area. He especially wanted to incorporate pressure into his idea as well. Conceptually, his principle is stated as:

• "If the velocity of a fluid increases, the pressure decreases and vice versa

Work in Moving a Volume of Fluid We will break it up into small sections: WORK is equal to FORCE

times DISPLACEMENT , (in this case, the length of the section the fluid travels. The grey area.)

FxW

WORK is done by the Section

One flowing FORWARD

111 lFW As well as Section Two working

AGAINST it, lFW 22

The work is negative due to Newton's Third Law and working

against gravity. Work must go into or onto moving it.

The Pressure formula can be inserted for the force and solving for F.

PAFA

FP

Substituting PA in for F, you will get the following two

equations:

lAPW 111 2222 lAPW and

Since WORK is energy we have to ask ourselves if there is any

other energy. Since the water rises it has HEIGHT and thus

POTENTIAL ENERGY.

)yy(mgW 123

Setting all three WORKS equal to the NET WORK.

.

,,

,,

12222111net

321net

321net

mgymgylAPlAPW

soWWWW

soWWWW

WHAT DOES THE NET WORK EQUAL TO?

A CHANGE IN KINETIC ENERGY!

12222111

2

2

2

1

12222111

2

2

2

1net

mgymgylAPlAPmv2

1mv

2

1

mgymgylAPlAPKE

mv2

1mv

2

1KEKEW

,

Consider that Density = Mass per unit Volume, AND

that VOLUME is equal to AREA time LENGTH , then substituting

and solving for m, we get:

lAm

We will now substitute for the MASS in the equation with energy.

12222111

2

2

2

1mgymgylAPlAPmv

2

1mv

2

1 To get:

12222111

2

o

2ρAlgyρAlgyΔlAPΔlAPρAΔlv

2

1ρAΔlv

2

1

We can now cancel out the AREA and LENGTH in all cases

because AREA times LENGTH equals VOLUME, and volume

remains constant in the pipe.

12222111

2

2

2

1gylAρgylAρlΔAPlΔAPvlΔAρ

2

1vlAρ

2

1

Leaving:

1221

2

2

2

1ρgyρgyPPρv

2

1ρv

2

1

Moving everything related, initials and finals together, to one

side results in:

2

2

2211ρgyρv

2

1Pρgyρv

2

1P 2

1

What this basically shows is that Conservation of Energy holds

true within a fluid and that if you add the PRESSURE, the

KINETIC ENERGY(in terms of density) and POTENTIAL

ENERGY(in terms of density) you get the SAME VALUE

anywhere along a streamline.

Constantρgyρv2

1P

1

2

11

Which is known a Bernoulli’s Equation!

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 2

1 2 2 1½ ½P P v v

h

Now, since the difference in pressure P = gh,

2 2

2 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

v2 h

h = 6 cm

2 2

2 1½ ½P gh v v

Bernoulli’s Equation (h1 = h2)

2gh = v22 - v1

2 Cancel ρ, then clear fractions:

2 2 2

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

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 2

1 1 1 2 2 2½ ½P gh v P gh v

P1 - P2 = gh2 - gh1 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

h2 h

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

2 2

1 1 1 2 2 2½ ½P gh v P gh v

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

Torricelli’s theorem:

2v gh

v2 0

Interesting Example of Torricelli’s

Theorem:

v

v

v

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 ghh

Torricelli’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/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.

Conceptual Applications

In general, things tend to move TOWARD areas of low pressure. When the velocity of a fluid increases, this creates a NET FORCE toward the low pressure area. Most of the time it happens when the AREA decreases. Example: Putting your thumb over a water hose.

Another example is an airfoil.

The shape of the top of the airfoil causes the streamlines to scrunch together and thus increasing their velocity. The airfoil then moves upward to the low pressure area created.

Example Problem

A large storage tank filled with water develops a small hole in its side at a point 16 m below the water level. If the rate of flow from the leak is 2.5 x 10-3 m3/min, determine (a) the speed at which the water leaves the hole and (b) the diameter of the hole.

Answer: a. 17.71 m/s; b. 0.00173 m or 1.73 mm

Example Problem

• Water flows through a fire hose of diameter 6.35 cm at a rate of 0.012 m3/s. The fire hose ends in a nozzle of inner diameter 2.2 cm. What is the velocity with which the water exits the nozzle?

• Answer: 31.6 m/s

Assignment

• Pages 353 – 354,

• #54, 55, 56, 62, 67, and 72