m48.dvi

ID Sheet: MISN-0-48

Title:

Introduction to Fluid Statics and Dynamics

Author: Jules Kovacs, Michigan State University

Version: 2/1/2000

Evaluation: Stage 2

Length: 1 hr; 16 pages

Input Skills

1. Draw a one-body force diagram, given the environment of an ob-

ject (MISN-0-10).

2. Set up and solve static equilibrium problems (MISN-0-6).
3. Use the work-energy principle (MISN-0-20).
4. Use the principle of conservation of energy for conservative forces

(MISN-0-21).

Output Skills (Knowledge)

K1. State Archimedes’ Principle, referring to an appropriate diagram.
K2. Apply the principle of energy conservation to derive Bernoulli’s

Theorem starting with a careful definition of each of the quantities
in the expression of the theorem.

Output Skills (Problem Solving)

S1. Start from the one-body diagram and Archimedes’ Principle and

determine the density of objects immmersed in liquids, given in-
formatin about the volume of the object, the depth of immersion
and the apparent weight. Also be able to determine all of the
forces on an immersed object.

S2. Use Bernoulli’s Equation and the continuity equation to solve

problems dealing with the flow of incompressible fluids.

External Resources (Required)

1. M. Alonso and E. Finn,

Physics

, Addison-Wesley (1970). For avail-

ability, see this module’s

Local Guide

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MISN-0-48

INTRODUCTION TO

FLUID STATICS AND DYNAMICS

J. Kovacs, Michigan State University

1. Introduction

The general principles that have been developed and applied to dis-

crete particles and rigid systems, consisting of distributions of particles,
also apply to systems which may be considered as continuous distribu-
tions and which are not part of a rigid system. The motion of fluids and
the static (stationary) behavior of rigid objects immersed in a fluid are
examples of the application of the general principles to such systems.

Archimedes’ Principle summarizes the effects expected when the princi-
ples associated with static equilibrium are applied to objects immersed in
fluids. Bernoulli’s Theorem results from the application of the principle
of energy conservation to the steady state motion of a fluid.

2. Archimedes’ Principle

Note: The designation AF below refers to M. Alonso and E. Finn,

Physics

, Addison-Wesley (1970). For availability, see this module’s

Local

Guide

AF: Archimedes’ Principle relates to static equilibrium of objects im-

mersed in fluids and is treated only very briefly in AF. It is stated
on page 115 between equations 7.22 and 7.23 where it arises in
conection with a discussion about the limiting speed of an object
falling through a viscous fluid. Its derivation appears as an exercise.

3. Bernoulli’s Theorem

AF: Section 10.13 (pp. 200-203). The basic equation of Bernoulli’s The-

orem is derived and applied to an example,

fluid flow

in a pipe.

The fluid movement of masses of air around the surface of the earth as a conse-

quence of the forces arising due to the earth’s rotation is examined briefly in (MISN-
0-18).

MISN-0-48

Acknowledgments

Preparation of this module was supported in part by the National

Science Foundation, Division of Science Education Development and
Research, through Grant #SED 74-20088 to Michigan State Univer-
sity.

A. Resource Summary

MOD: This module’s Problem Supplement.

AF: M. Alonso and E. Finn,

Physics

, Addison-Wesley (1970). For avail-

ability, see this module’s

Local Guide

Skill

Ref.

Items

p. 115; Problem 10.29

Sect. 10.13, Fig. 10.20

Skill

Ref.

Items

MOD

Probl. Suppl., Problem 1

MOD

Probl. Suppl., Problem 3

Sect. 10.13, Figs. 10.22, 10.23; Ex. 10.10; Question 14
(p. 204), Problems 10.26, 10.27, 10.28

MISN-0-48

LG-1

LOCAL GUIDE

The readings for this unit are on reserve for you in the Physics-Astronomy
Library, Room 230 in the Physics-Astronomy Building. Ask for them as
“The readings for CBI Unit 48.” Do

not

ask for them by book title.

MISN-0-48

PS-1

PROBLEM SUPPLEMENT

CAUTION: Carry along the dimensions of all of the quantites that go
into your solution to assure yourself that you are using a consistent set of
units. When you do the appropriate cancellations of units, each term in
your equation will have the same dimensions.

1. A large container, open at the top, is half-filled with water (density

00 gm/cm

). On top of this, filling the container, floats a layer of oil

(density 0

60 gm/cm

). Into this two-layer fluid is immersed a cube of

side length

of wood whose density is 0

90 gm/cm

a. Determine where the cube comes to rest when equilibrium is estab-

lished. Is is at the bottom of the container? At the air-oil interface?
At the oil-water interface? Wherever it is, determine the fraction
of the cube is in each medium (air, oil, water).

b. Determine the fraction of the cube that would be out of water if

the oil were removed.

1. Brief Answers (detailed assistance is given in the

Spec. Ass. Suppl.

a. One-fourth in oil, three-fourths in water, none in air. [S-1]

b. One-tenth in air, nine-tenths in water. [S-2]

2. AF: Problem 10.29.

3. An open water tank has its upper surface of water 2

meters

above an exit pipe. At the upper surface of the water, the pressure is
1.00 atm and the speed of the water is negligible.

a. Determine the pressure at the exit pipe when no water is flowing.

b. Determine how much the pressure at the exit pipe is diminished if

water flows there at a speed of 8.0 m/s.

c. Determine the maximum permissible speed of flow through this pipe

if the pressure at the exit pipe is to remain above 2.0 atm.

d. Determine the minimum diameter of the exit pipe is needed to main-

tain a pressure of 2.0 atm or more and the need for water may be as
great as 1

kg/sec. Numerical assistance: One atmosphere

of pressure may be taken as 1

N/m

, the density of water

as 1

kg/m

MISN-0-48

PS-2

Answers

(detailed assistance is given in the

Spec. Ass. Suppl.

a. 2.96 atm.

Help: [S-3]

b. diminished by 0.32 atm.

Help: [S-3]

c. 13.9 m/s

Help: [S-3]

d. 0.30 m diameter

4. AF: Question 14 (p. 204), Problems 10.26, 10.27, and 10.28.

Answers

10.26 a. 10 m/s, 2

N/m

b. 300 kg/min (or 0.3 m

/min)

+ (1

ρv

is the energy per unit volume This divided by

is the energy per kilogram Answer is 250 joules per kilogram

10.27 a. 10 m/s, 2

N/m

b. 300 kg/min

c. 250 J/kg

10.28 b. 11.2 s

MISN-0-48

AS-1

SPECIAL ASSISTANCE SUPPLEMENT

S-1

(from PS-problem 1a)

Because this is a static equilibrium problem, the condition that must be
satisfied is that the resultant force on the cube must be zero. That is,
the weight of the cube,

, is balanced by the upward bouyant force

on the cube exerted by the fluids.
The upward force

, according to Archimedes, is equal in magnitude to

the weight of the volume of fluid displaced by the object. If the cube
were immersed completely in the oil only, the weight of oil displaced
(the weight of an amount of oil whose volume equals the cube’s volume)
would be less than the cube’s weight so

would be less than

and

the cube would sink deeper. If the cube were completely immersed in
water only, it would not sink so deep that it would rise because

would

be larger than

and the net force on the cube would be upward, not

zero. Hence, in the oil-water situation, equilibrium would be occur with
the cube partially submerged in the oil and partially in the water.

Let

be the height of block in the oil, so

−

is the height of block in the

water (draw a sketch). The total weight of fluid displaced is the weight
of oil of volume

plus the weight of water of volume

(

−

Hence

water

(

−

) +

oil

and

wood

, the

weight of the cube. Equating and solving for

(

water

−

oil

)

water

−

wood

= 0

Thus one-fourth of the cube is in oil, three-fourths in water.

S-2

(from PS-problem 1b)

The bouyant force

is due only to the weight of that volume of water

that’s equal to the volume of the part of the cube that is submerged).

MISN-0-48

AS-2

S-3

(from PS-Problem 3)

Application of Bernoulli’s Theorem solves each of the parts of this prob-
lem.

1
2

ρv

ρgy

= constant

meaning that this quantity has the same value when evaluated at any
point in the fluid. Using the subscript “1” for the top surface, and
subscript “2” for a point in the fluid just at the exit pipe:

1
2

ρv

ρgy

1
2

ρv

ρgy

Now try to solve each part of the whole problem without further assis-
tance! If you find you just can’t do it, help for part (a) is given in [S-6],
for part (b) in [S-5], for part (c) in [S-4], and for part (d) in [S-7].

S-4

(from PS-Problem 3c)

If you want the pressure at the exit pipe to remain above 2 atmospheres,
then the maximum speed must be such that

(1/2)

ρv

,max

= 0

N/m

,max

= 13

9 m/s

S-5

(from PS-Problem 3b)

Again

= 0 (assume the tower is large enough so that the speed of a

point on the surface at the top is negligible).

= (1/2)

ρv

ρgy

N/m

= (1

2)(10

kg/m

)(8 m/s)

−

kg/(m s

)

= (10

+ 1

−

) N/m

= 2

64 atmospheres,

if water flows out of the exit pipe (diminished by 0

32 atmospheres with

the water flowing).

MISN-0-48

AS-3

S-6

(from PS-Problem 3a)

With the exit pipe closed

=0,

=0. Choosing the zero of altitude

to be at the top surface in the tank (choosing the reference level for
potential energy at the top surface), then

= 0,

−

20 meters.

= 1 atmosphere = 10

newtons/m

= ?,

= 10

kg/m

N/m

+ (10

kg/m

)(9

8 m/s

)(

−

20 m)

= 10

N/m

+ 1

kg/(m s

)

Note: 1 kg/(m s

) = 1 N/m

= 2

N/m

= 2

96 atmospheres.

S-7

(from PS-Problem 3d)

The rate of efflux of fluid (in kg/s) through opening of cross-sectional
area

ρ dV /dt

dm/dt

ρAv

, where

dv/dt

is the volume crossing

area

per unit time and

is the velocity of efflux.

For a given rate, a large value of

requires a small value of

and

vice-versa.
If

max

= 13

9 m/s then, as seen above, the pressure is greater than

2 atmospheres. With the efflux rate fixed, this puts a minimum restric-
tion on

Therefore,

kg/s = (10 kg/m

)(

)(13

9 m/s)

= (1

9) m

πR

= (1

9) m

= 0

15 meters, radius of outlet.

MISN-0-48

ME-1

MODEL EXAM

1. See Knowledge Skills K1-K2.

2. A 2.00 cubic centimeter cube of gold, whose den-

sity is 19.3 times that of water is suspended from
a spring scale and totally immersed in kerosene
(the cube hangs inside the fluid touching neither
sides nor bottom of the container). Kerosene has
a density 0.80 times that of water. The spring
scale is calibrated to read, in newtons, the value
of the tension in the cord connecting the scale to
whatever is suspended below it.

a. Draw a one-body force diagram showing all of the forces acting on

the cube. [O]

b. Evaluate the magnitude of the bouyant force on the cube. [K]

c. What would the scale read if the cube were hanging in the air? [C]

d. What does the scale read when the cube is immersed in the gasoline?

[H]

A very large reservoir is filled to a height

above a discharge outlet

which sends water through a vertical loop before discharging it at point
(3) into the atmosphere. Water weighs 62.4 lbf per cubic foot. At point
(1), the pipe has a cross-sectional area

= 1

f t

, at point (2)

= 1

f t

, while at (3),

= 0

f t

= 1

ft and

MISN-0-48

ME-2

= 7

ft. The outside pressure is 1.000 atmosphere (which is

equal to 2116 lbf ft

−

a. What is the pressure at the outlet (3)? [A]

b. What is the pressure at the top of the reservoir? [N]

c. What is the flow velocity of the fluid at the top of the reservoir?

[G]

d. What is the flow velocity at outlet (3)? [E]

e. What volume of water discharges from the pipe per second? [B]

f. What is the rate (volume per second) at which the water flows past

a cross-sction of pipe at points (1) and (2)? [M]

g. What is the flow velocity at (1)? [J]

h. What is the flow velocity at (2)? [D]

i. Find the pressure at (1). [L]

j. Find the pressure at (2). [F]

k. How high above the outlet can the level of point (2) be before the

pressure at (2) drops below zero? (If the pressure gets negative, the
water must support a tension, which it cannot. Hence the countin-
uous flow breaks up and the flow subsequently ceases.) [I]

Brief Answers

A. 1 atm.

B. 60 m

C. 0.388 newtons

D. 60 m/s

E. 80 ft/sec

F. 0.226 atmospheres

G. zero

H. 0.372 newtons

I. 77.7 ft