Southern Methodist University

Bobby B. Lyle School of Engineering

CEE 2342/ME 2342 Fluid Mechanics

Roger O. Dickey, Ph.D., P.E.

III. BASIC EQS. OF HYDRODYNAMICS

C. Integral Forms of the Basic Equations

3. Linear Momentum

Reading Assignments:

Chapter 5 Finite Control Volume Analysis;

Sections 5.2.1 Derivation of the Linear

Momentum Equation

Section 5.2.2 Application of the Linear

Momentum Equation

C. Integral Forms of the Basic Equations

Linear Momentum

When Isaac Newton originally formulated his

laws of motion, instead of expressing his second

law as F=ma he stated it in terms of the rate of

change of momentum; where momentum is

defined as mass times velocity, mV.

For a system of mass m, Newton stated his second law as:

Expressing this narrative equation mathematically:

FVDtmD

system theon actingforces external theof Sum

systema of momentumlinear theof change of rate Time

For a third time, reconsider Reynolds Transport

Theorem (RTT) for physical property B, where

b is the amount of the physical property per unit

mass of fluid :

CSCV

dAbVdbtDt

DB nV ˆ

Flux of B, i.e., transport rate of B per unit area

Amount of B per unit volume

For the linear momentum equation, i.e., principle

of conservation of linear momentum, B = mV,

and b = mV/m = V. Substituting into the RTT

expression:

CSCV

dAVdtDt

mD nVVVV ˆ

Substituting from Newton’s

Second Law and rearranging:

[Equation 5.22, p. 213]

Integral Form of Linear Momentum Equation

DtmD VF for

FnVVV

CSCV

dAVdt

ˆ

Expressing each term in words for clarity:

For steady flow : 0t

FnVVCS

dAˆ

Time rate of change of the linear momentum of the CV contents

FnVVV

CSCV

dAVdt

ˆ

Net rate of flow of linear momentum

across the CS

Sum of the forces acting on the contents of the CV

Integral Form of Steady Linear Momentum Equation

*Important Points –

(i) can be viewed in exactly the same

manner as the forces on a free-body. In this

case, the mass m inside the CV at time t is the

free-body under consideration, and the CS

delineates the free-body boundary.

F

(ii) Forces involved are usually placed into two categories;

• Body forces act on the entire contents of the CV, e.g., gravitational forces or weight

• Surface forces acting on or along the CS, e.g., fluid pressure forces, reaction forces from solid boundaries or immersed objects including both normal (pressure) and tangential (shear stress) forces

(iii) The linear momentum equation is a vector equation involving V on the left-hand side, and on the right; is a scalar. Thus, it may be written as three component, scalar equations. For steady flow and ;

F nV ˆ

xCS

FdAu nV ˆ

yCS

FdAv nV ˆ

zCS

FdAw nV ˆ

kjiV ˆˆˆ wvu

(iv) Very careful consideration must be given to algebraic signs for the forces, velocities, and the vector dot product in order to obtain correct results.

• is (+) for outflow from the CV, and () for inflow into the CV

• u, v, w, Fx , Fy , and Fz are (+) when

directed along the (+) x-, y-, and z-axes respectively, and negative when oppositely directed.

nV ˆ

nV ˆ

If the CV is fixed in space, it is of finite size, there are a finite number of inflow and outflow streams, and uniform flow prevails in the various inflow and outflow regions of the CS, then piecewise integration of the steady flow linear momentum equation over all inflow and all outflow regions of the CS yields finite summations. For example, along the x-axis:

All

xAll All

ininoutout mumu F

(+) because outflow regions have

0ˆ nV

() because inflow regions have

0ˆ nV

Consider a differential area element dA on the inflow region of a CS enclosing a fluid system, having pdA as the magnitude of the pressure force:

nF ˆpdApressure dA

V

n̂

In this scenario, velocity V is directed inward

while the unit normal vector, , is directed

outward yielding a negative dot product,

0cos 27090 because 0ˆ nV

n̂

Similarly, for a differential area element dA on the outflow region of a CS enclosing a fluid system, having pdA as the magnitude of the pressure force:

nF

ˆ pdApressure

dA

V

n̂

In this scenario, velocity V is directed

outward, likewise the unit normal vector, , is

directed outward yielding a positive dot

product,

0cos 9090 because 0ˆ nV

n̂

Applying similar conditions and reasoning along all

three coordinate axes, and substituting inQin for

and outQout for yields:

All

xAll All

inininoutoutout FQuQu

All

yAll All

inininoutoutout FQvQv

All

zAll All

inininoutoutout FQwQw

inm

outm

In most common engineering applications, the

various inflow and outflow streams have the same

constant density, , and the previous vector

component equations are slightly simplified to:

All

xAll All

ininoutout FQuQu

All

yAll All

ininoutout FQvQv

All

zAll All

ininoutout FQwQw

Refer to Handouts — III.C.3. Linear

Momentum Equation Examples for practical

examples dealing with application of the linear

momentum equation.

*Hydrodynamics Problem Solving Hints –

(1) Clues about when to apply the three Basic Equations of Hydrodynamics for problem solving:

• Continuity – virtually always

• Energy – when elevations, pressures, velocities, energy losses, energy input, or energy output are required

• Momentum – when forces are required

(2) The Continuity Equation is always

applicable, regardless of the presence or

absence of mechanical devices adding energy

to, or removing energy from the CV.

(3) For uniform flow, streamlines are straight

and parallel and the pressure distribution is

hydrostatic perpendicular to the direction of

flow. Stated another way,

(i.e., piezometric head is constant), along any

line or planar cross-section perpendicular to

the direction of flow.

Constant

pz

(4) Although all points on a plane perpendicular to

the flow have the same value of , it is

customary in uniform pipe flow to use the

elevation, z, and pressure, p, at the pipe

centerline as representative of a given cross-

section. Furthermore, it is often assumed that

pressure gauge readings give the pressure at the

pipe centerline (often a good approximation):

pz

p

Elevation Datum, z = 0

z

p = (z) 0z

z

Pressure GaugePipe Cross-section

(5) Although all points on a plane perpendicular to the direction of flow have the same value of , it is customary in uniform open

channel flow to use the elevation, zb , and

pressure, pb=γhb, at the channel bottom as

representative of a given cross-section. Further, it is often assumed that the angle, , that the sloping channel bottom makes with the horizontal is small, yielding cos 1.0 (often a good approximation for both natural and human-constructed channels):

pz

Elevation Datum, z = 0

zb

Free Surface, y = yfs

Channel Bottom, y = 0

pb = h

bpb = (y

fs cos)p

b yfs

y

x

z

hb

h

Open Channel Longitudinal-section

≈ 1

x-, y-coordinate axes parallel and normal to channel bottom

Substituting γyfs for pb , the Energy Equation for

open channel flow becomes,

Dividing out γ in the pressure head terms,

Lfs

bfs

b hg

Vγy

zg

Vγy

z 2

2

222

2

211

1

Lfsbfsb hg

Vyzg

Vyz 22

22

22

21

11

Finally, notice that for all points on a planar cross-

section perpendicular to the flow,

Therefore, drop the subscripts on zb and yfs

yielding the customary form of the Energy

Equation applied in open channel flow:

Lhg

Vyzg

Vyz 22

22

22

21

11

fsfsb yzyzyz z0Constant

(6) When applying the Linear Momentum Equation, the forces generated by momentum transfer may be visualized in two ways: (i) impact of inflow streams “crashing” into the fluid already inside the CV, and (ii) recoil from outflow streams “shooting” from the CV. This reveals that the pressure force component and the momentum transfer component in a given coordinate

direction at a given cross-section (e.g., Fpx1 and

u1Q1) are always additive.

That is, after solving the Linear Momentum Equation for resultant reaction force components,

either FRx or FRy, and substituting appropriate

numerical values for pressure force and velocity vector components, algebraic signs may be quickly checked for accuracy because each pair of pressure force and momentum transfer

components along the same coordinate axis—Fpx1

and u1Q1, Fpy1 and v1Q1, Fpx2 and u2Q2, Fpy2

and v2Q2—should have the same algebraic sign.