Cohen I.M., Kundu P.K. Fluid Mechanics

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22 Introduction

The pressure of the reference state decreases vertically as

=−ρ

g. (1.36)

In the ocean the speed of sound c is deﬁned by c

= ∂p/∂ρ, where the partial derivative

is taken at constant values of entropy and salinity. In the reference state these variables

are uniform, so that dp

= c

dρ

. Therefore, the density in the neutrally stable state

varies due to the compressibility effect at a rate

dρ

(−ρ

g) =−

ρg

, (1.37)

where the subscript “a” on ρ has been dropped because ρ

is nearly equal to the actual

density ρ.

The static stability of the ocean is determined by the sign of the potential density

gradient

dρ

pot

dρ

−

dρ

ρg

. (1.38)

The medium is statically stable if the potential density gradient is negative, and

so on. For a perfect gas, it can be shown that equations (1.30) and (1.38) are

equivalent.

Scale Height of the Atmosphere

Expressions for pressure distribution and “thickness” of the atmosphere can be

obtained by assuming that they are isothermal. This is a good assumption in the

lower 70 km of the atmosphere, where the absolute temperature remains within 15%

of 250 K. The hydrostatic distribution is

=−ρg =−

Integration gives

p = p

−gz/RT

where p

is the pressure at z = 0 . The pressure therefore falls to e

−1

of its surface

value in a height RT /g. The quantity RT/g, called the scale height, is a good measure

of the thickness of the atmosphere. For an average atmospheric temperature of T =

250 K, the scale height is RT/g = 7.3 km.

Exercises

1. Estimate the height to which water at 20

◦

C will rise in a capillary glass

tube 3 mm in diameter exposed to the atmosphere. For water in contact with glass the

wetting angle is nearly 90

◦

.At20

◦

C and water-air combination, σ = 0.073 N/m.

(Answer: h = 0.99 cm.)

Exercises 23

2. Consider the viscous ﬂow in a channel of width 2b. The channel is aligned

in the x direction, and the velocity at a distance y from the centerline is given by the

parabolic distribution

u(y) = U



1 −



In terms of the viscosity µ, calculate the shear stress at a distance of y = b/2.

3. Figure 1.10 shows a manometer, which is a U-shaped tube containing mer-

cury of density ρ

. Manometers are used as pressure measuring devices. If the ﬂuid

in the tank A has a pressure p and density ρ, then show that the gauge pressure in the

tank is

p − p

atm

= ρ

gh −ρga.

Note that the last term on the right-hand side is negligible if ρ  ρ

.(Hint: Equate

the pressures at X and Y .)

4. A cylinder contains 2 kg of air at 50

◦

C and a pressure of 3 bars. The air is

compressed until its pressure rises to 8 bars. What is the initial volume? Find the ﬁnal

volume for both isothermal compression and isentropic compression.

5. Assume that the temperature of the atmosphere varies with height z as

T = T

+ Kz.

Show that the pressure varies with height as

p = p



+ Kz



g/KR

where g is gravity and R is the gas constant.

Figure 1.10 A mercury manometer.

24 Introduction

6. Suppose the atmospheric temperature varies according to

T = 15 − 0.001z

where T is in degrees Celsius and height z is in meters. Is this atmosphere stable?

7. Prove that if e(T , v) = e(T ) only and if h(T , p) = h(T ) only, then the

(thermal) equation of state is equation (1.21) or pv = kT .

8. For a reversible adiabatic process in a perfect gas with constant speciﬁc

heats, derive equations (1.25) and (1.26) starting from equation (1.18).

9. Consider a heat insulated enclosure that is separated into two compartments

of volumes V

and V

, containing perfect gases with pressures and temperatures of

, p

, and T

, T

, respectively. The compartments are separated by an imperme-

able membrane that conducts heat (but not mass). Calculate the ﬁnal steady-state

temperature assuming each of the gases has constant speciﬁc heats.

10. Consider the initial state of an enclosure with two compartments as described

in Exercise 9. At t = 0, the membrane is broken and the gases are mixed. Calculate

the ﬁnal temperature.

11. A heavy piston of weight W is dropped onto a thermally insulated cylinder

of cross-sectional area A containing a perfect gas of constant speciﬁc heats, and

initially having the external pressure p

, temperature T

, and volume V

. After some

oscillations, the piston reaches an equilibrium position L meters below the equilibrium

position of a weightless piston. Find L. Is there an entropy increase?

Literature Cited

Taylor, G. I. (1974). The interaction between experiment and theory in ﬂuid mechanics. Annual Review of

Fluid Mechanics 6: 1–16.

Von Karman, T. (1954). Aerodynamics, New York: McGraw-Hill.

Supplemental Reading

Batchelor, G. K. (1967). “An Introduction to Fluid Dynamics,” London: Cambridge University Press,

(A detailed discussion of classical thermodynamics, kinetic theory of gases, surface tension effects,

and transport phenomena is given.)

Hatsopoulos, G. N. and J. H. Keenan (1981). Principles of General Thermodynamics. Melbourne, FL:

Krieger Publishing Co. (This is a good text on thermodynamics.)

Prandtl, L. and O. G. Tietjens (1934). Fundamentals of Hydro- and Aeromechanics, New York: Dover

Publications. (A clear and simple discussion of potential and adiabatic temperature gradients is given.)

Chapter 2

Cartesian Tensors

1. Scalars and Vectors .............. 25

2. Rotation of Axes: Formal Deﬁnition

of a Vector ....................... 26

3. Multiplication of Matrices ........ 29

4. Second-Order Tensor ............ 30

5. Contraction and Multiplication . . 32

6. Force on a Surface ............... 33

Example 2.1 .................... 35

7. Kronecker Delta and Alternating

Tensor........................... 36

8. Dot Product ..................... 37

9. Cross Product ................... 38

10. Operator ∇: Gradient, Divergence,

and Curl ........................ 38

11. Symmetric and Antisymmetric

Tensors .......................... 40

12. Eigenvalues and Eigenvectors of a

Symmetric Tensor ................ 41

Example 2.2 .................... 42

13. Gauss’ Theorem ................. 44

Example 2.3 .................... 45

14. Stokes’ Theorem ................. 47

Example 2.4 .................... 48

15. Comma Notation ................ 49

16. Boldface vs Indicial Notation..... 49

Exercises ........................ 50

Literature Cited ................. 51

Supplemental Reading ........... 51

1. Scalars and Vectors

In ﬂuid mechanics we need to deal with quantities of various complexities. Some

of these are deﬁned by only one component and are called scalars, some others are

deﬁned by three components and are called vectors, and certain other variables called

tensors need as many as nine components for a complete description. We shall assume

that the reader is familiar with a certain amount of algebra and calculus of vectors.

The concept and manipulation of tensors is the subject of this chapter.

A scalar is any quantity that is completely speciﬁed by a magnitude only, along

with its unit. It is independent of the coordinate system. Examples of scalars are

temperature and density of the ﬂuid. A vector is any quantity that has a magnitude

and a direction, and can be completely described by its components along three

speciﬁed coordinate directions. A vector is usually denoted by a boldface symbol,

for example, x for position and u for velocity. We can take a Cartesian coordinate

system x

, with unit vectors a

, a

, and a

in the three mutually perpendicular

directions (Figure 2.1). (In texts on vector analysis, the unit vectors are usually denoted

by i , j, and k. We cannot use this simple notation here because we shall use ij k to

DOI: 10.1016/B978-0-12-381399-2.50002-2

26 Cartesian Tensors

Figure 2.1 Position vector OP and its three Cartesian components (x

). The three unit vectors

are a

, a

, and a

denote components of a vector.) Then the position vector is written as

x = a

+ a

where (x

) are the components of x along the coordinate directions. (The

superscripts on the unit vectors a do not denote the components of a vector; the a’s

are vectors themselves.) Instead of writing all three components explicitly, we can

indicate the three Cartesian components of a vector by an index that takes all possible

values of 1, 2, and 3. For example, the components of the position vector can be

denoted by x

, where i takes all of its possible values, namely, 1, 2, and 3. To obey the

laws of algebra that we shall present, the components of a vector should be written

as a column. For example,

x =









In matrix algebra, one deﬁnes the transpose as the matrix obtained by interchanging

rows and columns. For example, the transpose of a column matrix x is the row matrix

=[x

2. Rotation of Axes: Formal Deﬁnition of a Vector

A vector can be formally deﬁned as any quantity whose components change similarly

to the components of a position vector under the rotation of the coordinate system.

2. Rotation of Axes: Formal Deﬁnition of a Vector 27

Figure 2.2 Rotation of coordinate systemO123toO1



Let x

be the original axes, and x



be the rotated system (Figure 2.2). The

components of the position vector x in the original and rotated systems are denoted

by x

and x



, respectively. The cosine of the angle between the old i and new j axes

is represented by C

. Here, the ﬁrst index of the C matrix refers to the old axes,

and the second index of C refers to the new axes. It is apparent that C

= C

little geometry shows that the components in the rotated system are related to the

components in the original system by



= x

+ x



i=1

. (2.1)

For simplicity, we shall verify the validity of equation (2.1) in two dimensions only.

Referring to Figure 2.3, let α

be the angle between old i and new j axes, so that

= cos α

. Then



= OD = OC + AB = x

cos α

+ x

sin α

. (2.2)

As α

= 90

◦

−α

,wehavesinα

= cos α

= C

. Equation (2.2) then becomes



= x

+ x



i=1

. (2.3)

In a similar manner



= PD = PB − DB = x

cos α

− x

sin α

28 Cartesian Tensors

Figure 2.3 Rotation of a coordinate system in two dimensions.

As α

= α

− 90

◦

(Figure 2.3), this becomes



= x

cos α

+ x

cos α



i=1

. (2.4)

In two dimensions, equation (2.1) reduces to equation (2.3) for j = 1, and to equa-

tion (2.4) for j = 2. This completes our veriﬁcation of equation (2.1).

Note that the index i appears twice in the same term on the right-hand side of

equation (2.1), and a summation is carried out over all values of this repeated index.

This type of summation over repeated indices appears frequently in tensor notation.

A convention is therefore adopted that, whenever an index occurs twice in a term, a

summation over the repeated index is implied, although no summation sign is explicitly

written. This is frequently called the Einstein summation convention. Equation (2.1)

is then simply written as



= x

, (2.5)

where a summation over i is understood on the right-hand side.

The free index on both sides of equation (2.5) is j, and i is the repeated or dummy

index. Obviously any letter (other than j ) can be used as the dummy index without

changing the meaning of this equation. For example, equation (2.5) can be written

equivalently as

= x

=···,

because they all mean x



= C

+ C

. Likewise, any letter can also

be used for the free index, as long as the same free index is used on both sides of

the equation. For example, denoting the free index by i and the summed index by k,

equation (2.5) can be written as



= x

. (2.6)

3. Multiplication of Matrices 29

This is because the set of three equations represented by equation (2.5) corresponding

to all values of j is the same set of equations represented by equation (2.6) for all

values of i.

It is easy to show that the components of x in the old coordinate system are

related to those in the rotated system by

= C



. (2.7)

Note that the indicial positions on the right-hand side of this relation are different

from those in equation (2.5), because the ﬁrst index of C is summed in equation (2.5),

whereas the second index of C is summed in equation (2.7).

We can now formally deﬁne a Cartesian vector as any quantity that transforms like

a position vector under the rotation of the coordinate system. Therefore, by analogy

with equation (2.5), u is a vector if its components transform as



= u

. (2.8)

3. Multiplication of Matrices

In this chapter we shall generally follow the convention that 3 × 3 matrices are repre-

sented by uppercase letters, and column vectors are represented by lowercase letters.

(An exception will be the use of lowercase τ for the stress matrix.) Let A and B be

two 3×3 matrices. The product of A and B is deﬁned as the matrix P whose elements

are related to those of A and B by



k=1

or, using the summation convention

= A

(2.9)

Symbolically, this is written as

P = A

•

B. (2.10)

A single dot between A and B is included in equation (2.10) to signify that a single

index is summed on the right-hand side of equation (2.9). The important thing to note

in equation (2.9) is that the elements are summed over the inner or adjacent index k.

It is sometimes useful to write equation (2.9) as

= A

= (A

•

where the last term is to be read as the “ij -element of the product of matrices A

and B.”

30 Cartesian Tensors

In explicit form, equation (2.9) is written as

























(2.11)

Note that equation (2.9) signiﬁes that the ij -element of P is determined by multiplying

the elements in the i-row of A and the j -column of B, and summing. For example,

= A

+ A

This is indicated by the dotted lines in equation (2.11). It is clear that we can deﬁne

the product A

•

B only if the number of columns of A equals the number of rows of B.

Equation (2.9) can be used to determine the product of a 3 × 3 matrix and a

vector, if the vector is written as a column. For example, equation (2.6) can be written

as x



= C

, which is now of the form of equation (2.9) because the summed index

k is adjacent. In matrix form equation (2.6) can therefore be written as







































Symbolically, the preceding is



= C

•

whereas equation (2.7) is

x = C

•



4. Second-Order Tensor

We have seen that scalars can be represented by a single number, and a Cartesian

vector can be represented by three numbers. There are other quantities, however, that

need more than three components for a complete description. For example, the stress

(equal to force per unit area) at a point in a material needs nine components for a

complete speciﬁcation because two directions (and, therefore, two free indices) are

involved in its description. One direction speciﬁes the orientation of the surface on

which the stress is being sought, and the other speciﬁes the direction of the force on

that surface. For example, the j -component of the force on a surface whose outward

normal points in the i-direction is denoted by τ

. (Here, we are departing from the

convention followed in the rest of the chapter, namely, that tensors are represented by

uppercase letters. It is customary to denote the stress tensor by the lowercase τ .) The

ﬁrst index of τ

denotes the direction of the normal, and the second index denotes

the direction in which the force is being projected.

This is shown in Figure 2.4, which gives the normal and shear stresses on an

inﬁnitesimal cube whose surfaces are parallel to the coordinate planes. The stresses

4. Second-Order Tensor 31

Figure 2.4 Stress ﬁeld at a point. Positive normal and shear stresses are shown. For clarity, the stresses

on faces FBCG and CDHG are not labeled.

are positive if they are directed as in this ﬁgure. The sign convention is that, on a

surface whose outward normal points in the positive direction of a coordinate axis,

the normal and shear stresses are positive if they point in the positive direction of

the axes. For example, on the surface ABCD, whose outward normal points in the

positive x

direction, the positive stresses τ

, τ

, and τ

point toward the x

and x

directions, respectively. (Clearly, the normal stresses are positive if they are

tensile and negative if they are compressive.) On the opposite face EFGH the stress

components have the same value as on ABCD, but their directions are reversed. This

is because Figure 2.4 shows the stresses at a point. The cube shown is supposed to be

of “zero” size, so that the faces ABCD and EFGH are just opposite faces of a plane

perpendicular to the x

-axis. That is why the stresses on the opposite faces are equal

and opposite.

Recall that a vector u can be completely speciﬁed by the three components u

(where i = 1, 2, 3). We say “completely speciﬁed” because the components of u in

any direction other than the original axes can be found from equation (2.8). Similarly,

the state of stress at a point can be completely speciﬁed by the nine components τ

(where i, j = 1, 2, 3), which can be written as the matrix

τ =









The speciﬁcation of the preceding nine components of the stress on surfaces parallel

to the coordinate axes completely determines the state of stress at a point, because