Fluid Mechanics, SG2214, HT2013September 11, 2013

Exercise 3: Conservation Equations and Stress Tensor

Example 1: conservative form of continuity equation (also in lecturenotes)

a) Consider a fixed, closed surface S in a fluid. Show that conservation of mass implies

t+

xi(ui) = 0 .

The change of mass in the volume is a balance between the mass flux in an out the domain and thechange in density. Conservation means that this should be zero

V

tdV +

S

ui ni dS = 0 .

Using the Gauss Theorem we can rewrite the surface integral as volume integrals,S

ui ni dS =

V

xi(ui) dV .

We then have V

{t

+

xi(ui)} dV = 0 .

This must be true for an arbitrary volume leading to the differential form

t+

xi(ui) = 0 .

b) Show that this can be written asD

Dt+

uixi

= 0 .

xi(ui) =

xiui +

uixi

= ui

xi+

uixi

This gives

t+ ui

xi D

Dt

+uixi

=D

Dt+

uixi

.

c) Explain what the following means

uixi

= 0 DDt

= 0 .

1

Remember the decomposition of uixj in its invariant parts

uixj

=1

2

(uixj

+ujxi

) 1

3

ukxk

ij deformation eij

+1

2

(uixj ujxi

)

rotation ij

+1

3

ukxk

ij volume change eij

.

The volume change eij = 0 due to the divergence-free condition. Therefore, if we are following a fluidparticle, the density and volume are constant. But at a fixed position, the density can change as

t= ui

xi.

Example 2: Reynolds Transport Theorem (also in lecture notes)

a) Use the Reynolds Transport Theorem to provide an alternative derivation of the conservation of massequation

t+

xi(ui) = 0.

Reynolds Transport Theorem:

D

Dt

V (t)

Tijk dV =

V (t)

(Tijkt

+

xl(ulTijk)

)dV

Put Tijk = ,

D

Dt

V (t)

dV =

V (t)

(

t+

xi(ui)

)dV

No flow through the surface, so the integrals are zero and valid for all V (t)

t+

xi(ui) = 0 .

b) Use this result to showD

Dt

V (t)

Fijk dV =

V (t)

DFijkDt

dV .

Put Tijk = Fijk in the Reynolds Transport Theorem

D

Dt

V (t)

Fijk dV =

V (t)

(

t(Fijk) +

xl(ulFijk)

)dV =

V (t)

(Fijkt

+ Fijk

t+ Fijk

xl(ul) + ul

Fijkxl

)dV =

V (t)

[Fijk

(

t+

xl(ul)

=0

)+

(Fijkt

+ ulFijkxl

=DF

ijkDt

)]dV =

V (t)

DFijkDt

dV .

2

The stress tensor Tij

The equation for the stress tensor was deduced by Stokes in 1845 from elementary hypotheses.Writing the stress tensor in the form

Tij = pij + ij ,the following statements should be true for the viscous stress tensor ij in a Newtonian fluid:

(i) ij should vanish if the flow involves no deformation of fluid elements: Tij = pij .(ii) ij is a function of the velocity gradients

uixj

. This is dependence is assumed local and at present

time, that is the stresses depends on the local deformation without any effect from past deformations.

(iii) the relationship between ij anduixj

should be isotropic as the physical properties of the fluid

are assumed to show no preferred direction.(iv) ij should be frame indifferent, so independent of the antisymmetric part ij

Finally linear dependence is assumed between ij anduixj

.

Tij = pij + 2eij .This proofed to be good!

Example 3: Derivation of the viscous stress tensor

If ij is a linear function of the components of eij it can then be written as

ij = cijkl ekl + dij ,

where dij = 0 from the first assumption above. It can be shown that the most general fourth-order isotropictensor is of the form

cijkl = Aijkl +Bikjl + Ciljk

where A,B and C are scalars.

a) Use this to show thatij = ekkij + 2eij

where and are scalars.

ij = cijkl ekl = (Aijkl +Bikjl + Ciljk)ekl = Aijekk +Beij + Ceji

Say that A = and that B = C = , then since eij = eji we have

ij = ekkij + 2eij

b) Show that if we have a Newtonian fluid where p = 13Tii then

Tij = (p+

2

3ukxk

)ij +

(uixj

+ujxi

)We have

Tij = pij + ij = pij + ekkij + 2eijThen

Tii = 3p+ 3eii + 2eii = 3p+ (3+ 2)eii

3

This gives = 23, and thusTij = p 2

3ekkij + 2eij

or

Tij = (p+

2

3ukxk

)ij +

(uixj

+ujxi

).

= 23 is assumed and shown to be valid for monoatomic gases. This implies that the mean pressure(normal stress) is the thermodynamic pressure. As consequence, there is no irreversible viscous loss inthe case of fluid expansion or compression (Work for expansion/compression is only provided by thepressure, not by viscosity)

Note that for incompressible fluid u = 0:

Tij = pij + 2eij .

Example 4

Calculate the stress vector at a surface through the origin of example 3, recitation 2 for a Newtonian fluid ifthe pressure at the origin is p0, and the unit normal is

a) in the x-direction b) in the y-direction c) at 45o in the x y planeAlso, for each case identify the normal stress and the shear stress, the viscous stress and the isotropic part

of the stress.Solution: Given the velocity the field u = x, v = y, (w = 0) and the definition of the stress tensor

Tij = p0ij + 2eij , the strain rate tensor and the stress tensor become respectively

eij =

u

x

1

2(u

y+v

x)

1

2(u

y+v

x)

v

y

= 0

0

Tij =

p0 + 2 00 p0 2

a) Unit normal n = ex and tangent t = ey, or equivalently

n =

10

; t =0

1

;Stress vector Ri = Tijnj

Ri =

p0 + 2 00 p0 2

10

=p0 + 2

0

Normal stress Rini (p0 + 2 0)

10

= p0isotropic part

+ 2viscous stress

4

Shear stress Riti (p0 + 2 0)0

1

= 0b) Unit normal n = ey and tangent t = ex, or equivalently

n =

01

; t =1

0

;Stress vector Ri = Tijnj

Ri =

p0 + 2 00 p0 2

01

= 0p0 2

Normal stress Rini (

0 p0 2)0

1

= p0isotropic part

2viscous stress

Shear stress Riti (0 p0 2

)10

= 0c)

n =12

11

; t = 12

11

;Stress vector Ri = Tijnj

Ri =

p0 + 2 00 p0 2

12

11

= 12

p0 + 2p0 2

Normal stress Rini

12

(p0 + 2 p0 2) 12

11

= p0isotropic part of stress

Shear stress Riti

12

(p0 + 2 p0 2) 12

11

= 2viscous stress

Example 5

Consider a control volume x [0, L], y [0, L] for the flow in the example 3 of recitation 2 and calculatethe net force resulting from the isotropic stress tensor on the control volume using the momentum theorem.

Momentum theorem in the absence of external body forces

d

dt

V

udV +

V

u(u n)dS =V

RdS

5

x-component for the steady flow u = x, v = y, (w = 0)V

u(u n)dS =V

RxdS =

V

pnxdS x-comp from isotropic stress tensor

+

V

xjnjdS x-comp from viscous stress tensor

y-componentV

v(u n)dS =V

RydS =

V

pnydS y-comp from isotropic stress tensor

+

V

yjnjdS y-comp from viscous stress tensor

Given the viscous stress tensor

ij =

2 00 2

we integrate long the boundary of our domain.Along x = L the normal is njdS = (ex)jdy and thus

ijnj =

2 00 2

10

=2

0

.Along the side x = 0 the normal is in the opposite direction njdS = (ex)jdy and

ijnj =

2 00 2

10

=2

0

.Hence we get no net contribution from the two surfaces of S at x = 0, and x = L

L0

ijnj(for x=0 and x=L) =

L0

20

+2

0

dy =0

0

We now consider the side along y = L the normal is njdS = (ey)jdx and thus

ijnj =

2 00 2

01

= 02

.Along the side y = 0 the normal is in the opposite direction njdS = (ey)jdx and

ijnj =

2 00 2

01

= 0

2

.Hence we get no net contribution from the two surfaces of S at y = 0, and y = L

L0

ijnj(for y=0 and y=L) =

L0

02

+ 02

dx =0

0

Therefore we have for the x-component

V

u(u n)dS =V

pnxdS x-comp from isotropic stress tensor

=

6

L0

u(u)dy =0 at x=0

+

L0

u(u)dy =2L2 at x=L

+

L0

u(v)dx =0 at y=0

+

L0

u(v)dx =x(L) at y=L

= 2L3 2L3

2=2L3

2

and the y-component V

v(u n)dS =V

pnydS x-comp from isotropic stress tensor

=

L0

v(u)dy =0 at x=0

+

L0

v(u)dy =y(L) at x=L

+

L0

v(v)dx =0 at y=0

+

L0

v(v)dx =2L2 at y=L

= 2L3

2+ 2L3 =

2L3

2

Note that you should obtain the same result using Bornoulllis theorem for irrotational flow and integratingthe pressure over the surface of the control volume.

Example 6

Show that the stress tensor Tij = pij for the flow

u = x,

where is constant.

Tensor notation:ui = iklkxl

The stress tensor then becomes:

Tij = pij + (uixj

+ujxi

)=

pij + (

xj(iklkxl) +

xi(jklkxl)

)=

pij + k(ikl

xlxj

+ jklxlxi

)=

pij + k(ikllj + jklli) =pij + k(ikj + jki) =

pij + k(ijk + ijk) = pij

7