CONSERVATION OF MASS AND BALANCE OF LINEAR...

J. N. Reddy Mass-Momenta - 1

CONSERVATION OF MASS AND BALANCE OF LINEAR MOMENTUM

Summary of integral theorems

Material time derivative

Reynolds’ transport theorem

Principle of conservation of mass

Principle of balance of linear momentum


Divergence Theorem

General Theorem

ˆ d d

n F F

ˆ * *ds d

n F F

Gradient Theorem

ˆ d d

nF F

Curl Theoremˆ d d

n F F

INTEGRAL THEOREMS


Gradient Theorem

n F d F d

INTEGRAL THEOREMS

The gradient of a function F represents the rate of change of F with respect to the coordinate directions.the partial derivative with respect to x, for example, gives the rate of change of F in the x direction.

F normal to the surfaceLevel surfaces


Gradient of a Scalar Function

a first- order tensor,that is, aF vector

in rectangular Cartesian system

in cylindrical coordinate system

e e e

e e e

ˆ ˆ ˆ

ˆ ˆ ˆ

x y z

x y x

x y z

F F FFx y z

1

1

e e e

e e e

ˆ ˆ ˆ

ˆ ˆ ˆ

r z

r z

r r z

F F FFr r z


Divergence Theorem

ˆ d d

n F F

INTEGRAL THEOREMS

The divergence represents the volume density of the outward flux of a vector field F from an infinitesimal volume around a given point. It is a local measure of its "outgoingness."

d

●

d

dFlux


Divergence of First-Order Tensors

F a zeroth tensor,that is, a scalar



e e e

F

ˆ ˆ ˆx y z

yx z

x y z

FF Fx y z

1

1

e e e

F

ˆ ˆ ˆ

( )

r z

zr

r r z

FrF F rr r z


Divergence of Second-order Tensors

S afirst- order tensor,that is, a vector

1

1

1

S e e

e

S e

e

ˆ ˆ

ˆ

( ) ˆ

( ) ˆ

yx xy yy zyxx zxx y

yzxz zzz

zrrr rr

zrr

S S S FS Fx y z x y z

SS Fx y z

SrS S r Sr r z

SrS S r Sr r z

e( ) ˆrz z zrz

rS S Srr r z




Curl Theorem (Stoke’s Theorem)

n F Fˆ d d

INTEGRAL THEOREMS

The curl of a vector F describes the infinitesimal rotation of F. A physical interpretation is as follows. Suppose the vector field describes the velocity field F = v of a fluid flow, say, in a large tank of liquid, and a small spherical ball is located within the fluid (the center of the ball being fixed at a certain point but free to rotate about an axis perpendicular to the plane of the flow). If the ball has a rough surface, the fluid flowing past the ball will make it rotate. The rotation axis (oriented according to the right hand rule) points in the direction of the curl of the field at the center of the ball, and the angular speed of the rotation is half the magnitude of the curl at this point.


Curl of a Vector Function in (x,y,z) system

A afirst- order tensor,that is, a vector

A e e e e e e

e e e e e e

e e e e e e

e e e

ˆ ˆ ˆ ˆ ˆ ˆ

ˆ ˆ ˆ ˆ ˆ ˆ

ˆ ˆ ˆ ˆ ˆ ˆ

ˆ ˆ ˆ

yx zx x x y x z

yx zy x y y y z

yx zz x z y z z

y yz x z xx y z

AA Ax x x

AA Ay y y

AA Az z z

A AA A A Ay z z x x y

0 eyez

ez 0

0

ex

ex

ey


Material Time DerivativeMaterial time derivative is the time derivative of a functionin spatial description with the material coordinate X held constant. We denote it with D/Dt (also called substantive derivative).

const

const const const

( , ) :

( , ) : i

i

DtDt t t

xDtDt t t t x

DDt t

X

X x X

X

x

v

Material (or Lagrange) description

Spatial (or Eulerian) description

Local change

Translational change


Suppose that a motion is described by the one-dimensional mapping, for t > 0. Determine (a) the velocities and accelerations in the spatial and material descriptions, and (b) the time derivative of a function

in the spatial and material descriptions.

The velocity can be expressed in the material and spatial coordinates as

The acceleration in the two descriptions is

EXAMPLE 5-1

1x t X

2( , )X t Xt

Problem statement:

Solution:/v Dx Dt

1( , ) , ( , ) ( , )Dx x xv X t X v x t X x t

Dt t t

/a Dv Dt

0 0( , ) ( , ),Dv X t v Dv x t v va a vDt t Dt t x


The material time derivative of in the material description is simply

The material time derivative of

in the spatial description is

which is the same as that calculated before, except that it is expressed in terms of the current coordinate, x.

EXAMPLE 5-1 (continued)( , )X t

2( , ) ( , )D X t X t XtDt t

2 2 1( , ) ( , ) / ( )x t X x t t xt t

2 2

2

2 21 1 1 1 1( )

D xt xt x t xtvDt t x t t t t t

J. N. Reddy

Consider the motion of a body described by the mapping

Determine the density as a function position x and time t.

Mass-Momenta - 13

11 2 2 3 3

11, ,Xx x X x X

tX

221

1 1 2 321

0 01

X X

x xv or

Thus, we have

, ,( )

ii

fixed fixed

xD vDt t t

Xv x v vtX

EXAMPLE 5-2Problem statement:

Solution:First, compute the velocity components


Next, compute

Integrating this equation with respect to t, we obtain

EXAMPLE 5-2 (continued)

11

1

2 21

v .D XxDt tX

1 1

1 1

1 2 21 1

ln ln lnX XD Dt ctX tX

where c is a constant of integration. If 0

0

at time = 0, then wehave ln ln ,and the density

becomest c c

2 00 1 2

1

11( )

tXtx


REYNOLDS’ TRANSPORT THEOREM

Since is changing with time, we cannot take the differentiation through the integral. However, if the integral were over the volume in the reference configuration (which is fixed), it is possible to interchange the integration and differentiation becauseD/Dt is differential with respect to time keeping Xconstant. The transformation and lets us to do exactly that.

Let each element of mass in the volume with closed boundary moves with the velocity . Let be any function. We are interested in the material time derivative of the integral

( , )tv x( )t

x( , )t

x x X( , )t

( , )D t dDt

x

( )t

0d J d


0 0

0

0 0

0v v

v v v

x X

n

( )

( )

( ) ( )

( ) ( )

( , ) [ ( , ),t]

ˆ

t

t

t t

t t

D D D DJt d t d dDt Dt Dt Dt

D DJ d dDt Dt

d d

t

J

t t

d

J

x

v d

REYNOLDS’ TRANSPORT THEOREMWe have

vn( ) ( ) ( )

ˆ( , )t t t

D t d dD t

dt

x

Thus


REYNOLDS’ TRANSPORT THEOREM

Thus the time rate of change of the integral of a function over a moving volume is equal to the integral of

the local time rate of plus the net outflow of over the surface of the moving volume. Here can be a scalar or a tensor of any order.

x( , )tx( , )t x( , )t

x( , )t

vn( ) ( ) ( )

ˆ( , )t t t

D t d dD t

dt

x

v

v

( ) ( )

( )

( , )t t

t

D t d dDt t

D dDt

x


The Principle of Conservation of Massin Spatial Description

ρdt

If a continuous medium of density ρ fills the volume at time t, the rate of increase of the total mass inside the volume is

( )t

The rate of mass outflow through the surface element is where ( is the outward normal). Hence, the rate of inflow through the entire surface is

nρv d nˆnv v n

v n vˆnρv d ρ d ρ d

d


The Principle of Conservation of Mass(continued)

0v vρ ρd ρ d ρ dt t

If no mass is created or destroyed inside the volume , this must be equal to rate of mass inflow through the surface. Equating the rate of mass inflow to the rate of increase of mass, we obtain

0 0v vorρ Dρρt Dt

In we shrink the volume to a point, we obtain

( )t


Conservation of Mass (continued)

This is the invariant form (i.e., valid in any coordinate system and dimension) of the statement of the principle of conservation of mass, also known as the continuity equation (this author does not prefer this name).

0 ( )ρ= ∇⋅ v

♠ Steady-state flows :

♠ Incompressible fluids :

0 = ∇⋅ v

0ρt

0DρDt


ˆ ˆ ˆ ˆˆ ˆ( )

( ) ( ) ( )

x y z

x y z

v v vx y z

v v vx y z

v i j k i j k

0

tyx zvv v

x y z

Cartesian component form

( ) 0=⋅∇+∂∂ vρ

tρ


J. N. Reddy22

The Principle of Conservation of Massin one dimension

x x +∆x∆x

A Infinitesimal volume, ∆V

ρvx

ρvx +∆x

x

If no mass is created or destroyed inside ∆V, the timerate of change of mass must equal to the rate of inflow of mass through the surface.

J. N. Reddy23

The Principle of Conservation of Massin one dimension (continued)

0

t t tx x x

t t t x x x

A x A xv A v A

t

A A v A v At x

vt x

vt x


1 1 0( )r zv r v vt r r r z

Cylindrical component form

( ) 0=⋅∇+∂∂ vρ

tρ

1e e eˆ ˆ ˆr zr r z


1v ( ) zr vrv v rr r z


Alternative form of REYNOLDS’ TRANSPORT THEOREM

Replace the function with in the Reynolds transport theorem and obtain

x( , )t x x( , ) ( , )t t

where the equation resulting from conservation of mass is used in arriving at the last step. We have

( , ) ( , )D Dt t d dDt Dt

x x

v

v

( ) ( )

( ) ( )

( , ) ( , )t t

t t

DD t t d dDt Dt

D D Dd dDt Dt Dt

x x


CONSERVATION OF MASS: Material Description

0

0 0ρ d ρd

Let be an arbitrary material volume occupied by the body in the reference configuration, and be the volume occupied by the body in the current configuration.

Conservation of mass states that if the mass is neither created nor destroyed during the motion from to , the mass of the material volume is conserved:

0

0

0d J d Since the volumes in the two configurations are related by

0ρ Jρ, we obtain


Consider a water hose with conical-shaped nozzle at its end, as shown in the figure.(a) Determine the pumping capacity required in order The velocity of the water (assuming incompressible for the present case) exiting the nozzle be 25 m/s. (b) If the hose is connected toa rotating sprinkler through its base, determine the average speed of the water Leaving the sprinkler nozzle.

AN EXAMPLEProblem statement:


AN EXAMPLESolution:(a) The principle of conservation of mass for steady one-dimensional flow requiresIf the exit of the nozzle is taken as the section 2, we cancalculate the flow at section 1 as (for an incompressible fluid, )

1 1 1 2 2 2A v A v

1 2 3 2

31 1 1 2 2

20 10 25 0 00254

( ) . m / s.Q A v A v

(b) The average speed of the water leaving the sprinkler nozzle can be calculated using the principle of conservation of mass for steady one-dimensional flow. We obtain

11 2 2 2 2 3 2

2 0 0052 3212 5 10

. m/s.( . )

QQ A v vd


BALANCE OF LINEAR MOMENTUMThe time rate of change of total linear momentum of a given continuum equals the vector sum of all external forces acting on the continuum. This also known as Newton’s Second Law.

Newton's First Law. Newton's First Law states that an object will remain at rest or in uniform motion in a straight line unless acted upon by an external force.

t d

f d

d

d

0

t f v

vn f

vf

ˆ

Dd d dDt

Dd d dDt

D dDt

ρ ρ

ρ ρ

ρ ρ


Conservation of Linear Momentum(continued)

Vector form of the equation of motion

vσ f Dρ ρDt

ˆ( )ˆ ˆ ˆ ˆ

ˆ ˆ ˆ

i ik ji j i i i

k

ji jii ii i i i i

j j

D vρf ρx Dt

σ σDv Dvρf ρ ρf ρx Dt x Dt

ee e e e

e e e

Cartesian Component Form


Conservation of Linear Momentum(continued)

3111 21 11

1 2 3

3212 22 22

1 2 3

13 23 33 33

1 2 3

ji ii

j

σ Dvρf ρx Dt

σσ σ Dvρf ρx x x Dt

σσ σ Dvρf ρx x x Dtσ σ σ Dvρf ρx x x Dt

Cartesian Component Form (expanded form)


2x

1x

3x

1dx2dx

3dx

3333 3

3dx

xσ

σ∂

+∂

3232 2

2dx

xσ

σ∂

+∂

2222 2

2dx

xσ

σ∂

+∂

1111 1

1dx

xσ

σ∂

+∂ 33σ

22σ

1212 2

2dx

xσ

σ∂

+∂

32σ

12σ

2121 1

1dx

xσ

σ∂

+∂

23σ13σ

1313 3

3dx

xσ

σ∂

+∂

2323 3

3dx

xσ

σ∂

+∂

3131 1

1dx

xσ

σ∂

+∂

31σ21σ

11σ

0 3fρ

0 1fρ

0 2fρ

0 1 0 2 0 3, , body force components (per unit mass)Origin is at the center of the parallelopiped

f f fρ ρ ρ =

EQUATIONS OF MOTIONin Rectangular Cartesian System

Alternative approach to the derivation: Sum all the forces on the infinitesimal block of dimensions 1 2 3, , anddx dx dx


Conservation of Linear MomentumSpecial Cases

0, for solid bodies

0, for steady state

Fluid Mechanics

Solid Mechanics

Viscous stress tensor

Hydrostatic pressure

vf v vt

ρ ρ

vf v v ,t

ρ ρ IP

vft

ρ ρ


EQUATIONS OF MOTIONFluid Mechanics

@¾11@x1

+@¾21

@x2+

@¾31

@x3+ ½f1 = ½

µ@v1@t

+ v1@v1@x1

+ v2@v1@x2

+ v3@v1@x3

¶

@¾12@x1

+@¾22

@x2+

@¾32

@x3+ ½f2 = ½

µ@v2@t

+ v1@v2@x1

+ v2@v2@x2

+ v3@v2@x3

¶

@¾13@x1

+@¾23

@x2+

@¾33

@x3+ ½f3 = ½

µ@v3@t

+ v1@v3@x1

+ v2@v3@x2

+ v3@v3@x3

¶

@¾xx

@x+

@¾yx

@y+

@¾zx

@z+ ½fx = ½

µ@vx@t

+ vx@vx@x

+ vy@vx@y

+ vz@vx@z

¶

@¾xy

@x+

@¾yy

@y+

@¾zy

@z+ ½fy = ½

µ@vy@t

+ vx@vy@x

+ vy@vy@y

+ vz@vy@z

¶

@¾xz

@x+

@¾yz

@y+

@¾zz

@z+ ½fz = ½

µ@vz@t

+ vx@vz@x

+ vy@vz@y

+ vz@vz@z

¶


@¾rr

@r+

1

r

@¾rµ

@µ+

@¾rz

@z+

1

r(¾rr ¡ ¾µµ) + ½fr

= ½

Ã@vr@t

+ vr@vr@r

+vµr

@vr@µ

+ vz@vr@z

¡ v2µr

!

@¾rµ

@r+

1

r

@¾µµ

@µ+

@¾µz

@z+

2¾rµ

r+ ½fµ

= ½

µ@vµ@t

+ vr@vµ@r

+vµvr

r+

vµr

@vµ@µ

+ vz@vµ@z

¶

@¾rz

@r+

1

r

@¾µz

@µ+

@¾zz

@z+

¾rz

r+ ½fz

= ½

µ@vz@t

+ vr@vz@r

+vµr

@vz@µ

+ vz@vz@z

¶

EQUATIONS OF MOTIONin cylindrical coordinates


dθ

eθ

ez

er

rrrr dr

rσ

σ∂

+∂

rr dr

rθ

θσ

σ∂

+∂

zrzr dr

rσ

σ∂

+∂

rr dθθ

σσ θ

θ∂

+∂

dθθθθ

σσ θ

θ∂

+∂

zz dθθ

σσ θ

θ∂

+∂

zzzz dz

zσ

σ∂

+∂

rzrz dz

zσ

σ∂

+∂

zz dz

zθ

θσ

σ∂

+∂

rzσ

zzσzθσ

rrσ

rθσ

zrσrθσ

θθσ

zθσdr

r

dz

z

r

θ

EQUATIONS OF MOTIONin cylindrical coordinates

Alternative approach to the derivation: Sum all the forces on the infinitesimal block of dimensions , , anddr rd dz


0

0 0

0 0

0 0

0 0 0

2

0 0 0 02

( ) ( ) ,

ˆ ,

d d

d d d

D d d dDt t t t

f x f X

σ P N P

u uv

Tn

0, (or ).d d dv J dV d J d σ a P A

2

0 0 0 2t

uS F fT T

2

0 0 0 2t

uP fT

EQUATIONS OF MOTIONin material description

J. N. Reddy Linear momentum 25

EXAMPLE: COUETTE FLOW

y

xz Flow

Assume:

steady state:

incompressible:

no body forces:

∇.v = 0

Flow of a viscous fluid between parallel plates (Couette Flow)

d

σzx

σyzσyy

σyx

σxx

σzz σzy

σxz

1 0xv v v

2 30 0,v v

0( )t

1

1

0vx

1 2( )( )x

v f xv f y

1 2 3 0f f f

J. N. Reddy Linear momentum 39

@¾xx

@x+

@¾yx

@y+

@¾zx

@z+ ½fx = ½

µ@vx@t

+ vx@vx@x

+ vy@vx@y

+ vz@vx@z

¶

@¾xy

@x+

@¾yy

@y+

@¾zy

@z+ ½fy = ½

µ@vy@t

+ vx@vy@x

+ vy@vy@y

+ vz@vy@z

¶

@¾xz

@x+

@¾yz

@y+

@¾zz

@z+ ½fz = ½

µ@vz@t

+ vx@vz@x

+ vy@vz@y

+ vz@vz@z

¶

For viscous fluids, the total stresses are given by

The viscous stresses are proportional to the gradient of the velocity field, and they are independent of x and z.

, ,, ,

xx xx yy yy zz zz

xy xy xz xz yz yz

P P P

ij

COUETTE FLOW (continued)

J. N. Reddy J.N. Reddy-ENGR214-L07(Feb. 8, 2000)

40

Thus the linear momentum equations become

This implies that the pressure P is only function of x. Thus, we have one equation

Thus we have

COUETTE FLOW (continued)

0 0 0, ,yxP P Px y y z

yxPx y

2 2

2 2yx x xv vPy y x y

J. N. Reddy

Problem statement:Given the following state of stress in a kinematicallyinfinitesimal deformation ,

determine the body force components for which the stress field describes a state of equilibrium.

Solution: The body force components are

AN EXAMPLE

( )ij ji 2

11 1 12 1 2 3 13 1 2

2 222 1 2 3 23 33 1 2 3

2 7 4 1 3

3 2 5 0 5 3 3

, , ,, ,

x x x x x xx x x x x x

3111 211 1 1

1 2 3

3212 222 2 2

1 2 3

13 23 333

1 2 3

4 4 0 0

4 4 0 0

1 0 3 4

[( ) ( ) ] ,

[( ) ( ) ] ,

[ ] .

f x xx x x

f x xx x x

fx x x

Mass-Momenta - 41


EXERCISESDetermine if the following velocity fields for an incompressible flow satisfies the continuity equation:

2 2 21 21 1 2 2 1 2 1 22 2( , ) , ( , ) wherex xv x x v x x r x x

r r

The velocity distribution between two parallel plates separated by distance b is

where y is measured from and normal to the bottom plate, x is taken along the plates, vx is the velocity component parallel to the plates, v0 is the velocity of the top plate in the x direction, and c is a constant. Determine if the velocity field satisfies the continuity equation and find the volume rate of flow and the average velocity.

1.

2.

0 1 0 0 0( ) , , , ,x y zy y yv y v c v v y bb b b


EXERCISES

For a cantilevered beam bent by a point load at the free end, the bending moment M about the y-axis is given by M = −Px. The axial stress is given by

where I is the moment of inertia of the cross section about the y-axis. Starting with this equation, use the two-dimensional equilibrium equations to determine stresses and and as functions of x and z.

,xxMz PxzI I

zz xz

If the stress field in a body has the following components in a rectangular Cartesian coordinate system

where a and b constants, determine the body force components necessary for the body to be in equilibrium.

2 2 21 2 2 1

2 2 2 22 1 2

23

2

01 3 03

0 0 2

( )

[ ] ( ) ( )

x x b x x

a b x x x b x

bx

3.

4.

CONSERVATION OF MASS AND BALANCE OF LINEAR...

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