Mass Momentum Energy Conservation

Module 2 : Convection

Lecture 12 : Derivation of conservation of energy

Objectives In this class:

Start the derivation of conservation of energy.Utilize earlier derived mass and momentum equations for simplificationShow that the viscous dissipation term is always positive

Conservation of Energy-1

Conservation of mass and momentum are complete and now the last conservation equation i.e.energy is derived.Again we start with the verbal form of the equation and then express this in mathematical termsNet rate of influx of energy into the Control Volume is equal to the rate of accumulation of energywithin the Control Volume.


Energy can enter the Control Volume in the form of Conduction, Convection due to mass enteringthe Control Volume or Work done on the fluid in the Control VolumeEnergy (per unit mass) for fluid consists of kinetic, potential and thermal components:

(12.1)


Consider a Control Volume of size dx X dy X dz where energy transfer due to mass entering inthe y direction is shown. Similar terms will exist for the x and z directions also but are not shownin the figure in the interest of clarity.


The conduction term has been seen alreadyThe rate of energy convected into the CV by virtue of mass entering in the ‘Y’ direction is shownin the figure on the y = 0 planeOn the y = dy plane the regular Taylor series expansion has been used and only the leading termhas been retained, as usual, for the conduction and convection terms


Net rate of heat conducted in:

(12.2)

Net rate of heat convected in:

− (12.2a)

Rate of storage of energy:

(12.3)


Subtract equn (12.2a) from equn (12.3) and group the appropriate terms to get:

(12.4)

Now look at energy transfer due to work. The rate of work done is computed as the dot productof the force and velocity:

(12.5)


Rate of work done is assumed positive if the force and velocity vectors are in the same direction.Only surface forces are used for work calculations and not body forces since the gravitationalpotential energy has already been included in the energy per unit mass of fluid term. Of course,this argument is valid only for gravitational body force term. Need to consider the work done ifother types of body forces exist.


Consider now the work terms on the Control Volume surfaces. Notice the signs of the work termson the different faces of the CV

Conservation of Energy-8a

Consider the top (pink), front (blue) and the right (red) face. The force due to the stress is in thedirection from left to right. The positive 'v' direction is also from left to right. The dot product is

therefore positive as shown.The left face (green), however, has force due to stress from right to left and 'v' in the left to rightdirection, making the work negative. It is not possible to show the bottom and back face whichalso have negative values.


The net work rate on the y = 0 and y = dy face due to the force in the ‘y’ direction istherefore:

Work rate done per unit volume is therefore:

(12.6)


Now let the volume ‘dxdydz’ be shrunk to zeroEqun 12.6 can be modified as:

(12.7)

The last term is zero since it is explicitly multiplied by ‘dy’ which tends to zero.There are two other terms due to forces in the ‘y’ direction on x = 0, x = dx and z = 0, z = dzplanes:


The total rate of work done due to forces in the y direction is therefore:

(12.8)

Work is a scalar and therefore there is an algebraic sum.Similarly there will be three terms each for the work due to forces in the ‘x’ and ‘z’ directionswhich will all be added together to the total work on the control volume.


Total rate of work is therefore:

(12.9)

Each of the product terms in equn (12.9) can be split into two terms and therefore a total of 18terms exist on the right hand side.


From the 18 terms in equn (12.9) there are 9 terms which also appear in the momentumequation (see equn (11.3) with Xz = −g). These 9 terms can therefore be simplified using the

same approximations as used earlier to derive the equn (11.3). Note that equn (11.3) wassimplified later using the continuity equation to obtain equn (11.10a)


The set of the 9 terms which can be simplified using the momentum equation is therefore: (with Xz = −g)

(12.10)


Simplify the RHS of equn (12.10). Note each coloured column adds to a single term below:

(12.11)


9 of the eighteen stress work terms therefore become:

(12.12)

The other 9 terms are:

(12.13)


Substitute the Stokes constitutive equn (11.5) in (12.13) gives:

(12.14)


The blue and yellow terms in the equation (12.14) are considered separately for convenience ofalgebraic manipulations. The terms coloured blue are first considered and simplified. The termscoloured yellow are then simplified and at the end the simplified equations are added together.


Consider the blue terms of (12.14) first. Terms coloured red, green and yellow on LHS arecombined algebraically together to become the same colour terms on RHS as below:

(12.15)


Consider equn (12.15). The terms with same colour on the RHS are grouped together to get thefinal compact form at the bottom:


Now consider the terms marked yellow in equn (12.14). Terms marked with the same colour arecombined together to obtain:


The total contribution from the equn (12.14) is therefore

The portion marked red in this term is always positive and is the viscous dissipation and wedenote this as ‘Q’.


The energy equation therefore becomes:

Need to convert this into a more usable form i.e. variables that are easily measurable. Controlvolume manipulations are complete and now we need some thermodynamic manipulations tocomplete the derivation.

Recap

In this class:

Start the derivation of conservation of energy.Utilize earlier derived mass and momentum equations for simplificationShow that the viscous dissipation term is always positive

DERIVATION OF THE NAVIER STOKES EQUATION

1. CAUCHY’S EQUATION

First we derive Cauchy’s equation using Newton’s second law.

We take a differential fluid element. We consider the element as a material element ( instead of a control volume) and apply Newton’s second law ∙ =

or since ( ) = ( ) ∙ = ( 1)

We express the total force as the sum of body forces and surface forces

∑ = ∑ + ∑ . Thus ( 1) can be written as

∙ = + ( 2)

We cosider the x-component of (Eq 2).

Since = and = ( , , ) we have

∙ = , + , ( 3) We denote the stress tensor ( pressure forces+ viscous forces)

= ,

Body forces: Gravity force Electromagnetic forse Centrifugal force Coriolis force

Surface forces: Pressure forces Viscous forces

the viscous stress tensor =

and strain ( deformation) rate tensor where

= = ⎥⎥⎥⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢⎢⎢⎢

⎣

⎡

∂∂

⎟⎟⎠

⎞⎜⎜⎝

⎛∂∂+

∂∂

⎟⎠⎞

⎜⎝⎛

∂∂+

∂∂

⎟⎟⎠

⎞⎜⎜⎝

⎛∂∂+

∂∂

∂∂

⎟⎟⎠

⎞⎜⎜⎝

⎛∂∂+

∂∂

⎟⎠⎞

⎜⎝⎛

∂∂+

∂∂

⎟⎟⎠

⎞⎜⎜⎝

⎛∂∂+

∂∂

∂∂

zw

zv

yw

zu

xw

yw

zv

yv

yu

xv

xw

zu

xv

yu

xu

21

21

21

21

21

21

Let = ( , , ) , = ( , , ), = ( , , ) be stress vectors on the planes perpendicular to the coordinate axes.

Then the stress vector at any point associated with a plane of unit normal vector = ( , , ) can be expressed as

= + + = ( , , ) .

yz- plane

xz-plane

xy-plane

zx

y

We consider the x-component of the net surface force ∑ , using the figure below.

Using Taylor’s formula we get

1 = −( − ) 2 = ( + )

3 = −( − 2 ) 4 = ( + 2 )

5 = −( − 2 ) 6 = ( + 2 )

Thus

, = 1 + 2 + 3 + 4 + 5 + 6 = ( + + )

If we assume that the only body force is the gravity force, we have

, = ∙ g = ∙ ∙ g

Now from ( 3) ∙ = , + , ( 3)

we have ∙ ∙ = ∙ ∙ g + ( + + )

We divide by and get the equation for the x-component:

∙ = g + + +

or ∙ ( + + + ) = g + + + eq x

In the similar way we derive the following equations for

y component: ∙ ( + + + ) = g + + + eq y

z component: ∙ ( + + + ) = g + + + eq z

Equations eq x,y,z, are called Cauchy’s equations.

THE NAVIER STOKES EQUATION

When considering ∑ , we can separate x components of pressure forces

and viscous forces:

= − + , = , =

In the similar way we can change y-component and z-component

Thus Cauchy’s equations become ∙ ( + + + ) = g − + + + eq A

In the similar way we derive the following equations for

y component: ∙ ( + + + ) = g − + + + eq B

z component: ∙ ( + + + ) = g − + + + eq C

According o the NEWTON’S LOW OF VISCOSITY the viscous stress components are related ( throw a linear combination) to the ( first) dynamic viscosity and the second viscosity . = 2 + , = ( + ) , = ( + ) (*)

= ( + ) , = 2 + , = ( + ) (**)

= ( + ) , = ( + ) = 2 + (***)

We substitute this values in to Cauchy’s equations eq A, B, C and get

THE NAVIER STOKES EQUATIONS for the compressible flow:

x-component:

∙ ( + + + )= g − + 2 + + ( + ) + ( + )

y-component:

∙ ( + + + )= g − + ( + ) + 2 + + ( + )

z-component:

∙ ( + + + )= g − + ( + ) + ( + ) + 2 +

Remark: For an incompressible flow we have = 0 and hence from (*), (**) and (***) = 2

where is the strain rate tensor for the velocity field ),,( wvuV =r

in Cartesian coordinates:

.

2

2

2

21

21

21

21

21

21

22

⎥⎥⎥⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢⎢⎢⎢

⎣

⎡

∂∂

⎟⎟⎠

⎞⎜⎜⎝

⎛∂∂+

∂∂

⎟⎠⎞

⎜⎝⎛

∂∂+

∂∂

⎟⎟⎠

⎞⎜⎜⎝

⎛∂∂+

∂∂

∂∂

⎟⎟⎠

⎞⎜⎜⎝

⎛∂∂+

∂∂

⎟⎠⎞

⎜⎝⎛

∂∂+

∂∂

⎟⎟⎠

⎞⎜⎜⎝

⎛∂∂+

∂∂

∂∂

=

⎥⎥⎥⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢⎢⎢⎢

⎣

⎡

∂∂

⎟⎟⎠

⎞⎜⎜⎝

⎛∂∂+

∂∂

⎟⎠⎞

⎜⎝⎛

∂∂+

∂∂

⎟⎟⎠

⎞⎜⎜⎝

⎛∂∂+

∂∂

∂∂

⎟⎟⎠

⎞⎜⎜⎝

⎛∂∂+

∂∂

⎟⎠⎞

⎜⎝⎛

∂∂+

∂∂

⎟⎟⎠

⎞⎜⎜⎝

⎛∂∂+

∂∂

∂∂

==

zw

zv

yw

zu

xw

yw

zv

yv

yu

xv

xw

zu

xv

yu

xu

zw

zv

yw

zu

xw

yw

zv

yv

yu

xv

xw

zu

xv

yu

xu

ijij

μμμ

μμμ

μμμ

μμετ

In the case when we consider an incompressible , isothermal Newtonian flow (density ρ =const,

viscosity μ =const), with a velocity field ))()()(( x,y,z, w x,y,z, vx,y,zuV =r

we can simplify the Navier-Stokes equations to his form:

x component:

)( 2

2

2

2

2

2

zu

yu

xug

xP

zuw

yuv

xuu

tu

x ∂∂+

∂∂+

∂∂++

∂∂−=⎟⎟

⎠

⎞⎜⎜⎝

⎛∂∂+

∂∂+

∂∂+

∂∂ μρρ

y- component:

)( 2

2

2

2

2

2

zv

yv

xvg

yP

zvw

yvv

xvu

tv

y ∂∂+

∂∂+

∂∂++

∂∂−=⎟⎟

⎠

⎞⎜⎜⎝

⎛∂∂+

∂∂+

∂∂+

∂∂ μρρ

z component:

)( 2

2

2

2

2

2

zw

yw

xwg

zP

zww

ywv

xwu

tw

z ∂∂+

∂∂+

∂∂++

∂∂−=⎟⎟

⎠

⎞⎜⎜⎝

⎛∂∂+

∂∂+

∂∂+

∂∂ μρρ

[ The vector form for these equations: VgPDt

VD rrr

2∇++−∇= μρρ ]

Gen

eral

Hea

t Con

duct

ion

Ei

Equ

atio

n

El

Δ&

&&

&&

&&

P.Ta

lukd

ar/M

ech-

IITD

tE

GQ

QQ

QQ

Qel

emen

tel

emen

tz

zy

yx

xz

yx

ΔΔ

=+

−−

−+

+Δ

+Δ

+Δ

+

)T

T.(z.y

.x.C

)T

T(m

CE

EE

tt

tt

tt

tt

tel

emen

t−

ΔΔ

Δρ

=−

=−

=Δ

Δ+

Δ+

Δ+

z.y

.x.g

VgG

elem

ent

elem

ent

ΔΔ

Δ=

=&

&&

tE

GQ

QQ

QQ

Qel

emen

tel

emen

tz

zy

yx

xz

yx

ΔΔ

=+

−−

−+

+Δ

+Δ

+Δ

+&

&&

&&

&&

tT

Tz

.y.x

.Cz

.y.x

.gQ

QQ

QQ

Qt

tt

zz

yy

xx

zy

xΔ−

ΔΔ

Δρ

=Δ

ΔΔ

+−

−−

++

Δ+

Δ+

Δ+

Δ+

&&

&&

&&

&

TT

Cg

QQ

1Q

Q1

QQ

1t

tt

zz

zy

yy

xx

x−

ρ=

+−

−−

Δ+

Δ+

Δ+

Δ+

&&

&&

&&

&

tC

gz

y.x

yz

.xx

z.y

Δρ

=+

ΔΔ

Δ−

ΔΔ

Δ−

ΔΔ

Δ− P.

Talu

kdar

/Mec

h-IIT

D

TT

CQ

Q1

QQ

1Q

Q1

tt

tz

zz

yy

yx

xx

−−

−−

Δ+

Δ+

Δ+

Δ+

&&

&&

&&

&

tC

gz

QQ

y.x

yz

.xx

QQ

z.y

tt

tz

zz

yy

yx

xx

Δρ

=+

ΔΔ

Δ−

ΔΔ

Δ−

ΔΔ

Δ−

Δ+

Δ+

Δ+

Δ+

⎞⎛

∂∂

⎞⎛

∂∂

∂T

T1

Q1

QQ

1&

&&

⎟ ⎠⎞⎜ ⎝⎛

∂∂−

∂∂=

⎟ ⎠⎞⎜ ⎝⎛

∂∂Δ

Δ−

∂∂Δ

Δ=

∂∂Δ

Δ=

Δ−

ΔΔ

Δ+→

ΔxT

kx

xTz

.y.k

xz

.y1xQ

z.y1

xQ

Qz

.y1lim

xx

xx

0x

⎟⎟ ⎠⎞⎜⎜ ⎝⎛

∂∂−

∂∂=

⎟⎟ ⎠⎞⎜⎜ ⎝⎛

∂∂Δ

Δ−

∂∂Δ

Δ=

∂∂

ΔΔ

=Δ

−

ΔΔ

Δ+

→Δ

yTk

yyT

z.x

.ky

zx1

yQz

x1y

QQ

zx1

limy

yy

y

0y

&&

&

⎠⎝

∂∂

⎠⎝

∂∂

ΔΔ

∂Δ

ΔΔ

ΔΔ

→Δ

yy

yy

z.x

yz

.xy

z.x

0y

⎟ ⎠⎞⎜ ⎝⎛

∂∂−

∂∂=

⎟ ⎠⎞⎜ ⎝⎛

∂∂Δ

Δ−

∂∂Δ

Δ=

∂∂Δ

Δ=

Δ−

ΔΔ

Δ+→

ΔzT

kz

zTy

.x.k

zy

.x1zQ

y.x1

zQ

Qy

.x1lim

zz

zz

0z

&&

&

tTC

gzT

kz

yTk

yxT

kx

∂∂ρ

=+ ⎟ ⎠⎞

⎜ ⎝⎛∂∂

∂∂+ ⎟⎟ ⎠⎞

⎜⎜ ⎝⎛∂∂

∂∂+ ⎟ ⎠⎞

⎜ ⎝⎛∂∂

∂∂&

Und

er w

hat c

ondi

tion?

T1

gT

TT

22

2∂

∂∂

∂&

P.Ta

lukd

ar/M

ech-

IITD

tT1

kgzT

yTxT

22

2∂∂

α=

+∂∂

+∂∂

+∂∂

0kg

TT

T2

2

2

2

2

2

=+

∂∂+

∂∂+

∂∂& k

zy

x2

22

∂∂

∂

tT1

zTyT

xT2

2

2

2

2

2

∂∂α

=∂∂

+∂∂

+∂∂

tz

yx

∂α

∂∂

∂

0zT

yTxT

2

2

2

2

2

2

=∂∂

+∂∂

+∂∂

y

P.Ta

lukd

ar/M

ech-

IITD

Documents

Mass Momentum Energy Conservation