Energy Conservation1

7/31/2019 Energy Conservation1

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Conservation of Energy-1

Derivation

Conservation of mass and momentum are

complete and now the last conservation

equation i.e. energy is derived.

equation and then express this in

mathematical terms

Net rate of influx of energy into the ControlVolume is equal to the rate of accumulation of

energy within the Control Volume.


2/21


Derivation

Energy can enter the Control Volume in the

form of Conduction, Convection due to mass

entering the Control Volume or Work done on

Energy(per unit mass) for fluid consists of

kinetic, potential and thermal components:

2

2

thermal kinetic potentialenergy e e e

Vu gh

= + +

= + + (12.1)


3/21


Derivation

Consider first, the energy interaction due to

conduction and convection due to massentering a Control Volume of size dx X dy X dz

dxdzy

Tk

dxdzdyy

Tk

ydxdz

y

Tk

vedxdz dxdzdyveyvedxdz )(

+

Z

Y

X


4/21


Derivation

The conduction term has been seen already

The rate of energy convected into the CV by

virtue of mass entering in the Y direction is

On the y=dy plane the regular Taylor series

expansion has been used and only the leading

term has been retained, as usual, for theconduction and convection terms


5/21


Derivation

Net rate of heat conducted in:

T T Tk k k

y y x x z z

+ +

(12.2)

Net rate of heat convected in:

Rate of storage of energy:

( ) ( ) ( )ue ve we

x y z

+ +

( )e dxdydzt

(12.2)

(12.3)


6/21


Derivation

Add equn (12.2) and equn (12.3) and group the

appropriate terms to get:

e e e eu v w e u v w

+ + + + + + +

Now look at energy transfer due to work.

Positive work rate if the force and velocity

vectors are in the same direction

continuity0

(12.4)

v.FW = (12.5)


7/21


Derivation

Note that we use only surface forces for work

calculations. Body forces are not used sincethe potential energy has already been

term. Of course, this argument is valid only for

gravitational body force term. Need to

consider the work done if other types of body

forces exist and can be added to the gravity

contribution.


8/21


Derivation

Consider now the work terms on the Control

Volume surfaces. Notice the signs of the workterms on the different faces of the CV

Positive work

dxdzvyy )(

(

)( )

yy

yy

dxdzv

v

dxdzdy v dyy y

+

+

( )( )zy zy vdz v dz dxdyz z

+ +

x

y

z

( )( )xy xyv

dx v dx dydzx x

+ +

Positive work

Positive work

Negative work


9/21


Derivation

The net work rate on the y=0 and y=dy face due

to the force in the y direction is therefore:

( )( )yy yy yyv

dxdz dxdzdy v dy dxdz vy y

+ +

Work rate done per unit volume is therefore:

( )

yy yy yy

yy yy

v vdxdzdy v dxdz dy dxdzdy dyy y y y

vv dxdzdy dxdzdy dy

y y y

= + +

= +

( )yy yyW v

v dy

dxdydz y y y

= +

(12.6)


10/21


Derivation

Now let the volume dxdydz be shrunk to zero

Equn 12.6 can be modified as:

( )yy yyW v

v dydxdydz y y y

= +

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=dz planes:

(12.7)= ( )yyvy

( ), ( )xy zyv vy y


11/21


Derivation

The total rate of work done due to forces in

the y direction is therefore:

x zxy yy zy

Wv v v

dxd d

= + +

(12.8)

Work is a scalar and therefore there is analgebraic sum.

Similarly there will be three terms each for thework due to forces in the x and z directionswhich will all be added together to the totalwork on the control volume.


12/21


Derivation

Total rate of work is therefore:

( ) ( ) ( )x y

xx yx zx

Wu u u

dxdydz z

= + +

Each of the product terms in equn (12.9) can

be split into two terms and therefore a total of

18 terms exist in equn (12.9)

( ) ( ) ( )x y z

( ) ( ) ( )y z

xy yy zy

xz yz zz

v v v

w w wx

+ + +

+ + +

(12.9)


13/21


Derivation

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)

x y

x y z

y z

xx yx zx

xy yy zy

xz yz zz

u u u uu u u v wz t x y z

v v v vv v u v w

t x y z

w w w ww w u v w

x t x y z

+ + = + + +

+ + + + + + +

+ + + + + + + +

gw

(12.10)


14/21


Derivation Simplify the RHS of equn (12.10). Note each

coloured column adds to a single term below:

2 2 2 21 1 1 1

2 2 2 2

u u u u u u u uu u v w u v w

t x y z t x y z

+ + + + + +

2 2 2 21 1 1 1+2 2 2 2

v v v v v v v vv u v w u v wt x y z t x y z

w w ww u v

t x

+ + + + + +

+ +

2 2 2 2

2

1 1 1 1+

2 2 2 2

1 1

2 2

w w w w ww u v w

y z t x y zgw gw

Vu

t

+ + + +

+

+

2 2 21 1

2 2

V V Vv w

y z

gw

+ +

+2 2 2 2where V u v w= + + (12.11)


15/21


16/21


Derivation

Substitute the Stokes constitutive equn (11.5)

in (12.13) gives:

2

u u u v u w u u .

3

2( . ) 2

3

2( . ) 2

3

ux x y x y x z z

v v u v v v w vP u

y y y x x z y z

w w u w w vP u

z z z x x z

+ + + + +

+ + + + + +

+ + + + + +

w w

y y

(12.14)


17/21


Derivation

Consider terms marked in blue in equn (12.14)

Red, green, yellow terms on LHS become thecorresponding colour terms on RHS purely

2 2

2 2

2 22 2

3 3

2 22 2

3 3

u u v w u u u u v u u wP P

x x y z x x x x y x x z

v u v w v v v v v u v wP P

y x y z y y y y y x y z

wP

+ + +

+ + +

2 22 2

2 23 3

u v w w w w w v w u wP

z x y z z z z z y z x z

+ + +

(12.15)


18/21


Derivation

Consider equn (12.15). The terms with same

colour are grouped to together:2 2 2

2 22

u u v w u u u u v u u wP

+ + + +

2 2 2

2 22

3 3

v u v w v v v v v u v wP

y x y z y y y y y x y z

+ + + +

2 2 2

2 22

2 2

23 3

2Add together to get -P .u

3

w u v w w w w w v w u w

P z x y z z z z z y z x z

u v u w w v

x y x z z y

+ + + +

+ + +


19/21


Derivation

Now consider the terms marked yellow in

equn (12.14). They can be combined as:

u v u w u u

+ + +

2 22

u v v v w v

y x x z y z

u w w v w w

z x x z y y

u v w u v w

y x x z z y

+ + + +

+ + + +

= + + + + +


20/21


Derivation

The total contribution from the equn (12.14)

+

+

222

3

2

y

v

z

w

z

w

x

u

y

v

x

u

This term is always positive and is the viscous

dissipation we denote this as Q.

+

+

+

+

+

+

222

y

w

z

v

z

u

x

w

x

v

y

u


21/21


Derivation

The energy equation therefore becomes:

QuPgwV

Dt

D

z

Tk

zy

Tk

yx

Tk

x+++

+

+

= .

2Dt

De 2

Need to convert this into a more usable formi.e. variables that are easily measurable.

Control volume manipulations are complete

and now we need some thermodynamicmanipulations to complete the derivation.

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Energy Conservation1