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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.
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Conservation of Energy-2
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)
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Conservation of Energy-3
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
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Conservation of Energy-4
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
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Conservation of Energy-5
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)
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Conservation of Energy-6
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)
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Conservation of Energy-7
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.
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Conservation of Energy-8
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
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Conservation of Energy-9
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)
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Conservation of Energy-10
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
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Conservation of Energy-11
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.
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Conservation of Energy-12
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)
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Conservation of Energy-13
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)
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Conservation of Energy-14
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)
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Conservation of Energy-16
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)
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Conservation of Energy-17
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)
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Conservation of Energy-18
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
+ + + +
+ + +
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Conservation of Energy-19
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
+ + + +
+ + + +
= + + + + +
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Conservation of Energy-20
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
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Conservation of Energy-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.