Download Equations of Motion and Control Volume Analysis in Fluid Dynamics and more Slides Acting in PDF only on Docsity!
bjc 6.1 4/15/
C HAPTER 6
THE CONSERVATION EQUATIONS
6.1 LEIBNIZ ’ RULE FOR DIFFERENTIATION OF INTEGRALS
6.1.1 D IFFERENTIATION UNDER THE INTEGRAL SIGN
According to the fundamental theorem of calculus if is a smooth function and
the integral of is
(6.1)
then the derivative of is
. (6.2)
Similarly if
(6.3)
then
(6.4)
Suppose the function depends on two variables and the integral is a definite integral.
(6.5)
where and are constant. The derivative with respect to is
f f
I x( ) f ( x') dx' cons tant
x
I x( )
d x
dI (^) = f ( x)
I x( ) f ( x') dx' x
cons tant
dx
dI (^) = –f ( x)
f
I t( ) f ( x' ,t) dx' a
b
a b t
Leibniz’ rule for differentiation of integrals
4/15/13 6.2 bjc
(6.6)
The order of the operations of integration and differentiation can be exchanged and so it is permissible to bring the derivative under the integral sign.
We are interested in applications to compressible flow and so from here on we
will interpret the variable as time. Now suppose that both the kernel of the inte- gral and the limits of integration depend on time.
(6.7)
This situation is shown schematically below with movement of the boundaries indicated.
Figure 6.1 Integration with a moving boundary. The function is shown at one instant in time.
Using the chain rule the substantial derivative of (6.7) is
(6.8)
Now make use of the results in (6.2), (6.4) and (6.6). Equation (6.8) becomes,
dI t( ) dt
,t
, (^) f ( x' ,t) dx' a
b
t
I t a t( , ( ) ,b t( )) f ( x' ,t) dx' a t( )
b t( )
f ( x t, )
x a t( ) (^) b t( )
f ( x t, 0 )
f ( x t, )
DI
Dt
,t
,I
,a
,I da dt
,b
,I db dt
= + +^ ------
Leibniz’ rule for differentiation of integrals
4/15/13 6.4 bjc
Leibniz’ rule extended to three dimensions describes the time rate of change of
the amount of contained inside.
. (6.10)
Note that when (6.9) is generalized to three dimensions the boundary term in (6.9) becomes a surface integral. Equation (6.10) can be expressed in words as follows.
(6.11)
The Leibniz relationship (6.10) is fundamental to the development of the transport
theory of continuous media. The velocity vector is that of the control volume
surface itself. If the medium is a moving fluid the surface velocity is specified
independently of the fluid velocity. Consider a fluid with velocity vector which is a function of space and time.
F V
D
Dt
------ F Vd V t( )
,F
,t
------- Vd V t( )
F U (^) A •n dA A t( )
Rate of change of the total amount of F ª« inside V ®«
« «
©
« «
« «
¨ ¬ Rate of change due to changes ª« of F within V ®«
©
« «
¨ ¬
Rate of change due to movement of the surface A enclosing ª« more or less F within V®«
« «
©
« «
« «
¨ ¬
= +
U A
U A
U U
UA
U (^) A
V (t)
x 1
x
x
mij Q
G
n(t)
dA
UA
UA
U(x1,x2,x3,t)
heat transfer
body force
surface stress
Conservation of mass
bjc 6.5 4/15/
Figure 6.3 Control volume de fi ned in a fl ow fi eld subject to sur- face stresses, body forces and heat conduction.
Now let the velocity of each surface element of the control volume be the same
as the velocity of the flow,. In effect, we assume that the surface is
attached to the fluid and therefore the control volume always contains the same set of fluid elements. This is called a Lagrangian control volume. In this case Leibnitz’ rule becomes
. (6.12)
Use the Gauss theorem to convert the surface integral to a volume integral. The result is the Reynolds transport theorem.
. (6.13)
6.2 C ONSERVATION OF MASS
Let where is the density of the fluid. The Reynolds transport theorem gives
. (6.14)
The left-hand-side is the rate of change of the total mass inside the control volume. If there are no sources of mass within the control volume, the left-hand-side must be zero. Since the choice of control volume is arbitrary, the kernel of the right- hand-side must therefore be zero at every point in the flow.
Thus the continuity equation in the absence of mass sources is
. (6.15)
U = U A
D
Dt ------ F Vd V t( )
,F
,t ------- Vd V t( )
F U •n dA A t( )
D
Dt ------ F Vd V t( )
,F
,t ¤£ -------+^ ¢^ •(^ FU)¦¥^ dV V t( )
F = l l
D
Dt ------ l dV V t( )
,l ,t ¤£ ------+^ ¢^ •(^ lU)¦¥^ dV V t( )
,l ,t
------+ ¢ •( lU) = 0
Conservation of momentum
bjc 6.7 4/15/
and is the viscous stress tensor. The net force acting on the control volume is
the integral of the stress tensor, , over the surface plus the integral of any body
force vectors per unit mass, (gravitational acceleration, electromagnetic accel- eration, etc.), over the volume.
The isotropy of the pressure implies that it acts normal to any surface element in the fluid regardless of how it is oriented. The viscous part of the stress can take on many different forms. In Aeronautics and Astronautics we deal almost exclu- sively with Newtonian fluids discussed in Chapter 1 such as air or water where the viscous stress is linearly proportional to the rate-of-strain tensor of the flow.
The general form of the stress-rate-of-strain constitutive relation in Cartesian coordinates for a compressible Newtonian fluid is
(6.20)
where,
. (6.21)
Recall that. The stress components in cylindrical and spherical polar
coordinates are given in Appendix 2.
Interestingly, there are actually two viscosity coefficients that are required to account for all possible stress fields that depend linearly on the rate-of-strain ten-
sor. The so-called shear viscosity arises from momentum exchange due to
molecular motion. A simple model of is described in Appendix I. The bulk vis-
cosity is a little more mysterious. It contributes only to the viscous normal
force and seems to arise from the exchange of momentum that can occur between colliding molecules and the internal degrees of freedom of the molecular system. Some typical values of the bulk viscosity are shown in Figure 6.4.
o (^) ij m (^) ij
G
o (^) ij 2 μS (^) ij^2 3
= – ¤£ --- μ – μ (^) v¦¥b (^) ij S (^) kk
S (^) ij^1 2
,U (^) i ,x (^) j
,U (^) j ,x (^) i
S (^) kk = ¢ •U
μ μ μ (^) v
Conservation of momentum
4/15/13 6.8 bjc
Figure 6.4 Physical properties of some common fl uids at one atmo- sphere and 298.15°K.
For monatomic gases that lack such internal degrees of freedom,. For
some polyatomic gases such as CO 2 the bulk viscosity is much larger than the
shear viscosity.
Recall the discussion of elementary flow patterns from Chapter 4. Any fluid flow can be decomposed into a rotational part and a straining part. According to the Newtonian model (6.20) only the straining part contributes to the viscous stress.
Although is called the shear viscosity it is clear from the diagonal terms in
(6.20) that there are viscous normal force components proportional to. However they make no net contribution to the mean normal stress defined as
. (6.22)
This is not to say that viscous normal stresses are unimportant. They play a key role in many compressible flow phenomena we will study later especially shock waves.
μ (^) v = 0
μ μ
m (^) mean = ( 1 3 ) m (^) ii = –P+ μ (^) v S (^) kk
Conservation of energy
4/15/13 6.10 bjc
The stress tensor acting over the surface does work on the control volume as do the body force vectors. In addition, there may be conductive heat flux, , through the surface. There could also be sources of heat within the flow due to chemical reactions, radiative heating, etc. For a general fluid the internal energy per unit mass is a function of temperature and pressure.
The rate of change of the energy inside the control volume in Figure 6.3 is
. (6.27)
The Reynolds transport theorem and Gauss’ theorem lead to
. (6.28)
Since the equality holds over an arbitrary volume, the kernel must be zero and we have the differential equation for conservation of energy.
. (6.29)
The sum of enthalpy and kinetic energy
(6.30)
is called the stagnation or total enthalpy and plays a key role in the transport of energy in compressible flow systems. Take care to keep in mind that the flow
energy is purely the sum of internal and kinetic energy,. Some typical gas transport properties at 300K and one atmosphere are shown in Figure 6.4.
According to Fick’s law, the heat flux vector in a linear heat conducting medium is:
(6.31)
where is the thermal conductivity.
The rightmost column in Figure 6.4 is the Prandtl number
Q
e = f ( T ,P)
D Dt ------ l ( e +k) dV V t( )
0 (^ (^ – PI+^ o)^ •U^ – Q)^ •^ n^ dA+ (^ lG^ •U)^ dV
V t( )
A t( )
,l ( e +k) ,t ------------------------- ¢ lU e P l ¤£ +^ • ¤£ ¤£ +^ ---^ +k¦¥–o^ •U^ +Q¦¥–lG^ •U¦¥^ dV V t( )
0 =^0
,l ( e +k) ,t
------------------------ ¢ lU e P l
- • ¤£ ¤£ + --- +k¦¥–o •U +Q¦¥ –lG •U = 0
h (^) t e P l
- --- + k h 1 2 = = +---U (^) i U (^) i
e +k
Q (^) i = – g ,( T ,x (^) i)
g
Summary - differential form of the equations of motion
bjc 6.11 4/15/
(6.32)
The Prandtl number can be thought of as comparing the rate at which momentum is transported by viscous diffusion to the rate at which temperature diffuses
through conductivity. For most gases the Prandtl number is around. This num- ber is close to one due to the fact that heat and momentum transport are accomplished by the same basic mechanism of molecular collision with lots of space between molecules. Liquids are often characterized by large values of the Prandtl number and the underlying mechanisms of heat and momentum transport in a condensed fluid are much more complex.
6.5 S UMMARY - DIFFERENTIAL FORM OF THE EQUATIONS OF
MOTION
The coordinate-independent form of the equations of motion is
. (6.33)
Using index notation the same equations in Cartesian coordinates are
Pr
μC (^) p g
,l ,t ------+ ¢ •( lU) = 0
,lU ,t
-----------+ ¢ •( lU U +PI– o)– lG = 0
,l ( e +k) ,t
------------------------ ¢ lU e P l
- • ¤£ ¤£ + --- +k¦¥–o •U +Q¦¥ –lG •U = 0
Integral form of the equations of motion
bjc 6.13 4/15/
6.6.1 I NTEGRAL EQUATIONS ON AN EULERIAN CONTROL VOLUME The simplest case to consider is the Eulerian control volume used in Chapter 1
where. This is a stationary volume fixed in space through which the
fluid moves. In this case the Leibniz rule reduces to
. (6.36)
Note that, since the Eulerian volume is fixed in space and not time dependent, the
lower case form of the derivative is used in (6.36). See for comparison (6.6) and (6.9).
Let in Equation (6.36)
(6.37)
Use (6.15) to replace in (6.37).
(6.38)
Now use the Gauss theorem to convert the volume integral on the right-hand-side of (6.38) to a surface integral. The integral form of the mass conservation equation valid on a finite Eulerian control volume of arbitrary shape is
(6.39)
The integral equations for conservation of momentum and energy are derived in a similar way using (6.25), (6.29), and (6.36). The result is
U A = 0
d dt ----- F Vd V
,F
,t ------- Vd V
d dt
F = l
d dt ----- l dV V
,l ,t ------ Vd V
,l ,t
d dt ----- l dV V
¢ • ( lU)dV V
d dt ----- l V lU •n dA A
d + V
Integral form of the equations of motion
4/15/13 6.14 bjc
(6.40)
6.6.2 MIXED EULERIAN-LAGRANGIAN CONTROL VOLUMES More general control volumes where part of the surface may be at rest and other parts may be attached to the fluid are of great interest especially in the analysis of
propulsion systems. Now use the general form of the Liebniz rule with.
(6.41)
Use (6.15) to replace in (6.41) and use the Gauss theorem to convert the volume integral to a surface integral. The result is the integral form of mass con- servation on an arbitrary moving control volume.
(6.42)
The integral equations for conservation of momentum and energy on a general, moving, finite control volume are derived in a similar way using (6.25), (6.29), and (6.35). Finally, the most general integrated form of the conservation equations is
d
dt-----V^0 l^ dV+ A^0 lU^ •n^ dA =^0
d
dt-----V^0 lU^ dV+ A^0 (^ lU U^ +PI–^ o)^ •n^ dA–V^0 lG Vd =^0
d dt-----^ l^ (^ e^ +k)^ V^ lU e^
P ¤ +^ l^ ---^ +k¦
£ ¥–o •U +Q ¤ ¦
£ ¥ (^) n dA ( lG •U) dV V
A
d + 0
V
0 =^0
F = l
D Dt ------ l dV V t( )
,l ,t ------ Vd V t( )
l U (^) A •n dA A t( )
,l ,t
D
Dt ------ l dV V t( )
lU •n dA A t( )
l U (^) A •n dA A t( )
Applications of control volume analysis
4/15/13 6.16 bjc
. (6.44)
Momentum fluxes integrated on are directly related to the lift and drag forces
exerted on the body. The integrated momentum equation gives
(6.45)
. (6.46)
where the integrals are the drag and lift by the fl ow on the body. The integral momentum balance in the streamwise direction is
. (6.47)
and in the vertical direction
. (6.48)
( lU) •n dA A (^1)
A 1
( lU U +PI– o) n A ( PI– o) •n dA A (^2)
• d + A (^1)
Drag ( PI– o) •n dA A (^2)
x (^1)
; Lift ( PI– o) •n dA A (^2)
x (^2)
( lU U +PI– o) •n dA A (^1)
x (^1)
( lU U +PI– o) •n dA A (^1)
x (^2)
Applications of control volume analysis
bjc 6.17 4/15/
6.7.2 EXAMPLE 2 - C HANNEL FLOW WITH HEAT ADDITION
Heat addition to the compressible flow shown above occurs through heat transfer
through the channel wall,. There is no net mass addition to the control volume.
. (6.49)
The flow is steady with no body forces. In this case, the energy balance is,
. (6.50)
The contribution of the viscous stresses to the energy balance is zero along the
wall because of the no-slip condition, and, as long as the streamwise
velocity gradients are not large, the term is very small on the upstream
and downstream faces, ,. In this approximation, The energy balance
becomes,
. (6.51)
A 1 A^2
bQ (^) w
Aw
n n
Q
( lU) n A ( lU) •n dA A (^2)
• d + A (^1)
( lU e( +k) +PU– o •U +Q) •n dA A
U A
w
lU e P l ¤£ +^ ---^ +k¦¥^ •n^ dA A
Q • ndA A
Applications of control volume analysis
bjc 6.19 4/15/
6.7.3 EXAMPLE 3 - S TATIONARY FLOW ABOUT A ROTATING FAN.
This is the prototype example for propellers, compressors and turbines. In contrast to a steady flow, a stationary flow is one where time periodic motions such as the rotation of the fan illustrated above do occur, but the properties of the flow aver- aged over one fan rotation period or one blade passage period are constant. This will be the case if the fan rotation speed is held constant.
Here we make use of a mixed Eulerian-Lagrangian control volume. The Lagrangian part is attached to and moves with the fan blade surfaces and fan axle. Remember the fluid is viscous and subject to a no slip condition at the solid sur- face. The Eulerian surface elements are the upstream and downstream faces of the control volume as well as the cylindrical surrounding surface aligned with the axis
of the fan. We will assume that the fan is adiabatic, and there is no mass injection through the fan surface. The integrated mass fluxes are zero.
(6.55)
Momentum fluxes integrated on are equal to the surface forces exerted by the
flow on the fan.
(6.56)
Flow
A (^) e
A (^) f
A (^) e
Q = 0
( lU) •n dA A (^) e
A (^) e
( lU U +PI– o) n A ( lU U( – U (^) A) +PI– o) •n dA A (^) f
• d + A (^) e
Applications of control volume analysis
4/15/13 6.20 bjc
or
(6.57)
where the vector force by the fl ow on the fan is
(6.58)
Note that the flow and fan velocity on are the same due to the no-slip condition
. (6.59)
For stationary flow, the integrated energy fluxes on are equal to the work/sec
done by the fl ow on the fan.
. (6.60)
(6.61)
If the flow is adiabatic, and viscous normal stresses are neglected on and
the energy equation becomes,
(6.62)
( lU U +PI– o) •n dA A (^) e
+ F = 0
F ( PI– o) •n dA A (^) f
A (^) f
( U – U (^) A) A (^) f
A (^) e
( lU e( +k) +PU– o •U) •n dA A (^) e
( PU– o •U) •n dA A (^) f
Work ( PU– o •U) •n dA A (^) f
= = bW
A (^) e A (^) f
l e P l ¤£ +^ ---^ +k¦¥^ U^ •n^ dA A (^) e
