# ����������ѧ(CFD)�ĵ�����2. Governing equations

2. FLUID-FLOW EQUATIONS 2.1 Control-volume approach 2.2 Conservative integral and differential equations 2.3 Non-conservative differential equations 2.4 Non-dimensionalisation Summary Examples

SPRING 2011

Fluid dynamics is governed by conservation equations for: mass; momentum; energy; (for a non-homogenous fluid) individual constituents. The continuum equations for these can be expressed mathematically in many different ways. In this section we shall show that they can be written as equivalent: integral (i.e. control-volume) equations; differential equations; In addition, the differential equations may be either conservative or non-conservative. The focus will be the integral equations because they are physically more fundamental and form the basis of the finite-volume method. However, the equivalent differential equations are often easier to write down, manipulate and, in some cases, solve analytically. Although there are different fluid variables, most of them satisfy the same generic equation, which can therefore be solved computationally by the same subroutine. This is called the scalar-transport or advection-diffusion equation.

2.1 Control-Volume Approach The rate of change of some quantity within an arbitrary control volume is determined by: the net rate of transport across the bounding surface ("flux"); the net rate of production within that control volume ("source"). FLUX RATE OF CHANGE + SOURCE (1) V out of boundary = inside V inside V The flux across the bounding surface can be divided into: advection: transport with the flow; diffusion: net transport by molecular or turbulent fluctuations. (Some authors �C but not this one �C prefer the term convection to advection.)
RATE OF CHANGE + ADVECTION + DIFFUSION = SOURCE inside V inside V through boundary

(2)

The finite-volume method is a natural discretisation of this.

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2.2 Conservative Integral and Differential Equations 2.2.1 Mass (Continuity)

u
Physical principle (mass conservation): mass is neither created nor destroyed.

V

un

A For an arbitrary control volume (aka "cell") with volume V: mass of fluid: V For a typical cell face with area A and component of velocity un along the (outward) normal: mass flux through cell face: C = Q = u n A = u A
d (mass ) + net outward mass flux = 0 dt

d ( V)+ dt

��
faces

u A = 0

(3)

1 u1

2

u2

e.g. steady, 1-d flow: 2 u 2 A2 (mass flux ) out

1u1 A1 (mass flux) in

=0 =0

t

A corresponding conservative differential equation for mass conservation can be derived by considering a (fixed) control volume comprised of a cuboid with sides x, y, z as shown.

z

w
y

n s b
x

e

Using (3),
d( V ) 1dt3 2
rate of change of mass

+ ( uA) e ( uA) w + ( vA) n ( vA) s + ( wA) t ( wA) b = 0 1444444444442444444444443
net outward mass flux

where density and velocity are averages over the volume or the cell faces. Noting that V = x y z and Aw = Ae = y z etc., d( x y z ) + [( u ) e ( u ) w ] y z + [( v) n ( v) s ] z x + [( w) t ( w) b ] x y = 0 dt Dividing by the volume, xyz: ( u ) e ( u ) w ( v) n ( v) s ( w) t ( w) b d + + + =0 x y z dt Proceeding to the limit x, y, z �� 0:

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( u ) ( v) ( w) + + + =0 t x y z

(4)

This analysis is analogous to the finite-volume procedure, except that in the latter the control volume does not shrink to zero; i.e. it is a finite-volume not infinitesimal-volume approach.

(**** MSc course only ***) More mathematically, for an arbitrary volume V with closed surface V: d dV + u dA = 0 dt V V

(5)

For a fixed volume, take d/dt under the integral sign and apply the divergence theorem to the surface integral: t + ( u) dV = 0 V Since V is arbitrary, the integrand must be identically zero. Hence, + ( u) = 0 t

(6)

Incompressible Flow For incompressible flow volume as well as mass is conserved, so that for a fixed control volume and steady (i.e., time-independent) flow: (uA) e (uA) w + (vA) n (vA) s + ( wA) t ( wA) b = 0 1444444444 24444444444 4 3
net outward VOLUME flux

Substituting for the face areas, dividing by volume and proceeding to the limit as above produces u v w (7) + + =0 x y z which is usually taken as the incompressibility equation. In fact, this continues to hold for time-dependent incompressible flows, although this is harder to show by this method when the density is non-uniform.

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2.2.2 Momentum Physical principle (Newton's Second Law): rate of change of momentum = force The total rate of change of momentum for fluid passing through a control volume consists of: time rate of change of total momentum inside the control volume; plus net momentum flux (difference between rate at which momentum leaves and enters). For a cell with volume V and a typical face with area A: = ( V )u momentum in cell = mass �� u = ( u A)u momentum flux = mass flux �� u d (momentum) + net outward momentum flux = force dt 144444444 244444444 3 4 4
total rate of change of momentum

u

V A

un

d (mass �� u) + dt

�� (mass
faces

flux �� u) = F

(8)

(Note that momentum, velocity and force are vectors; u and F have 3 components). e.g. for steady flow through a conduit: (momentum flux)out �C (momentum flux)in = force Q(u 2 u1 ) = F

1 u1 F

2

u2

Fluid Forces There are two types: surface forces (proportional to area; act on control-volume faces) body forces (proportional to volume) (i) Surface forces are usually expressed in terms of stress (= force per unit area): force stress = or force = stress �� area area The main surface forces are: pressure p: always acts normal to a surface; viscous stresses : frictional forces arising from relative motion. For a simple shear flow there is only one non-zero stress component: u �� 12 = y but, in general, is a symmetric tensor (the components of stress imparted by external fluid on individual faces of a volume of fluid are shown right) and has a more complex expression for its �� components. In incompressible flow: �� u j u = ( i + ) ij x j xi
11

y

��

��

U
��22 ��12 ��21
x

y

��11

21

��12 ��22

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(ii) Body forces The main body forces are: gravity: the force per unit volume is
g = (0,0, g )

z

��g

(For constant-density fluids pressure and weight can be combined in the governing equations as a piezometric pressure p* = p + gz) centrifugal and Coriolis forces (apparent forces in a rotating reference frame)

centrifugal force: �� ( �� r) =

2

R

r

Coriolis force: 2 ��u

u

In inertial frame In rotating frame
2 2 (Because the first of these can be written as the gradient of 1 R it can also be 2 absorbed into a modified pressure and hence is usually ignored �C see the Examples).

Equivalent Differential Equation Once again, a conservative differential equation can be derived by considering a fixed Cartesian control volume with sides x, y and z.
z

x For the x-component: d + ( uA) e u e ( uA) w u w + ( vA) n u n ( vA) s u s + ( wA) t u t ( wA) b u b ( Vu ) 1444444444444442444444444444443 t 23 d4 4 1

net outward momentum flux

rate of change of momentum

= ( p w Aw pe Ae ) 14 244 4 3
pressure force in x direction

+ viscous and other forces

Substituting the cell dimensions: d ( x y z u ) + [( u ) e u e ( u ) w u w ] y z + [( v) n u n ( v) s u s ] z x + [( w) t u t ( w) b u b ] x y dt = ( p w p e ) y z + viscous and other forces Dividing by the volume x y z (and changing the order of pe and pw): ( p pw ) d( u ) ( uu ) e ( uu ) w ( vu ) n ( vu ) s ( wu ) t ( wu ) b + + + = e dt x y z x + viscous and other forces

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2

R

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In the limit as x, y, z �� 0: ( u ) ( uu ) ( vu ) ( wu ) p + + + = + 2 u + other forces t x y z x (9)

Notes. (1) The viscous term is given without proof (but MSc students read the notes below). 2 is 2 2 2 the Laplacian operator 2 + 2 + 2 . x y z (2) The pressure force per unit volume in the x direction is given by (minus) the pressure gradient in that direction. (3) You should be able to derive the y and z-momentum equations by inspection / patternmatching.

(**** MSc course only ***) Separating surface forces (determined by a stress tensor ij) and body forces (fi per unit volume), the control-volume equation for the i component of momentum may be written d (10) u i dV + u i u j dA j = ij dA j + f i dV dt V V V V where the stress tensor has pressure and viscous parts: u i u j 2 u k (11) + ) ij = ( ij ij = p ij + ij , x j x i 3 x k For a fixed volume, take d/dt under the integral sign and apply the divergence theorem to the surface integrals: ( u i ) ( u i u j ) ij f i dV = 0 + t x j x j V Since V is arbitrary, the integrand must vanish identically. Hence, for arbitrary surface and body forces: ( u i ) ( u i u j ) ij (12) + = + fi t x j x j Splitting the stress tensor into pressure and viscous terms: ( ui ) ( ui u j ) p ij (13) + + + fi = t xi x j x j
If the fluid is incompressible and viscosity is uniform then the viscous term simplifies to give ( ui ) ( u i u j ) p + + 2ui + f i = t xi x j

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2.2.3 General Scalar
A similar equation may be derived for any physical quantity that is advected or diffused by a fluid flow. For each such quantity an equation is solved for the concentration (i.e. amount per unit mass) ��: for example, the concentration of salt, sediment or a chemical constituent. Diffusion occurs when concentration varies with position. It typically involves transport from regions of high concentration to regions of low concentration, at a rate proportional to area and concentration gradient. For many scalars it may be quantified by Fick's diffusion law: rate of diffusion = diffusivity �� gradient �� area �� = A n This is often referred to as gradient diffusion. A common example is heat conduction. For an arbitrary control volume: V�� amount in cell: advective flux: diffusive flux: source
C��

(mass �� concentration) (mass flux �� concentration) (�Cdiffusivity �� gradient �� area) (source density �� volume)

u

�� A n SV

V A

un

Balancing the rate of change against the net flux through the boundary and rate of production yields the scalar-transport or (advection-diffusion) equation:
rate of change + net outward flux = source d ( V ��) + dt

�� (C ��
faces

�� A) = S V n

(14)

( ��) �� �� �� + ( u�� ) + ( v�� ) + ( w�� )=S t x x y y z z

(15)

(**** MSc course only ****) This may be expressed more mathematically as: d �� dV + ( u�� ����) dA = S dV dt V V V

(16)

For a fixed control volume, taking the time derivative under the integral sign and using Gauss's divergence theorem as before gives a corresponding differential equation ( ��) (17) + ( u�� ��) = S t

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2.2.4 Momentum Components as Transported Scalars
In the momentum equation, if the viscous force A = (u / y ) A is transferred to the LHS it looks like a diffusive flux: e.g. for the x-component of momentum: u d ( Vu ) + �� (Cu A) = other forces dt n faces (The viscous force has been simplified a bit! �C see 2.2.5 below �C but the essence is correct.) Compare this with the generic scalar-transport equation: d �� ( V ��) + �� ( C �� A) = S V dt n faces
Each component of momentum satisfies its own scalar-transport equation with concentration velocity (u, v or w) diffusivity viscosity source other forces Consequently, only one generic scalar-transport equation need be considered. Computationally, the same subroutine may be used to solve the general scalar-transport equation for each variable (but with different diffusivities and source density S).

The analysis above assumes that all non-advective flux is simple gradient diffusion: �� A n Actually, the real situation is a little more complex. For example, in the u-momentum equation the full expression for the 1-component of viscous stress through the 2-face is u v 12 = y + x The u/ y part is gradient diffusion of u, but the v/ y term is not. In general, non-advective fluxes F �� that can't be represented by gradient diffusion are discretised conservatively (i.e. worked out for particular cell faces) but are transferred to the RHS as a source term: d �� ( V �� ) + �� [C �� A + F ��] = S dt n faces

2.2.6 Moving Control Volumes
The control-volume equations are equally applicable to moving control volumes, provided the normal velocity component un is taken relative to the mesh; i.e. u n = (u u mesh ) n This enables the finite-volume method to be used for calculating flows with moving boundaries1: for example, surface waves or internal combustion engines.

See, for example: Apsley, D.D. and Hu, W., 2003, CFD Simulation of two- and three-dimensional free-surface flow, International Journal for Numerical Methods in fluids, 42, 465-491.

1

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2.3 Non-Conservative Differential Equations
Two equivalent differential forms of the flow equations may be derived from the controlvolume equations in the limit as the control volume shrinks to a point. From fixed control volumes we obtain governing equations in conservative form as above; this is called the Eulerian approach. Using control volumes moving with the fluid we obtain the governing equations in non-conservative form; this is called the Lagrangian approach. Both forms can, however, be obtained by mathematical manipulation of the other and that is what we shall do here. For the conservative differential equations derived earlier all terms involving derivatives of dependent variables have differential operators "on the outside". Hence, they can be integrated directly to give an equivalent integral form. For example, in one dimension:
df = g ( x) dx (differential form)

f ( x 2 ) f ( x1 ) = g ( x) dx x1 (integral form) fluxout fluxin = source

x2

This is the 1-d version of: (*** MSc course only ***)

The three-dimensional version uses partial derivatives and the divergence theorem to change the differentials to surface integrals. As an example of how essentially the same equation can appear in conservative and nonconservative forms consider a simple 1-d example: d 2 ( y ) = g ( x) (conservative form �C can be integrated directly) dx dy 2y = g ( x) (equivalent non-conservative form) dx To describe how a particular element of fluid changes as it moves with the flow we need a material derivative.

Material Derivatives The rate of change of some property in a fluid element moving with the flow is called the material (or substantive) derivative. It is denoted by D��/Dt and worked out as follows. Every field variable �� is a function of both time and position; i.e. �� = ��(t , x, y, z ) As one follows a path through space �� will change with time because: it changes with time t at each point, and it changes with position (x, y, z) as it moves with the flow.

(x(t), y(t), z(t))

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Thus, the total time derivative following an arbitrary path (x(t), y(t), z(t)) is d�� �� �� dx �� dy �� dz = + + + dt t x dt y dt z dt The material derivative is obtained when the path follows the flow (i.e., dx/dt = u, etc.): �� �� �� D�� �� �� +u +v +w (18) x y Dt t z Using this definition, it is possible to write a non-conservative but more compact form of the governing equations. For a general scalar �� the sum of time-dependent and advective terms in its transport equation is ( ��) ( u��) ( v��) ( w��) + + + t x y z �� �� ( w) �� ( v) �� ( u ) = ��+ + x �� + u x + y �� + v y + z �� + w z t t (by the product rule) �� �� �� �� ( u ) ( v) ( w) + + = + t + u x + v y + w z �� + z y x 4444 4444 t 144444 44444 2 3 1 2 3
= 0 by continuity = D�� / Dt by definition

(on collecting terms)
D�� (19) Dt Hence, using the definition of the material derivative, the combined time-dependent and advective terms in a scalar transport equation can be written as the much shorter (but nonconservative) form D��/Dt. =

For example, the material derivative of velocity, i.e. Du/Dt, is acceleration and the momentum equation can be written Du p (20) = + 2 u + other forces Dt x 1 3 2
mass�� acceleration

This form is shorter and quicker to write and is used both for notational convenience and to derive theoretical results in special cases (see the Examples, Q2). However, in the finitevolume method it is the conservative form which is actually approximated directly. (*** MSc course only ***) The material derivative can be defined more compactly as �� D�� �� D�� �� = + ui = + u �� or x Dt t Dt t Simplify the derivation of (19) above using the summation convention or vector notation.

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2.4 Non-Dimensionalisation
Although it is possible to work entirely in dimensional quantities, there are good theoretical reasons for working in non-dimensional variables. These include the following. All dynamically-similar problems (same Re, Fr etc.) can be solved with a single computation. The number of relevant parameters (and hence the number of graphs) is reduced. It indicates the relative size of different terms in the governing equations and, in particular, which might conveniently be neglected. Computational variables are the same order of magnitude, yielding better numerical accuracy.

2.4.1 Non-Dimensionalising the Governing Equations
For incompressible flow the governing equations are: u v w + + =0 continuity: x y z Du p x-momentum: = + 2u Dt x

(21) (22)

Adopting reference scales U0, L0 and 0 for velocity, length and density, respectively, and 2 derived scales L0/U0 for time and 0U 0 for pressure, each fluid quantity can be written as a product of a dimensional scale and a non-dimensional variable (indicated by an asterisk *): L x = L0 x * , t = 0 t*, u = U 0u* , = 0 *, p p ref = ( 0U 02 ) p * , etc. U0 (Note: In incompressible flow it is differences in pressure that are important, not absolute values. Since these differences are usually much smaller than the absolute pressure it is numerically more accurate to work in terms of the difference from a reference pressure pref). Substituting these into mass and momentum equations, (21) and (22), yields: u * x *
*

+

v * y *

+

w* z * +

=0

(23)

U L 1 *2 * (24) u where Re = 0 0 0 Dt * x * Re From this, it is seen that the key dimensionless group is the Reynolds number Re. If Re is large then viscous forces would be expected to be negligible in much of the flow.

Du *

=

p *

Having derived the non-dimensional equations it is usual to drop the asterisks and simply declare that you are "working in non-dimensional variables".

2.4.2 Common Dimensionless Groups
If other types of fluid force are included then each introduces another non-dimensional group. For example, gravitational forces lead to a Froude number (Fr) and Coriolis forces to a

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Rossby number (Ro). Some of the most important dimensionless groups are given below. (See also the Section 3 examples.) If U and L are representative velocity and length scales, respectively, then: Re �� UL �� UL Reynolds number (viscous flow; = dynamic viscosity)

Fr ��

U gL
U c U L

Froude number (open-channel flow; g = gravity)

Ma �� Ro ��

Mach number (compressible flow; c = speed of sound)

Rossby number (rotating flows;

= angular rotation rate)

We ��

U 2L

Weber number (free-surface flows;

= surface tension)

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Summary
Fluid dynamics is governed by conservation equations for mass, momentum, energy and, for a non-homogeneous fluid, the amount of individual constituents. The governing equations can be expressed in equivalent integral (control-volume) or differential forms. The finite-volume method is a direct discretisation of the control-volume approach. Differential forms of the flow equations may be conservative (i.e. can be integrated directly to give something of the form "fluxout �C fluxin = source") or non-conservative. A particular control-volume equation takes the form: rate of change + net outward flux = source There are really just two canonical equations to discretise and solve: mass conservation (continuity): d ( V ) + ��C = 0 (C = uA is outward mass flux) dt faces scalar-transport (or advection-diffusion) equation: d �� ( V��) + ��( C�� A ) = SV dt n faces rate of change advection diffusion source Each Cartesian velocity component (u, v, w) satisfies its own scalar-transport equation. However, these equations differ from those for a passive scalar because they are non-linear and strongly coupled through the advective fluxes and pressure forces. Non-dimensionalising the governing equations, allows dynamically-similar flows (those with the same values of Reynolds number, etc.) to be solved with a single calculation, reduces the overall number of parameters, indicates when certain terms in the governing equations are significant or negligible and ensures that the main computational variables are of similar magnitude.

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Examples
Q1. The continuity and x-momentum equations can be written for 2-d flow in conservative form (and with compressibility neglected in the viscous forces) as + ( u ) + ( v) = 0 t x y p ( u ) + ( uu ) + ( vu ) = + 2 u t x y x respectively. (a) By expanding derivatives of products show that these can be written in the equivalent non-conservative forms: D u v + ( + )=0 Dt x y Du p = + 2u Dt x D = +u +v . where the material derivative is given (in 2 dimensions) by Dt t x y Define what is meant by the statement that a flow is incompressible. To what does the continuity equation reduce in incompressible flow? Write down conservative forms of the 3-d equations for mass and x-momentum. Write down the z-momentum equation, including gravitational forces. Show that, for constant-density flows, pressure and gravity can be combined in the momentum equations as the piezometric pressure p + gz.

(b)

(c) (d) (e)

**** Remaining parts of this question for MSc course only ****) (f) Write the conservative mass and momentum equations in vector notation. (g) Write the conservative mass and momentum equations in suffix notation using the summation convention. In a rotating reference frame there are additional apparent forces (per unit volume): 2 �� ( �� r ) or R centrifugal force: Coriolis force: 2 ��u R where is the angular velocity vector (with magnitude and direction along the axis of rotation), u is the fluid velocity in the rotating reference frame, r is the position vector (relative to a point r on the axis of rotation) and R is a vector perpendicularly outward from the axis of rotation to the point. By writing the centrifugal force as the gradient of some quantity show that it can be subsumed into a modified pressure. Also, find the components of the Coriolis force if rotation is about the z axis.

(h)

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2

R

Q2. The x-component of the momentum equation is given by Du p = + 2u Dt x Using this equation derive the velocity profile in fully-developed, laminar flow for: (a) pressure-driven flow between stationary parallel planes ("Poiseuille flow"); (b) constant-pressure flow between stationary and moving planes ("Couette flow").

Q3. (**** MSc Course only ****) By applying Gauss's divergence theorem deduce the conservative and non-conservative differential equations corresponding to the integral scalar-transport equation d �� dV + ( u�� ��) dA = S dV dt V V V

Q4. In each of the following cases, state which of (i), (ii), (iii) is a valid dimensionless number. Carry out research to find the name and physical significance of these numbers. (L = length scale; U = velocity scale; z = height; P = pressure; = density; = dynamic viscosity; = kinematic viscosity; g = gravitational acceleration; = angular velocity). (a) (b) (i)
P P0 ; U UL ; (i)
1/ 2

(ii)

P P0 ; 2 2 U UL (ii) ;
1

(iii) (iii)

U 2 ( P P0 ) UL

(c)

(d)

gd dz (i) dU dz U (i) ; L

;

(ii)

U ; gL
gL ; U

(iii)

P P0 g U L

(ii)

(iii)

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Q5. (Computational Hydraulics Examination, May 2008) The momentum equation for a viscous fluid in a rotating reference frame is
Du = p + 2u 2 Dt ��u

(*) is the

where is density, u = (u,v,w) is velocity, p is pressure, is dynamic viscosity and angular-velocity vector. (The symbol �� denotes a vector product.) (a) (b) (c) If
= (0,0, ) write the x and y components of the Coriolis force ( 2 �� u ).

Hence write the x- and y-components of equation (*). Show how Equation (*) can be written in non-dimensional form in terms of a Reynolds number Re and Rossby number Ro (both of which should be defined). Define the terms conservative and non-conservative when applied to the differential equations describing fluid flow. Define (mathematically) the material derivative operator D/Dt. Then, noting that the continuity equation can be written ( u ) ( v) ( w) + + + = 0, t x y z show that the x-momentum equation can be written in an equivalent conservative form. If the x-momentum equation were to be regarded as a special case of the general scalar-transport (or advection-diffusion) equation, identify the quantities representing: (i) concentration; (ii) diffusivity; (iii) source. Explain why the three equations for the components of momentum cannot be treated as independent scalar equations. Explain (briefly) how pressure can be derived in a CFD simulation of: (i) high-speed compressible gas flow; (ii) incompressible flow.

(d)

(e)

(f)

(g)

(h)

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