02 Derivation of Rans

3rd year fluids and engineering analysis

Tristan Robinson

Department of Civil, Environmental and Geomatic EngineeringUniversity College London

8th October 2013

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Table of contents

1 Reynolds Averaged Navier-Stokes EquationsExamplesContinuityFluid accelerationForcesReynolds shear stressesEquation of motion

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Reynolds Averaged Navier-Stokes Equations

Incrompressible flow of a Newtonian fluid

ut + u

ux + v

uy + w

uz = 1 px + gx + 2u

(uux +

uv y +

uw z

)vt + u

vx + v

vy + w

vz = 1 py + gy + 2v

(v ux +

v v y +

v w z

)wt + u

wx + v

wy + w

wz = 1 pz + gz + 2w

(w ux +

w v y +

w w z

) Continuity

ux

+vy

+wz

= 0

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Numerical applications

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Exact Solutions: Laminar & steady

Flow between parallel plates Flow on smooth surfaces

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Example: Oil Skimmer

An oil skimmer uses a 5 m widex 6 m long moving belt above afixed platform ( = 60) to skimoil off rivers. The belt travels at 3m/s. The distance between thebelt and the fixed platform is 2mm. The belt discharges into anopen container on the ship.

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Example: Oil Skimmer

Assume that the fluid is crude oil ( = 860 kg/m3 and = 1 102 Ns/m2).

Find the discharge Q?

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Calculations

Assume laminar flow

u =Uya

+y2 ay

2

(gx +

dpdx

) In this case dpdx = 0, gx = g cos(60) = 9.81 0.5 = 4.905

m/s2 and = 860 q =

a0 udy

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Calculations

q =Ua2 a

3

12 4218.3

a = 0.002 m and U = 3 m/s

q =3 0.0002

2 0.002

3

12 1 102 4218.3 = 0.0027

Hence Q = 0.0027 5 = 0.0136 m3/s

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Incrompressible flow of a Newtonian fluid

ut + u

ux + v

uy + w

uz = 1 px + gx + 2u

(uux +

uv y +

uw z

)vt + u

vx + v

vy + w

vz = 1 py + gy + 2v

(v ux +

v v y +

v w z

)wt + u

wx + v

wy + w

wz = 1 pz + gz + 2w

(w ux +

w v y +

w w z

) Continuity

ux

+vy

+wz

= 0

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Continuity Equation

Derived from the principle of conservation of mass Consider flow into and out of a small control volume

(xyz) within the fluid Since the volume of incompressible fluid inside the block

remains constant Volume of flow in must equal volume of flow out.

ux

+vy

+wz

= 0

Hence fluid motion is subject to constraints linking thevelocity components.

It is an important component in the description of fluidmotion.

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Derivation of the continuity equation

Consider flow through an infinitesimally small fluid element

The flow into surface ABCD x = ut . The mass flowing into surface ABCD m = xyz Hence mxin = utyz.

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Continuity equation

The mass flow out of surface ABCD in t is

mxout =(u +

ux

x)tyz

The mass flow into the whole cubic element in t is thesum of the flow into each of the surfaces ABCD, ABBA

and ADDA

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Continuity equation

Mass in (min)

uyzt + vxzt + wxyt

Mass out (mout )(u +

ux

x)yzt +

(v +

vy

y)xzt +

(w +

wz

z)xyt

Change of mass (m)

m =

ttxyz

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Continuity equation

Change of mass is the difference in mass out minus massin

m = mout min Cancelling xyzt in every term. We obtain the general form of the continuity equation

(u)x

+(v)y

+(w)z

=

t

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Steady and incompressible two-dimensionalcontinuity equation

Incompressible ( is constant throughout the fluid) Two-dimensional (w = 0) Steady flow (t = 0)

ux

+vy

= 0

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Acceleration of a fluid particle

In the Eulerian system, the velocity field is specified bycomponents u, v ,w in directions x , y , z.

At a particular instant in time: (u, v ,w) = f (x , y , z) For unsteady flows: (u, v ,w) = f (x , y , z, t) If the corresponding components of acceleration are:

ax =dudt, ay =

dvdt, az =

dwdt

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As u is a function of x , y , z, t Use the chain rule of partial differentiation

du(x , y , z, t)dt

=ux

dxdt

+uy

dydt

+uz

dzdt

+ut

dtdt

We can write this asdudt

= uux

+ vuy

+ wuz

+ut

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Hence

dudt = u

ux

+ vuy

+ wuz

+ut

dvdt = u

vx

+ vvy

+ wvz

+vt

dwdt

substantive acc.

= uwx

+ vwy

+ wwz

convective acc.

+wt

local acc.

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Acceleration

ax =U(x2, t2) U(x1, t1)

t

Separate change in time and displacement

ax =U(x1, t2) U(x1, t1)

t+U(x2, t2) U(x1, t2)

t

Re-write as

ax =U(x , t2) U(x , t1)

t+

xt

U(x2, t) U(x1, t)x

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Henceax =

Ut

+ UUx

Temporal acceleration: at =Ut

Convective acceleration: ac = UUx

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Forces on fluid particles

II

nFFF

nF II

I

G=mg

I

Volume forces: acts throughout the entire body G. Surface forces: acts as reaction, Fn, and viscous force,F .

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Total force

Most important areF

total force

=

F gravity body force

+

F pressure +

F friction surface forces

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Pressure Force

Consider aninfinitesimally smallrectangular block offluid of density .

The net pressureforce in the xdirection:

Fx =(P

(P +

Px

x))

zy

Mass of block of fluid = xzy25 / 66

Pressure Forces

Pressure force in x-direction /unit mass = 1

Px

Pressure force in y -direction /unit mass = 1

Py

Pressure force in z-direction /unit mass = 1

Pz

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Body forces

For engineering purposes, these are nearly always due togravity (g).

Adopting the convention that z is positive verticallyupwards.

Consider the force acting on a block of mass = xyz Total vertical force on fluid block = gxyz Gravity force in z-direction / unit mass = g ( 9.81 m/s2)

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Body forces

More generally: Gravity force in x direction / unit mass = gx Gravity force in y direction / unit mass = gy Gravity force in z direction / unit mass = gz

Where gs is the component of g in direction s.

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Friction Forces

For a turbulent flow field, the total average frictional force isthe sum of the average viscous friction and the averageturbulent friction.

For incompressible flow, the viscous frictional force / unitmass can be determined by considering the frictionalforces acting on a small element of fluid (xyz)

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Friction Forces

The net force acting on the topand bottom faces in thex-direction is:

Fx =((

+

zz) )xy

As = uz

Fx = 2uz2

xyz

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Friction Forces

Giving a force /unit mass:

F =

2uz2

= v2uz2

Similar terms can be derived to describe the net forces onthe other four faces of the element

Giving a total viscous friction force in the x direction /unitmass:

Fx =

(2ux2

+2uy2

+2uz2

)=

2u

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Friction Forces

The corresponding terms in the y and z directions are:

Fy =

(2vx2

+2vy2

+2vz2

)=

2v

Fz =

(2wx2

+2wy2

+2wz2

)=

2w

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Shear stress

Shear stress: = FA

Shear strain: = xy

Newtonian fluid: =

d

dt

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Equation of viscosity

Shear strain

=(u(y + y) u(y))

yt

Rate of shear straind

dt=

dudy

Equation of viscosity

= dudy

=xy

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Navier-Stokes Equations

So far we have derivedut + u

ux + v

uy + w

uz = 1 px + gx + 2u

vt + u

vx + v

vy + w

vz = 1 py + gy + 2v

wt + u

wx + v

wy + w

wz = 1 pz + gz + 2w

Continuityux

+vy

+wz

= 0

Missing the Reynolds and Averaged parts

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Turbulent flow

The velocity is composedof a mean part and afluctuating part

u = u + u

v = v + v

w = w + w

Similarly for the pressureterm

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Average of mean and fluctuatingcomponents

Average is given as

u + u = u + u = u

As u = u, u = 0, uu = u2, and uu 6= 0 Hence

(u1 + u1)(u2 + u2) = u1u2 + u

1u2

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Continuity

Substitute into the continuity equation

(u + u)x

+(v + v )

y+(w + w )

z= 0

Multiply out

ux

+vy

+wz

+u

x+v

y+w

z= 0

Henceux

+vy

+wz

= 0

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Acceleration

Acceleration is the only part that contributes to theturbulence force:

uux

+ vuy

+ wuz

uvx

+ vvy

+ wvz

uwx

+ vwy

+ wwz

The product rule and continuity can be used to re-writethese equations

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Contribution from the acceleration term

Re-write as

uux

+ vuy

+ wuz

(uu)x

+(vu)y

+(wu)z

uvx

+ vvy

+ wvz

(uv)x

+(vv)y

+(wv)z

uwx

+ vwy

+ wwz

(uw)x

+(vw)y

+(ww)z

Is this correct? Multiply it out again to test!

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We get

(uux

+ uux

)+

(vuy

+ uvy

)+

(wuz

+ uwz

)

(uvx

+ vux

)+

(vvy

+ vvy

)+

(wvz

+ vwz

)

(uwx

+ wux

)+

(vwy

+ wvy

)+

(wwz

+ wwz

) Use continuity

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Re-write as

u(ux

+vy

+wz

)+ u

ux

+ vuy

+ wuz

v(ux

+vy

+wz

)+ u

vx

+ vvy

+ wvz

w(ux

+vy

+wz

)+ u

wx

+ vwy

+ wwz

We get back to our original form

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Average acceleration

Consider only the horizontal component

uux

+ vuy

+ wuz

=uux

+uvy

+uwz

Let u = u + u, v = v + v and w = w + w

(uu + uu)x

+(uv + uv )

y+(uw + uw )

z

Re-write as

(uu)x

+(uv)y

+(uw)z

+(uu)x

+(uv )y

+(uw )z

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Average acceleration

Hence

uux

+ vuy

+ wuz

+(uu)x

+(uv )y

+(uw )z

In the other directions we get

uvx

+ vvy

+ wvz

+(v u)x

+(v v )y

+(v w )z

uwx

+ vwy

+ wwz

+(w u)x

+(w v )y

+(w w )z

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Reynolds shear stresses

The resulting force in the x-direction / unit mass is given by:

Fx = ((uu

)x

+(uv

)y

+(uw

)z

)

The resulting force in the y -direction / unit mass is given by:

Fy = (v u

x+v v

y+v w

z

) The resulting force in the z-direction / unit mass is given by:

Fz = (w u

x+w v

y+w w

z

)45 / 66

Reynolds shear stresses

Caused by turbulent transfer of momentum across thefluid, with sweeps of high momentum fluid moving intoregions of slow-moving fluid and ejections of lowmomentum fluid out into regions of high velocity.

The effect is characterised by the turbulent fluctuations(u, v ,w ) that occur in the three velocity componentsabout their average values (U,V ,W ). These turbulentresisting forces are often described in terms of Reynoldsshear stresses

uu, uv , uw

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Momentum transfer Reynolds or turbulent stresses are due to momentum

transfer between layers of flow with mean velocity U(y)

Ix = x z v mass transfer

(U + u) xvelocity

t

The corresponding force

F =Ixt

= x z v (U + u) = x z uv

As U does not depend on time and v = 0.v (U + u) = v U + v u = v U + uv = uv .

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Turbulent stresses

Turbulent stress (or Reynolds stress = Rxy )

turb = Fx z

= uv .

These are named after Osborne Reynolds who derivedthem in 1895.

They are sometimes summarised as a 2nd order stresstensor.

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Equation of motion

Use Newtons second Law

F = ma

Per unit massF = a

We balance out the accleration with the forces

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Equations of motion for incompressible flow


ut + u

ux + v

uy + w

uz = 1 px + gx + 2u

(uux +

uv y +

uw z

)vt + u

vx + v

vy + w

vz = 1 py + gy + 2v

(v ux +

v v y +

v w z

)wt + u

wx + v

wy + w

wz = 1 pz + gz + 2w

(w ux +

w v y +

w w z

) Continuity

ux

+vy

+wz

= 0

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Application of Navier-Stokes Equations

The equations are nonlinear partial differential equations No full analytical solution exists The equations can be solved for several simple flow

conditions Numerical solutions to Navier-Stokes equations are

increasingly being used to describe complex flows

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Apply RANS to static fluid

There is no motion: u = v = w = 0 (ax = ay = az = 0) No motion implies no friction Let z correspond to vertical direction: gx = gy = 0 andgz = g

RANS reduces to

0 = 1

px

; 0 = 1

py

; 0 = 1

pz g

Hence p is not dependent on x and y . Integrate wrt z: p = gz + constant .

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Apply RANS to frictionless flow For steady flow along a streamline, velocity Us is a function

of position s on the curve but is constant with time. RANS for flow in the s direction reduces to:

UsUss

= 1

ps

+ gs

Use product rule and as gs = g in the vertical direction:12(U2s)

s+

1

ps

+ gzs

= 0

Integrate wrt s12U2s +

p

+ gz = constant

Bernoullis equation (frictionless flow along a streamline)53 / 66

Apply RANS to laminar flow between twoparallel plates

The flow is steady, uniform and fully developed The flow is is 2-dimensional (x , z) with velocity

components (u,w) The gap between plates is 2b and no-slip condition at the

boundaries. Flow is in the x direction (w = 0). From RANS: p

z= g

From continuity: ux

= 0

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Apply RANS to laminar flow between twoparallel plates

The pressure term is given as

p = gz + C(x)

No change in velocity u/x = 0 implies

1 p/x = constant

2uz du

dz

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Reduced RANS

RANS reduces to

0 = 1

px

+

d2udz2

We can integrate twice wrt z:

px

z + dudz

= C

12px

z2 + u = Cz + D

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Reduced RANS

RANS reduces to

0 = 1

px

+

d2udz2

We can integrate twice wrt z:

px

z + dudz

= C

12px

z2 + u = Cz + D

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Solve flow between two parallel plates

Using the no-slip boundary conditions that u = 0 at z = band z = b, we find that:

C = 0, and D = 12px

b2

Henceu =

12

px

(z2 b2

) Which is a parabolic velocity distribution.

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Mean velocity

We can determine the mean velocity:

Q = bb

u dz = bb

12

px

(z2 b2

)dz

=1

2Px

(13z3 b3z

)bb

= 23

Px

b3

We know that Q = 2bU:

U = b2

3px

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Piezometric pressure

The piezometric pressure: p + gz In open channel p is atmospheric hence only the second

term gz is significant Let p = gh:

U = b2

3 (gh)x

= b2g

3hx

This is analogous to groundwater flow (Darcy equation)

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Infinite Horizontal Plates: Laminar Flow

Derive the equation for the laminar, steady, uniform flowbetween infinite horizontal parallel plates.

ut + u

ux + v

uy + w

uz = 1 px + gx + 2u

(uux +

uv y +

uw z

)vt + u

vx + v

vy + w

vz = 1 py + gy + 2v

(v ux +

v v y +

v w z

)wt + u

wx + v

wy + w

wz = 1 pz + gz + 2w

(w ux +

w v y +

w w z

) Continuity

ux

+vy

+wz

= 0

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Derive the equation for the laminar, steady, uniform flowbetween infinite horizontal parallel plates.

0 = 1

px

+

(2uy2

)0 = 1

py

+ gy

Continuityux

= 0

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gy =py p = gy + A(x)

0 = px

+

(2uy2

) dp

dx=

(d2udy2

)

dpdx

dy =

(d2udy2

)dy

ydpdx

+ A = (dudy

)

ydpdx

+ Ady =

(dudy

)dy

y2

2dpdx

+ Ay + B = u63 / 66


No slip condition u = 0 at y = 0 and y = a Hence B = 0 and A = a2

dpdx

u =y(y a)

2dpdx

Discharge per unit width

q = a

0

y(y a)2

dpdx

dy =1

2dpdx

[y3

3 ay

]a0

=1

2dpdx

(a3

3 a2

)

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Inclined plane

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Infinite Inclined Plates: Laminar Flow Derive the equation for the laminar, steady, uniform flow

between infinite inclined parallel plates One fixed and theother moving with velocity U.

ut + u

ux + v

uy + w

uz = 1 px + gx + 2u

(uux +

uv y +

uw z

)vt + u

vx + v

vy + w

vz = 1 py + gy + 2v

(v ux +

v v y +

v w z

)wt + u

wx + v

wy + w

wz = 1 pz + gz + 2w

(w ux +

w v y +

w w z

) Continuity

ux

+vy

+wz

= 0

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Reynolds Averaged Navier-Stokes EquationsExamplesContinuityFluid accelerationForcesReynolds shear stressesEquation of motion

Documents

02 Derivation of Rans