Jan 06, 2016
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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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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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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Acceleration of a fluid particle
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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Acceleration of a fluid particle
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 of a fluid particle
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Acceleration of a fluid particle
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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Acceleration of a fluid particle
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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Contribution from the acceleration term
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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Contribution from the acceleration term
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
Reynolds Averaged Navier-Stokes Equations
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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Infinite Horizontal Plates: Laminar Flow
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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Infinite Horizontal Plates: Laminar Flow
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
Infinite Horizontal Plates: Laminar Flow
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