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CIVE 1400: Fluid Mechanics. www.efm.leeds.ac.uk/CIVE/FluidsLevel1 Lectures 16-19 (^178)
CIVE1400: An Introduction to Fluid Mechanics
Dr P A Sleigh [email protected]
Dr CJ Noakes [email protected]
January 2008 Module web site: www.efm.leeds.ac.uk/CIVE/FluidsLevel
Unit 1: Fluid Mechanics Basics 3 lectures Flow Pressure Properties of Fluids Fluids vs. Solids Viscosity Unit 2: Statics 3 lectures Hydrostatic pressure Manometry/Pressure measurement Hydrostatic forces on submerged surfaces Unit 3: Dynamics 7 lectures The continuity equation. The Bernoulli Equation. Application of Bernoulli equation. The momentum equation. Application of momentum equation. Unit 4: Effect of the boundary on flow 4 lectures Laminar and turbulent flow Boundary layer theory An Intro to Dimensional analysis Similarity
CIVE 1400: Fluid Mechanics. www.efm.leeds.ac.uk/CIVE/FluidsLevel1 Lectures 16-19 (^179)
Real fluids
Flowing real fluids exhibit
viscous effects, they:
x “stick” to solid surfaces
x have stresses within their body.
From earlier we saw this relationship between
shear stress and velocity gradient:
W v
du dy
The shear stress, W , in a fluid
is proportional to the velocity gradient
- the rate of change of velocity across the flow.
For a “Newtonian” fluid we can write:
W P
du dy
where P is coefficient of viscosity
(or simply viscosity).
Here we look at the influence of forces due to
momentum changes and viscosity
in a moving fluid.
Unit 4 Laminar and turbulent flow
Injecting a dye into the middle of flow in a pipe,
what would we expect to happen?
This
this
or this
Unit 4
All three would happen -
but for different flow rates.
Top: Slow flow
Middle: Medium flow
Bottom: Fast flow
Top: Laminar flow
Middle: Transitional flow
Bottom: Turbulent flow
Laminar flow:
Motion of the fluid particles is very orderly
all particles moving in straight lines
parallel to the pipe walls.
Turbulent flow:
Motion is, locally, completely random but the
overall direction of flow is one way.
But what is fast or slow?
At what speed does the flow pattern change?
And why might we want to know this?
CIVE 1400: Fluid Mechanics. www.efm.leeds.ac.uk/CIVE/FluidsLevel1 Lectures 16-19 (^182)
The was first investigated in the 1880s
by Osbourne Reynolds
in a classic experiment in fluid mechanics.
A tank arranged as below:
CIVE 1400: Fluid Mechanics. www.efm.leeds.ac.uk/CIVE/FluidsLevel1 Lectures 16-19 (^183)
After many experiments he found this
expression
U P
ud
U = density, u = mean velocity, d = diameter P = viscosity
This could be used to predict the change in
flow type for any fluid.
This value is known as the
Reynolds number, Re:
Re
U P
ud
Laminar flow: Re < 2000
Transitional flow: 2000 < Re < 4000
Turbulent flow: Re > 4000
Unit 4
What are the units of Reynolds number?
We can fill in the equation with SI units:
U
P
U
P
kg m u m s d m Ns m kg m s
ud kg m
m s
m m s kg
Re
3 2
It has no units!
A quantity with no units is known as a
non-dimensional (or dimensionless) quantity.
(We will see more of these in the section on
dimensional analysis.)
The Reynolds number, Re,
is a non-dimensional number.
Unit 4
At what speed does the flow pattern change?
We use the Reynolds number in an example:
A pipe and the fluid flowing
have the following properties:
water density U = 1000 kg/m^3
pipe diameter d = 0.5m
(dynamic) viscosity, P = 0.55x10^3 Ns/m^2
What is the MAXIMUM velocity when flow is
laminar i.e. Re = 2000
Re
. .
. /
u u u
U P
P U
ud
u d
u m s
2000
2000 2000 0 55 10 1000 0 5
0 0022
CIVE 1400: Fluid Mechanics. www.efm.leeds.ac.uk/CIVE/FluidsLevel1 Lectures 16-19 (^190)
Attaching a manometer gives
pressure (head) loss due to the energy lost by
the fluid overcoming the shear stress.
L
Δ p
The pressure at 1 (upstream)
is higher than the pressure at 2.
How can we quantify this pressure loss
in terms of the forces acting on the fluid?
CIVE 1400: Fluid Mechanics. www.efm.leeds.ac.uk/CIVE/FluidsLevel1 Lectures 16-19 (^191)
Consider a cylindrical element of
incompressible fluid flowing in the pipe,
area A
τ ο τ ο
τw
τw
W w is the mean shear stress on the boundary
Upstream pressure is p ,
Downstream pressure falls by ' p to ( p - ' p )
The driving force due to pressure
driving force = Pressure force at 1 - pressure force at 2
pA p p A p A p
d ' ' '
S 2 4
The retarding force is due to the shear stress
u
u
shear stress area over which it acts
= area of pipe wall
=
w
w
W
W S dL
Unit 4
As the flow is in equilibrium,
driving force = retarding force
'
'
p
d dL
p
L d
w
w
S W S
W
4 4
Giving pressure loss in a pipe in terms of:
x pipe diameter
x shear stress at the wall
Unit 4
What is the variation of shear stress in the flow?
τw
τw
r
R
At the wall
W w
R p 2 L
'
At a radius r
W
W W
r p L r wR
2
'
A linear variation in shear stress.
This is valid for:
x steady flow
x laminar flow
x turbulent flow
CIVE 1400: Fluid Mechanics. www.efm.leeds.ac.uk/CIVE/FluidsLevel1 Lectures 16-19 (^194)
Shear stress and hence pressure loss varies
with velocity of flow and hence with Re.
Many experiments have been done
with various fluids measuring
the pressure loss at various Reynolds numbers.
A graph of pressure loss and Re look like:
This graph shows that the relationship between
pressure loss and Re can be expressed as
laminar
turbulent or
'
'
p u
p u
v
v 1 7.^ (^ 2 0. ) CIVE 1400: Fluid Mechanics. www.efm.leeds.ac.uk/CIVE/FluidsLevel1 Lectures 16-19 (^195)
Pressure loss during laminar flow in a pipe
In general the shear stress W w****. is almost
impossible to measure.
For laminar flow we can calculate
a theoretical value for
a given velocity, fluid and pipe dimension.
In laminar flow the paths of individual particles
of fluid do not cross.
Flow is like a series of concentric cylinders
sliding over each other.
And the stress on the fluid in laminar flow is
entirely due to viscose forces.
As before, consider a cylinder of fluid, length L,
radius r, flowing steadily in the centre of a pipe.
Unit 4
r
δ r r R
The fluid is in equilibrium,
shearing forces equal the pressure forces.
W S S
W
2
2
r L p A p r^2
p L
r
' '
'
Newtons law of viscosity says W P
du dy
We are measuring from the pipe centre, so
W P
du dr
Giving:
'
'
p L
r du dr du dr
p L
r
2
2
P
P
Unit 4
In an integral form this gives an
expression for velocity,
u
p L
³ r dr
' 1
2 P
The value of velocity at a
point distance r from the centre
u
p L
r r ^ C
' 2
4 P
At r = 0 , (the centre of the pipe), u = umax , at
r = R (the pipe wall) u = 0 ;
C
p L
' R^2
4 P
At a point r from the pipe centre when the flow is
laminar:
u
p L
r R^ r
' 1 4
2 2
P
This is a parabolic profile
(of the form y = ax^2 + b )
so the velocity profile in the pipe looks similar to
v
CIVE 1400: Fluid Mechanics. www.efm.leeds.ac.uk/CIVE/FluidsLevel1 Lectures 16-19 (^202)
Understand this Boundary layer growth diagram.
CIVE 1400: Fluid Mechanics. www.efm.leeds.ac.uk/CIVE/FluidsLevel1 Lectures 16-19 (^203)
Boundary layer thickness:
G = distance from wall to where u = 0.99 umainstream
G increases as fluid moves along the plate. It reaches a maximum in fully developed flow.
The G increase corresponds to a drag force increase on the fluid.
As fluid is passes over a greater length:
*** more fluid is slowed
- by friction between the fluid layers
- the thickness of the slow layer increases.**
Fluid near the top of the boundary layer drags the fluid nearer to the solid surface along.
The mechanism for this dragging may be one of two types:
Unit 4 First: viscous forces (the forces which hold the fluid together)
When the boundary layer is thin: velocity gradient du/dy, is large
by Newton’s law of viscosity
shear stress, W = P (du/dy) , is large.
The force may be large enough to drag the fluid close to the surface.
As the boundary layer thickens velocity gradient reduces and shear stress decreases.
Eventually it is too small to drag the slow fluid along.
Up to this point the flow has been laminar.
Newton’s law of viscosity has applied.
This part of the boundary layer is the laminar boundary layer
Unit 4 Second: momentum transfer
If the viscous forces were the only action the fluid would come to a rest.
Viscous shear stresses have held the fluid particles in a constant motion within layers. Eventually they become too small to hold the flow in layers;
the fluid starts to rotate.
The fluid motion rapidly becomes turbulent. Momentum transfer occurs between fast moving main flow and slow moving near wall flow. Thus the fluid by the wall is kept in motion. The net effect is an increase in momentum in the boundary layer. This is the turbulent boundary layer.
CIVE 1400: Fluid Mechanics. www.efm.leeds.ac.uk/CIVE/FluidsLevel1 Lectures 16-19 (^206)
Close to boundary velocity gradients are very large. Viscous shear forces are large. Possibly large enough to cause laminar flow. This region is known as the laminar sub-layer.
This layer occurs within the turbulent zone it is next to the wall. It is very thin – a few hundredths of a mm.
Surface roughness effect
Despite its thinness, the laminar sub-layer has vital role in the friction characteristics of the surface.
In turbulent flow: Roughness higher than laminar sub-layer: increases turbulence and energy losses.
In laminar flow: Roughness has very little effect
Boundary layers in pipes Initially of the laminar form. It changes depending on the ratio of inertial and viscous forces; i.e. whether we have laminar (viscous forces high) or turbulent flow (inertial forces high).
CIVE 1400: Fluid Mechanics. www.efm.leeds.ac.uk/CIVE/FluidsLevel1 Lectures 16-19 (^207)
Use Reynolds number to determine which state.
Re^ U
P
ud
Laminar flow: Re < 2000 Transitional flow: 2000 < Re < 4000 Turbulent flow: Re > 4000
Laminar flow: profile parabolic (proved in earlier lectures) The first part of the boundary layer growth diagram.
Turbulent (or transitional), Laminar and the turbulent (transitional) zones of the boundary layer growth diagram. Length of pipe for fully developed flow is the entry length. Laminar flow | 120 u diameter Turbulent flow | 60 u diameter
Unit 4 Boundary layer separation
Divergent flows: Positive pressure gradients. Pressure increases in the direction of flow.
The fluid in the boundary layer has so little momentum that it is brought to rest, and possibly reversed in direction. Reversal lifts the boundary layer.
u 1 p (^1)
u 2 p 2
p 1 < p 2 u 1 > u (^2)
This phenomenon is known as boundary layer separation.
Unit 4 **Boundary layer separation:
- increases the turbulence
- increases the energy losses in the flow.**
Separating / divergent flows are inherently unstable
Convergent flows:
x Negative pressure gradients
x Pressure decreases in the direction of flow.
x Fluid accelerates and the boundary layer is thinner.
u (^1)
p (^1)
u (^2) p (^2)
p 1 > p (^2) u 1 < u (^2)
x Flow remains stable
x Turbulence reduces.
x Boundary layer separation does not occur.
CIVE 1400: Fluid Mechanics. www.efm.leeds.ac.uk/CIVE/FluidsLevel1 Lectures 16-19 (^214)
Fluid accelerates to get round the cylinder Velocity maximum at Y. Pressure dropped.
Adverse pressure between here and downstream. Separation occurs
CIVE 1400: Fluid Mechanics. www.efm.leeds.ac.uk/CIVE/FluidsLevel1 Lectures 16-19 (^215)
Aerofoil Normal flow over a aerofoil or a wing cross-section.
(boundary layers greatly exaggerated)
The velocity increases as air flows over the wing. The pressure distribution is as below so transverse lift force occurs.
Unit 4 At too great an angle boundary layer separation occurs on the top Pressure changes dramatically. This phenomenon is known as stalling.
All, or most, of the ‘suction’ pressure is lost. The plane will suddenly drop from the sky!
Solution: Prevent separation.
1 Engine intakes draws slow air from the boundary layer at the rear of the wing though small holes
2 Move fast air from below to top via a slot.
3 Put a flap on the end of the wing and tilt it.
Unit 4 Examples: Exam questions involving boundary layer theory are typically descriptive. They ask you to explain the mechanisms of growth of the boundary layers including how, why and where separation occurs. You should also be able to suggest what might be done to prevent separation.
CIVE 1400: Fluid Mechanics. www.efm.leeds.ac.uk/CIVE/FluidsLevel1 Lectures 16-19 (^218)
Lectures 18 & 19: Dimensional Analysis
Unit 4: The Effect of the Boundary on Flow
Application of fluid mechanics in design makes use of experiments results. Results often difficult to interpret. Dimensional analysis provides a strategy for choosing relevant data. Used to help analyse fluid flow Especially when fluid flow is too complex for mathematical analysis.
Specific uses: x help design experiments x Informs which measurements are important x Allows most to be obtained from experiment: e.g. What runs to do. How to interpret.
It depends on the correct identification of variables Relates these variables together Doesn’t give the complete answer Experiments necessary to complete solution
CIVE 1400: Fluid Mechanics. www.efm.leeds.ac.uk/CIVE/FluidsLevel1 Lectures 16-19 (^219)
Uses principle of dimensional homogeneity It gives qualitative results which only become quantitative from experimental analysis.
Dimensions and units
Any physical situation can be described by familiar properties.
e.g. length, velocity, area, volume, acceleration etc.
These are all known as dimensions.
Dimensions are of no use without a magnitude. i.e. a standardised unit e.g metre, kilometre, Kilogram, a yard etc.
Dimensions can be measured. Units used to quantify these dimensions.
In dimensional analysis we are concerned with the nature of the dimension i.e. its quality not its quantity.
Unit 4 The following common abbreviations are used:
length = L mass = M time = T force = F temperature = 4
Here we will use L, M, T and F (not 4 ).
We can represent all the physical properties we are interested in with three:
L, T
and one of M or F
As either mass (M) of force (F) can be used to represent the other, i.e. F = MLT- M = FT^2 L-
We will mostly use LTM:
Unit 4 This table lists dimensions of some common physical quantities:
Quantity SI Unit Dimension velocity m/s ms -1^ LT- acceleration m/s 2 ms -2^ LT- force N kg m/s 2 kg ms^ -2^ M LT- energy (or work) Joule J N m, kg m 2 /s 2 kg m^
(^2) s -2 ML 2 T-
power Watt W N m/s kg m 2 /s 3
Nms - kg m 2 s -3^ ML^
(^2) T-
pressure ( or stress) Pascal P, N/m^2 , kg/m/s 2
Nm- kg m -1s -2^ ML^
-1T-
density kg/m^3 kg m -3^ ML - specific weight N/m 3 kg/m 2 /s 2 kg m^
-2s - ML -2T- relative density a ratio no units
1 no dimension viscosity N s/m 2 kg/m s
N sm- kg m -1s -1^ M L^
-1T-
surface tension N/m kg /s 2
Nm- kg s -2^ MT-
CIVE 1400: Fluid Mechanics. www.efm.leeds.ac.uk/CIVE/FluidsLevel1 Lectures 16-19 (^226)
2 nd^ S theorem
Each S group is a function of n governing or repeating variables plus one of the remaining variables.
Choice of repeating variables
Repeating variables appear in most of the S groups. They have a large influence on the problem. There is great freedom in choosing these.
Some rules which should be followed are x There are n ( = 3) repeating variables. x In combination they must contain all of dimensions (M, L, T) x The repeating variables must not form a dimensionless group. x They do not have to appear in all S groups. x The should be measurable in an experiment. x They should be of major interest to the designer.
It is usually possible to take U, u and d This freedom of choice means: many different S groups - all are valid. There is not really a wrong choice. CIVE 1400: Fluid Mechanics. www.efm.leeds.ac.uk/CIVE/FluidsLevel1 Lectures 16-19 (^227)
An example
Taking the example discussed above of force F induced on a propeller blade, we have the equation
0 = I ( F, d, u, U , N, P ) n = 3 and m = 6
There are m - n = 3 S groups, so I ( S 1 , S 2 , S 3 ) = 0
The choice of U , u, d satisfies the criteria above.
They are: x measurable, x good design parameters x contain all the dimensions M,L and T.
Unit 4
We can now form the three groups according to the 2nd theorem,
S 1 U a^1^^ u db^1^^ c^1 F
S 2 U a^2^^ u b^2^^ d c^2 N
S 3 U a^3^^ u b^3^^ dc^3 P
The S groups are all dimensionless, i.e. they have dimensions M^0 L^0 T^0
We use the principle of dimensional homogeneity to equate the dimensions for each S group.
Unit 4
For the first S group, S 1 U a^1^^ u db^1^^ c^1 F
In terms of dimensions
M L T M L L T L M L T 0 0 0 3 a^^1 1 b^^1 c 1 2
The powers for each dimension (M, L or T), the powers must be equal on each side.
for M: 0 = a 1 + 1 a 1 = -
for L: 0 = -3a 1 + b 1 + c 1 + 1 0 = 4 + b 1 + c 1
for T: 0 = -b 1 - 2 b 1 = - c 1 = -4 - b 1 = -
Giving S 1 as
S U
S
U
1
1 2 2
(^1 2 )
(^) u (^) d F
F u d
CIVE 1400: Fluid Mechanics. www.efm.leeds.ac.uk/CIVE/FluidsLevel1 Lectures 16-19 (^230)
And a similar procedure is followed for the other S groups.
Group S 2 U a^2^^ u db^2^^ c^2 N
M L T M L L T L T
0 0 0 3 a^1^^ 1 b^^1 c 1 1
for M: 0 = a 2
for L: 0 = -3a 2 + b 2 + c 2
0 = b 2 + c 2
for T: 0 = -b 2 - 1
b 2 = - c 2 = 1
Giving S 2 as
S U
S
2
0 1 1
2
u d N^ Nd u
CIVE 1400: Fluid Mechanics. www.efm.leeds.ac.uk/CIVE/FluidsLevel1 Lectures 16-19 (^231)
And for the third, S 3 U a^3^^ u db^3^^ c^3 P
M L T M L L T L ML T
0 0 0 3 a^^3 1 b^^3 c 3 1 1
for M: 0 = a 3 + 1 a 3 = -
for L: 0 = -3a 3 + b 3 + c 3 - b 3 + c 3 = -
for T: 0 = -b 3 - 1 b 3 = - c 3 = -
Giving S 3 as
S U P
S
P
U
3
1 1 1
3
(^) u d
ud
Unit 4 Thus the problem may be described by
I ( S 1 , S 2 , S 3 ) = 0
I
U
P
U
F
u d
Nd (^2 2) u ud
©¨^
This may also be written:
F
u d
Nd
U u ud
I
P
2 2 U
Wrong choice of physical properties.
If, extra, unimportant variables are chosen :
Extra S groups will be formed
Will have little effect on physical performance
Should be identified during experiments
If an important variable is missed: x A S group would be missing. x Experimental analysis may miss significant behavioural changes.
Initial choice of variables should be done with great care.
Unit 4 Manipulation of the S groups
Once identified the S groups can be changed. The number of groups does not change. Their appearance may change drastically.
Taking the defining equation as:
I ( S 1 , S 2 , S 3 ……… S m-n ) = 0
The following changes are permitted: i. Combination of exiting groups by multiplication or division to form a new group to replaces one of the existing.
E.g. S 1 and S 2 may be combined to form S1a = S 1 / S 2 so the defining equation becomes
I ( S 1a , S 2 , S 3 ……… S m-n ) = 0 ii. Reciprocal of any group is valid. I ( S 1 ,1/ S 2 , S 3 ……… 1/ S m-n ) = 0 iii. A group may be raised to any power. I ( ( S 1 ) 2 , ( S 2 ) 1/2, ( S 3 ) 3 ……… S m-n ) = 0 iv. Multiplied by a constant. v. Expressed as a function of the other groups S 2 = I ( S 1 , S 3 ……… S m-n )
In general the defining equation could look like
I ( S 1 , 1/ S 2 ,( S 3 ) i……… 0.5 S m-n ) = 0
CIVE 1400: Fluid Mechanics. www.efm.leeds.ac.uk/CIVE/FluidsLevel1 Lectures 16-19 (^238)
Dynamic similarity
If geometrically and kinematically similar and the ratios of all forces are the same.
Force ratio F F
M a M a
L
L
m m p p
m m p p
L T
L
L T
L u
m p
u
U
U
O
O
O O
O
O
U O O OU
3 3 2
2
2 2 2
This occurs when the controlling S group is the same for model and prototype.
The controlling S group is usually Re. So Re is the same for model and prototype:
U P
U P
m m m m
p p p p
u d u d
It is possible another group is dominant. In open channel i.e. river Froude number is often taken as dominant.
CIVE 1400: Fluid Mechanics. www.efm.leeds.ac.uk/CIVE/FluidsLevel1 Lectures 16-19 (^239)
Modelling and Scaling Laws
Measurements taken from a model needs a scaling law applied to predict the values in the prototype.
An example:
For resistance R, of a body moving through a fluid. R , is dependent on the following:
U ML -3^ u: LT -1^ l:(length) L P : ML -1^ T - So
I (R, U , u, l, P ) = 0
Taking U , u, l as repeating variables gives:
R
u l
ul
R u l
ul
U
I
U
P
U I
U
P
2 2
2 2
This applies whatever the size of the body i.e. it is applicable to prototype and a geometrically similar model.
Unit 4 For the model
R u l
m u l m m m
m m m
U m
I
U
2 2 P
and for the prototype
R u l
p u l p p p
p p p
U p
I
U
2 2 P
Dividing these two equations gives
R u l R u l
u l u l
m m m m p p p p
m m m m p p p p
U
U
I U P
I U P
2 2 2 2
W can go no further without some assumptions. Assuming dynamic similarity, so Reynolds number are the same for both the model and prototype:
U
P
U
P
m m m m
p p p p
u d u d
so
R R
u l u l
m p
m m m p p p
U
U
2 2 2 2
i.e. a scaling law for resistance force:
O R O UO^2 u^ O L^2
Unit 4 Example 1 An underwater missile, diameter 2m and length 10m is tested in a water tunnel to determine the forces acting on the real prototype. A 1/20th^ scale model is to be used. If the maximum allowable speed of the prototype missile is 10 m/s, what should be the speed of the water in the tunnel to achieve dynamic similarity?
Dynamic similarity so Reynolds numbers equal:
U
P
U
P
m m m m
p p p p
u d u d
The model velocity should be
u u
d m p d
p m
p m
m p
U
U
P
P
Both the model and prototype are in water then,
P m = P p and U m = U p so
u u
d d m p p m^ s m
This is a very high velocity. This is one reason why model tests are not always done at exactly equal Reynolds numbers. A wind tunnel could have been used so the values of the
U and P ratios would be used in the above.
CIVE 1400: Fluid Mechanics. www.efm.leeds.ac.uk/CIVE/FluidsLevel1 Lectures 16-19 (^242)
Example 2
A model aeroplane is built at 1/10 scale and is to be tested in a wind tunnel operating at a pressure of 20 times atmospheric. The aeroplane will fly at 500km/h. At what speed should the wind tunnel operate to give dynamic similarity between the model and prototype? If the drag measure on the model is 337.5 N what will be the drag on the plane?
Earlier we derived an equation for resistance on a body moving through air:
R u l
ul u l
U I ¸
U
P
2 2 U 2 2 I Re
For dynamic similarity Rem = Rep , so
u u
d m p d
p m
p m
m p
U
U
P
P
The value of P does not change much with pressure so
P m = P p
For an ideal gas is p = U RT so the density of the air in the
model can be obtained from
p p
RT
RT
p p
m p
m p
m p p p
m p m p
U
U
U
U
U
U
U U
CIVE 1400: Fluid Mechanics. www.efm.leeds.ac.uk/CIVE/FluidsLevel1 Lectures 16-19 (^243)
So the model velocity is found to be
u u u
u km h
m p p
m
And the ratio of forces is
R R
u l u l
R R
m p
m p
m p
U
U
2 2 2 2
2 2 20 1
So the drag force on the prototype will be
R (^) p^1 R (^) m u N 0 05
Unit 4 Geometric distortion in river models
For practical reasons it is difficult to build a geometrically similar model.
A model with suitable depth of flow will often be far too big - take up too much floor space.
Keeping Geometric Similarity result in: x depths and become very difficult to measure; x the bed roughness becomes impracticably small; x laminar flow may occur - (turbulent flow is normal in rivers.)
Solution: Abandon geometric similarity.
Typical values are 1/100 in the vertical and 1/400 in the horizontal.
Resulting in: x Good overall flow patterns and discharge x local detail of flow is not well modelled.
The Froude number (Fn) is taken as dominant. Fn can be the same even for distorted models.
Unit 4
