FUNDAMENTALS OF
FLUID MECHANICS
Chapter 7 Dimensional Analysis
Modeling, and Similitude
Jyh-Cherng Shieh
Department of Bio-Industrial Mechatronics Engineering
National Taiwan University
12/24/2007
1
MAIN TOPICS
™ Dimensional Analysis
™ Buckingham Pi Theorem
™ Determination of Pi Terms
™ Comments about Dimensional Analysis
™ Common Dimensionless Groups in Fluid Mechanics
™ Correlation of Experimental Data
™ Modeling and Similitude
™ Typical Model Studies
™ Similitude Based on Governing Differential Equation
2
Dimensional Analysis 1/4
™ A typical fluid mechanics problem in which
experimentation is required, consider the steady flow of an
incompressible Newtonian fluid through a long, smoothwalled, horizontal, circular pipe.
™ An important characteristic of this system, which would
be interest to an engineer designing a pipeline, is the
pressure drop per unit length that develops along the pipe
as a result of friction.
3
Dimensional Analysis 2/4
™ The first step in the planning of an experiment to study
this problem would be to decide on the factors, or
variables, that will have an effect on the pressure drop.
™ Pressure drop per unit length
∆p l = f ( D , ρ, µ , V )
9 Pressure drop per unit length depends on FOUR variables:
sphere size (D); speed (V); fluid density (ρ); fluid viscosity
(m)
4
Dimensional Analysis 3/4
™ To perform the experiments in a meaningful and
systematic manner, it would be necessary to change on of
the variable, such as the velocity, which holding all other
constant, and measure the corresponding pressure drop.
™ Difficulty to determine the functional relationship between
the pressure drop and the various facts that influence it.
5
Series of Tests
6
Dimensional Analysis 4/4
™ Fortunately, there is a much simpler approach to the
problem that will eliminate the difficulties described
above.
DCollecting these variables into two nondimensional
combinations of the variables (called dimensionless
product or dimensionless groups)
D∆p l
ρV 2
⎛ ρVD ⎞
⎟⎟
= φ⎜⎜
⎝ µ ⎠
9 Only one dependent and one
independent variable
9 Easy to set up experiments to
determine dependency
9 Easy to present results (one graph)
7
Plot of Pressure Drop Data Using …
D ∆p l
ρV 2
=
L( F / L3 )
( FL− 4 T 2 )( FT −1 ) 2
=& F 0 L0T 0
ρVD ( FL−4 T 2 )( LT −1 )( L)
0 0 0
=&
=
F
LT
−
2
µ
( FL T )
dimensionless product or
dimensionless groups
8
Buckingham Pi Theorem 1/5
™ A fundamental question we must answer is how many
dimensionless products are required to replace the original list of
variables ?
™ The answer to this question is supplied by the basic theorem of
dimensional analysis that states
™If an equation involving k variables is dimensionally
homogeneous, it can be reduced to a relationship among
k-r independent dimensionless products, where r is the
minimum number of reference dimensions required to
describe the variables.
Buckingham Pi Theorem
Pi terms
9
Buckingham Pi Theorem 2/5
™ Given a physical problem in which the dependent variable
is a function of k-1 independent variables.
u 1 = f ( u 2 , u 3 ,....., u k )
DMathematically, we can express the functional relationship
in the equivalent form
g ( u 1 , u 2 , u 3 ,....., u k ) = 0
Where g is an unspecified function, different from f.
10
Buckingham Pi Theorem 3/5
™ The Buckingham Pi theorem states that: Given a relation
among k variables of the form
g ( u 1 , u 2 , u 3 ,....., u k ) = 0
™ The k variables may be grouped into k-r independent
dimensionless products, or Π terms, expressible in
functional form by
Π 1 = φ( Π 2 , Π 3 , , , , Π k − r )
or
φ( Π 1 , Π 2 , Π 3 , , , , Π k − r ) = 0
r ?? Π??
11
Buckingham Pi Theorem 4/5
™ The number r is usually, but not always, equal to the
minimum number of independent dimensions required to
specify the dimensions of all the parameters. Usually the
reference dimensions required to describe the variables
will be the basic dimensions M, L, and T or F, L, and T.
™ The theorem does not predict the functional form of φ or
ϕ . The functional relation among the independent
dimensionless products Π must be determined
experimentally.
™ The k-r dimensionless products Π terms obtained
from the procedure are independent.
12
Buckingham Pi Theorem 5/5
™ A Π term is not independent if it can be obtained from a
product or quotient of the other dimensionless products of
the problem. For example, if
2Π1
Π5 =
Π 2Π 3
or
Π6 =
Π1
3/ 4
Π 32
then neither Π5 nor Π6 is independent of the other
dimensionless products or dimensionless groups.
13
Determination of Pi Terms 1/12
™ Several methods can be used to form the dimensionless
products, or pi term, that arise in a dimensional analysis.
™ The method we will describe in detail is called the
METHOD of repeating variables.
™ Regardless of the method to be used to determine the
dimensionless products, one begins by listing all
dimensional variables that are known (or believed) to
affect the given flow phenomenon.
™ Eight steps listed below outline a recommended
procedure for determining the Π terms.
14
Determination of Pi Terms 2/12
™Step 1 List all the variables. 1
>List all the dimensional variables involved.
>Keep the number of variables to a minimum, so that we
can minimize the amount of laboratory work.
>All variables must be independent. For example, if the
cross-sectional area of a pipe is an important variable,
either the area or the pipe diameter could be used, but
not both, since they are obviously not independent.
γ=ρ×g, that is, γ,ρ, and g are not independent.
15
Determination of Pi Terms 3/12
™Step 1 List all the variables. 2
>Let k be the number of variables.
>Example: For pressure drop per unit length, k=5. (All
variables are ∆pl, D,µ,ρ, and V )
∆p l = f ( D, ρ, µ, V )
16
Determination of Pi Terms 4/12
™Step 2 Express each of the variables in terms of
basic dimensions. Find the number of reference
dimensions.
>Select a set of fundamental (primary) dimensions.
>For example: MLT, or FLT.
>Example: For pressure drop per unit length , we choose
FLT.
∆p l =& FL−3
D =& L
ρ =& FL−4 T 2
r=3
−2
−1
µ =& FL T
V =& LT
17
Determination of Pi Terms 5/12
™Step 3 Determine the required number of pi terms.
>Let k be the number of variables in the problem.
>Let r be the number of reference dimensions (primary
dimensions) required to describe these variables.
>The number of pi terms is k-r
>Example: For pressure drop per unit length k=5, r = 3,
the number of pi terms is k-r=5-3=2.
18
Determination of Pi Terms 6/12
™Step 4 Select a number of repeating variables,
where the number required is equal to the number
of reference dimensions.
>Select a set of r dimensional variables that includes all
the primary dimensions ( repeating variables).
>These repeating variables will all be combined with
each of the remaining parameters. No repeating
variables should have dimensions that are power of the
dimensions of another repeating variable.
>Example: For pressure drop per unit length ( r = 3)
select ρ , V, D.
19
Determination of Pi Terms 7/12
™Step 5 Form a pi term by multiplying one of the
nonrepeating variables by the product of the
repeating variables, each raised to an exponent that
will make the combination dimensionless. 1
>Set up dimensional equations, combining the variables
selected in Step 4 with each of the other variables
(nonrepeating variables) in turn, to form dimensionless
groups or dimensionless product.
>There will be k – r equations.
>Example: For pressure drop per unit length
20
Determination of Pi Terms 8/12
™Step 5 (Continued) 2
a b c
Π1 = ∆p l D V ρ
( FL−3 )( L) a ( LT −1 ) b ( FL−4 T 2 ) c =& F 0 L0T 0
F :1 + c = 0
L : −3 + a + b − 4 c = 0
T : − b + 2c = 0
⇒ a = 1, b = −2, c = −1
Π1 =
∆p l D
ρV
2
21
Determination of Pi Terms 9/12
™Step 6 Repeat Step 5 for each of the remaining
nonrepeating variables.
a
b c
Π 2 = µD V ρ
( FL−2 T )( L) a ( LT −1 ) b ( FL−4 T 2 ) c =& F 0 L0T 0
F :1 + c = 0
L : −2 + a + b − 4 c = 0
T : 1 − b + 2c = 0
⇒ a = −1, b = −1, c = −1
µ
Π2 =
DVρ
22
Determination of Pi Terms 10/12
™Step 7 Check all the resulting pi terms to make
sure they are dimensionless.
>Check to see that each group obtained is dimensionless.
>Example: For pressure drop per unit length .
Π1 =
∆p l D
2
=& F0 L0T 0 =& M 0 L0T 0
ρV
µ
0 0 0
0 0 0
&
&
=F L T =M LT
Π2 =
DVρ
23
Determination of Pi Terms 11/12
™Step 8 Express the final form as a relationship
among the pi terms, and think about what is means.
>Express the result of the dimensional analysis.
Π 1 = φ( Π 2 , Π 3 , , , , Π k − r )
>Example: For pressure drop per unit length .
∆p l D
ρV 2
⎛ µ ⎞
⎟⎟
= φ⎜⎜
⎝ DVρ ⎠
Dimensional analysis will not provide
the form of the function. The function
can only be obtained from a suitable set
of experiments.
24
Determination of Pi Terms 12/12
™ The pi terms can be rearranged. For example, Π2, could
be expressed as
ρVD
Π2 =
µ
∆p l D
ρV
2
⎛ ρVD ⎞
⎟⎟
= φ⎜⎜
⎝ µ ⎠
25
Example 7.1 Method of Repeating
Variables
z A thin rectangular plate having a width w and a height h is
located so that it is normal to a moving stream of fluid.
Assume that the drag, D, that the fluid exerts on the plate
is a function of w and h, the fluid viscosity, µ ,and ρ,
respectively, and the velocity, V, of the fluid approaching
the plate. Determine a suitable set of pi terms to study this
problem experimentally.
26
Example 7.1 Solution1/5
™ Drag force on a PLATE
D = f ( w , h , ρ, µ , V )
>Step 1:List all the dimensional variables involved.
D,w,h, ρ,μ,V k=6 dimensional parameters.
>Step 2:Select primary dimensions M,L, and T. Express
each of the variables in terms of basic dimensions
D =& MLT −2
µ =& ML−1T −1
w =& L
ρ =& ML−3
h =& L
V =& LT −1
27
Example 7.1 Solution2/5
>Step 3: Determine the required number of pi terms.
k-r=6-3=3
>Step 4:Select repeating variables w,V,ρ.
>Step 5~6:combining the repeating variables with each
of the other variables in turn, to form dimensionless
groups or dimensionless products.
28
Example 7.1 Solution3/5
Π 1 = Dw a V b ρ c = ( MLT − 2 )( L ) a ( LT − 1 ) b ( ML − 3 ) c = M 0 L0 T 0
M :1 + c = 0
L : 1 + a + b − 3c = 0
T : −2 − b = 0
>> a = − 2 , b = − 2 , c = − 1
Π1 =
D
2
2
w V ρ
Π 2 = hw a V b ρ c = L ( L ) a ( LT − 1 ) b ( ML − 3 ) c = M 0 L0 T 0
M :c = 0
L : 1 + a + b − 3c = 0
T:b=0
>> a = − 1, b = 0, c = 0
h
Π2 =
w
29
Example 7.1 Solution4/5
Π 3 = µw a V bρ c = ( ML−1T −1 )( L) a ( LT −1 ) b ( ML−3 ) c = M 0 L0T 0
M :1 + c = 0
L : −1 + a + b − 3c = 0
T : −1 − b = 0
>> a = −1, b = −1, c = −1
µ
Π3 =
wVρ
30
Example 7.1 Solution5/5
>Step 7: Check all the resulting pi terms to make sure
they are dimensionless.
>Step 8: Express the final form as a relationship among
the pi terms.
The functional relationship is
Π 1 = φ ( Π 2 , Π 3 ), or
⎛ h
µ ⎞
⎜
⎟
=
φ
,
⎜
⎟
2 2
w V ρ
⎝ w wV ρ ⎠
D
31
Selection of Variables 1/4
™ One of the most important, and difficult, steps in applying
dimensional analysis to any given problem is the selection
of the variables that are involved.
™ There is no simple procedure whereby the variable can be
easily identified. Generally, one must rely on a good
understanding of the phenomenon involved and the
governing physical laws.
™ If extraneous variables are included, then too many pi
terms appear in the final solution, and it may be difficult,
time consuming, and expensive to eliminate these
experimentally.
32
Selection of Variables 2/4
™ If important variables are omitted, then an incorrect result
will be obtained; and again, this may prove to be costly
and difficult to ascertain.
™ Most engineering problems involve certain simplifying
assumptions that have an influence on the variables to be
considered.
™ Usually we wish to keep the problems as simple as
possible, perhaps even if some accuracy is sacrificed
33
Selection of Variables 3/4
™ A suitable balance between simplicity and accuracy is an
desirable goal.~~~~~
™ Variables can be classified into three general group:
BGeometry: lengths and angles.
BMaterial Properties: relate the external effects and the
responses.
BExternal Effects: produce, or tend to produce, a change
in the system. Such as force, pressure, velocity, or
gravity.
34
Selection of Variables 4-1/4
™ Points should be considered in the selection of variables:
BClearly define the problem. What’s the main variable
of interest?
BConsider the basic laws that govern the phenomenon.
BStart the variable selection process by grouping the
variables into three broad classes.
Continued
35
Selection of Variables 4-2/4
™ Points should be considered in the selection of variables:
BConsider other variables that may not fall into one the
three categories. For example, time and time dependent
variables.
BBe sure to include all quantities that may be held
constant (e.g., g).
BMake sure that all variables are independent. Look for
relationships among subsets of the variables.
36
Determination of Reference Dimension 1/3
™ When to determine the number of pi terms, it is important
to know how many reference dimensions are required to
describe the variables.
™ In fluid mechanics, the required number of reference
dimensions is three, but in some problems only one or two
are required.
™ In some problems, we occasionally find the number of
reference dimensions needed to describe all variables is
smaller than the number of basic dimensions. Illustrated in
Example 7.2.
37
Example 7.2 Determination of Pi Terms
z An open, cylindrical tank having a diameter D is
supported around its bottom circumference and is filled to
a depth h with a liquid having a specific weight γ. The
vertical deflection, δ , of the center of the bottom is a
function of D, h, d, γ, and E, where d is the thickness of
the bottom and E is the modulus of elasticity of the bottom
material. Perform a dimensional analysis of this problem.
38
Example 7.2 Solution1/3
™ The vertical deflection
δ = f (D, h, d , γ , E )
For F,L,T. Pi terms=6-2=4
For M,L,T Pi terms=6-3=3
δ=L
D=L
h=L
d=L
γ = FL−3 , γ = ML− 2 T − 2
E = FL− 2 , E = ML−1T − 2
39
Example 7.2 Solution2/3
™ For F,L,T system, Pi terms=6-2=4
D and γ are selected as repeating variables
a1 b1
a 2 b2
Π1 = δD γ
Π 2 = hD γ
Π 3 = dD a 3 γ b3
Π 4 = ED a 4 γ b4
⎛h d E ⎞
δ
h
d
E
δ
⎟⎟
Π1 = , Π 2 = , Π 3 = , Π 4 =
⇒ = Φ ⎜⎜ , ,
D
D
D
Dγ
D
⎝ D D Dγ ⎠
40
Example 7.2 Solution3/3
™ For M,L,T system, Pi terms=6-3=3 ?
A closer look at the dimensions of the variables listed
reveal that only two reference dimensions, L and MT-2
are required.
41
Determination of Reference Dimension 2/3
EXAMPLE ∆h = f (D, γ , σ )
MLT SYSTEM
∆h D
L
L
γ
M
2 2
LT
Pi term=4-3=1
FLT SYSTEM
σ
M
T
2
∆h D
L
L
γ
F
L3
σ
F
L
Pi term=4-2=2
42
Determination of Reference Dimension 3/3
Set Dimensional Matrix
MLT SYSTEM
∆h D
M
0
0
L
T
1
0
1
0
γ
σ
1
1
−2 0
−2 −2
Rank=2 Pi term=4-2=2
FLT SYSTEM
∆h D
F
L
0
1
0
1
T
0
0
γ
σ
1
1
− 3 −1
0
0
∆h
Π1 =
D
⎛ σ ⎞
∆h
Π2 = 2 ⇒
= Φ⎜ 2 ⎟
⎜D γ⎟
D
D γ
⎝
⎠
σ
43
Uniqueness of Pi Terms 1/4
™ The Pi terms obtained depend on the somewhat arbitrary
selection of repeating variables. For example, in the
problem of studying the pressure drop in a pipe.
∆p l = f ( D, ρ, µ, V )
Selecting D,V, and ρ as
repeating variables:
∆p l D
Selecting D,V, and µ as
repeating variables:
⎛ ρVD ⎞
∆p l D 2
⎟⎟
= Φ 2 ⎜⎜
Vµ
⎝ µ ⎠
ρV 2
⎛ ρVD ⎞
⎟⎟
= Φ1 ⎜⎜
⎝ µ ⎠
44
Uniqueness of Pi Terms 2/4
∆p l D
ρV
2
⎛ ρVD ⎞
⎟⎟
= Φ1 ⎜⎜
⎝ µ ⎠
⎛ ρVD ⎞
∆p l D 2
⎟⎟
= Φ 2 ⎜⎜
Vµ
⎝ µ ⎠
Both are correct, and both would
lead to the same final equation for
the pressure drop. There is not a
unique set of pi terms which
arises from a dimensional analysis.
The functions Φ1 and Φ2 are will
be different because the dependent
pi terms are different for the two
relationships.
45
Uniqueness of Pi Terms 3/4
EXAMPLE Π1 = Φ (Π 2 , Π 3 )
Form a new pi term
'
Π2
(
=
a b
Π2Π3
'
Π1 = Φ1 Π 2 , Π 3
)
(
'
= Φ2 Π2 , Π2
)
All are correct
46
Uniqueness of Pi Terms 4/4
∆p l D
ρV 2
⎛ ρVD ⎞
⎟⎟
= Φ1 ⎜⎜
⎝ µ ⎠
∆p l D
ρV 2
Selecting D,V, and ρ as
repeating variables:
2
∆
p
D
ρVD
l
×
=
µ
Vµ
⎛ ρVD ⎞
∆p l D 2
⎟⎟
= Φ 2 ⎜⎜
Vµ
⎝ µ ⎠
47
Common Dimensionless Groups 1/2
™ A list of variables that
commonly arise in fluid
mechanical problems.
™ Possible to provide a
physical interpretation to
the dimensionless groups
which can be helpful in
assessing their influence
in a particular application.
48
Froude Number 1/2
Fr =
2
2 2
ρ
V
V
L
2
>> Fr =
=
gL
ρ gL3
gL
V
™ In honor of William Froude (1810~1879), a British civil engineer,
mathematician, and naval architect who pioneered the use of towing
tanks for the study of ship design.
™ Froude number is the ratio of the forces due to the acceleration of a
fluid particles (inertial force) to the force due to gravity (gravity
forces).
™ Froude number is significant for flows with free surface effects.
™ Froude number less than unity indicate subcritical flow and values
greater than unity indicate supercritical flow.
49
Froude Number 2/2
as =
2
*
s
*
dVs V
dV
dVs
= Vs
=
Vs*
dt
ds
l
ds
Vs
V =
V
*
s
V 2 * dVs*
FI =
Vs
m
*
l
ds
s
s =
l
*
FI
V 2 * dVs* V 2
V
Fr =
Vs
=
≡
≡
*
FG
gl
ds
gl
gl
50
Reynolds Number 1/2
ρVl Vl
=
Re =
µ
υ
™ In honor of Osborne Reynolds (1842~1912), the British engineer
who first demonstrated that this combination of variables could be
used as a criterion to distinguish between laminar and turbulent flow.
™ The Reynolds number is a measure of the ration of the inertia forces
to viscous forces.
™ If the Reynolds number is small (Re<<1), thi sis an indication that
the viscous forces ar dominant in the problem, and it may be
possible to neglect the inertial effects; that is, the density of the fluid
will no be an important variable.
51
Reynolds Number 2/2
™ Flows with very small Reynolds numbers are commonly referred to
as “creeping flows”.
™ For large Reynolds number flow, the viscous effects are small
relative to inertial effects and for these cases it may be possible to
neglect the effect of viscosity and consider the problem as one
involving a “nonviscous” fluid.
™ Flows with “large” Reynolds number generally are turbulent. Flows
in which the inertia forces are “small” compared with the viscous
forces are characteristically laminar flows.
52
Euler number
p
∆p
≡
Eu =
2
ρV
ρV 2
™ In honor of Leonhard Euler (1707~1783), a famous Swiss
mathematician who pioneered work on the relationship between
pressure and flow.
™ Euler’s number is the ratio of pressure force to inertia forces. It is
often called the pressure coefficient, Cp.
53
Cavitation Number
pr − pv
Ca =
1
ρV 2
2
™ For problems in which cavitation is of concern, the dimensionless
group ( p r − p v ) / 12 ρ V 2 is commonly used, where pv is the vapor
pressure and pr is some reference pressure.
™ The cavitation number is used in the study of cavitation phenomena.
™ The smaller the cavitation number, the more likely cavitation is to
occur.
54
Cauchy Number & Mach Number 1/2
ρV 2
Ca =
Eυ
V
=
Ma =
c
ρ
V
=V
Eυ
Eυ
ρ
ρV 2
= Ca
Ma =
Eυ
2
™ The Cauchy number is named in honor of Augustin Louis de
Cauchy (1789~1857), a French engineer, mathematician, and
hydrodynamicist.
™ The Mach number is named in honor of Ernst Mach (1838~1916),
an Austrian physicist and philosopher.
™ Either number may be used in problems in which fluid
compressibility is important.
55
Cauchy Number & Mach Number 2/2
™ Both numbers can be interpreted as representing an index of the
ration of inertial force to compressibility force, where V is the flow
speed and c is the local sonic speed.
™ Mach number is a key parameter that characterizes compressibility
effects in a flow.
™ When the Mach number is relatively small (say, less than 0.3), the
inertial forces induced by the fluid motion are not sufficiently large
to cause a significant change in the fluid density, and in this case the
compressibility of the fluid can be neglected.
™ For truly incompressible flow, c=∞ so that M=0.
56
Strouhal Number 1/2
ωl
St =
V
™ In honor of Vincenz Strouhal (1850~1922), who used this parameter
in his study of “singing wires.” The most dramatic evidence of this
phenomenon occurred in 1940 with the collapse of the Tacoma
Narrow bridges. The shedding frequency of the vortices coincided
with the natural frequency of the bridge, thereby setting up a
resonant condition that eventually led to the collapse of the bridge.
™ This parameter is important in unsteady, oscillating flow problems
in which the frequency of the oscillation is ω .
57
Strouhal Number 2/2
™ This parameter represents a measure of
the ration of inertial force due to the
unsteadiness of the flow (local
acceleration) to the inertial forces due to
change in velocity from point to point in
the flow field (convective acceleration).
This type of unsteady flow may develop
when a fluid flows past a solid body (such
as a wire or cable) placed in the moving
stream.
For example, in a certain Reynolds number range, a periodic flow will
develop downstream from a cylinder placed in a moving stream due to
a regular patterns of vortices that are shed from the body.
58
Weber Number 1/2
2
ρV l
We =
σ
™ Named after Moritz Weber (1871~1951), a German professor of
naval mechanics who was instrumental in formalizing the general
use of common dimensionless groups as a basis for similitude
studies.
™ Weber number is important in problem in which there is an interface
between two fluids. In this situation the surface tension may play an
important role in the phenomenon of interest.
59
Weber Number 2/2
™ Weber number is the ratio of inertia forces to surface tension forces.
™ Common examples of problems in which Weber number may be
important include the flow of thin film of liquid, or the formation of
droplets or bubbles.
™ The flow of water in a river is not affected significantly by sureface
tension, since inertial and gravitational effects are dominant
(We>>1).
60
Correlation of Experimental Data
™ Dimensional analysis only provides the dimensionless groups
describing the phenomenon, and not the specific relationship
between the groups.
™ To determine this relationship, suitable experimental data must be
obtained.
™ The degree of difficulty depends on the number of pi terms.
61
Problems with One Pi Term
™ The functional relationship for one Pi term.
Π1 = C
where C is a constant. The value of the constant must still be
determined by experiment.
62
Example 7.3 Flow with Only One Pi Term
z Assume that the drag, D, acting on a spherical particle that falls very
slowly through a viscous fluid is a function of the particle diameter,
d, the particle velocity, V, and the fluid viscosity, μ. Determine,
with the aid the dimensional analysis, how the drag depends on the
particle velocity.
63
Example 7.3 Solution
™ The drag
D = f (d, V , µ )
D =& F
ρ =& ML−3
d =& L
µ =& FL−2T
V =& LT −1
D
D = CµVd
Π1 =
=C
µVd
For a given particle and fluids, the drag varies
D∝V
directly with the velocity
64
Problems with Two or More Pi Term 1/2
™ Problems with two pi terms
Π1 = Φ( Π 2 )
the functional relationship
among the variables can the
be determined by varying
Π2 and measuring the
corresponding value of Π1.
The empirical equation
relating Π2 and Π1 by using
a standard curve-fitting
technique.
An empirical relationship is
valid over the range of Π2.
Dangerous to
extrapolate beyond
valid range
65
Problems with Two or More Pi Term 2/2
™ Problems with three pi terms.
Π1 = Φ(Π 2 ,Π 3 )
To determine a suitable empirical equation
relating the three pi terms.
To show data correlations on simple graphs.
Families curve of curves
66
Modeling and Similitude
To develop the procedures for designing
models so that the model and prototype
will behave in a similar fashion…….
67
Model vs. Prototype 1/2
™ Model ? A model is a representation of a physical system that may
be used to predict the behavior of the system in some desired respect.
Mathematical or computer models may also conform to this
definition, our interest will be in physical model.
™ Prototype? The physical system for which the prediction are to be
made.
™ Models that resemble the prototype but are generally of a different
size, may involve different fluid, and often operate under different
conditions.
™ Usually a model is smaller than the prototype.
™ Occasionally, if the prototype is very small, it may be advantageous
to have a model that is larger than the prototype so that it can ve
more easily studied. For example, large models have been used to
study the motion of red blood cells.
68
Model vs. Prototype 2/2
™ With the successful development of a valid model, it is possible to
predict the behavior of the prototype under a certain set of
conditions.
™ There is an inherent danger in the use of models in that predictions
can be made that are in error and the error not detected until the
prototype is found not to perform as predicted.
™ It is imperative that the model be properly designed and tested and
that the results be interpreted correctly.
69
Similarity of Model and Prototype
™ What conditions must be met to ensure the similarity of model and
prototype?
~ Geometric Similarity
>Model and prototype have same shape.
>Linear dimensions on model and prototype correspond within
constant scale factor.
~ Kinematic Similarity
>Velocities at corresponding points on model and prototype differ
only by a constant scale factor.
~ Dynamic Similarity
>Forces on model and prototype differ only by a constant scale
factor.
70
Theory of Models 1/5
™ The theory of models can be readily developed by using
the principles of dimensional analysis.
™ For given problem which can be described in terms of a
set of pi terms as
Π1 = φ( Π 2 , Π 3 , , , Π n )
This relationship can be formulated
with a knowledge of the general
nature of the physical phenomenon
and the variables involved.
This equation applies to any system that is
governed by the same variables.
71
Theory of Models 2/5
™ A similar relationship can be written for a model of this
prototype; that is,
Π 1m = φ( Π 2 m , Π 3 m , , , , Π nm )
where the form of the function will be the same as long
as the same phenomenon is involved in both the
prototype and the model.
The prototype and the model must have
the same phenomenon.
72
Theory of Models 3/5
™ Model design (the model is designed and operated)
conditions, also called similarity requirements or modeling
laws.
Π 2 = Π 2m
Π 3 = Π 3 m ..... Π n = Π nm
™ The form of Φ is the same for model and prototype, it
follows that
This is the desired prediction equation and
Π 1 = Π 1m
indicates that the measured of Π1m obtained
with the model will be equal to the
corresponding Π1 for the prototype as long as
the other Π parameters are equal.
73
Theory of Models 4/5 – Summary1
™ The prototype and the model must have the same
phenomenon.
For prototype Π1 = φ( Π 2 , Π 3 , , , Π n )
For model
Π 1m = φ( Π 2 m , Π 3 m , , , , Π nm )
74
Theory of Models 5/5 – Summary2
™ The model is designed and operated under the following
conditions (called design conditions, also called similarity
requirements or modeling laws)
Π 2 = Π 2m
Π 3 = Π 3 m ..... Π n = Π nm
™ The measured of Π1m obtained with the model will be
equal to the corresponding Π1 for the prototype.
Π 1 = Π 1m
Called prediction equation
75
Theory of Models EXAMPLE 1
™ Example: Considering the drag force on a sphere.
⎛ ρ VD ⎞
F
⎜
⎟
=
f
1
F = f ( D , V , ρ, µ )
⎜
⎟
µ
ρV 2 D 2
⎝
⎠
BThe prototype and the model must have the same phenomenon.
⎛ ρ VD ⎞
F
⎛ ρ m Vm D m ⎞
Fm
⎟⎟
⎟⎟
= f1 ⎜⎜
= f1 ⎜⎜
2
2
2
2
µm
ρ m Vm D m
⎝
⎠
ρV D
⎝ µ ⎠ prototype
BDesign conditions.
⎛ ρ VD
⎜⎜
⎝ µ
⎛ ρ VD
⎞
⎟⎟
= ⎜⎜
⎠ mod el ⎝ µ
⎞
⎛
F
⎟
⎜
=
2 2
⎜
⎜ ρV 2 D 2 ⎟
⎠ mod el ⎝ ρ V D
⎝
BThen … ⎛⎜
F
⎞
⎟⎟
⎠ prototype
⎞
⎟
⎟
⎠ prototype
76
Theory of Models EXAMPLE 2
™ Example: Determining the drag force on a thin rectangular plate (w
× h in size)
D
D = f (w , h , µ , ρ , V )
w 2ρV 2
⎛ w ρ Vw
= Φ ⎜⎜ ,
µ
⎝ h
⎞
⎟⎟
⎠
BThe prototype and the model must have the same phenomenon.
Dm
w m 2 ρ m Vm 2
⎛ w m ρ m Vm w m
,
= Φ ⎜⎜
h
µm
⎝ m
BDesign conditions.
BThen …
D
2
w ρV
2
=
⎞
⎟⎟
⎠
D
w 2ρV 2
⎛ w ρ Vw
= Φ ⎜⎜ ,
µ
⎝ h
⎞
⎟⎟
⎠ prototype
2
2
wm
w ρ V w
ρ Vw
= , m m m =
hm
h
µm
µ
Dm
w m 2 ρ m Vm 2
⎛ w
>> D = ⎜⎜
⎝ wm
⎞
⎟⎟
⎠
⎛ ρ
⎜⎜
⎝ ρm
⎞⎛ V
⎟⎟ ⎜⎜
⎠ ⎝ Vm
⎞
⎟⎟ D m
⎠
77
Example 7.5 Prediction of Prototype
Performance from Model Data 1/2
z A long structural component of a bridge has the cross section
shown in Figure E7.5. It is known that when a steady wind blows
past this type of bluff body, vortices may develop on the downwind
side that are shed in a regular fashion at some definite frequency.
Since these vortices can create harmful periodic forces acting on the
structure, it is important to determine the shedding frequency. For
the specific structure of interest, D=0.1m, H=0.3m, and a
representative wind velocity 50km/hr. Standard air can be assumed.
The shedding frequency is to be determined through the use of a
small-scale model that is to be tested in a water tunnel. For the
model Dm=20mm and the water temperature is 20℃.
78
Example 7.5 Prediction of Prototype
Performance from Model Data 2/2
Determine the model dimension, Hm, and the velocity at which the
test should be performed. If the shedding frequency ω for the
model is found to be 49.9Hz, what is the corresponding frequency
for the prototype?
−5
3
~ For air at standard condition µ = 1 .79 × 10 kg / m − s, ρ = 1 .23 kg / m
~ For water at 20℃, µ water = 1 × 10 − 3 kg / m − s, ρ water = 998 kg / m 3
79
Example 7.5 Solution1/4
9 Step 1:List all the dimensional variables involved. ω
D,H,V,ρ,μ k=6 dimensional variables.
9 Step 2:Select primary dimensions F,L and T. List the
dimensions of all variables in terms of primary dimensions.
r=3 primary dimensions
−1
&
ω=T
V =& LT −1
D =& L
H =& L
ρ =& FL− 4 T 2
µ =& ML − 2 T
80
Example 7.5 Solution2/4
9 Step 3: Determine the required number of pi terms.
k-r=6-3=3
9 Step 4:Select repeating variables D,V, μ.
9 Step 5~6:combining the repeating variables with each of
the other variables in turn, to form dimensionless groups.
a1
b1 c 1
Π 1 = ωD V µ
Π 3 = ρ D a 3 V b3 µ c3
ωD
=
V
ρ VD
=
µ
Π 2 = HD
a2
b2
V µ
c2
H
=
D
81
Example 7.5 Solution3/4
9 The functional relationship is
⎛ H ρ VD
ωD
= φ⎜⎜ ,
V
µ
⎝D
⎞
⎟⎟
⎠
Strouhal number
9 The prototype and the model must have the same
phenomenon.
⎛ H m ρ m Vm D m
ωm D m
= φ⎜⎜
,
µm
Vm
⎝ Dm
⎞
⎟⎟
⎠
82
Example 7.5 Solution4/4
9 The Design conditions.
H Hm
=
D Dm
ρ VD ρ m V m D m
=
µ
µm
Then….
Hm =
H
D m = ... = 60 mm
D
Vm =
µm ρ D
V = ... = 13 .9 m / s
µ ρm Dm
V Dm
ω=
ω m = ... = 29 .0 Hz
Vm D
83
Model Scales
™ The ratio of a model variable to the corresponding
prototype variable is called the scale for that variable.
Length Scale
l1
l
l
l
= 1m ⇒ λ l = 1m = 2 m
l 2 l 2m
l1
l2
Velocity Scale
Vm
λV =
V
Density Scale
ρm
λρ =
ρ
Viscosity Scale
λµ =
µm
µ
84
Validation of Models Design
™ The purpose of model design is to predict the effects of
certain proposed changes in a given prototype, and in this
instance some actual prototype data may be available.
Validation of model design ?
™ The model can be designed, constructed, and tested, and
the model prediction can be compared with these data. If
the agreement is satisfactory, then the model can be
changed in the desired manner, and the corresponding
effect on the prototype can be predicted with increased
confidence.
85
Distorted Models
™ In many model studies, to achieve dynamic similarity
requires duplication of several dimensionless groups.
™ In some cases, complete dynamic similarity between
model and prototype may not be attainable. If one or more
of the similarity requirements are not met, for example,
if Π 2 ≠ Π 2 m , then it follows that the prediction equation
Π 1 = Π 1m is not true; that is, Π 1 ≠ Π 1m
™ MODELS for which one or more of the similarity
requirements are not satisfied are called DISTORTED
MODELS.
86
Distorted Models EXAMPLE-1 1/3
™ Determine the drag force on a surface ship, complete
dynamic similarity requires that both Reynolds and Froude
numbers be duplicated between model and prototype.
Fr m =
Re m
Vm
( gl m )
1/ 2
= Fr p =
Vp
( g l p )1 / 2
Vp l p
Vm l m
=
= Re p =
υm
υp
Froude numbers
Reynolds numbers
™ To match Froude numbers
between model and prototype
Vm ⎛⎜ l m
=
Vp ⎜⎝ l p
⎞
⎟
⎟
⎠
1/ 2
87
Distorted Models EXAMPLE-1 2/3
™ To match Reynolds numbers between model and
prototype
1/ 2
3/ 2
υm
V l
= m m
υp
Vp l p
υ m ⎛⎜ l m
=
υ p ⎜⎝ l p
⎞
⎟
⎟
⎠
l m ⎛⎜ l m
=
l p ⎜⎝ l p
⎞
⎟
⎟
⎠
If lm/ lp equals 1/100(a typical length scale for ship
model tests) , then υm/υp must be 1/1000.
>>> The kinematic viscosity ratio required to
duplicate Reynolds numbers cannot be attained.
88
Distorted Models EXAMPLE-1 3/3
™ It is impossible in practice for this model/prototype scale
of 1/100 to satisfy both the Froude number and Reynolds
number criteria; at best we will be able to satisfy only
one of them.
™ If water is the only practical liquid for most model test of
free-surface flows, a full-scale test is required to obtain
complete dynamic similarity.
89
Distorted Models EXAMPLE-2 1/2
™ In the study of open channel or free-surface flows.
Typically in these problems both the Reynolds number
and Froude number are involved.
Fr m =
Re m
Vm
(g m l m )
1/ 2
= Fr p =
Vp
( g p l p )1 / 2
ρ p Vp l p
ρ m Vm l m
=
= Re p =
µm
µp
™ To match Froude numbers
between model and prototype
Froude numbers
Reynolds numbers
Vm ⎛⎜ l m
=
Vp ⎜⎝ l p
⎞
⎟
⎟
⎠
1/ 2
=
λl
90
Distorted Models EXAMPLE-2 2/2
™ To match Reynolds numbers between model and
prototype
Vm µ m ρ p l p
=
Vp
µp ρm l m
µm /µp lp
lm
=
lp
ρp / ρm l m
Vm
=
Vp
λl
3/2
=
µm /µp
ρp /ρm
=
υm
υp
If lm/ l p equals 1/100(a typical length scale for ship
model tests) , then υm/υp must be 1/1000.
>>>The kinematic viscosity ratio required to
duplicate Reynolds numbers cannot be attained.
91
Typical Model Studies
™ Flow through closed conduits.
™ Flow around immersed bodies.
™ Flow with a free surface.
92
Flow Through Closed Conduits 1/5
™ This type of flow includes pipe flow and flow through
valves, fittings, and metering devices.
™ The conduits are often circular, they could have other
shapes as well and may contain expansions or contractions.
™ Since there are no fluid interfaces or free surface, the
dominant forces are inertial and viscous forces so that the
Reynolds number is an important similarity parameter.
93
Flow Through Closed Conduits 2/5
™ For low Mach numbers (Ma<0.3), compressibility effects
are usually negligible for both the flow of liquids or gases.
™ For flow in closed conduits at low Mach numbers, and
dependent pi term, such as pressure drop, can be expressed
as
⎛ l i ε ρVl ⎞
Dependent pi term= φ⎜⎜ l , l , µ ⎟⎟
⎝
⎠
Where l is some particular length of the system and li
represents a series of length terms, ε/ l is the relative
roughness of the surface, and ρVl/µ is the Reynolds number.
94
Flow Through Closed Conduits 3/5
™ To meet the requirement of geometric similarity
l im l i
=
lm
l
εm ε
l im l m ε m
= ⇒
=
=
= λl
lm l
li
l
ε
™ To meet the requirement of Reynolds number
ρ m Vm l m ρVl
Vm µ m ρ l
=
⇒
=
µm
µ
µ ρm l m
V
If the same fluid is used, then
µm
l
=
=1
µ
lm
Vm
l
=
⇒ Vm = V / λ l
V
lm
95
Flow Through Closed Conduits 4/5
™ The fluid velocity in the model will be larger than that in
the prototype for any length scale less than 1. Since length
scales are typically much less than unity.
™ Reynolds number similarity may be difficult to achieve
because of the large model velocities required.
Vm
l
=
⇒ Vm = V / λ l
V
lm
Vm > V
λl < 1
96
Flow Through Closed Conduits 5/5
™ With these similarity requirements satisfied, it follows that
the dependent pi term will be equal in model and
prototype. For example,
Dependent pi term Π1 =
∆p
ρV 2
™ The prototype pressure drop
ρ
∆p =
ρm
⎛ V
⎜⎜
⎝ Vm
2
⎞
⎟⎟ ∆p m In general ∆p ≠ ∆p m
⎠
97
Example 7.6 Reynolds Number Similarity
™ Model test are to be performed to study the flow through a large
valve having a 2-ft-diameter inlet and carrying water at a flowrate of
30cfs. The working fluid in the model is water at the same
temperature as that in the prototype. Complete geometric similarity
exits between model and prototype, and the model inlet diameter is
30 in. Determine the required flowrate in the model.
98
Example 7.6 Solution
™To ensure dynamic similarity, the model tests should be
run so that
Re m = Re
Vm D m VD
V
D
=
⇒ m =
νm
ν
V Dm
Same fluid used in the model and prototype
Q m Vm A m
Dm
=
= ... =
Q
VA
D
(3/12ft)
3
Qm =
( 30 ft / s ) = 3 .75 cfs
(2ft)
99
Flow Around Immersed Bodies 1/7
™ This type of flow includes flow around aircraft,
automobiles, golf balls, and building.
™ For these problems, geometric and Reynolds number
similarity is required.
™ Since there are no fluid interfaces, surface tension is not
important. Also, gravity will not affect the flow pattern, so
the Froude number need not to be considered.
™ For incompressible flow, the Mach number can be omitted.
100
Flow Around Immersed Bodies 2/7
™ A general formulation for these problems is
⎛ l i ε ρ Vl ⎞
⎟⎟
Dependent pi term= φ⎜⎜ , ,
⎝ l l µ ⎠
Where l is some characteristic length of the system and li
represents other pertinent lengths, ε/ l is the relative
roughness of the surface, and ρVl/µ is the Reynolds number.
Model of the National Bank of Commerce, San
Antonio, Texas, for measurement of peak, rms,
and mean pressure distributions. The model is
located in a long-test-section, meteorological
wind tunnel.
101
Flow Around Immersed Bodies 3/7
™ Frequently, the dependent variable of interest for this type
of problem is the drag, D, developed on the body.
The dependent pi terms would usually be expressed in the
form of a drag coefficient
CD =
D
1
ρV 2 l 2
2
™ To meet the requirement of geometric similarity
l im l i
=
lm
l
εm ε
l im l m ε m
= ⇒
= λl
=
=
lm l
li
l
ε
102
Flow Around Immersed Bodies 4/7
™ To meet the requirement of Reynolds number similarity
Vm µ m ρ l
υm l
ρ m Vm l m ρVl
⇒
=
=
=
µm
µ
V
µ ρm l m
υ lm
D
1
ρV 2 l 2
2
=
Dm
1
ρ m Vm 2 l m 2
2
ρ
⇒D=
ρm
⎛ V
⎜⎜
⎝ Vm
⎞
⎟⎟
⎠
2
⎛ l
⎜⎜
⎝ lm
2
⎞
⎟⎟ D m
⎠
Vm
l
=
⇒ Vm = V / λ l
The same fluid is used, then
lm
V
103
Flow Around Immersed Bodies 5/7
™ The fluid velocity in the model will be larger than that in
the prototype for any length scale less than 1. Since length
scales are typically much less than unity.
™ Reynolds number similarity may be difficult to achieve
because of the large model velocities required.
104
Flow Around Immersed Bodies 6/7
™How to reduce the fluid velocity in the model ?
A different fluid is used in the model such that υ m / υ < 1
For example, the ratio of the kinematic viscosity of water to that
of air is approximately 1/10, so that if the prototype fluid were
air, test might be run on the model using water.
Vm < V This would reduce the required model velocity, but
it still may be difficult to achieve the necessary
velocity in a suitable test facility, such as a water
tunnel.
105
Flow Around Immersed Bodies 7/7
™How to reduce the fluid velocity in the model ?
Same fluid with different density.. ρm>ρ
An alternative way to reduce Vm is to
increase the air pressure in the tunnel so that
ρm>ρ. The pressurized tunnels are obviously
complicated and expensive.
106
Example 7.7 Model Design Conditions and
Predicted Prototype Performance
™ The drag on an airplane cruising at 240 mph in standard air is to be
determined from tests on a 1:10 scale model placed ina pressurized
wind tunnel. To minimize compressibility effects, the airspeed in the
wind tunnel is also to be 240 mph. Determine the required air
pressure in the tunnel (assuming the same air temperature for model
and prototype), and the drag on the prototype corresponding to a
measured force of 1 lb on the model.
107
Example 7.7 Solution1/2
™The Reynolds numbers in model and prototype are the
same. Thus,
ρ m Vm l m ρVl
=
µ
µm
Vm=V, lm/ l=1/10
µm
ρm µ m V l
=
= 10
µ
ρ
µ Vm l m
The same fluid with ρm=ρ and μm=μ cannot be
used if Reynolds number similarity is to be
maintained.
108
Example 7.7 Solution2/2
We assume that an increase in pressure does not
significantly change the viscosity so that the required
increase in density is given by the relationship
ρm
= 10
ρ
For ideal gas
pm ρm
=
= 10
p
ρ
D
Dm
=
2
2
2
2
1
1
2 ρV l
2 ρ m Vm l m
D = ...
109
Flow Around Immersed Bodies at High
Reynolds Number 1/3
™ Unfortunately, in many situations the flow characteristics
are not strong influenced by the Reynolds number over the
operating range of interest.
™ Consider the variation in the drag coefficient with the
Reynolds number for a smooth sphere of diameter d
placed in a uniform stream with approach velocity, V.
™ At high Reynolds numbers the drag is often essentially
independent of the Reynolds number.
110
Flow Around Immersed Bodies at High
Reynolds Number 2/3
The effect of Reynolds number on the drag coefficient, CD for a
smooth sphere with CD = D/ ½ AρV2, where A is the projected area
of sphere, πd2/4.
111
Flow Around Immersed Bodies at High
Reynolds Number 3/3
™ For problems involving high velocities in which the Mach
number is greater than about 0.3, the influence of
compressibility, and therefore the Mach number (or
Cauchy number), becomes significant.
™ In this case complete similarity requires not only
geometric and Reynolds number similarity but also Mach
number similarity so that
Vm V
c υ lm
= ⇒
=
cm
c
cm υ m l
112
Flow with a Free Surface 1/8
™ This type of flow includes flow in canals, rivers, spillways,
and stilling basics, as well as flow around ship.
™ For this class of problems, gravitational, inertial forces,
and surface tension are important and, therefore, the
Froude number and Weber number become important
similarity parameters.
™ Since there is a free surface with a liquid-air interface,
forces due to surface tension may be significant, and the
Weber number becomes another similarity parameter that
needs to be considered along with the Reynolds number.
113
Flow with a Free Surface 2/8
™ A general formulation for these problems is
⎛ l ε ρVl V ρV 2 l ⎞
⎟
Dependent pi term= φ⎜ i , ,
,
,
⎜ l l µ
σ ⎟
gl
⎝
⎠
Where l is some characteristic length of the system and li
represents other pertinent lengths, ε/ l is the relative
roughness of the surface, and ρVl/µ is the Reynolds number.
™ To meet the requirement of Froude number similarity
Vm
gml m
=
V
gl
gm = g
Vm
=
V
lm
= λl
l
114
Flow with a Free Surface 3/8
™ To meet the requirement of Reynolds number and Froude
number similarity
ρ m Vm l m ρVl
Vm µ m ρ l
υm l
=
⇒
=
=
µ
V
µ ρm l m
υ lm
µm
υm
3/ 2
= (λ l )
υ
The working fluid for the prototype is normally either freshwater
or seawater and the length scale is small.
It is virtually impossible to satisfy υ m / υ = (λ l )3 / 2 , so models
involving free-surface flows are usually distorted.
115
Flow with a Free Surface 4/8
™ The problem is further complicated if an attempt is made
to model surface tension effects, since this requires the
equality of Weber numbers, which leads to the condition
σ m / ρ m Vm 2 l m
ρ m Vm 2 l m ρV 2 l
2
(
)
⇒
=
=
λ
=
l
2
σ
σ/ρ
σm
V l
For the kinematic surface tension σ/ρ. It is evident that the same
fluid cannot be used in model and prototype if we are to have
similitude with respect to surface tension effects for λ l ≠ 1.
116
Flow with a Free Surface 5/8
™ Fortunately, in many problems involving free-surface
flows, both surface tension and viscous effect are small
and consequently strict adherence to Weber and Reynolds
number similarity is not required.
™ Certainly, surface tension is not important in large
hydraulic structures and rivers.
™ Our only concern would be if in a model the depths were
reduced. to the point where surface tension becomes an
important factor.
How to overcome this problem ?
117
Flow with a Free Surface 6/8
™ Different horizontal and vertical length scales, which
introduce geometric distortion, are often used to eliminate
surface tension effects in the model.
™ Although this approach eliminates surface tension effects
in the model, it introduces geometric distortion that must
be accounted for empirically, usually by increasing the
model surface roughness.
)Verification
test in the model must be made. Model
.
roughness can be adjusted to give satisfactory
agreement between model and prototype.
118
Flow with a Free Surface 7/8
™ For large hydraulic structures, such as dam spillways, the
Reynolds numbers are large so that viscous forces are
small in comparison to the force due to gravity and inertia.
™ In this case Reynolds number similarity is not maintained
and models are designed on the basis of Froude number
similarity.
119
Flow with a Free Surface 8/8
The model Reynolds
numbers are large so that
they are not required to
equal to those of the
prototype.
A scale hydraulic model (1:197) of the Guri Dam in Venezuela
which is used to simulate the characteristics of the flow over
and below the spillway and the erosion below the spillway.
120
Example 7.8 Froude Number Similarity
™ A certain spillway for a dam is 20 m wide and is designed to carry
125 m3/s at flood stage. A 1:15 model is constructed to study the
flow characteristics through the spillway. Determine the required
model width and flowrate. What operating time for the model
corresponds to 1 24-hr period in the prototype? The effects of
surface tension and viscosity are to be neglected.
121
Example 7.8 Solution1/2
™The width, wm, of the model of spillway is obtained from
the length scale
wm
1
= λl =
w
15
wm
20m
=
= 1.33m
15
™With the neglect of surface tension and viscosity, the
dynamic similarity will be achieved if the Froude numbers
are equal between model and prototype
Vm
gml m
=
V
gl
gm = g
Vm
=
V
lm
= λl
l
122
Example 7.8 Solution2/2
™ The flowrate
Q m Vm A m
=
=
Q
VA
Q m = (λ l )
5/ 2
lm ⎛ lm ⎞
⎜
⎟
l ⎝ l ⎠
2
3
Q = ... = 0.143m / s
V
l tm
=
Vm l m t
tm
V lm
=
= λl
t
Vm l
123
Similitude Based on Governing Differential
Equations 1/5
™ For a steady incompressible two-dimensional flow of a
Newtonian fluid with constant viscosity.
DThe mass conservation equation is
∂u ∂v
+
=0
∂x ∂y
Has dimensions of 1/time.
DThe Navier-Stokes equations are
⎛ ∂2u ∂2u ⎞
⎛ ∂u
∂p
∂u ⎞
⎟
⎟⎟ = −
+ µ⎜ 2 +
ρ ⎜⎜ u
+v
2 ⎟
⎜ ∂x
∂x
∂y ⎠
∂
y
⎝ ∂x
⎝
⎠
⎛ ∂2v ∂2v ⎞
⎛ ∂v
∂v ⎞
∂p
⎟
⎟⎟ = −
ρ ⎜⎜ u
+v
− ρg + µ ⎜ 2 +
⎜ ∂x
∂y ⎠
∂y
∂ y 2 ⎟⎠
⎝ ∂x
⎝
Has dimensions of
force/volume
124
Similitude Based on Governing Differential
Equations 2/5
™ How to non-dimensionalize these equations ?
x* =
x
l
y* =
y
l
∂u V ∂u *
=
∂x
l ∂x *
u* =
u
V
∂2u
∂x 2
v* =
v
V
∂
=
∂x
The mass conservation equation
p* =
p
p0
t* =
t
τ
2 *
∂
u
V
∂
u
⎛
⎞
⎜
⎟= 2
⎝ ∂ x ⎠ l ∂ x *2
∂u ∗
∂x ∗
+
∂v ∗
∂y∗
=0
125
Similitude Based on Governing Differential
Equations 3/5
The Navier-Stokes equations
Reynolds number
∗
∗
∗
⎡ p 0 ⎤ ∂p∗ ⎡ µ ⎤⎛⎜ ∂ 2 u ∗ ∂ 2 u ∗ ⎞⎟
⎡ l ⎤ ∂u
∗ ∂u
∗ ∂u
+v
= −⎢ 2 ⎥ ∗ + ⎢
⎥⎜ ∗2 + ∗2 ⎟
⎢ τV ⎥ ∗ + u
∗
∗
∂x
∂y
⎣ ⎦ ∂t
⎢⎣ ρ V ⎥⎦ ∂x
⎣ ρ Vl ⎦⎝ ∂x
∂y ⎠
∗
∗
∗
⎡ p 0 ⎤ ∂p∗ ⎡ gl ⎤ ⎡ µ ⎤⎛⎜ ∂ 2 v ∗ ∂ 2 v ∗ ⎞⎟
⎡ l ⎤ ∂v
∗ ∂v
∗ ∂v
+v
= −⎢ 2 ⎥ ∗ − ⎢ 2 ⎥ + ⎢
⎥⎜ ∗2 + ∗2 ⎟
⎢ τV ⎥ ∗ + u
∗
∗
∂x
∂y
⎣ ⎦ ∂t
⎣ V ⎦ ⎣ ρ Vl ⎦⎝ ∂x
⎢⎣ ρ V ⎥⎦ ∂y
∂y ⎠
Strouhal number
Euler number
Reciprocal of the square
of the Froude number
126
Similitude Based on Governing Differential
Equations 4/5
™ From these equations it follows that if two systems are
governed by these equations, then the solutions (in terms
of u*,v*,p*,x*,y*, and t*) will be the same if the four
parameters are equal for the two systems.
™ The two systems will be dynamically similar. Of course,
boundary and initial conditions expressed in
dimensionless form must also be equal for the two systems,
and this will require complete geometric similarity.
127
Similitude Based on Governing Differential
Equations 5/5
™ These are the same similarity requirements that would be
determined by a dimensional analysis if the same variables
were considered. These variables appear naturally in the
equations.
™ All the common dimensionless groups that we previously
developed by using dimensional analysis appear in the
governing equations that describe fluid motion when these
equations are expressed in term of dimensionless variables.
™ The use of governing equations to obtain similarity laws
provides an alternative to conventional dimensional
analysis.
128