CHAPTER 5
DIMENSIONAL ANALYSIS AND SIMILITUDE
5.1 DIMENSIONAL ANALYSIS
One approach to solve fluid mechanics problems is by using dimensional
analysis, a mathematical technique that makes use of the study of dimensions.
Dimensional analysis is related to similitude; however, the approach is quite
different. In dimensional analysis, the prediction of physical parameters will
influence the flow, and then group these parameters into dimensionless
combinations for better understanding of flow phenomena. Dimensional
analysis is particularly helpful in experimental work because it provides a guide
to the things that significantly influence the phenomena; thus it indicates the
direction in which experiment work is important.
The significant of dimensional analysis are;
1. Useful for research study especially in design work by reducing the
number of variables.
2. To express in dimensionless equation to find the significant of each
parameters.
3. To simplify the analysis of complex phenomenon in systematic order.
5.2 UNITS AND DIMENSIONS All physical phenomena are expressible in terms of a set of basic or
fundamental dimensions as shown in Table 6.1.
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Table 5.1
Dimension SI US
Mass (M) Kg Lb
Length (L) M ft
Time (T) S S
Temperature (θ) oC oF
All equations related to a physical phenomenon must be dimensionally
homogeneous. This is known as Principle of Dimensional Homogeneity.
5.3 DIMENSIONAL HOMOGENEITY
An equation which expresses the proper relationship between the variables in a
physical phenomenon will be dimensionally homogenous. This means that
each of additive terms in an equation should have the same dimension. For
example; P (kg/ms) = ρgh (kg/ms), where both sides are in same units.
There are 2 types of system which are MLTθ and FLTθ. In a MLTθ system, the
fundamental dimensions are mass M, length L, time T, and temperatureθ .
While for FLTθ system, mass M is replaced by force F. For instance; Area for
rectangular, A = Length, L x width, b = m2 (SI unit). But in dimensional analysis
principle, value is not important. Thus, Area = Length, L x width, L = L2. Table
6.2 shows quantities of fluid mechanics and hydraulic in MLTθ system.
DIMENSIONAL ANALYSIS AND SIMILARITY
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Table 5.2 QUANTITY SYMBOL SI UNIT DIMENSION
FUNDAMENTAL Mass Length Time GEOMETRIC Area Volume Angle Strain Moment of Inertia DYNAMIC Force Weight Specific weight Density Pressure Shear stress Modulus of elasticity Momentum Angular momentum Moment of momentum Torsion Energy Work Power Dynamic viscosity Surface tension KINEMATIC Linear velocity Angular velocity Rotational speed Acceleration Angular acceleration Gravity Discharge Kinematics viscosity Stream function Circulation Vorticity
m L t
A V θ e I
F W γ ρ P ι
E, K M
T E W P μ σ
U,v,u ω N a α g Q υ Ψ Γ Ω
kg m s
m2 m3
M4
N (kgm/s2) N
N/m3 kg/m3
Pa (N/m2) Pa
kgm/s
Nm J
W (J/s) Pa.s
(N.s/m2) (kg/ms)
N/m
m/s rad/s (s-1)
s-1 m/s2 s-2
ms-2 m3/s m2/s m2/s m2/s s-1
M L T
L2 L3
M0L0T0 L0 L4
MLT-2 MLT-2
ML-2T-2 ML-3
ML-1T-2 ML-1T-2 ML-1T-2 MLT-1 ML2T-1 ML2T-1 ML2T-2 ML2T-2 ML2T-2 ML2T-3 ML-1T-1
MT-2
LT-1 T-1 T-1
LT-2 T-2
LT-2 L3T-1 L2T-1 L2T-1 L2T-1 T-1
Source: Rajput (1998)
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5.4 METHODS There are a lot of methods can be used to reduce into a smaller number of
dimensionless parameters such as Bridgman method, Matrix-tenor etc. Two of
the commonly used methods are Raleigh method (basic principle) and
Theorem Pi Buckingham.
5.4.1 RAYLEIGH’s METHOD This method is a basic for a small number of parameters; it becomes rather
cumbersome when a large number of parameters are involved.
ncba AAACAA .....4321 =
Where
A1 = dependent variables
nAAAA .....432 = independent variables
C = a dimensionless constant
The dimensions of each quantities, nAAAA .....432 are written and the sum
exponents of each, which are M, L and T on both sides are equated. Solution of
equations on simplification yields dimensionless groups controlling the
phenomenon.
Example 1: Express dimensionless equation for the speed V with a wave pressure travels
through a fluid. Consider the physical factors probably influence the speed are
compressibility, K density, ρ dan kinematics viscosity, ν.
Answer: 1. Write the fundamental dimension for all dimensions given using MLTθ.
⎥⎦
⎤⎢⎣
⎡=⎥⎦
⎤⎢⎣⎡=⎥⎦
⎤⎢⎣⎡=⎥⎦
⎤⎢⎣⎡=
TL
LM
LTMK
TLV
2
32 ,,, νρ
DIMENSIONAL ANALYSIS AND SIMILARITY
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2. Let write the equation like this:
dbaCKV νρ=
3. Insert fundamental dimension into the equation while C is dimensionless
constant.
dba
TL
LM
LTM
TL
⎟⎟⎠
⎞⎜⎜⎝
⎛⎟⎠⎞
⎜⎝⎛
⎟⎠⎞
⎜⎝⎛=
2
32
4. To satisfy dimensional homogeneity, net power of each dimension must be
identical on both sides of this equation. Thus,
For M : 0 = a + b
For L : 1 = -a + (-3b) + 2d
For T : -1 = -2a + (-d)
5. Solve those three equations, we obtain a = ½, b = -½ dan d = 0,
6. So the equation will be,
02/12/1 νρ −= CKV or ρ
KCV =
∴ Wave speed is not affected by the fluid’s kinematics viscosity, v
Example 2:
The drag force DF on a sphere in laminar flow is known to depend on its
diameter D, velocity of flow V, density of fluid ρ and coefficient of viscosityμ .
Obtain an expression for DF using Rayleigh’s method.
Answer: Using M, L, and T as primary units,
Since
dcbaD VCDF μρ= where, C = dimensionless constant.
Thus, [ ] [ ] [ ] [ ]dcba TMLMLLTLMLT 11312 −−−−− =
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Equating the powers of M, L, and T on both sides,
For M: 1 = c + d (1)
For L: 1 = a + b + (-3c) + (-d) (2)
For T: -2 = -b + (-d) (3)
Since there are three equations and four unknowns, three variables can be
expressed in terms of the fourth using substitution method.
From (1) c = 1 - d
From (3) b = 2 - d = 2 – d
From (2) a = 1 - b + 3c + d = 1 - (2 - d) + 3 (1 - d) + d = 2 – d
∴ ddddD VDF μρ ⋅⋅= −−− 122
d
VDVD ⎟⎟
⎠
⎞⎜⎜⎝
⎛=
ρμρ 22
⎟⎟⎠
⎞⎜⎜⎝
⎛ρ
μρ=
VDfnVDF 22
D
5.4.2 Pi Buckingham Theorem When a large number of variables are involved, Raleigh’s method becomes
lengthy. In such circumstances, the Buckingham's method is useful. This
method expressed the variables related to a dimensional homogenous
equation as:
( )nXXXfX ....., 321 =
where, the dimension at each section is the same.
The Buckingham Phi Theorem can also be expressed in terms of ∏ as shown
in on the right.
)....,,()( 321 knphi −ΠΠΠΠ=φ
where, m = the primary dimensions which are M, L, T and θ
n = dimensional variables such as velocity, discharge and
density.
k = reduction
DIMENSIONAL ANALYSIS AND SIMILARITY
7
Example 3: Drag force FD exerted on a submerged sphere as it moves through a viscous
fluid. Certainly parameters involve are diameter, D, velocity, V, dynamic
viscosity, μ and density of fluid, ρ. Express dimensionless equation by using Pi
Buckingham theorem.
Answer:
1. List the influence factors and count n.
FD = fn (D, V, μ, ρ)
where FD= dependent variable
D,V, μ and ρ = independent variables
n = 5 (FD, D, V, μ, ρ)
2. Choose dimensional system (MLTθ or FLTθ) & list the dimensions of each
variable. Find m.
Choose MLTθ, )LM
LTM,
TL(L,fn
TML
32 =
3. Find k. It is usually equal to m which is cannot exceed but rarely less than m.
Then find n-k or n-m (the number of dimensionless Π groups needed).
n - m = 5 – 3 = 2 so we can write f (Π1, Π2) = 0
4. Choose repeating variables which is 3 (number must be same as m). These
variables must contain the entire fundamental dimensions which are L, T and
M. In this question, we choose
D (L), V (L/T), ρ (M/L3)
Notes:
Number of repeating variables must same with number of fundamental
dimension and choose from independent variables (in this example choose
3)
m = 3 (M, L and T)
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To choose these variables must relate to mass, geometry and kinematics.
Make sure that the entire fundamental dimensions (M, L & T) include in the
chosen variable (at least 1).
5. Form the phi group, which are;
π 1 = ρa1 Db1 Vc1 FD
π 2 = ρa2 Db2 Vc2 μ
The repeating variables need to be inserted in the phi group (π 1 and π 2).
Equate the exponents on both sides, solve and form dimensionless groups.
6. Since the π s are dimensionless, we can replace with M0L0T0 in π 1
π 1 = ρa1 Db1 Vc1 FD
M0 L0 T0 = ⎟⎠⎞
⎜⎝⎛
⎟⎠⎞
⎜⎝⎛
⎟⎠⎞
⎜⎝⎛
2
11
1
3 TML
TLL
LM c
ba
M: 0 = a1 + 1
L : 0 = -3a1 + b1 + c1 + 1
T : 0 = - c1 – 2
Solve a1, b1 & c1 a1 = -1, b1 = -2, c1 = -2
7. So π 1 = ρa1 Db1 Vc1 FD
π 1= ρ-1 D-2 V-2 FD
8. Then solve π 2
π 2 = ρa1 Db1 Vc1 μ
M0 L0 T0 = ⎟⎠⎞
⎜⎝⎛
⎟⎠⎞
⎜⎝⎛
⎟⎠⎞
⎜⎝⎛
LTM
TLL
LM c
ba 1
11
3
M: 0 = a1 + 1
L : 0 = -3a1 + b1 + c1 - 1
T : 0 = - c1 – 1
DIMENSIONAL ANALYSIS AND SIMILARITY
9
Solve a1, b1 & c1 a1 = -1, b1 = -1, c1 = -1
9. So π 2 = ρa1 Db1 Vc1 μ
π 2 = ρ-1 D-1 V-1 μ
10. Rearrange the pi groups as desired and expressed as
( )nf ππππ ......., 321 =
)( 22VDFf
DVD
ρρμ
=
Example 4: Show that
⎟⎟
⎠
⎞
⎜⎜
⎝
⎛=
2/32/3
HgPHgq φ
when flow rate, q (m3/s/m) is over the spillway and assume that would affected
by height of water over weir, H (m), height of weir, P (m) and gravity, g (m/s2)
Answer:
1. List the influence factors and count n.
q = fn (H, P, g) n = 4
2. Choose dimensional system (MLTθ or FLTθ) & list the dimensions of each
variable. Find m.
),,(2
3
TLLLf
TLL
=
3. Find k and then find n-k
So k = 2 same as m
m = 2 (L and T)
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n - m = 4 – 2 = 2 so there are 2 groups of phi; π 1, π 2 = 0
4. Choose repeating variables which is 2 (same as k). These variables must
contain L and T.
Choose: H and g
In this example, the repeating variables can be found easily in the equation
given.
⎟⎟⎠
⎞⎜⎜⎝
⎛=
2/32/3
HgPHgq φ ⎟
⎟⎠
⎞⎜⎜⎝
⎛=
2/32/3 HgH
Hgq φ
5. Form phi group which is 2 groups:
π 1 = Ha1 gb1 q
π 2 = Ha2 gb2 P
Equate the exponents on both sides, solve its and form dimensionless groups.
6. Since the phi is dimensionless, we can replace with M0L0T0 in π1
π1 = Ha1 gb1 q
L0 T0 = ( ) ⎟⎟⎠
⎞⎜⎜⎝
⎛⎟⎠⎞
⎜⎝⎛
LTL
TLL
ba
31
21
L : 0 = a1 + b1 + 2 T : 0 = - 2b1 - 1
Solve a1, b1 & c1 a1 = -3/2, b1 = -1/2
7. So π1 = Ha1 gb1 q π1= H-3/2 g-1/2 q
8. Repeat step 6 – 7 for π2
9. Rearrange the pi groups as desired and expressed as
( )21 ππ fn=
⎟⎟⎠
⎞⎜⎜⎝
⎛=
gHPfn
gHq
2/32/3
⎟⎟⎠
⎞⎜⎜⎝
⎛=
gHPfngHq2/3
2/3
DIMENSIONAL ANALYSIS AND SIMILARITY
11
5.5 SIMILARITY In hydraulic and aeronautical engineering valuable results are obtained at
relatively small cost by performing tests on small scale models of full size
systems (prototypes). Similarity laws help us to interpret the results of model
studies. The relation between model and prototype is classified into three:
1. Geometry Similarity
2. Kinematics Similarity
3. Dynamic Similarity
5.5.1 MODEL (m) Model is similar with object/structure required in certain scale ratio. It is need to
be tested in laboratory with similar condition in real phenomenon. The size of
model is not necessary smaller than prototype.
5.5.2 PROTOTYPE (p) Prototype is an object/actual structure in full size. It is need properly tested in
actual phenomenon, example: spillway structure in open channel, ship etc
5.5.3 GEOMETRY SIMILARITY The prototype and model have identical shapes but differ only in size. Ratio of
corresponding length in prototype and model show as,
Length, rm
p LLL
=
Area, 2
2
2
rm
p
m
p LL
LAA
==
Volume, 3
3
3
rm
p
m
p LLL
VV
==
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5.5.4 KINEMATICS SIMILARITY In addition to geometric similarity, ratio of velocities at all corresponding points
in flow are the same.
Velocity scale ratio, m
pr V
VV =
As time, T is dimensionally L/V. By that
Time scale ratio, rm
p TTT
= and r
rr V
LT =
While, for discharge and acceleration scale ratio
Discharge scale ratio, r
r
m
p
m
p
m
m
p
p
m
pr T
L
TTL
L
TL
TL
Q33
3
3
3
====
Acceleration scale ratio, 2
2
2
2
2
r
r
m
p
m
p
m
m
p
p
m
pr T
L
T
TLL
TLT
L
aa
a ====
5.5.5 DYNAMIC SIMILARITY Two systems have dynamic similarity if, in addition to dynamic similarity,
corresponding forces are in the same ratio in both. The force scale ratio is
rm
p FFF
=
Basically, if the geometric and kinematics similarities exist, it shows two
systems are dynamically similar. The ratios of these systems of all
corresponding forces are the same. The respective forces includes;
a) Gravity = GF b) Viscosity = vF c) Elasticity = EF d) Surface tension = TF e) Inertia = IF
DIMENSIONAL ANALYSIS AND SIMILARITY
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Resultant=++++=∑ TEVPG FFFFFF and FI = - Resultant
Then, strict dynamic similarity means;
=====Ip
Im
Tp
Tm
Ep
Em
vp
vm
Gp
Gm
FF
FF
FF
FF
FF
Constant
5.5.6 ADVANTAGES USING SIMILARITY The advantages are;
1. Performances of object can be predicted.
2. Economy and easy to build, where design of model can be done many
times until reach a certain values.
3. Nonfunctional structure also can be measured such as dam.
5.5.7 NON-DIMENSIONAL PARAMETER
By using Raleigh’s Method or the Buckingham Phi Theorem, the number of
dimensional variables such as mass, length and time used in an analysis of
flow is reduced to a few non-dimensional variables.
There are the five non-dimensional parameters that represent the ratio of
forces per unit volume.
1. Reynolds Number
2. Froude Number
3. Mach Number
4. Euler Number
5. Weber Number
thus 0=+++++ ITEVPG FFFFFF
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Table 5.2
5.5.7.1 Reynolds Number Reynolds number is a non-dimensional parameter that is used when viscous
force is dominant. Reynold’s number represents the ratio between inertia force
FI and viscosity force FV.
The Reynolds number can be expressed as:
Re =V
I
FF
=ForceViscosity
Force Inertia
νμ
ρμρ LVLV
LVVL
===22
DIMENSIONAL ANALYSIS AND SIMILARITY
15
Below is the equation for dynamic similarity where viscous forces are
predominant.
ppm
m
LVLV⎟⎠⎞
⎜⎝⎛===⎟
⎠⎞
⎜⎝⎛
ννReRe
Reynolds number is used for the following types of flow:
Completely submerged flow
Completely enclosed flow through pipes and plates
Viscous flow as in settling of particles in fluids
Flow in flow meter in pipes, venturi meter, or orifice meter
Example 5: An oil (density = 917 kg/m3, dynamic viscosity = 0.29 Pa.s) flows in a 15 cm
diameter pipe at a velocity of 2 m/s. What would be the velocity of water flow in
a 1 cm diameter pipe, to make the two flows dynamically similar? The density
and viscosity of water can be taken as 998 kg/m3 and 1.31 x 10-3 Pa
respectively.
Answer: Reynolds similarity law is applicable,
p
pmm
LVLV⎟⎠⎞
⎜⎝⎛===⎟
⎠⎞
⎜⎝⎛
ννReRe
rr
r
r
r
p
mr LLV
VVρμν
===
0623.0
917998
151
1.29.01031.11 3
=⎟⎠⎞
⎜⎝⎛⎟⎠⎞
⎜⎝⎛
=
⎟⎟⎠
⎞⎜⎜⎝
⎛⎟⎟⎠
⎞⎜⎜⎝
⎛=
−x
LLV
V
p
m
p
mp
m
p
m
ρρμ
μ
smxVm /1246.00623.02 ==
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Example 6: A 1:6 scale model of a passenger car is tested in a wind tunnel. The prototype
velocity is 60 km/h. If the model drag is 250 N, what is the drag force and
power are required for prototype. The air in the model and prototype can be
assumed to have the same properties.
Answer: Reynolds similarity law is applicable
p
pmm
LVRRLV⎟⎠⎞
⎜⎝⎛===⎟
⎠⎞
⎜⎝⎛
νν then
r
rr L
V ν=
If ),..(1 pmpmr ei μμρρν === then r
r LV 1
=
100m/skm/h3606/1
60====
r
pm L
VV
5.5.7.2 Froude Number
The Froude number represents the ratio between inertia force and gravity
force. Froude number is expressed as:
Fr = gLV
The Froude number is applied where gravity forces are predominant. The
number is used in the analysis of:
Wave action such as breakwaters and ships
Free surface flow in open-channels
Hydraulic structures, such as spillways, stilling basins, weirs, and notches
Forces on bridge piers and offshore structures
DIMENSIONAL ANALYSIS AND SIMILARITY
17
The equation for dynamic similarity where gravity forces is predominant is given
below.
ppm
mgLVFrFr
gLV
⎟⎟⎠
⎞⎜⎜⎝
⎛===⎟
⎟⎠
⎞⎜⎜⎝
⎛)()(
Example 7: A spillways model with 1:50 scale can manage discharge of 1.25 m3/s, find the
discharge of prototype? If flood scenario takes only 12 hours in prototype, how
long should it take in the model?
Answer:
( ) ( )m
mm
p
pp gL
VFr
gL
VFr ===
Graviti g is same for model and prototype.
The length scale ratio, Lr =Lp/Lm, when rr LV =
Discharge scale ratio, Qr = VrLr2 = Lr
5/2
As Lr =1/50 , 5.2
150
25.1⎟⎠⎞
⎜⎝⎛== p
m
p QQQ
smQp /22097 3=
Time ratio, rr
rr L
VLT ==
50112
== m
p
m TTT
hours697.150
12==mT
Example 8: A model boat 1/100 size of its prototype has 0.12 N of resistance when
simulating a speed of 5 m/s of the prototype. What is the corresponding
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resistance in the prototype? Water is the fluid in both cases and frictional forces
can be neglected.
Answer:
( ) ( )m
mm
p
pp gL
VFrgL
VFr ===
If Lp/Lm = Lr , rr LV = and ( ) ( )322
)()(
rrrm
pr LVL
DayaDaya
F ρρ ===
The density of fluid is same, by that,
ρp = ρm and ρr = 1,
Maka, 3r
LFF
m
p =
Fp = Prototype force = (100/1)3 x (0.12) = 120 kN 5.5.7.3 Mach Number The Mach number represents the ratio between inertia force and
compressibility or elastic force. The ratio is mathematically presented as
below.
Mach Number = 1/2
1/2
force)bility (Compressi
force) (Inertia
The Mach number is used where compressibility effects are predominant in the
flow and can be expressed as:
M = CV
/EV
=ρ
Where:
C = velocity of sound in medium
DIMENSIONAL ANALYSIS AND SIMILARITY
19
The equation for dynamic similarity where compressibility effects are
predominant is given on the right.
ppm
m CVMM
CV
⎟⎠⎞
⎜⎝⎛===⎟
⎠⎞
⎜⎝⎛
Example 9: An airfoil moves at 650 km/h through still air at 20oC. If the elastic stress and
density of air at this temperature is 21 kg/cm2 and 0.216 kg/m3, find Mach’s
number.
Answer: V = 650 km/h = 180.6 m/s K = 21 kg/cm2 = 21 x 104 kg/m2 ρ = 0.216 kg/m3
14.0
126.010216.180
4===
xKVMρ
5.5.7.4 Euler Number
The Euler number represents the ratio between inertia force and pressure force
as shown on the right. When pressure dominates flow, dynamic similarity is
obtained using the Euler number for both prototype and model.
Example 10: A prototype spillway has a characteristic velocity of 2 m/s, 996 kg/m3 of density
and surface tension is 0.0712. What is corresponding length of prototype if
Weber number of model is 5.03 x 105 ?
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Answer:
mmp
m
LVWWLV⎟⎟⎠
⎞⎜⎜⎝
⎛===⎟
⎟⎠
⎞⎜⎜⎝
⎛
σρ
σρ 22
522
1003.50712.0
)2(996 xLLVWp ===σ
ρ
5.5.7.5 Weber Number
The Weber number represents the ratio between inertia force and surface
tension as displayed on the right. When surface tension dominates flow,
dynamic similarity is obtained using the Weber number for both prototype and
model.
m
mp
m
LVWWLV⎟⎠⎞
⎜⎝⎛===⎟
⎠⎞
⎜⎝⎛
σρ
σρ 22
mLp 9=
DIMENSIONAL ANALYSIS AND SIMILARITY
21
EXERCISE 1. The capillary rise, h of a fluid of density, ρ and surface tension σ in a
tube of diameter D depends upon the contact angle θ and gravity g.
Obtain an expression for h by Rayleigh’s method.
2. The stagnation pressure, ps in an air flow depends upon the static
pressure, po , the velocity, V of the free stream and density, ρ of the air.
Derive a dimensionless expression for ps by Rayleigh’s method.
3. The velocity of flow, u is very near to the rotating disk depends on the
angular velocity, ω of the disk, the radial distance r, vertical distance z
and kinematics viscosity of the fluid, ν. Derive a relationship for u in
dimensionless form by using Pi Theorem.
4. The shear stress, ιo at the bed of a rough channel depends upon the
depth of flow, y, velocity of the fluid, V, roughness of the bed, ε and fluid
density, ρ and viscosity, μ. Derive an expression for ιo in dimensionless
form by using Pi Theorem.
5. Obtain expressions for the velocity and force ratio similitude for a model
which obeys Mach’s law similarity.
6. If a 1.0 m diameter of pipe carrying air at a velocity 3.8 m/s is to be
modeled for dynamic similarity by 10 cm diameter of water pipe, what
would be the velocity of water?
7. The resistance offered to the movement of a 2.0 m long ship model in a
towing tank full of fresh water while moving with a speed of 1.5 m/s was
450 N.
(a) If the prototype is 60 m in length, what will be the corresponding
velocity?
(b) What would be the force required to drive at a corresponding
velocity for a prototype of 80 m length in seawater (relative
density 1.025)?
REFERENCES:
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1. Franzini J.B. and Finnermore E.J. 2006. Fluid Mechanics. Mc Graw Hill,
10th Edition.
2. Subramanya, K (1993). Theory and Application of Fluid Mechanics. Mc
Graw Hill. New Delhi.
3. Zarina Md Ali and Ishak Baba (2003) E-module Hidraulics. UTHM
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