sab2513 hydraulic chapter 6
TRANSCRIPT
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What is Dimensional Analysis???
A powerful & useful tool that can be used to investigate
& obtain solutions to real problems.
To condense the number of separate variables involved
in a particular type of physical system into smaller
number of non-dimensional groups of variables.
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Application of Dimensional Analysis???
Form an equation from dimensional Analysis.
By using any - term/grouping to solve the
problem.
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Units and Dimension
Three (3) fundamental units and dimensions: (m = 3)
All physical parameters can be expressed in terms of
number of fundamental dimension:
Examples; Velocity, v = LT-1
Discharge, Q = L3T-1
Force, F = MLT-2
Quantity Unit Dimension
Length m L
Time s T
Mass kg M
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Fundamental quantities
(1) Geometrical
(2) Kinematic
(3) Dynamic
***For more details please refer to lecture note in TABLE6.1 page 86.
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Methods to determine the non-dimensionalgroup of variables
(1) The Buckingham Theorem
(2) The Repeating Variable Method
(3) The Rayleigh Method
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Buckingham Pi Theorem Known as the - theorem
Number of physical quantities or variables equals three
or more, (n)
Provide an excellent tool by which these quantities can
be organized into smallest number of significant, non-
dimensional groupings from which an equation can be
evaluated.
Number of fundamental variable (m)
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The drag force (N) exerted by a flowing fluid
on a body is a function of the density (kg/m3
),dynamic viscosity (kg/ms), velocity of the fluid
(m/s) and a characteristics length of the body
(m). Develop a general equation?
Example 1
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Solution
(1) List the variables, units and dimensions
Variable Unit Dimension
F N MLT-2
kg/m3 ML-3
kg/ms ML-1T-1
v m/s LT-1
L m L
n = 5 m = 3
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(2) Number of group n m = 5 3= 2
[1 , 2] = 0
(3) Select the repeating variables
(depend on number of fundamental dimension m)
L , v and
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(4) The first - term can be expressed at the product of thechosen repeating variables with unknown exponent and
one other variable/quantity with power of 1.
1 = La vb c F1
(5) Repeat for next - term with the same repeating variables
plus one other variable
2 = La vb c 1
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(6) For each - solve for the unknown exponents bydimensional analysis.
1 = La
vb
c
F1
M0L0T0 = (L)a (LT-1)b (ML-3)c (MLT-2)1
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T0 0 = - b - 2
b = - 2
M0 0 = c + 1
c = - 1
M0
L0
T0
= (L)a
(LT-1
)b
(ML-3
)c
(MLT-2
)1
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L0 0 = a + b 3c + 1
= a 2 3(-1) + 1
a = -2
1 = L-2 v-2 -1 F1
=
22Lv
F
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2 = La vb c 1
M0L0T0 = (L)a (LT-1)b (ML-3)c (ML-1T-1)1
T0 0 = - b - 1
b = - 1
M0 0 = c + 1
c = - 1
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L0 0 = a + b 3c - 1
= a 1 3(-1) - 1
a = -1
2 = L-1 v-1 -1 1
=
vL
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[1 , 2 ] = 0
0,22
vLvL
F
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Repeating Variables Method The simplest method
Same as Buckingham Pi theorem but has a different way
to solve - term.
Number of fundamental variable (m)
Procedure:
(1) ---- (3) Follow Buckingham Pi Theorem
- List of variables- - term
- Repeating variable
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(4) From eqns for M, L and T from selected repeating variables
L = L ------------ (4.1)
v = L/T T = L/v -------- (4.2)
= M/L3
M = L3
--------(4.3)
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(5) Replace dimensions for each other variables with (4)
Example;
1
2
T
MLF
2
3
v
L
LL
2
2
4
vL
L 22vL
1 22
vL
F
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The force F(N) to water flow which acted on
the sluice gate are dependent on flow
velocity, v(m/s), density of water, (kg/m3),dynamic viscosity, (kg/ms), wetted section
area of sluice gate, A(m2) and elasticity
modulus, E(kg/ms2). Develop a general
equation to the other variables as a function offorce, F.
Example 2
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Solution
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(1) List the variables, units and dimensions
Variable Unit Dimension
F N MLT-2
kg/m3 ML-3
kg/ms ML-1T-1
v m/s LT-1
A m2 L2
E kg/ms2 ML-1T2
n = 6 m = 3
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(2) Number of group n m = 6 3= 3
[1
, 2
, 3] = 0
(3) Select the repeating variables
(depend on number of fundamental dimension m)
A , v and
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(4) The first - term can be expressed at the product of thechosen repeating variables with unknown exponent and
one other variable/quantity with power of 1.
1 = Aa1
vb1
c1
F
(5) Repeat for next - term with the same repeating variables
plus one other variable
2 = Aa2 vb2 c2 E
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(7) For each - solve for the unknown exponents by
dimensional analysis.
1
= Aa1 vb1 c1 F
M0L0T0 = (L2)a1 (LT-1)b1 (ML-3)c1 (MLT-2)1
(6) Repeat for next - term with the same repeating variables
plus one other variable
3 = Aa3 vb3 c3
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M0 0 = c1 + 1
c1 = - 1
T0 0 = - b1 - 2
b1 = - 2
(i) M0
L0
T0
= (L2
)a1
(LT-1
)b1
(ML-3
)c1
(MLT-2
)1
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L0 0 = 2a1 + b1 3c1 + 1
= 2a1 2 3(-1) + 1
a = -1
1 = A-1 v-2 -1 F
=
2vA
F
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(ii) 2 = Aa2
vb2
c2
E
M0L0T0 = (L2)a2 (LT-1)b2 (ML-3)c2 (ML-1T2)
M0 0 = c2 + 1
c2 = - 1
T0 0 = - b2 - 2
b2 = - 2
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L0 0 = 2a2 + b2 3c2 - 1
= 2a2 2 3(-1) - 1
a2 = 0
2 = A0 v-2 -1 E
=
2v
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(iii) 3 = Aa3
vb3
c3
M0L0T0 = (L2)a3 (LT-1)b3 (ML-3)c3 (ML-1T-1)
M0 0 = c3 + 1
c3 = - 1
T0 0 = - b3 - 1
b3 = - 1
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L0 0 = 2a3 + b3 3c3 - 1
= 2a3 1 3(-1) - 1
a3 = 1/2
3 = A1/2 v-1 -1
=
vA
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[1 , 2 , 3] = 0
0,,22
vAv
E
vA
F
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Hydraulic Similitude & Model Studies
-full size structure - geometric reduction of prototype in 2
(actual) - the model studies are based comprises
the theory of hydraulic similitude
The relationship between model & prototype is
determined by the law of hydraulic similitude/similarity
Research
Prototype Model
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Hydraulic similitude
3 types
(1) Geometric similarity
- the similarity of shape
- any model length is related to equivalent length in the
prototype(i) Length,
(ii) Area,
(iii) Volume,
LrL
L
m
p
2
2
2
LrL
L
m
p
3
3
3
LrL
L
m
p
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Hydraulic similitude
(2) Kinematic similarity
- the similarity of motion
- at similar points at similar times, the model must
reproduce to scale the velocity and direction of flow
experienced within the prototype. Example;
Velocity,
m
m
p
p
m
p
TL
T
L
v
v
p
m
m
p
T
T
L
L
TrLr 1
Tr
Lr
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Hydraulic similitude
(3) Dynamic similarity
- the similarity of force
- at similar points, the model must reproduce to scale all
of the forces experienced within the prototype.
Example;
Fr
F
F
m
p
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You as an engineer wants to study the ability of a prototypesluice gate using a small scale model. The ratio of geometric
model to prototype scale is 1:150.
Water density for both model and prototype are same, but
the dynamic viscosity of water for prototype is 1.2 timesdynamic viscosity of water in the model because the
temperature difference. The velocity scale ratio vp/vm=0.008.
Determine the force acting on the prototype sluice gate ifforce acting on the model sluice gate is 30 Newton. Use
dimensional analysis from Example 2.
.
Example 3
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Solution
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- Dimensional analysis from Example 2
or
or
21
vA
F
22v
vA
3
22
vL
F
vL
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Solution
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- Given:
- To get the force on the prototype sluice gate, similarity 1 is
used because got the variable of Force.
- So
150 ;1
p
m
LL
30mF N1.2 ;p m
mp 11
mpvL
F
vL
F
2222
0.008p
m
vv
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Solution
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- or
m
p
m
p
m
p
mpvv
LLFF
22
2
215030.0 0.008 1
1
N2.43