SAB2513 Hydraulic Chapter 6

7/31/2019 SAB2513 Hydraulic Chapter 6

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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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8

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



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



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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(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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(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

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SAB2513 Hydraulic Chapter 6