Testing of hydraulic design of structures through physical hydraulic model studies

Testing of Hydraulic Design of Structures
through Physical Hydraulic Model Studies
Oct 22, 2020
Resource Person :Dr.Muhammad Mubashir Qureshi
General Manager (Technical), Associated Consulting Engineers
ACE (RON)
Associated
Consulting
Engineers
(ACE.Ltd) One Day In House Training

Session – 1 (Before Tea)
 Discuss questions like; What are hydraulic structures?, What is hydraulic
design procedure, Why to test?, What are testing options? Why to conduct
physical testing of structures?
 Conduct SWOT Analysis of Physical Model Testing
Session – 2 (Before Lunch)
 Theoretical Background of Physical Modeling
 Practical Exercise to Compute Model Dimensions
Session – 3 (After Lunch)
 Procedure for Designing the Physical Model
 Case Study : Physical Hydraulic Model Testing of Diamer Basha Dam Project
TRAINING OUTLINE

SESSION – 1
 What are hydraulic structures?,
 What is hydraulic design procedure,
 Why to test?,
 What are testing options?
 Why to conduct physical testing of structures?
 Conduct SWOT Analysis of Physical Model Testing
4

What are hydraulic structures?
5

6
What are hydraulic structures?,

TYPICAL SPILLWAY CROSS-SECTION
Reservoir

CHUTE SPILLWAYS
Auxiliary Spillway of Tarbela Dam Service Spillway of Tarbela Dam

SHAFT SPILLWAY
Shaft spillway at Jatiluhur Dam

STEPPED SPILLWAY
Stepped Spillway at Gold Creek Dam

Hydraulic Structures of a Hydropower Scheme

Hydraulic Structures
 Storage and Diversion Structures & Their Components
 Spillways, Tunnels, Diversion during Construction structures, Barrages, Navigational locks, Fish Ladder
 Conveyance Structures
 Canals, Aqueducts, Flumes, Outlets, Head regulators, Cross Regulators
 Hydropower
 Power Intake, headrace tunnel, forebay, pressure tunnel, surge tanks, Turbines, tailrace channels/tunnels
 Energy Dissipation Structures
 Stilling Basin, flip bucket, roller bucket, plunge pool
 Drainage
 Ditches and Drains, Open Drains, Tubewells, Tile Drains
 Flood Control Structures
 Drinking Water Supply
 Water Treatment

What is hydraulic design
procedure?
19

20
Conventional procedure for hydraulic design and analysis (modified) (Mays and Tung, 1992)
Proceed for Physical
Hydraulic Model Testing
Yes
Proceed for Structural
Design
No
Change Design
Revised
Design

What is Physical Model and Why to
Test Structures?
21

What is Physical Model and Why to Test Structures?
 A physical hydraulic model is a small-scale version of the
area where flow and pressure is to be analyzed and includes
both topography and structures.
 The reasons for using these kinds of models are,
 natural flowing water is complicated to analyze by
theory as it includes irregular boundaries such as
variable slope, expansions, constrictions etc.
 the hydraulic structures often need to function for a
number of flow situations and conditions which cannot
be investigated by analyzing a full-scale prototype.
22

SWOT Analysis of Physical Model
Testing
23

24
Strengths
• Physical scale modelling has been used successfully in the
design and investigation of hydraulic structures for over 100
years.
• Provided the right similitude law is used, they give access to
reliable and uncontested results.
• They help in detecting and overcoming the defects in the
hydraulic structures.
Weakness
• The model scale depends on the laboratory
facilities (space, pump capacity, measuring tools)
• High cost for building and measurement
equipment, low geometrical flexibility once the
model has been constructed, etc.
Opportunities
• The hydraulic performance and the working details of
hydraulic structures can be visualized and be predicted.
• Boundary Conditions can be easily applied
• A lot of alternative designs can be simulated and studied.
So, that the most economical, accurate and applicable
design could be selected.
Threats
• If the physical model is not designed or conducted
properly, the results will be vague and inaccurate.
• The modification of the model to trial alternative
arrangements or to optimize features can add
weeks to a testing program.

SESSION – 2
 Theoretical Background of Physical Modeling
 Practical Exercise to Compute Model Dimensions
25

Model Analysis
 Model: is a small scale replica of the actual structure.
 Prototype: the actual structure or machine.
Note: It is not necessary that the models should be
smaller than the prototype, they may be larger than
prototype.
Prototype Model
Lp3
Lp1
Lp2
Fp1
Fp3
Fp2
Lm
3
Lm
1
Lm
2 Fm
1
Fm
3
Fm
2

Similitude-Type of Similarities
 Similitude: is defined as similarity between the model
and prototype in every respect, which mean model and
prototype have similar properties or model and
prototype are completely similar.
 Three types of similarities must exist between model
and prototype.
 Geometric Similarity
 Kinematic Similarity
 Dynamic Similarity

 Geometric Similarity: is the similarity of shape. It is said to exist
between model and prototype if ratio of all the corresponding
linear dimensions in the model and prototype are equal. E.g.
p p p
r
m m m
L B D
L
L B D
  
 Where: Lp, Bp and Dp are Length, Breadth, and diameter of
prototype and Lm, Bm, Dm are Length, Breadth, and diameter of
model.
 Lr= Scale ratio
 Note: Models are generally prepared with same scale ratios in
every direction. Such a model is called true model. However,
sometimes it is not possible to do so and different convenient
scales are used in different directions. Such a models is call
distorted model

 Kinematic Similarity: is the similarity of motion. It is said to exist
between model and prototype if ratio of velocities and
acceleration at the corresponding points in the model and
prototype are equal. E.g.
1 2 1 2
1 2 1 2
;
p p p p
r r
m m m m
V V a a
V a
V V a a
   
 Where: Vp1& Vp2 and ap1 & ap2 are velocity and accelerations at
point 1 & 2 in prototype and Vm1& Vm2 and am1 & am2 are velocity
and accelerations at point 1 & 2 in model.
 Vr and ar are the velocity ratio and acceleration ratio
 Note: Since velocity and acceleration are vector quantities, hence
not only the ratio of magnitude of velocity and acceleration at the
corresponding points in model and prototype should be same; but
the direction of velocity and acceleration at the corresponding
points in model and prototype should also be parallel.

 Dynamic Similarity: is the similarity of forces. It is said to exist
between model and prototype if ratio of forces at the
corresponding points in the model and prototype are equal. E.g.
 
 
 
 
 
 
g
i v
p p p
r
i v g
m m m
F
F F
F
F F F
  
 Where: (Fi)p, (Fv)p and (Fg)p are inertia, viscous and gravitational
forces in prototype and (Fi)m, (Fv)m and (Fg)m are inertia, viscous
and gravitational forces in model.
 Fr is the Force ratio
 Note: The direction of forces at the corresponding points in model
and prototype should also be parallel.

Types of forces encountered in fluid Phenomenon
 Inertia Force, Fi: It is equal to product of mass and acceleration
in the flowing fluid.
 Viscous Force, Fv: It is equal to the product of shear stress due to
viscosity and surface area of flow.
 Gravity Force, Fg: It is equal to product of mass and acceleration
due to gravity.
 Pressure Force, Fp: it is equal to product of pressure intensity
and cross-sectional area of flowing fluid.
 Surface Tension Force, Fs: It is equal to product of surface
tension and length of surface of flowing fluid.
 Elastic Force, Fe: It is equal to product of elastic stress and area
of flowing fluid.

Dimensionless Numbers
 These are numbers which are obtained by dividing the
inertia force by viscous force or gravity force or
pressure force or surface tension force or elastic force.
 As this is ratio of once force to other, it will be a
dimensionless number. These are also called non-
dimensional parameters.
 The following are most important dimensionless
numbers.
 Reynold’s Number
 Froude’s Number
 Euler’s Number
 Weber’s Number

 Reynold’s Number, Re: It is the ratio of inertia force to the viscous force
of flowing fluid.
. .
Re
. .
. . .
. . .
Velocity Volume
Mass Velocity
Fi Time Time
Fv Shear Stress Area Shear Stress Area
QV AV V AV V VL VL
du V
A A A
dy L

   
  
 
  
    
2
. .
. .
. .
. .
Velocity Volume
Mass Velocity
Fi Time Time
Fe
Fg Mass Gavitational Acceleraion Mass Gavitational Acceleraion
QV AV V V V
Volume g AL g gL gL

 
 
  
   
 Froude’s Number, Re: It is the ratio of inertia force to the gravity
force of flowing fluid.

 Eulers’s Number, Re: It is the ratio of inertia force to the pressure force
of flowing fluid.
2
. .
Pr . Pr .
. .
. . / /
u
Velocity Volume
Mass Velocity
Fi Time Time
E
Fp essure Area essure Area
QV AV V V V
P A P A P P

 
 
  
   
2 2
. .
. .
. .
. . .
Velocity Volume
Mass Velocity
Fi Time Time
We
Fg Surface Tensionper Length Surface Tensionper Length
QV AV V L V V
L L L
L

  
   

  
   
 Weber’s Number, Re: It is the ratio of inertia force to the surface
tension force of flowing fluid.

 Mach’s Number, Re: It is the ratio of inertia force to the elastic force of
flowing fluid.
2 2
2
. .
. .
. .
. . /
: /
Velocity Volume
Mass Velocity
Fi Time Time
M
Fe Elastic Stress Area Elastic Stress Area
QV AV V L V V V
K A K A KL C
K
Where C K

  


  
    


Model Laws or similarity Laws
 We have already read that for dynamic similarity ratio of
corresponding forces acting on prototype and model should be equal.
i.e  
 
 
 
 
 
 
 
 
 
 
 
g p
v s e I
p p p p p p
v s e I
g p
m m m m
m m
F F
F F F F
F F F F
F F
    
   
 
 
 
 
Thus dynamic similarity require that
v g p s e I
v g p s e I
p p
I
v g p s e m
m
F F F F F F
F F F F F F
F
F F F F F
    
   

   
 Force of inertial comes in play when sum of all other forces is not
equal to zero which mean
 In case all the forces are equally important, the above two
equations cannot be satisfied for model analysis

Model Laws or similarity Laws
 However, for practical problems it is seen that one
force is most significant compared to other and is
called predominant force or most significant force.
 Thus for practical problem only the most significant
force is considered for dynamic similarity. Hence,
models are designed on the basis of ratio of force,
which is dominating in the phenomenon.
 Finally the laws on which models are designed for
dynamic similarity are called models laws or laws of
similarity. The followings are these laws
 Reynold’s Model Law
 Froude’s Model Law
 Euler’s Model Law
 Weber’s Model Law

Froude’s Model Law
 It is based on Froude’s number and states that Froude’s
number for model must be equal to the Froude’s
number for prototype.
 Froude’s Model Law is used in problems where gravity
forces is only dominant to control flow in addition to
inertia force. These problems include:
 Free surface flows such as flow over spillways, weirs, sluices,
channels etc.
 Flow of jet from orifice or nozzle
 Waves on surface of fluid
 Motion of fluids with different viscosities over one another
   
e e
/ 1; : ,
m m
P P
P m
P P m m P m
P P P
r r r r
m m
P
m
m
V V
V V
F F or or
g L g L L L
V V L
V L where V L
V L
L
V
L
  
   
 
 
 

 The Various Ratios for Froud’s Law are obtained as
r
P P P r
m m m
P P
r
m m
2 2 5/2
r
sin
Velocity Ratio: V
T L /V L
Time Ratio: Tr=
T L /V
V / Vr
Acceleration Ratio: a = 1
V / Tr
Discharge Ratio: Q
Force Ratio: Fr=
m
P
P m
p
P
r
m m
r
r
r
P
m r
P P
r r r r r
m m
r r
V
V
ce
L L
L
V
L
V L
L
L
L
a T
a T L
A V
L V L L L
A V
m a

  
  
   
   

 
2 2 2 2 3
3
2 2 2 3 2 7/2
Power Ratio: Pr=Fr.Vr=
r r r r r r r r r r r r r r r
r r r r r r r r r r r r
Q V L V V L V L L L
L V V L V L L L
    
   
   
  

 Q. In the model test of a spillway the discharge and velocity of flow
over the model were 2 m3/s and 1.5 m/s respectively. Calculate the
velocity and discharge over the prototype which is 36 times the
model size.
   
 
2.5 2.5
p
m
2.5 3
For Discharge
Q
36
Q
36 2 15552 /sec
r
p
L
Q m
 
  
p
m
For Dynamic Similarity,
Froude Model Law is used
V
36 6
V
6 1.5 9 /sec
r
p
L
V m
  
  
 Solution: Given that
 For Model
 Discharge over model, Qm=2 m3/sec
 Velocity over model, Vm = 1.5
m/sec
 Linear Scale ratio, Lr =36
 For Prototype
 Discharge over prototype, Qp =?
 Velocity over prototype Vp=?

Classification of Models
 Undistorted or True Models: are those which are
geometrically similar to prototype or in other words if the scale
ratio for linear dimensions of the model and its prototype is same,
the models is called undistorted model. The behavior of prototype
can be easily predicted from the results of undistorted or true
model.
 Distorted Models: A model is said to be distorted if it is not
geometrically similar to its prototype. For distorted models
different scale ratios for linear dimension are used.
 For example, if for the river, both horizontal and vertical scale
ratio are taken to be same, then depth of water in the model of
river will be very very small which may not be measured
accurately.
 The followings are the advantages of distorted models
 The vertical dimension of the model can be accurately measured
 The cost of the model can be reduced
 Turbulent flow in the model can be maintained
 Though there are some advantage of distorted models, however the results of such models
cannot be directly transferred to prototype.

Classification of Models
 Scale Ratios for Distorted Models
 
 
 
r
r
P
P
Let: L = Scale ratio for horizontal direction
L =Scale ratio for vertical direction
2
Scale Ratio for Velocity: Vr=V /
2
Scale Ratio for area of flow: Ar=A /
P P
H
m m
P
V
m
P
m r V
m
P P
m
m m
L B
L B
h
h
gh
V L
gh
B h
A
B h


 
     
         
3/2
P
Scale Ratio for discharge: Qr=Q /
V
r r
H V
P P
m r r r r r
H V V H
m m
L L
A V
Q L L L L L
A V
  

Distorted model
 Q. The discharge through a weir is 1.5 m3/s. Find the discharge
through the model of weir if the horizontal dimensions of the
model=1/50 the horizontal dimension of prototype and vertical
dimension of model =1/10 the vertical dimension of prototype.
 
 
   
3
p
r
r
3/ 2
P
3/ 2
Solution:
Discharge of River= Q =1.5m /s
Scale ratio for horizontal direction= L =50
Scale ratio for vertical direction= L =10
Since Scale Ratio for discharge: Qr=Q /
/ 50 10
V
P
H
m
P
V
m
m r r
H
p m
L
L
h
h
Q L L
Q Q



  
3
1581.14
1.5/1581.14 0.000948 /
m
Q m s

  

Distorted model
 Q. A river model is to be constructed to a vertical scale of 1:50 and a
horizontal of 1:200. At the design flood discharge of 450m3/sec, the
average width and depth of flow are 60m and 4.2m respectively.
Determine the corresponding discharge in model.
𝐷𝑖𝑠𝑐ℎ arg 𝑒 𝑜𝑓 𝑅𝑖𝑣𝑒𝑟 = 𝑄𝑝 = 450𝑚3
/𝑠
𝑊𝑖𝑑𝑡ℎ = 𝐵𝑝 = 60𝑚 𝑎𝑛𝑑 𝐷𝑒𝑝𝑡ℎ = 𝑦𝑝 = 4.2 𝑚
Horizontal scale ratio= Lr 𝐻 =
𝐵𝑃
𝐵𝑚
= 200
Vertical scale ratio= Lr 𝑉 =
𝑦𝑃
𝑦𝑚
=50
Since Scale Ratio for discharge: Qr=QP/𝑄𝑚
= 𝐿𝑟 𝐻 𝐿𝑟 𝑉
3/2
∴ 𝑄𝑝/𝑄𝑚 = 200 × 503/2
= 70710.7
⇒ 𝑄𝑚 = 450/70710.7 = 6.365 × 10−3
𝑚3
/𝑠

SESSION – 3
 Procedure for Designing the Physical Model
 ToRs of a Typical Physical Hydraulic Model Testing Project
 Case Study 1: Physical Hydraulic Model Testing of Diamer Basha Dam Project
 Case Study 2 : Physical Hydraulic Model Testing of Ubrug Spillway, Jatiluhur Dam
Project, Indonesia
5

Procedure for Designing the
Physical Model
4
6
Selecting the Suitable scale for the physical model
Constructing the model
Testing the model
Re-testing the modified model

ToRs of a Typical Physical Hydraulic
Model Testing Project
4
7

1
KOHALA HYDROPOWER PROJECT
Terms of Reference for Hydraulic Model Study (Modified)
1 INTRODUCTION
The Kohala hydropower project is planned on Jhelum River in Azad Jammu & Kashmir. The purpose of the project is to produce electric power of 1100 MW for connection to the national grid of Pakistan. The major structures of the pro
proposed as under:
 A 58.50m high gravity Dam near Siran village about 27 km u/s of Muzaffarabad city on road to Srinagar
 4 nos. Intakes on left bank of Jhelum river
 4 connection tunnels from Intake to desanders
 4 underground desanders
 4 connection tunnels from desanders to low pressure headrace tunnel(s)
 About 18 km. Low pressure headrace tunnel
 Headrace surge tunnel
 High pressure headrace tunnel
 4 steel penstocks
 Underground Powerhouse near Barsala village about 30 km from Muzaffarabad on road to Kohala
 Tailrace surge tunnel
 Tailrace tunnel
The Jhelum River at the dam site has mean annual flow of 312 m3/s and the Probable Maximum Flood (PMF) at the dam site is calculated as 11800 m3/s. The reservoir of the Siran will be located in a narrow gorge of 6.5 km. length
pond only volume of about 15 Mill. m3.
The gravity dam of the project consists of 11 blocks. The openings for flood discharge include three spillway bays in the river channel, two sluices aside of both ends of spillway. As well as providing flood discharges, the sluices will be
flush reservoir sediment from the reservoir – sufficiently far from the dam to maintain deposition-free conditions at the power intake.
The civil works proposed for the river diversion for construction of dam comprises of one diversion tunnel, an upstream rock-fill cofferdam with clay core, a downstream RCC cofferdam and rock-fill cofferdam. The upstream rock-fill co
and downstream rock-fill cofferdam are only used in first dry season.
The rock formation along the Jhelum River in the dam u/s and d/s areas consists of the inter-bedded sandstone, shale. Consideration will be given to the geology for the purposes of estimating its erosion characteristics. This will b
together with the hydraulics of the jet impacts, to determine the extent of design for energy dissipation downstream of the dam and spillway.
Hydraulic Model studies of the Kohala Hydropower Project are planned to be carried out for its following structures:

Case Study : Physical Hydraulic
Model Testing of Diamer Basha
Dam Project
4
9

DESIGN MEMORANDUM
Subject : River Diversion during Construction and Reservoir Filling
1. General
The present memorandum summaries the criteria applied and the design of the diversion works for the Diamer Basha Dam Project including definition of design floods, freeboard margins and the technical key parameters of the diversion work structures. Aiming on an economic design, the se-
lection of the design flood in conjunction with the consequences of dam failure during construction is taken into account in the design.
The period required for construction of the about 270 m high (225 m above riverbed) Diamer Basha Dam is estimated to be approximately 8 to 9 years.
1.1 Design Criteria
1.1.1 Selection of Design Floods
For the selected project layout with an RCC dam at dam site C d/s and the corresponding construction period of 9 years, the design flood for the diversion works was defined as given in Table 1 below in accordance with the feasibility study design. In case of the Diamer Basha Dam Project
floods are generated by
- Glacier and snowmelt from June to August;
- Dam Break flood waves generated by failure of a natural dam.
With reference to the corresponding hydrological studies, design floods resulting from glacier and snowmelt with reoccurrence periods of 25 and 100 years have been taken into account in the present design of the diversion works. In the August 2004 Feasibility Study Vol. III the dry season de-
sign flood is determined for the period from October to March based on the monthly maximum 1-day flood presented in Table 3.4.13. These estimates were reviewed and finally substituted by the design hydrographs and the corresponding instantaneous peak discharges. The instantaneous
peak discharges of the design floods with reoccurrence probabilities of 25 and 100 years are:
HQ25 = 12,560 m³/s HQ100 = 14,330 m³/s (114.1 % of HQ25)
The incremental cost for maintaining the same level of flood protection during diversion is obviously low for consideration of a design flood with a return period of 100 years compared to that of 25 years in relation to the potential damages resulting from overtopping the cofferdams in the event of
a flood. In view of the difference in the flood peaks of less than 15 % and the rather long construction period of 9 years, in accordance with the recommendations of the panel of experts, the diversion design flood is defined as the peak discharge of a flood with a return period of 100 years.
Detailed information on the variation of maximum observed discharges during low and high flow season is summarized in ANNEX A.
Table 1: Diversion Design Floods for the Diamer Basha Dam Project
Design Period Probability Design Discharge

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