Storage type of Hydropower plant

.
Component parts & Layout of hydropower project:-
 Layout of Hydropower Project:-
• Generally, every hydropower project starts with intake and ends at tailrace. But almost every plant has
some different features with others either in component parts or in constructional features. Mainly it
depends upon the hydrology, topography, location etc.

a. Intake: - The arrangement of structure situated at the entrance of the canal or tunnel or pipes through
which withdraws from the river, pond or reservoir and conveyed to the powerhouse to produce electricity.
b. Weir/Dam: - A weir is a solid obstruction put across the river to raise its water level and divert the water
into the canal. A dam and weir has only one difference in height. Dam stores the water comparatively for
longer duration.
c. Barrage: - The function of barrage is similar to that of weir but the heading up of water is effected by the
gate alone.
d. Reservoir: - The main purpose of reservoir is to store water during raining season and supply the same
water during dry season and thus it helps in supplying water to the turbines accordingly to the load.
e. Spillway: - Whatever may be the type of dam it is absolutely necessary to provide and safe passage for the
flood water downstream so as to avoid danger of the dam being overtopped. The part of the dam which
discharges the flood to the d/s side is called spillway. It must be hydraulically and structurally efficient
design.
f. Headrace canal: - Head race canal conveys the water diverted from intake to fore bay tank with suitable
velocity and required flow of water.
 g. De-silting Basin: -
De-silting basin removes silt and sand particles bigger
than 0.3 mm size which shorten the life of the penstock
pipe and turbine. Normally it is built in pipe
combination with fore bay tank to reduce cost of the
scheme.
h. Fore bay: - A fore bay or Head pond is a regulating
reservoir at the downstream end of the canal just at the
upstream of penstocks to act as a small balancing
reservoir. Water is temporarily stored in the fore bay in
the event of a rejection of load by the turbine and there
is with drawl from it when load is increased. If the
power house is located at the toe of dam a separate fore
bay is not required
i. Surge tank: - when conveyance system consists of pressure tunnel a surge tank is usually required at
the upstream of penstock. The surge tank helps in stabilizing the velocity and pressure in the penstock. It
also used to reduce the water hammer effect.

j. Penstock: - it is the pipe that supply water from head pond reservoir, fore bay or surge tank to the
turbine of power house at high pressure. It is a part of water conveyance system where pressurized
flow occurs.
k. Tail race: - The channel through which water is discharged back to the river after running the
turbines. The outlet level of the tail race channel should be at higher than the flood level of the
stream. There should avoid excessive silting and scouring of the tailrace channel bed. If the
powerhouse is close to the stream, the outflow may be discharged directly into the stream.
l. Power house: - The main function of the power house is to protect the electro-mechanical units
from rain and other weather effects.
m. Draft tube:
It is a pipe, which connects the turbine and the tail race or outlet channel
through which the water exhausted from the runner flows to the outlet
channel. The draft tubes are either conical draft tubes with a circular
section or they are elbow shaped tubes with gradually increasing area. A
draft tube is an essential part of reaction turbine. Draft tube decreases
the pressure at runner exit to a value less than atmospheric pressure and
thereby increasing working head.
 The function of these tubes are:-
• To achieve the recovery of the velocity head at the runner outlet which otherwise
would have gone to waste as exit loss.
• Permits the turbine to be installed at higher level than the tail water level which
facilitate the maintenance of the turbine.
• A conduit which connects the outlet of a reaction turbine runner to the tailrace.
n. Casing: - It is a structural arrangement provided in around the turbine runner to
control and guide the flow with minimum head loss and to control the splashing of
flow after striking the runner. Casing depends on the types of turbines used in power
house.
o. Turbine: - The hydraulic machines which convert hydraulic energy into mechanical
energy are called turbines. This mechanical energy is used in running an electric
generator, which is directly coupled to the shaft of the tube. Thus mechanical energy
is converted into electrical energy.
 Dam
• Dam is an obstruction or a barrier built across a stream or a river to raise
the water level or store water at the upstream side in the form of
reservoir or pond. The water stored behind the dam is used for various
purpose such as navigation, flood control, water supply, irrigation,
power generation. It is most important hydraulic structure built on river.
It’s cost is about 20% of total cost.
 Function of dam
1. Develop a reservoir which has a capacity to store water.
2. Built up head for power generation.
3. Divert flow from the river.
Classification of dam
Dam may be classified by various ways. They are:-
• Classification Based on its function.
1. Storage or conservation dam:-Such dam is use to store surplus water during rainy
season to be used during the period of deficit supply (dry season). Storage dams may be
further classified accordingly to the purpose of the storage. Such as water supply,
irrigation, navigation, hydro electric power generation, fish farming etc. storage dam are
usually high height and they may be constructed by concrete, stone, earth and rock fill.
• Advantage:-
• Dam height is very high; hence it can be stores enough water can be used for multipurpose
projects.
• The dam is usually permanent as they are mostly made of concrete or masonry.
• Provision of overall spillway at crest is possible as it is made of concrete. Cost of separate spillway
construction is saved.
• Disadvantage:-
• Cost of construction is very high.
• Strong foundation for such heavy structure is not always available in dam site.
• Risk of the foundation failure due to earthquake or overturning during reservoir fail
condition is always there.
 2.Diversion dam
• Diversion dam are ordinarily constructed to provide head (Raise water level) for carrying
water into canal, ditches or other conveyance systems. The height of dam is relatively small so
no reservoir is formed to store water. It is mainly used in irrigation and in simple ROR type
hydropower development.
• Advantage
• No possibility of submergence upstream as the dam has no storage.
• Cost of construction is low.
• Risk of flooding is nil as the overflow is regulated.
• Disadvantage
• It is not suitable for Non-perennial River because seasonal storage is required for such river.
• It cannot be used for multi-purpose projects.
3. Detention dam:- Detention dam is constructed to store and detain flood water temporarily
and to release it when flood subsided. i.e. it retard flood runoff and minimize the effect of
sudden floods. Detention dam are also constructed to trap sediment and big boulders or
cobbles carried by flood water known as debris. These are also called debris dams.
4.Coffer dam:- It is temporary nature dam to make the dry space for the construction of project
or main dam.
Classification Based On head:-
I. Low head Dam(H<=15m)
II. Medium head Dam(15m to 70m)
III. High head Dam(71m to 250m)
IV. Very high Dam(more than 250m)
A. Classification Based on Material of Construction
i. Earth Dam
ii. Rock Fill Dam
iii.Concrete Dam
iv. Steel Dam
v. Timber Dam
vi. Masonry Dam etc.
Site selection for Dam:-
• Topographical Factor:- Dam is suitable at neck of river is narrow and valley which has large storage capacity
• Geological factor:- foundation having sound rock in river bed, concrete gravity dam are preferable where as
arc dam is preferable when foundation is not so strong but abutment are good to take up thrust. If rocks are
available but they have faults or fissures, they will be grouted first before founding gravity dam over them.
Availability of construction material and skilled labor:-
• Accessibility:- Dam site should be easily accessible. It economizes the transportation cost of construction
material, equipment and manpower required which highly reduces cost of dam.
• Sociological factor:- Land inundated by the reservoir after the dam construction also affects the economy of the
dam. Resettlement and rehabilitation cost is more for reservoir and dam construction so the dam should be
located in such a place where it minimizes the adverse impact of the society.
i. According to the climatic condition
ii.According to the Diversion condition
iii.According to the environmental and ecological condition.
Mainly following points should be taken while
selecting the dam site:-
• Dam site should be selected in a narrow valley to reduce length of dam.
• For safe and cheap construction, good foundation at moderate depth
should be available at the dam site.
• Bed and sides of the river should be fairly water tight.
• Construction material should be available in local site.
• A good communication to dam site is essential.
• Location of suitable spillway should be available.
• A healthy climate at dam site is necessary.
• The site should have lower percolation loss.
• Dam site should be such that it avoids which carry a high percentage of silt.
• The area will get submerged after the construction should be minimum.
• Catchment should be non-erodible upstream of dam which is essential to
reduce silting of the reservoir.
Force acting on dam
The forces that act on the dam are the following:-
i. Weight of dam (Self weight)(W)
ii. Horizontal hydrostatic pressure due to water(P)
iii.Uplift pressure due to water percolated under the dam(U)
iv.Wind and waves forces(𝑷𝒘 )
v. Earthquake forces(𝑷𝒆 )
vi.Pressure due to ice formation at water surface(𝑷𝒊 )
vii.Silt pressure due to deposited silt in the reservoir(𝑷𝒔 )
• For analyzing the structural stability of dam, individual above
mentioned loads have to be classified into consideration of the relatively
importance of the loads is as follows:-
a. Primary loads
b. Secondary loads
c. Exceptional loads
Primary loads: -
a. Primary loads are identified as universally applicable and of
prime importance to all dams e.g. Water and related seepage
loads, self weight load etc.
i. Horizontal force acting at the dam body of unit length:-
• Water load due to hydrostatic pressure:-
Water load due to hydrostatic pressure

• This is the largest force on the dam. It has the largest capacity for
disturbing the stability of the dam. It is the horizontal force acting at the
C.G.
1 𝛾𝑤 𝐻 2
• Horizontal pressure (𝑃𝐻 ) = ∗ 𝛾𝑤 𝐻 ∗ 𝐻 =
2 2
• If there is tail water of height 𝐻′ on D/S side, horizontal pressure
2
1 𝛾𝑤 𝐻 ′
force(𝑃𝐻 ′ = ∗ 𝛾𝑤 𝐻′ ∗ 𝐻′ =
2 2
• Where, 𝛾𝑤 = 𝑢𝑛𝑖𝑡 𝑤𝑒𝑖𝑔ℎ𝑡 𝑜𝑓 𝑤𝑎𝑡𝑒𝑟 = 𝜌 ∗ 𝑔 = 1000 ∗ 9.81 =
9810 𝑁/𝑚3=9.81 KN/m3
Vertical force acting at the dam body of the unit length

• Vertical component of water force (𝑃𝑉 ) 𝑖𝑛 𝑢/𝑠 = weight of water in

ABCD= 𝛾𝑤 ∗ 𝐴𝑟𝑒𝑎 𝑜𝑓 𝐴𝐵𝐶𝐷 ∗ 1m
• ⇒ 𝑃𝑉 = 𝛾𝑤 ∗ 𝐴𝑟𝑒𝑎 𝑜𝑓 𝐴𝐵𝐶𝐷
• Vertical component of water force per unit length of dam (𝑃𝑉 ′ ) 𝑖𝑛 𝑑/𝑠 =
weight of water in EFG = 𝛾𝑤 ∗ 𝐴𝑟𝑒𝑎 𝑜𝑓 𝐸𝐹𝐺 ∗ 1m
• ⇒𝑃𝑉 ′ = 𝛾𝑤 ∗ 𝐴𝑟𝑒𝑎 𝑜𝑓 𝐸𝐹𝐺
Self Weight of Dam:-

• It is the most important force, particularly for gravity dam, stability of

dam largely depends on this force.
• Considering unit length of dam.
• Therefore, total weight of dam (w)= W1+W2+W3
W1= Weight of a triangular portion ABK act through C.G. = 𝛾𝐶 ∗
𝐴𝑟𝑒𝑎 𝑜𝑓 𝑡𝑟𝑖𝑎𝑛𝑔𝑢𝑙𝑎𝑟 𝑝𝑜𝑟𝑡𝑖𝑜𝑛 𝐴𝐵𝐾 ∗ 1
W2=Weight of a rectangle portion IJKL act through C.G. = 𝛾𝐶 ∗
𝐴𝑟𝑒𝑎 𝑜𝑓 𝑟𝑒𝑐𝑡𝑎𝑛𝑔𝑢𝑙𝑎𝑟 𝑝𝑜𝑟𝑡𝑖𝑜𝑛 𝐼𝐽𝐾𝐿 ∗ 1
W3=Weight of a triangular portion GHL act through C.G. = 𝛾𝐶 ∗
𝐴𝑟𝑒𝑎 𝑜𝑓 𝑡𝑟𝑖𝑎𝑛𝑔𝑢𝑙𝑎𝑟 𝑝𝑜𝑟𝑡𝑖𝑜𝑛 𝐺𝐻𝐿 ∗ 1
Where, 𝛾𝐶 = 𝑢𝑛𝑖𝑡 𝑤𝑒𝑖𝑔ℎ𝑡 𝑜𝑓 𝐶𝑜𝑛𝑐𝑟𝑒𝑡𝑒
Uplift Pressure due to seeping water under the dam:-

• Uplift force (U) is represented by the resultant effective vertical

components of water pressure. Uplift pressure is equal to the product
of unit weight of water and area of uplift diagram. There are three
different possibilities namely,
• Full uplift acting but with tail water, Without Tail water and Uplift
acting with drainage gallery:-
• Ref fig.page no 187
B. Secondary loads:-
• a. Horizontal forces acting at the dam body of unit length.
• After some months or years silt is deposited at the bottom, which exerts
additional horizontal pressure on the dam.
𝟏
• Silt pressure force (𝑷𝑺 )𝑯 = ∗ 𝑲𝒂 ∗ 𝜸𝑺 ′ ∗ 𝑯𝒔 𝟐
𝟐
1−sin ∅𝑆
• Where,𝐾𝑎 = 𝑎𝑐𝑡𝑖𝑣𝑒 𝑙𝑎𝑡𝑒𝑟𝑎𝑙 𝑝𝑟𝑒𝑠𝑠𝑢𝑟𝑒 𝑐𝑜𝑒𝑓𝑓𝑖𝑐𝑖𝑒𝑛𝑡 =
1+sin ∅𝑠
• ∅𝑠 = 𝑎𝑛𝑔𝑙𝑒 𝑜𝑓 𝑠ℎ𝑒𝑎𝑟𝑖𝑛𝑔 𝑟𝑒𝑠𝑖𝑠𝑡𝑎𝑛𝑐𝑒 𝑠𝑒𝑑𝑖𝑚𝑒𝑛𝑡 = 300
• 𝛾𝑆 ′ = 𝛾𝑠 − 𝛾𝑤 and 𝛾𝑠 = 𝑠𝑒𝑑𝑖𝑚𝑒𝑛𝑡 𝑠𝑎𝑡𝑢𝑟𝑎𝑡𝑒𝑑 𝑢𝑛𝑖𝑡 𝑤𝑒𝑖𝑔ℎ𝑡(18 𝑡𝑜 20 𝐾𝑁/𝑚3
• After some months or years silt is deposited at the bottom, which exerts
additional horizontal pressure on the dam.
• Silt pressure is sometime not considered:-
• Initially silt is not deposited.
• It had already been consider under the weight and it behaves more like solid less
like liquid.
• It is somewhat impervious and helps in reducing uplift pressure in dam.
Hydrodynamic water wave load:-
• Waves develop on the surface of the reservoir due to wind. The waves exert pressure on the
dam. The magnitude of wave pressure depends upon the height of the waves, developed in the
reservoirs.
• An empirical formula, 𝑃𝑤𝑎𝑣𝑒 = 2 ∗ 𝛾𝑤 ∗ 𝐻𝑤 2
• 𝑤ℎ𝑒𝑟𝑒, 𝐻𝑤 =
𝑤𝑎𝑣𝑒 ℎ𝑒𝑖𝑔ℎ𝑡 𝑐𝑎𝑛 𝑏𝑒 𝑓𝑜𝑢𝑛𝑑 𝑖𝑛 𝑡ℎ𝑒 𝑓𝑢𝑛𝑐𝑡𝑖𝑜𝑛 𝑜𝑓 𝒇𝒆𝒕𝒄𝒉 𝑭 𝑑𝑖𝑠𝑡𝑎𝑛𝑐𝑒 𝑎𝑛𝑑 𝑤𝑖𝑛𝑑 𝑣𝑒𝑙𝑜𝑐𝑖𝑡𝑦.
• 𝑪𝒂𝒔𝒆 − 𝑰 ∶ − 𝑰𝒇 𝑭 ≤ 𝟑𝟐 𝑲𝒎
• 𝐻𝑤 = 0.032 𝐹𝑉 + 0.763 − 0.271 ∗ 𝐹 0.25
• 𝑪𝒂𝒔𝒆 − 𝑰𝑰: −𝑰𝒇 𝑭 > 32 𝑲𝒎
• 𝐻𝑤 = 0.032 𝐹𝑉 , where, V= velocity of wind in Km/hr
Vertical forces acting at the dam body of unit
length :-
• Silt vertical pressure ((𝑷𝑺 )𝑽 = Weight of silt *𝑲𝒂
• But in nature, silt deposition is not of the isotropic in nature, so the
analysis is done with the assumption of the equivalent liquid of the
sediment and water mixture.
C.Exceptional Load:-
• Earthquake pressure
Design criteria for concrete gravity dam and principle
(Modes of failure of gravity dam)
σ𝑀 σ
• Overturning of the dam: FOS= σ +𝑣𝑒 = σ 𝑀𝑀𝐴𝑛𝑡𝑖𝑐𝑙𝑜𝑐𝑘𝑤𝑖𝑠𝑒 (𝑅𝑒𝑠𝑖𝑠𝑡𝑖𝑛𝑔 𝑚𝑜𝑚𝑜𝑚𝑡)
, the value of (𝐹𝑂 ) in excess
𝑀−𝑣𝑒 𝐶𝑙𝑜𝑐𝑘𝑤𝑖𝑠𝑒 (𝑜𝑣𝑒𝑟 𝑡𝑢𝑟𝑛𝑖𝑛𝑔 𝑚𝑜𝑚𝑒𝑛𝑡)
of 1.25 Max generally acceptable. But (𝐹𝑂 ) ≥1.5 is desirable.
• Sliding of the dam (low masonary dam)
𝜇σ𝑉
• Factor of safety (FS) = σ𝐻
> 1 Where, 𝜇 = 𝑐𝑜𝑒𝑓𝑓𝑖𝑐𝑖𝑒𝑛𝑡 𝑜𝑓 𝑓𝑟𝑖𝑐𝑡𝑖𝑜𝑛 it varies from 0.65 to 0.75
• σ 𝑉 = 𝑁𝑒𝑡 𝑣𝑒𝑟𝑡𝑖𝑐𝑎𝑙 𝑓𝑜𝑟𝑐𝑒𝑠
• σ 𝐻 = 𝑠𝑢𝑚 𝑜𝑓 𝑡ℎ𝑒 ℎ𝑜𝑟𝑖𝑧𝑜𝑛𝑡𝑎𝑙 𝑓𝑜𝑟𝑐𝑒𝑠 𝑐𝑎𝑢𝑠𝑖𝑛𝑔 𝑠𝑙𝑖𝑑𝑖𝑛𝑔
• Sliding factor of safety can be expressed in the following definition:-
• Sliding Factor (S.F)
σ𝑯
• , Sliding Factor (S.F.)= σ𝑽
= 𝒕𝒂𝒏𝜽
• Where, σ 𝑉 = 𝑁𝑒𝑡 𝑣𝑒𝑟𝑡𝑖𝑐𝑎𝑙 𝑓𝑜𝑟𝑐𝑒𝑠, σ 𝐻 = 𝑠𝑢𝑚 𝑜𝑓 𝑡ℎ𝑒 ℎ𝑜𝑟𝑖𝑧𝑜𝑛𝑡𝑎𝑙 𝑓𝑜𝑟𝑐𝑒𝑠 𝑐𝑎𝑢𝑠𝑖𝑛𝑔 𝑠𝑙𝑖𝑑𝑖𝑛𝑔
• F.S >1 is desirable. For safety against sliding the sliding factor (S.F) should be less than coefficient
of friction (𝜇). 𝑆. 𝐹 < 𝜇.
𝝁
• FS= , 𝑭𝑶𝑹 𝑭𝑺 < 1, 𝑆. 𝐹 < 𝜇
𝑺.𝑭
Shear Friction Factor (S.F.F)

In case of high dams,

• The FOS against sliding when the shear strength of the joint is also
considered is known as shear friction factor (S.F.F).
𝝁 σ 𝑽+𝑩𝒒
• Therefore, Safety Factor(F.S)= σ𝑯
≥𝟒
• Where,
𝜇 = 𝑐𝑜𝑒𝑓𝑓𝑖𝑐𝑖𝑒𝑛𝑡 𝑜𝑓 𝑓𝑟𝑖𝑐𝑡𝑖𝑜𝑛
B= width of the joint or section
• q= Average Shear strength of the joint which is usually taken as 14 kg/ m2
• σ 𝑉 = 𝑁𝑒𝑡 𝑣𝑒𝑟𝑡𝑖𝑐𝑎𝑙 𝑓𝑜𝑟𝑐𝑒𝑠
• σ 𝐻 = 𝑠𝑢𝑚 𝑜𝑓 𝑡ℎ𝑒 ℎ𝑜𝑟𝑖𝑧𝑜𝑛𝑡𝑎𝑙 𝑓𝑜𝑟𝑐𝑒𝑠 𝑐𝑎𝑢𝑠𝑖𝑛𝑔 𝑠𝑙𝑖𝑑𝑖𝑛𝑔
The value of Shear friction factor (S.F.F) should lie between 4 to 5.
Crushing of the dam or foundation

• Crushing failure occurs when the compressive failure occurs when the
compressive stress in the dam or foundation exceeds the safe limit.
Maximum compressible stress in dam or foundation is determined by
applying the theory of bending combined with thrust or eccentrically
loaded short column.
𝑅𝑣 6∗𝑅𝑣 ∗𝑒 𝑅𝑣 6𝑒
• 𝜎 = 𝜎𝑛 + 𝜎𝑏 = ± = 1± … … . (1)
𝐵 𝐵2 𝐵 𝐵
𝐵
𝑒 =< e should be less than or equal to B/6 for safe dam.
6
Tension Failure of the dam:-
• Until resultant of the forces lies with in the middle third with of the base tension can not developed
any where in the dam.
𝑅𝑣 6𝑒 6𝑒 𝐵
• 0= 𝐵
1± 𝐵
⇒1= 𝐵
⇒𝑒= 6
, This rule is called middle third rule.
• Hence, analysis of middle third rule following key point can be drawn:-
i. In case of direct compression, compressive stress is identical in all section.
ii. In case of eccentric loading, compressive stress varies in section.
iii. In case of eccentric loading, one of the part of the considered section, there will be compressive
stress more than direct. Compressive stress and in another section less than direct compressive
section.
iv. Addition of bending stress will cause more compressive stress at one part and decrease in direct
compressive stress in another part.
v. Variation of stress depends upon the position of eccentricity of the resultant force.
vi. If compressive stress exceeds allowable compressive strength of concrete than, there will be
crushing failure of the dam.
vii.Negative compressive stress i.e. tensile stress, as the concrete weak in tension so its very
sensitive in the case of concrete dam.
 Foundation Treatment:-
• Foundation of the dam must be strong enough to withstand the foundation
pressure exerted on it as the entire load acting on the dam are ultimately
transmitted to the foundation. So it should consists of solid and sound rock
but in actual practice such as ideal foundation is not available so it is
necessary to carry out geological investigation to determine the suitability of
the dam foundation . If defects are observed then suitable measures can be
adopted. Foundation treatment commonly adopted for all foundation is
carried out in three steps:-
a. Preparing the surface.
removing the entire loose rock till a sound bed rock is exposed. The excavation of loose rock laying over
the bed rock should be carried out carefully so as not to damage bed rock and final surface obtain
above is stepped so as to increase the frictional resistance of dam against the sliding.
b. Grouting the foundation.
C. Foundation Drainage.
 Numerical
• A concrete dam as shown in figure of given profile is proposed by a designer
for implementation. The unit shear resistance and angle of shearing
resistance is 500 KN/m2 and 350 respectively. Unit weight of concrete is 24
KN/ m3. You are the person to implement it so you want to check the stability
of dam against flotation overturning and sliding before implementation.
Based upon your findings do you implement it? What measures do you
recommend to implement it ?
6m

H=50m
40m

B=26m
Solution:-
From question, unit shear resistance (q) = 500 KN/m2
• Angle of shear resistance (∅) = 350
• Unit weight of concrete (𝛾𝑐 ) = 24 𝐾𝑁/𝑚3
• Considering 1 m length of dam for analysis. The pressure diagram and
their location are shown in figure below and calculation is done in
below.
Name of forces Designation Magnitude of forces Lever Arm (m) Moment about toe(KN-m)
KN (- ) (+) Clockwise(-ive) Anticlockwise (+ive)
a.
1.
.
Vertical force
Weight of dam W1 24*55*6=7920 23 (+) 182160
W2 24*1/2*20*40=9600 13.33 (+)127698
Weight of water above W3 9.81*1/2*20*40=3924 6.67 (+) 26173.08

σ 𝑉1 (+) 21444
1. Uplift pressure force U1 -26*392.4= -10202.4 13 (-) 132631.2

U2 1 2*26/3=17.33 (-) 22100.95

− ∗ 98.1 ∗ 26 = −1275.3
2
(-) 11477.7
෍𝑈

Net vertical forces

෍𝑉 ෍ 𝑉1 − ෍ 𝑈 = + 9966.3

a. Horizontal forces
1. U/S water pressure force 𝑃𝐻1 ½*490.5*50=12262.5 50/3=16.67 (-) 204415.88

1. D/S water pressure force 𝑃𝐻2 -1/2*392.4*40= -7848 40/3= 13.33 (+)104613.84

= (+) 4414.5
෍𝐻

440914.92
෍ 𝑀𝑅 +

359148.03
෍ 𝑀𝑂 −

81766.85 (MR-MO)
෍𝑀 =
 W1= 𝛾𝐶 ∗ 𝐴𝑟𝑒𝑎 𝑜𝑓 𝑟𝑒𝑐𝑡𝑎𝑛𝑔𝑙𝑒 = 24 ∗ 55 ∗ 6 = 7920𝐾𝑁 , Lever arm from toe = 20+6/2 =23 m
1
W2= 𝛾𝐶 ∗ 𝐴𝑟𝑒𝑎 𝑜𝑓 𝑇𝑟𝑖𝑎𝑛𝑔𝑙𝑒 = 24 ∗ ∗ 20 ∗ 40 = 9600𝐾𝑁 , Lever arm from toe =2*20/3 =13.33 m
2
1
W3= 𝛾𝑤 ∗ 𝐴𝑟𝑒𝑎 𝑜𝑓 𝑇𝑟𝑖𝑎𝑛𝑔𝑙𝑒 = 9.81 ∗ ∗ 20 ∗ 40 = 3924𝐾𝑁 , Lever arm from toe = 20/3 =6.67 m
2

• Now,
σ𝑀 81766.89
• 𝑋ത = σ = = 8.20𝑚
𝑉 9966.3
𝐵 26
• Eccentricity (e) = − 𝑋ത = − 8.2 = 4.80 𝑚
2 2

𝐵 26
• = = 4.33
6 6
𝐵
• Here, e > , hence the dam is unstable.
6
σ 𝑉1 21444
1. Flotation Factor (𝐹𝐹 ) =σ𝑈
= = 1.87 > 1.0 𝐻𝑒𝑛𝑐𝑒 𝑠𝑎𝑓𝑒.
11477.7
σ𝑀 440914.92
2. Overturning Factor (𝐹𝑂 ) = σ 𝑅 = = 1.23 < 1.25 𝐻𝑒𝑛𝑐𝑒 𝑢𝑛𝑠𝑡𝑎𝑏𝑙𝑒
𝑀𝑂 359148.03
𝜇σ𝑉 𝑡𝑎𝑛35∗9966.3
3. Sliding factor = σ = = 1.58 > 1.0 𝐻𝑒𝑛𝑐𝑒 𝑠𝑎𝑓𝑒
𝐻 4414.5
𝜇 σ 𝑉+𝐵𝑞 (𝑡𝑎𝑛35∗9966.3)+(26∗500)
• Shear friction factor = σ𝐻
= = 4.53 > 4 𝐻𝑒𝑛𝑐𝑒 𝑠𝑡𝑎𝑏𝑙𝑒.
4414.5
• Conclusion:- This dam have safe against sliding, flotation but unsafe against tension and overturning.
• To safe over turning, the positive moment (anticlockwise) moment should be increased. i.e. increased by self weight of dam
and downstream water depth. Rod anchoring for overturning to make safe. Use high grade of concrete for tension to make safe.
 Design of earthen dam (Embankment Dam):-
• Earth dam are constructed by locally available earthwork in trapezoidal section.
These dams have been constructed since early days of civilization. The following are
the main reasons due to which these dams have been and still continue:-
1. The basic material used in the construction is the local soil or rock filled or both.
This aspect makes these dams economical.
2. No specific type of foundation is required.
3. No skill of very high standard is required.
4. No materials have to be imported from outside as locally available soil is the main
material of their construction.
5. Earth dam are more suitable than gravity dam if strong foundation at reasonable
depth exists.
6. Earthen dam are more economical than gravity dam if construction material are
locally available.
 Design consideration:-
• If an earth has to safe and stable during construction and in its operation, it must be
designed considering the following part:-
1. The earth dam should never be allowed to be over topped during heavy flood. This
object can be achieved by providing spillways of adequate capacity.
2. Sufficient free board should be provided.
3. The dam portion laying downstream of the impervious core should be properly
drained.
4. The U/S and D/S slopes should be such that they are not damaged by pore pressure
during and after construction.
5. The upstream face of the dam should remain stable during rapid drawdown.
6. In order to prevent internal erosion of the soil the seepage flow through the dam and
foundation should be suitably controlled.
7. Upstream and downstream slopes of the dam should be flat. This makes safety against
the shear failure of the foundation.
8. Free passage of water, through holes made by piping.
9. The dam should be defined by considering earthquake resistant measure.
 Design of earth dam:-
• Design of the section of an earth dam evolves fixing the sizes of the following
element:-
1. Height of dam:- The height of the dam is governed by the high flood level (HFL) and free
board.
• i.e. Height of dam (H)= High flood level (HFL) + Free Board(FB)
2. Top width of the dam:- The height of the dam is the most important factor to fix the
width of the dam. top width of the dam can be found by the following empirical formula.
𝐻
• Top width (B)=( + 3) 𝑓𝑜𝑟 𝑣𝑒𝑟𝑦 𝑙𝑜𝑤 𝑑𝑎𝑚(𝐻 ≪ 30 𝑚)
3
𝐻
• B= 0.55*𝐻 + for H < 30 m
0.5
5
• B= 1.65*(𝐻 + 1.5)0.33 m for H > 30 m
• Also, B= 1.67 𝐻 is commonly used formula to calculate the width of dam. Width of dam
should not be less than 3 m.
Free Board:-
1. The vertical distance between the top of the dam and maximum reservoir level is known as free
board. Free board may further be stated as normal free board and minimum free board.
• Difference between top of the dam and normal reservoir level is known as normal free board.
Difference between top of the dam and HFL when all the spillways and other outlets are
working as planned, is known as minimum free board.
• The free board is at least 1.5 times the wave height. The wave height can be found by :-

• 𝑪𝒂𝒔𝒆 − 𝑰 ∶ − 𝑰𝒇 𝑭 ≤ 𝟑𝟐 𝑲𝒎
• 𝑯𝒘 = 𝟎. 𝟎𝟑𝟐 𝑭𝑽 + 𝟎. 𝟕𝟔𝟑 − 𝟎. 𝟐𝟕𝟏 ∗ 𝑭 𝟎.𝟐𝟓
• 𝑪𝒂𝒔𝒆 − 𝑰𝑰: −𝑰𝒇 𝑭 > 32 𝐾𝑚
• 𝑯𝒘 = 𝟎. 𝟎𝟑𝟐 𝑭𝑽 , where, V= velocity of wind in Km/hr, 𝑯𝒘 = 𝑾𝒂𝒗𝒆 𝒉𝒆𝒊𝒈𝒉𝒕, 𝑭 =
𝑭𝒆𝒕𝒄𝒉 𝒊𝒏 𝑲𝒎
• USBR suggested For free board: - Nature of spillway Height of dam Free Board
Free Any 2 to 3 m
Controlled < 60 m 2.5 m
Controlled  60 m 3m
Central impervious core of the dam:- The thickness of the impervious core
depends upon the following factors:-
 Rate of seepage through core.
 Type of material available for core.
 Design of proposed filter that will permit proper construction.
• Usually Top width of core = 0.80 to 1.0m
𝐻𝑀𝑎𝑥
• Width of core at bottom should be equal to 1.25 * where, 𝐼𝑐𝑟 =
𝑖𝑐𝑟
𝑐𝑟𝑖𝑡𝑖𝑐𝑎𝑙 𝑔𝑟𝑎𝑑𝑖𝑒𝑛𝑡 𝑣𝑎𝑙𝑢𝑒.
1. U/S and D/S slopes:-
• The U/S and D/S slopes of earth dam depends upon following factors:-
 Type of soil used in the construction of the dam.
 Foundation conditions. Type of Material U/S Slope D/S Slope
 Height of the dam. 1. Homogeneous well graded 2:1 2:1
 U/s slope, 2:1 to 4:1 for stability 1. Homogeneous silty clay 3:1 2.5:1
 D/S slope , 2:1 to 2.5 :1 1. Height < 15 m 2.5:1 2:1
1. Height > 15 m 3:1 2.5:1
1. Sand or sand and gravel with 3:1 2.5:1
clay core
 Seepage analysis:-
• Seepage takes place through and under all dam both earth and concrete. The
problem is to minimize and control seepage so that it will have no harmful effect.
While analyzing the seepage flow, the following assumption must be kept in
mind:-
1. Water is incompressible ( density remains constant).
2. Pore spaces do not change with time regardless of water pressure. No
consolidation or expansion.
3. The hydraulic boundary condition at entry and exit are known.
4. The quantity of water entering and leaving the soil element is same.
5. The flow of water through porous medium follows Darcy’s law (Q= KiA). In other
words the seeping water flows under the hydraulic gradient due to gravity head
loss only.
6. No change in the degree of saturation.
 Computation of seepage rate from flow net :-
• A network of flow line and equipotential lines is known as a flow net. A portion between any two
successive flow line is known as flow channel. The area enclosed between two successive flow lines
and successive equipotential lines is known as field.
• Let B and L be the width and length of the field.
• ∆ℎ = ℎ𝑒𝑎𝑑 𝑑𝑟𝑜𝑝 𝑡ℎ𝑟𝑜𝑢𝑔ℎ 𝑡ℎ𝑒 𝑓𝑖𝑒𝑙𝑑.
• ∆𝑞 = 𝐷𝑖𝑠𝑐ℎ𝑎𝑟𝑔𝑒 𝑝𝑎𝑠𝑠𝑖𝑛𝑔 𝑡ℎ𝑟𝑜𝑢𝑔ℎ 𝑓𝑙𝑜𝑤 𝑐ℎ𝑎𝑛𝑛𝑒𝑙.
• H = total head causing the flow.
• According to the Darcy’s law, considering unit thickness
∆ℎ 𝐻 1 𝐻
• ∆𝑞 = 𝐾𝑖𝐴 = 𝑘 ∗ ∗ 𝐵∗1 =𝐾∗ ∗ ∗ 𝐵 ∗ 1 , 𝑏𝑒𝑐𝑎𝑢𝑠𝑒: − ∆ℎ = , 𝑤ℎ𝑒𝑟𝑒 𝑁𝑑 = 𝑛𝑜 𝑜𝑓 𝑑𝑟𝑜𝑝
𝐿 𝑁𝑑 𝐿 𝑁𝑑
𝐻 𝐵 𝑁𝑓 𝐵
• 𝐻𝑒𝑛𝑐𝑒, 𝑡𝑜𝑡𝑎𝑙 𝑑𝑖𝑠𝑐ℎ𝑎𝑟𝑔𝑒 𝑡ℎ𝑟𝑜𝑢𝑔ℎ 𝑓𝑙𝑜𝑤𝑛𝑒𝑡 𝑄 = σ ∆𝑞 = ∆𝑞 ∗ 𝑁𝑓 = 𝐾 ∗ ∗ 𝐿 ∗ 𝑁𝑓 = 𝐾 ∗ 𝐻 ∗𝐿
𝑁𝑑 𝑁𝑑
• Where, 𝑁𝑓 = 𝑁𝑜. 𝑜𝑓 𝑓𝑙𝑜𝑤 𝑐ℎ𝑎𝑛𝑛𝑒𝑙 𝑖𝑛 𝑡ℎ𝑒 𝑓𝑙𝑜𝑤𝑛𝑒𝑡.
• In flow net usually taken as square hence B=L.
𝑁𝑓
• Therefore, 𝑄 = 𝐾 ∗ 𝐻 .
𝑁𝑑
Phreatic Line:-
 Phreatic line or seepage line is a line in the body of the dam below which
there are positive hydrostatic pressures. On the line itself the hydrostatic
pressure is zero. Above the line, there is a cone of capillary saturation. The
effect of capillary is however neglected in dam.
 Characteristic of Phreatic Line:-
1. The phreatic line must be normal to the slope at the point of entry. This is
essentially because u/s face represents 100% equipotential line.
2. Pressure along phreatic line is atmospheric.
3. The presence of pervious foundation below the dam does not affect the
location of phreatic line.
4. Focal point of base parabola is taken slightly with the length of horizontal
filter.
5. In the case of zoned dams having impervious core at the centre the outer
pervious shell may be neglected.
.
Seepage discharge through a dam with impervious foundation:-
• Case –I:- Phreatic Line emerging with the down stream water.
𝐾
• ⇒q= ∗ (𝐻1 2 − 𝐻2 2 )
2∗𝐿0
• This is a required equation for calculation of specific seepage
discharge from saturated soils.
Case-II :- Seepage discharge for the dam with phreatic line emerging at the
downstream face of the dam above the downstream water :-

𝐾∗(𝐻1 2 −(𝐻2 +∆)2 ) 𝐾∗∆ 𝐻0

• = {1+ln( )}
2∗(𝑙−∆𝑚𝑑 ) 𝑚𝑑 ∆
The value of ‘∆′ 𝑎𝑟𝑒 𝑑𝑒𝑡𝑒𝑟𝑚𝑖𝑛𝑒𝑑 𝑏𝑦 𝑡𝑟𝑖𝑎𝑙 𝑎𝑛𝑑 𝑒𝑟𝑟𝑜𝑟.
Case –III:- Seepage discharge for the dam with phreatic line
emerging at the downstream face of the dam if there is no
downstream water.
2 𝐿 𝐿2
, ∆= ±∗ ( 2 − 𝐻1 2 )
𝑚𝑑 𝑚𝑑
The location of the emerging of phreatic line are
important because of :-
 The location of phreatic line inside the dam body while help in
finding the seepage discharge through the dam and finding the
water pressure any point inside the dam body.
 It is required to check the possible failure of the dam by sloughing
or other process of failure.
 The demarcation line of the saturated and dry soil which is used
to calculation of specific weight of soil mass for stability analysis.
Seepage discharge through a dam with impervious foundation
Intake
• It is a hydraulic structure constructed at the head of an irrigation canal, tunnel conduit or power
canal to obtain the required quantity of water from the river to reservoir for the different
engineering purposes..
 Purpose of Intake
The main purpose of intake structure as a regulating structure are as follows:-
1. To admit and regulate water from the source and possibly to meter the flow rate.
2. To control the sediment entry into the canal.
3. To admit smooth, easy and turbulence free entry of water into the canal.
4. To prevent the clogging of the entrance with floating debris.
 Location for intake.
 The river reach upstream of the intake should be well established with stable banks.
 The best location for the intake is the outer (Concave) bank to avoid bed load, with the intake
located toward the downstream end of the bend.
 The location of the pond level and their variations, navigation hazards and location of the
diversion structure must be considered.
 The structure should be aligned to produce a suitable curvature of flow into the intake and a
diversion angle of around 300 𝑡𝑜 450 is usually recommended to produce curvature of flow.
 The entry to the tunnel and penstocks taking from the reservoir has to be placed lower than the
Dead storage level.
Types of intake:-
• Types of intake structure depend upon the types of power plant as well as its layout.
• According to the location it can be classified into two types:-
Surface intake Sub surface intake

• Side intake Power or fore bay intake

• Frontal intake Tower intake
• Drop intake
• Himalayan intake
• According to the hydraulic features it can be classified as :-
• Run-off river intake
• Canal intake
• Dam intake
• Tower intake
• Shaft intake
Side intake:-
• A structure built along a river bank and in front of a canal/conduit end for
diverting the required water supply. Side intakes are simple, less expensive,
easy to built and maintain. It is normally be sited immediately upstream of
diversion structure.

• Frontal intake:- Frontal intake are constructed

in front of the river at the dam body of the river.
It is normally used at low head plants where the
intake and the powerhouse is an integral part of
the dam. Under sluice should be constructed
where the sediment load transported by the
river contains large particle.
Drop intakes:-
• The drop intakes are located in small stream and they are adding water to either a reservoir feeder tunnel or
directly to a power tunnel. The main advantage of drop intake are:-
• It has minimum impact on the flow pattern of the river.
• It is favorable for small and steep river plant.
• Simple and economical structure.
Disadvantage:-
• Difficult to access for handling during flooding and maintenance.
• Intake chamber remains almost inaccessible throughout monsoon season.
• Himalayan intake:- Himalayan intake is provided for ROR project with daily peaking requirements in relatively steep
river. The objective of the Himalayan intake are:-
• Debris control during the monsoon season.
• Sediment by pass during the monsoon season.
• Daily peaking during the low flow season.
• Sediment control of intake.
• Run Of River Intake:- Run of river intake is provided for run of river plant and form an integral part of the power
house. In silty rivers special arrangement for the desilting are usually constructed upstream of the intake.
Fig drop intake
Canal Intake:-
• It is similar to the run off river intake but it admits water into the canal not directly through the
penstock. The invert level of the intake is raised to form a sill to control bed load and trash rack is
provided to prevent the entry of debris.
Dam intake:- Generally for the power house located at the toe of the dam, such type of intake
structure is suitable. In case of valley dam plants, this type of intake is suited in the body of the dam.
it is provided in the body of the dam and is used in high head hydroelectric plant.
Tower intake:-
• Tower intake are normally on hill side not far off from the dam. When it is not convenient to
provide other type of intake directly on the u/s face of the dam. They are constructed when there
is a large discharge or wide fluctuation of water level. There are two types :-
• Wet tower intake
• Dry tower intake
• Wet tower intake is preferable if a fluctuation of water level or wave is generated quite often in
reservoir level.
Shaft intake:-
• It is vertical or near vertical shaft driven in the bed which carries
water to penstock for power house to power production.
Basic design Consideration (Guide Line) for surface intake:-
The necessary consideration to be taken toward the design of intake are following :-
• Hydraulic consideration
• Minimum head loss
• Minimize the flow contraction
• Minimize the vortex formation.
• Structural consideration
• Concrete structure:- (Weir, intake, Energy Dissipater, Divide wall) should be structurally stable
for the various loading condition.
• Metallic Structure:- (Trash rack, Gate) structurally stable for the various loading condition.
• Operational consideration:-
• Smooth operation of the gate and trash racks without failure, sediment exclusion.
• Environmental consideration:-
• Fish ladder in head works to enhance the fish migration.
• Minimum flow on downstream
Losses in the intake
• The intake losses include entrance loss, rack losses and head losses.
• Entrance loss:-
𝑉2 𝑉𝑓 2
• Loss due to change in direction of flow is given by , (ℎ)𝐿1 = −𝐶∗
2𝑔 2𝑔
• Where, V = velocity in canal
• 𝑉𝑓 = 𝑣𝑒𝑙𝑜𝑐𝑖𝑡𝑦 𝑜𝑓 𝑓𝑙𝑜𝑤 𝑖𝑛 𝑡ℎ𝑒 𝑚𝑎𝑖𝑛 𝑠𝑡𝑟𝑒𝑎𝑚
• C= a constant depending on the angle of diversion and ranging between 0.8
to 0.40
• 𝛼 = 𝑎𝑛𝑔𝑙𝑒 𝑜𝑓 𝑑𝑖𝑣𝑒𝑟𝑠𝑖𝑜𝑛
𝑉2
• The loss due to sudden contraction of area at the diversion, (ℎ)𝐿2 = 𝐾 ∗
2𝑔
• Where, K varies between 0.03 for round mouthed entry to 1.3 for sharp
entry.
Head Losses:-
• The gate contract the flow, the coefficient 𝐶𝑑 in the following formula,
• Q= 𝐶𝑑 ∗ 𝐴 ∗ 2𝑔ℎ
• Where, Q= flow in conduit, A= area of gate opening, h=Difference in
head on the gate
• 𝐶𝑑 = 𝑐𝑜𝑒𝑓𝑓𝑖𝑐𝑖𝑒𝑛𝑡 𝑜𝑓 𝑑𝑖𝑠ℎℎ𝑎𝑟𝑔𝑒(0.62 𝑡𝑜 0.83)
Rack losses:- widely used formula is
𝑡 4/3 𝑉𝑏 2
• Kirchmer’s formula (ℎ𝐿 )𝑟 = 𝑘𝑟 ( ) ∗ ∗ sin ∅
𝑏 2𝑔
• Where, (ℎ𝐿 )𝑟 = 𝑅𝑎𝑐𝑘 𝐿𝑜𝑠𝑠, 𝑘𝑟 = 𝑎 𝑓𝑎𝑐𝑡𝑜𝑟 𝑑𝑒𝑝𝑒𝑛𝑑𝑠 𝑜𝑛 𝑐𝑟𝑜𝑠𝑠 −
𝑠𝑒𝑐𝑡𝑖𝑜𝑛 𝑜𝑓 𝑏𝑎𝑟
• b= spacing between bars
• 𝑉𝑏 = 𝑣𝑒𝑙𝑜𝑐𝑖𝑡𝑦 𝑜𝑓 𝑓𝑙𝑜𝑤 𝑖𝑛𝑓𝑟𝑜𝑛𝑡 𝑜𝑓 𝑏𝑎𝑟
• ∅ = 𝑎𝑛𝑔𝑙𝑒 𝑜𝑓 𝑏𝑎𝑟 𝑤𝑖𝑡ℎ ℎ𝑜𝑟𝑖𝑧𝑜𝑛𝑡𝑎𝑙
Gates
• Gates in the controlled spillways, different types of gates are provided on the
crest of the spillway specially on concrete dam. The gate provides additional
storage on the dam during dry season. But during flood time, these gates are
lifted or opened to allow full original spillways capacity as designed. Vertical
and radial gates are commonly used in hydropower projects.
The design of gates shall involves following
• 1. The gates shall be designed for the hydrostatic and hydrodynamic forces
taking into consideration forces arising from wave effects, seismic loads and ice
formation wherever applicable.
• 2. The additional water head to the static head to account for the sub-
atmospheric pressure downstream of gates located in conduits/sluices should be
specified to the designer
• 3. The gate is normally designed to close under its own weight with or without
addition of ballast but sometimes it may require a positive thrust for closing, in
which case hoist shall be suitable for that purpose.
Design criteria
The gate, in general, shall satisfy the following criteria:
1. It shall be reasonably watertight, the maximum permissible leakage being not
more than 5 litres/min/m length of seal in case of crest gates and medium head
conduit gates. The figure of permissible leakage is the upper limit before which
remedial measures shall be required to rectify defects:
2. It shall be capable of being raised or lowered by the hoist at the specified speed;
3. Power operated gates shall normally be capable of operation by alternate means in
case of power supply failure;
4. If meant for regulation, it shall be capable of being held in position within the
range of travel to pass the required discharge without cavitation and undue
vibration, and
5. Whenever necessary, model studies may be carried out for high head regulating
gates.
Types of gates
The common types of spillway gates are
• Vertical lift gate
• Radial gate
• Flap Gate
• Flash Board
• Stop logs gale
• Needle gate

a) Vertical lift gate:

Radial gate
Flap Gate Flash Board

Flap Gate
Stop logs gate Needle gate
Components of ROR Hydropower project
Settling Basin:-
• During the collection of water through the intake, it is not possible to extract only
the clear water. Certain percentage of suspended as well as bed loads will also
enter along with water.
• If the sediment particles along with the water enter into the turbine, it will cause
the damage to the turbine due to the abrasion. So it is necessary to supply as much
as clear water into the turbine or only the allowable size of the sediment is
accepted. In order to exclude the sediment particles that are harmful to the turbine.
Settling basin is generally constructed along the diversion channel.
• Settling basin has wider size and has the much depth of diversion channel. So the
velocity will be grately reduced causing settling of incoming sediment. Length and
width of the settling tank is designed such that the minimum required size of the
sediment is settled down along its length.
Types of settling Basin:-
• Periodic flushing type:-
• Closed the upper gate.
• Open the gallery gate.
• After the empty of the chamber, the upper gate is opened with the flushing discharge of (10 to
15%) of the designed discharge. This jet type velocity will flush the remaining sediment.
• Power plant shut down during flushing.
• Manual removal.
• Continuous flushing type:-
• To abstract water continuously during operation to prevent any deposition to scour.
• It is usually 20 to 30 % of the design flow is used for flushing.
• Flushing gate is located in the downstream of canal.
• Common flushing system has longitudinal hoppers with a flushing canal running along the bottom
of the hoppers.
• Intermittent flushing system:-
• No loss of water during the time between flushing and operation.
• The main advantage of this type is power plant can be operated during flushing.
Purpose of settling basin:-
• To remove of suspended particles and minimize the wear and tear of the
nozzle and runners of the turbine and other hydro-mechanical
components.
• To remove the unwanted sediment particles entering into the
conveyance system.
• To remove sediment as it cause abrasion and erosion of civil structures,
gates, valves.
• To remove the fine grained suspended matter from the water drawn
from intake to ensure that the water entering the conveyance system is
free of sediments that can damage the penstock and turbine runners
due to abrasion.
Design Criteria of the settling basins:-
The design of the settling basin aims to meet the following
criteria:-
• To determine maximum size of particles 2 mm escaped from the
gravel trap.
• To determine total sediment load to the turbine.
• To optimize sediment exclusion with respect to cost parameters.
• To secure high generation regularity.
• To use the available length and to manage with as minimum
width as possible depending on the topography and intake
location.
 Fall Velocity:-
• If the sediment falls in the quiescent liquid (No turbulence) then it will start to sink due to
its gravity. The sediment will be descending with retardation due to varying resultant
force, until the frictional resistance is fully developed. After the frictional resistance force
is fully developed and then the sediment will be falling with constant velocity due to
constant frictional resistance is called fall velocity.
• The force due to self weight of the particle is,
1
• 𝐹 𝐷𝑟𝑎𝑔 𝑟𝑒𝑠𝑖𝑠𝑡𝑎𝑛𝑐𝑒 = 𝜌𝑠 − 𝜌 ∗ 𝑔 ∗ 𝑉𝑜𝑙𝑢𝑚𝑒 𝑜𝑓 𝑡ℎ𝑒 𝑝𝑎𝑟𝑡𝑖𝑐𝑙𝑒𝑠 = 𝜌𝑠 − 𝜌 ∗ 𝑔 ∗ ∗ 𝜋 ∗
6
𝑑3 … … … . . 1
• The force due to motion of the particle in horizontal plane inside the fluid is given by ,
𝑉𝑆 2
• 𝐹𝐷𝑟𝑎𝑔 = 𝐶𝐷 ∗ 𝐴 ∗ 𝜌 ∗ ……………. 2
2
.• The particle will only move when equation (1) and (2) are equal ( Because,
Drag force just greater than frictional force due to self weight of the
sphere)
𝑉𝑆 2 1
• Therefore, 𝐶𝐷 ∗ 𝐴 ∗ 𝜌 ∗ = 𝜌𝑠 − 𝜌 ∗ 𝑔 ∗ ∗ 𝜋 ∗ 𝑑 3
2 6
1 3 1 3
2 𝜌𝑠 −𝜌 ∗𝑔∗ ∗𝜋∗𝑑 𝜌𝑠 −𝜌 ∗𝑔∗ ∗𝜋∗𝑑 4 𝑔 𝜌𝑠
• ⇒𝑉𝑆 = 3
= 3
𝜋∗𝑑2
= ∗ ∗ −1 ∗𝑑
𝐶𝐷 ∗𝐴∗𝜌 𝐶𝐷 ∗ ∗𝜌 3 𝐶𝐷 𝜌
4
4 𝑔 𝜌𝑠
• ⇒𝑉𝑆 = ∗ ∗ 𝐺 − 1 ∗ 𝑑 Because, G=
3 𝐶𝐷 𝜌
• Where, G= specific Gravity of the particle
• 𝜌𝑠 = 𝑠𝑝𝑒𝑐𝑖𝑓𝑖𝑐 𝑤𝑒𝑖𝑔ℎ𝑡 𝑜𝑓 𝑡ℎ𝑒 𝑝𝑎𝑟𝑡𝑖𝑐𝑙𝑒
• 𝜌 = 𝑢𝑛𝑖𝑡 𝑤𝑒𝑖𝑔ℎ𝑡 𝑜𝑓 𝑤𝑎𝑡𝑒𝑟
• d= diameter of particle
• 𝐶𝐷 = 𝑐𝑜𝑒𝑓𝑓𝑖𝑐𝑖𝑒𝑛𝑡 𝑜𝑓 𝑑𝑟𝑎𝑔
•
Ideal Settling Basin:-
• Following assumptions are made for an ideal horizontal settling basin:-
• The flow is steady and uniform.
• The flow is quiescent (no turbulence).
• Solid entering in deposition zones are not re-suspended.
𝐷
• Settling Time(𝑇𝑆 ) = … … … … … … … … … . (1)
𝑤
Where, w= settling or fall velocity.
𝐿
• Detention time (𝑇𝐷 ) = … … … … … … … . 2 , 𝑤ℎ𝑒𝑟𝑒, 𝑉 = ℎ𝑜𝑟𝑖𝑧𝑜𝑛𝑡𝑎𝑙 𝑓𝑙𝑜𝑤 𝑣𝑒𝑙𝑜𝑐𝑖𝑡𝑦
𝑣
• For quiescent liquid all particles of settling velocity(w) are removed when detention time
( time it takes for a particle to travel across the basin.)equal to settling time.
𝐷 𝐿
• Therefore, =
𝑤 𝑣

𝐿 𝐿 𝐿∗𝐴 𝐿∗𝐷∗𝐵 𝐷
• Also, = = = =
𝑣 𝑄/𝐴 𝑄 𝑄 𝑤
𝑄 𝑄 𝑄
• ⇒w= = , the ratio is termed as surface loading. As is surface area
𝐿∗𝐵 𝐴𝑆 𝐴𝑆
Real settling basin:- (Component of basin and their
design)
• Real settling basin:- (Component of basin and their design)
• The objective of design of a real basin is to achieve conditions most closely
relating to the ideal flow. But it is really very difficult to achieve. The following
factor mainly influence the settling capacity of a real basin.
• Turbulence of water in the basin
• Short circuiting and current within the basin caused by,
Wind induced the surface current.
Density current induced by thermal effect.
Boundary friction.
Separation and contraction zones which generate re-circulating eddy.
• The basin is divided into three main zones:-
Inlet zones
Settling zones
Outlet zones
a) Inlet zones:-
Gradually decrease micro turbulence and avoid secondary current by decreasing the flow velocity.
Needs to distribute the flow uniformly over the vertical X-Section.
• Method for good flow distribution:-
 Gradually channel expansion 110 )
 Entrance vertical slope(1:1)
 Baffles and screen
b) Settling Zone:- Main part of the basin. Effective length is considered to be 85% of its uniform
length. Minimum basin length = (4 to 6) of width.
• Critical velocity of flow = 0.25 to 0.50 m/sec.
• Depth of settling tank = 3 to 5.5 m
• Longitudinal slope of the tank = 1:50 to 1:20
• Dead storage depth = (20 to 25%) of whole depth of flow.
c) Outlet Zone:-
• It is used to facilitate getting back of the flow into the conveyance system. It control the water
level in the basin.
• The outlet transition = 260 .
• Velocity of flushing gallery= 2 to 3.5 m/sec
• Minimum flushing frequency = 3 to 4 days
• Flushing discharge = 10 to 30% of design discharge
Value of CD (𝒄𝒐𝒆𝒇𝒇𝒊𝒄𝒊𝒆𝒏𝒕 𝒐𝒇 𝒅𝒓𝒂𝒈)
For spherical particle moving in the water.
24
• Reynold’s number (Re) < 1, 𝐶𝐷 = (Linear Relationship)
𝑅𝑒
24 3
• For Re between 1 to 1000 , 𝐶𝐷 = + 𝑅𝑒 + 0.34 𝑁𝑜𝑛 − 𝑙𝑖𝑛𝑒𝑎𝑟 𝑅𝑒𝑙𝑎𝑡𝑖𝑜𝑛 𝑠ℎ𝑖𝑝
𝑅𝑒
• For Re > 1000, 𝐶𝐷 = 0.40 , Constant Value.
• For Laminar flow,
• Drag force due to viscosity (Stroke’s Law)𝑭𝑫 = 𝟑𝟐 ∗ 𝝁 ∗ 𝑽𝒔 ∗ 𝒅 … … … … … … 𝟏
𝒈
• And , 𝑽𝒔 = ∗ 𝑮 − 𝟏 ∗ 𝒅𝟐 , 𝒘𝒉𝒆𝒓𝒆 Ѵ = 𝑲𝒊𝒏𝒆𝒎𝒂𝒕𝒊𝒄 𝑽𝒊𝒔𝒄𝒐𝒔𝒊𝒕𝒚
𝟏𝟖∗Ѵ
𝟐 𝟑𝒕+𝟕𝟎
• For temperature, 𝑽𝒔 = 𝟒𝟏𝟖 ∗ 𝑮 − 𝟏 ∗ 𝒅 *( )
𝟏𝟎𝟎
• ‘Ѵ’ depends upon the temperature,

Temp(t)𝑡 𝑜 𝑐 0 5 10 15 20 25
Ѵ(Centistokes) 1.79 1.52 1.31 1.15 1.01 0.90
Design Principle of settling Basin:-
1. Flushing Discharge:- The flushing discharge should be capable of
inducing bed load movement usually flushing discharge range
from (10 to 30%) of the intake discharge.
𝐾∗𝑄
• { Calculate surface area {𝐴𝑠 = ).
𝑤
• Assume L/B = 4 to 10 .
• If depth is not given check width(B) ≤ 4.75 𝑄),
• 𝑇𝑎𝑘𝑒, 𝐾 = 𝑐𝑜𝑒𝑓𝑓𝑖𝑐𝑖𝑒𝑛𝑡 𝑎𝑐𝑐𝑜𝑢𝑛𝑡 𝑡𝑜 𝑡𝑢𝑟𝑏𝑢𝑙𝑒𝑛𝑐𝑒 (1.2 𝑡𝑜 1.5)}
 2.Velocity of Basin:-
• Limiting flow velocity can be determined by the relation.
• V= a 𝑑 … … … … … … . . (1)
• Where, v= flow through velocity (m/sec)
• d= diameter of particles upto which sediment load is desired to be removed in (mm)
• a = constant which is , a= 36 for d> 1 mm
a= 44 for 1 mm>d>0.1 mm
a= 51 for d< 0.1 mm
• However, R.S. Varshney (1985) recommends the following coefficient for Himalayan
rivers,
a= 55 for d> 1 mm
a= 66 for 1 mm>d>0.1 mm
a= 77 for d< 0.1 mm
3.Size of sediment load to be removed:-
• The efforts should be made to removes much as silt load and sediment load as
economically and hydraulically. The largest particle is to eliminate sediment particles
up to 0.15 to 0.20 mm size.
4. Dimension of settling Basin:- If the depth of Basin is D, width B, flow velocity V, then
discharge passing through the basin is
• Q= B*D*V………………….(2)
′ 𝐷
• If ‘w’’ is settling velocity then the settling time ‘𝑡𝑠 = … … … . 3
𝑤
𝐿
• Also, detention period (𝑡𝐷 ) = ……………..(4)
𝑉
• We know, to remove all sediment particle, detention period should be equal or
greater than the settling time.
• Now, equating equation(3) and (4)
𝐷 𝐿
• = ⇒ 𝐿𝑤 = 𝐷𝑉 … … … … … . 5
𝑤 𝑉
. If turbulence is neglected then, 𝐿𝑆𝑒𝑡𝑡𝑙𝑖𝑛𝑔 𝐵𝑎𝑠𝑖𝑛 = 1.5 ∗ 𝐿
 If turbulence is considered then, Settling Velocity = w-w’
• Where, w’ = reduction due to turbulence = 𝛼 ∗ 𝑣
0.132
And, 𝛼 =
𝐷
𝐷∗𝑉 𝐷∗𝑉 𝐷∗𝑉 𝐷∗𝑉∗𝐷1/2 𝐷 1.5 ∗𝑉
 Settling Length(L)= = = 0.132 = =
𝑤−𝑤′ 𝑤−𝛼∗𝑉 𝑤− ∗𝑉 𝑤∗ 𝐷−0.132∗𝑉 𝑤∗ 𝐷−0.132∗𝑉
𝐷

ƛ2 ∗𝑉 2 ∗( 𝐷−0.20)2
Velikanov’s FormulaL=
7.51∗𝑤 2
• Where,
• ƛ = 𝑉𝑒𝑙𝑖𝑘𝑎𝑛𝑜𝑣 ′ 𝑠𝐶𝑜𝑛𝑠𝑡𝑎𝑛𝑡 𝑑𝑒𝑝𝑒𝑛𝑑𝑠 𝑢𝑝𝑜𝑛 % 𝑜𝑓 𝑡𝑟𝑎𝑝𝑝𝑒𝑑 𝑠𝑒𝑑𝑖𝑚𝑒𝑛𝑡
[for 95% removal ƛ= 1.2]
 .
 Find out the dimension of settling basin for a high head project in Himalayan
River which utilize a discharge of 60 m3/sec and gross head of 300m. The
sediment particles larger than 0.15 mm (w= 1.5 cm /sec) had to be trap in the
basin. Consider effect of turbulence as well as. Draw plan a section of the basin
showing major component.(Note:- Here depth is not given)
• Solution:-
𝐾∗𝑄
1. Calculate surface area(𝐴𝑆 ) =
𝑤
• Where, K= coefficient to account for turbulence ( 1.2 to 1.5)
• Consider, purpose of two chamber in a settling basin because the flow is
greater.
• i.e flow in one chamber = 60/2 = 30 m3/sec.
𝐾∗𝑄 1.5∗30
• Therefore, 𝐴𝑆 ) = = 1.5 = 3000 𝑚2
𝑤
100
. 𝐿
2. Assume , = 4 𝑡𝑜 10
𝐵
𝐿
• Assume , = 8, 𝑇ℎ𝑒𝑟𝑒𝑓𝑜𝑟𝑒 𝐿 =8∗𝐵
𝐵
• Now, 𝐴𝑆 = 𝐿 ∗ 𝐵 = 8𝐵 ∗ 𝐵 = 3000
3000
•⇒𝐵= = 19.36 𝑚
8

• Check width (B)= 4.75 𝑄 = 4.75 30 = 26.01 𝑚 > 19.36 𝑚 𝑜𝑘

• Adopt, B= 20 m
• Therefore, L= 8*20 = 160 m
3. Calculate the limiting velocity(V):-
• V= 𝑎 𝑑 here, d= 0.15 mm and a= 66 for Himalayan River
• V= 66 0.15 = 25.56 𝑐𝑚/𝑠𝑒𝑐 =0.2556 m/sec
 𝑄 30
4.Depth of settling Basin (D)= = = 5.868𝑚
𝑉∗𝐵 0.2556∗20
• Adopt, , D ≅ 6.0 𝑚
• Therefore, size of the settling basin without considering turbulence, 160m*20m*6m.
• Also, With considering Turbulent effect { Velikanov’s Formula}
𝐷1.5 ∗𝑉 61.5 ∗0.2556
• Settling Length (L) = = = 1250.87 ≅ 1255 𝑚
𝑤∗ 𝐷−0.132∗𝑉 0.015∗ 6−0.132∗0.2556
ƛ2 ∗𝑉 2 ∗( 𝐷−0.20)2
• Velikanov’s Formula, L=
7.51∗𝑤 2
• Where, ƛ = 𝑉𝑒𝑙𝑖𝑘𝑎𝑛𝑜𝑣 ′ 𝑠𝐶𝑜𝑛𝑠𝑡𝑎𝑛𝑡 𝑑𝑒𝑝𝑒𝑛𝑑𝑠 𝑢𝑝𝑜𝑛 % 𝑜𝑓 𝑡𝑟𝑎𝑝𝑝𝑒𝑑 𝑠𝑒𝑑𝑖𝑚𝑒𝑛𝑡. [Take, ƛ= 1.5]
ƛ2 ∗𝑉 2 ∗( 𝐷−0.20)2 1.52 ∗(0.2556)2 ∗( 6−0.20)2
• Therefore, Length of Settling basin(L)= =
7.51∗𝑤 2 7.51∗(0.015)2
=440.20 m ≅ 445 𝑚
• Therefore, size of the basin with considering turbulent is 1255m*20m*6m
Remember
95% removal ƛ= 1.2
88%, removal ƛ = 0.8