HYDROELECTRIC POWER PLANTS - … · LECTURE NOTES – I « HYDROELECTRIC POWER PLANTS » Prof. Dr. Atıl BULU Istanbul Technical University College of Civil Engineering Civil Engineering

LECTURE NOTES – I

« HYDROELECTRIC POWER PLANTS »

Prof. Dr. Atıl BULU

Istanbul Technical University College of Civil Engineering

Civil Engineering Department Hydraulics Division

Chapter 1

Potential Energy of Surface Water

Every particle of water that appear, either as precipitation or as ground water, on the surface or in the ground disposes of a definite potential energy, the magnitude of which depends upon the elevation above sea level of the ground where it reaches the surface or where it emerges from ground. The flowing water has a kinetic energy as well which is V2/2g, which is comparably high at the steep areas usually in uplands around 5 – 6 m/sec and decreases to 1 – 2 m/sec in lowland areas, or at the mouth of the river. When descending from the high elevations to the sea or the lake, the surface flow as rivers, potential and kinetic energies are converted to non-usable energy, the kinetic energy loss being negligible as compared to the potential energy. The potential energy is dissipated to overcome internal friction of turbulent flow, to supply energy to whirls, eddies and spiral flows, to scour the material of the river bed and to transport bed load, while the water is descending to a body of water. The mechanical work, wasted in overcoming frictional resistance (bed resistance), is converted entirely into heat and lost. The fundamental principle of water power development is to reduce the amount of energy dissipated as heat without paralyzing the flow of water. Friction (energy) losses can be reduced in principle by three methods:

1. In any stretch of the river, the energy loss required for the conveyance of water is reduced by decreasing the velocity of flow.

LShR

nVS

SRn

V

loss 034

22

0

210

321

=→=

=

↓↓→↓→ losshVS0

This can be achieved by increasing the depth of water through building dams. Thus at the dam, between backwater level and the natural surface of the stream, a certain head H is formed. For creating head H, backwater should be produced with a non-uniform flow curve (M1) extending to distance L0. Out of the total water level difference H0, on a river section L0, with channel slope S0, the head available for utilization is H, while the potential energy H1 is dissipated in overcoming the reduced frictional resistance in the stream. The degree of utilization of potential energy in any section L0 relating to an assumed head H, for any given discharge Q, is expressed by,

Prof. Dr. Atıl BULU 1

0HH

m =η

If a second dam is constructed at the downstream of the same backwater curve, the utilization is around 50%. By erecting a series of dams and weirs, a considerable portion of potential energy in any stream or in an entire river basin can be utilized. Let us examine two dams at an arbitrary distance x from each other, which does not exceed L0 and staying in the same backwater curve.

It is assumed that the backwater can be substituted by a quadratic parabola. The head created at the upper dam at distance x from the lower one can be expressed as;

yxLHyxS−

=−0

00

When taking into consideration that,

20

0

2cLH

=


For the rise y of the backwater parabola, we get,

220

02

2x

LHcxy ==

The total head created by both dams is,

220

0

0

000

222x

LHx

LHHyHH −+=+=

As the utilization of the above head refers to section (L0+x), the mean hydraulic slope available for utilization,

xL

xL

HxLHH

Su +

−+=

0

220

0

0

00

22

The maximum value of Ju is determined by,

( )

( )( ) ( )( )

( ) 0000

20

200

2,1

200

2

20

20

2000

0

20

20

20

0

020

2000

200

414.012222

442

02

0222

22

22

0

LLLLx

LLLx

LxLx

xLxxLLxLxL

dxdS

xLxxLL

LH

dxd

dxdS

xLLxHxLHLHS

dxdS

u

u

u

u

=−=+−=

+±−=

=+−−

=+

−+−+−=

⎥⎦

⎤⎢⎣

⎡⎟⎟⎠

⎞⎜⎜⎝

⎛ −+=

+−+

=

=

( )

0

0

0

0max

00

202

0

00

0

00

max

586.0414.1828.0

414.0

414.02

414.02

LH

LHS

LL

LL

HLLHH

S

u

u

=×=

+

×−×+=

The degree of utilization is,


586.0=mη It can be proved that, by increasing the number of dams, the degree of head utilization can be raised hydraulically, but it is questionable whether or not this solution is economical as far as construction costs are concerned. If three dams are installed on the same backwater curve, the degree of utilization is 0.670 while in case of four dams this value is increased to 0.725. Whereas the ratio between developments with two and four dams amounts to 0.725/0.586= 0.24 increase in the degree of utilization, the costs of investment go up excessively and may reach almost the double. It can be concluded that by a series of dams, not more than 60% of the total head can be developed. B) Another method of reducing the head required for the conveyance of water is to divert the whole part of the flow into an artificial bypass, usually termed power canal, the slope of which is considered flatter than that of the original water course. The difference in elevation between the power canal and the original watercourse is thus gradually increased and a head available for power generation at the most suitable site is created.

How can this gain in elevation be obtained? Again through the reduction of frictional losses, that is, through improving conveyance conditions by shaping the canal according to a design of a suitable cross-section, by removing aquatic growth from the bed and by lining the canal. The resulting head can be especially high if there are rapids in the original water course. The diversion of water into power canal is affected by a diversion dam or weir built across the river bed. The head created for the purposes of utilization on any river L0 + L is,

( )LSSHH d 1−+=

Out of which,


( )LSS 1−

Is the gain in head attributable to the power canal of length L. In case of developments with power canals, the degree of utilization varies very wide limits and under average conditions a value ηm = 0.80 can be assumed. C) If the river is sinuous and especially if the valley itself is characterized by sharp or horseshoe bends, they can readily cut by a channel or a tunnel. With this of development, a difference in elevation capable of being utilized is obtained. In power utilization the above mentioned possibilities of developing the available fall are usually combined.


LECTURE NOTES – II





CHAPTER 2

MEASURES

In the field of hydroelectric development, work and energy (output of a power plant) are expressed generally in kilowatt-hours (kWh). Power (capacity of a power plant) is usually expressed in kilowatts (kW) and sometimes the power of hydraulic machinery is in horsepower (HP).

1 megawatt (MW) = 1000 kW 1 megawatt-hours (MWh) = 1000 kWh

1 gigawatt-hours (GWh) = 106 kWh

1 HP = 75 kgm/sec = 736 watts = 0.736 kW 75 kgm/sec = 75×9.81 ≈ 736 Nm/sec = Joule/sec = Watt

1 kW = 1.36 HP

1 kW = 102736.075

= kgm/sec

1 HP-hour = 0.736 kWh

1 HP-hour = 75×3600 = 270000 kgm 1 HP-hour = 75×9.81×3600 = 2648700 Nm (Joule)

1Kwh = =736.0

270000 367000 kgm = 3.6 × 106 Joule

1 kgm = 9.81 joules (Nm) 1 kgm/sec = 9.81 joules/sec = 9.81 watt

In the field of thermal power generation, work and energy are also measured in kilogram-calories (Cal).

1 Cal = 427 kgm

1kWh = =427

367000 860 Cal

Example 2.1: Calculate the quantity in kWh of the energy generated from 1 kg of coal of 4000 calories by a thermal power plant having an overall efficiency of 24%. Solution: Considering that η = 0.24, from 1 kg of coal of 4000 calories, the thermoelectric plant generates a quantity of electric energy that corresponds to,

960400024.0 =× Calories

As, 1 kWh = 860 Cal

The electric energy generated from 1 kg of coal is,


kWh12.1860960

=

Example 2.2: Compute in kg weight of coal saved per annum by a hydroelectric plant operating at an annual average capacity of 8000 kW, supposing the fuel consumption of the substituting thermal plant is 3500 Cal/kWh, and the quality of coal is characterized by 4000 Cal/kg. Solution: The hydroelectric produces an annual output of,

kWh61070243658000 ×=××

When generating the same quantity of energy, the consumption of the substituting thermal plant would be,

CalkWhkWhCal 106 105.2410703500 ×=××

Accordingly, the annual saving in coal attained by operating the hydroelectric plant amounts to,

tonkg 612001012.6104105.24 7

3

10

=×=××

Example 2.3: How long does it take a 100 W bulb to consume the same quantity of energy as is required for a tourist of 70 kg in weight carrying an outfit of 35 kg to climb a mountain of 970 m in height? The tourist is assumed to set out on his way to the top of the 970 m mountain from a hostel situated 340 m above sea level. Solution: The work done by the tourist is,

( ) ( ))(64893281.966150

661503570340970JouleNm

kgm=×

=+×−

The hourly consumption of a 100 W bulb is 0.10 kWh,

kgmKwh 36700081.9360010001 ≅

×=

The electric energy consumed per hour by the bulb is equivalent to a mechanical work of 0.1×367000 = 36700 kgm. Accordingly, the electric energy equivalent to the work done by the tourist is consumed in,

hours80.13670066150

=


Example 2.4: A laborer working at an average capacity shovels 8 m3 of earth a day up to a vertical distance of 1.60 m from a material having specific weight of 1.8 ton/m3. Compute in kg the quantity of coal of 4200 Cal required for obtaining the same work if the thermal station operates at an efficiency of 24%. Solution: The work done daily by the laborer is,

)(22600081.9230402304004.236.18.18

NmJoulekgmtm

≅×==××

Cal

kgmCal

54427

230404271

≅

=

In case of a 24% efficiency, from the coal having a calorific value of 4200,

grkg 54054.0420024.0

54==

×

Coal is required to substitute the work done in 1 day by the laborer. Example 2.5: Determine the quantity of heat generated by braking and stopping a goods train consisting of 50 wagons and traveling at a velocity of 36 km/hour. Calculate the time required for a small hydroelectric power plant of 15 kW capacity to generate an equivalent amount of electric energy. The average weight of each wagon is 20 ton. Solution: The mass of goods train amounts to,

mkg 2sec10200081.9

2000050≅

×

In the process of braking the goods train running at a speed of 36 km/hour = 10 m/sec, the kinetic energy converted into heat equals,

kWh

kgmmV

9.13367000

101.5

101.52

101020002

6

622

=×

×=×

=

Consequently, the small hydroelectric plant of 15 kW capacity is capable of producing an equivalent electric energy in,

min56926.015

9.13== hour


LECTURE NOTES – III





CHAPTER 3

Power in Flowing Water

Power is the proportion of performed work in unit time. N = Power, E = Energy,

dtdEN =

p0/γ

h2=p2/γ

h1

H.D.

G V

The total energy of any water particle of weight G, under pressure p, and moving at a velocity V, at an elevation h1 above a horizontal datum is, according to Bernoulli theorem,

⎟⎟⎠

⎞⎜⎜⎝

⎛++=γp

gVhGE2

2

1 (kgm, Nm, Joule)

With the substitution,

⎟⎟⎠

⎞⎜⎜⎝

⎛++=

+=

γ

γγ

02

20

2p

gVhGE

hpp

Working with gage pressure,

⎟⎟⎠

⎞⎜⎜⎝

⎛+=

gVhGE2

2

Total mechanical energy of any water mass of 1 kg in weight amounts to,


gVh2

2

+ (m)

The total potential energy of the volume of water V stored in the reservoir is,

0VHE γ=

While it’s inherent energy is, Ci VHE γ=

HC = the height of the center of gravity of the volume of stored water above the chosen datum. The energy consumed in emptying the reservoir is exactly equal to inherent energy of the entire mass as imagined to be in the center of gravity. The total energy of water discharging from the reservoir is subject to constant change and equals the inherent energy pertaining to the momentary water surface. Accordingly, the work performed during emptying the reservoir is not the sum of the total energy of water particles stored in the full reservoir, but is the inherent potential energy of the water mass. If the surface of water in the reservoir is kept constant level by balanced inflow and outflow, both total energy and inherent energy in any volume of effluent water of weight G are equal for any arbitrary time.

00 GHVHEEi === γ (kgm, Nm, Joule)


If the discharge at a velocity V1 through any section of a stream amounts to Q (m3/sec), the power of the flowing mass per second, with reference to any datum at a depth H1 below the surface,

⎟⎟⎠

⎞⎜⎜⎝

⎛+=

gVHQ

dtdE

2

21

11 γ (kgm/sec, watt)

The energy per second of the body of water flowing through section 2 located downstream from section 1 and having a height of water level H2 above the datum is,

⎟⎟⎠

⎞⎜⎜⎝

⎛+=

gVHQ

dtdE

2

22

22 γ

Accordingly, with any discharge Q flowing steadily between constant levels by a difference in elevation H = H1 – H2, the energy dissipated in section 1-2 in every second,

⎟⎟⎠

⎞⎜⎜⎝

⎛ −+=⎟⎟

⎠

⎞⎜⎜⎝

⎛+−⎟⎟

⎠

⎞⎜⎜⎝

⎛+

gVVHQ

gVHQ

gVHQ

222

22

21

22

2

21

1 γγγ

The overall potential power for the considered stretch is,

⎟⎟⎠

⎞⎜⎜⎝

⎛ −+=

gVVHQN p 2

22

21γ (kgm/sec, watt)

The portion of power that originates from changes in velocity is generally negligible as against the potential power from the differences in elevation. The theoretical potential power in any river stretch with a difference in elevation H is,

QHN p γ= (kgm/sec, watt)


LECTURE NOTES – IV





CHAPTER 4

Potential Water Power

For any stretch of a watercourse, characterized by a difference in level of H meters, conveying a discharge of Q (m3/sec), the theoretical (potential) power,

( )

( )

( )kWQHQHN

HPQHQHN

kgmQHQHN

p

p

p

8.9736.03.13

3.1375

1000

sec1000

=×=

==

== γ

If the rate of flow changes along a stretch, the mean value of the discharges pertaining to the two terminal sections of the stretch is to be substituted in the equation, , ( ) 2/21 QQQ += The theoretical power resources of any river or river system are given by the total of the values computed for the individual stretches,

( )kWQHQHN p ∑ ∑=×= 8.9736.075

1000

Potential water power resources can be characterized by different values according to the discharge taken as basis of computation. The conventional discharges are,

Figure. Discharges used for characterising potential water power resources


1. Minimum potential power, or theoretical capacity of 100%, is the term for the value computed from the minimum flow observed. Np100

2. Small potential power. The theoretical capacity of 95% can be derived from the discharge of 95% duration as indicated by the average flow duration curve. Np95

. 3. Median or average potential power. The theoretical capacity of 50% can be

computed from the discharge of 50% duration as represented by the average flow duration curve. Np50.

4. Mean potential power. The value of theoretical mean capacity can be ascertained by taking into account the average of mean flow. The average of mean flow is understood as the arithmetic mean of annual mean discharges for a period of 10 to 30 years. The annual mean discharge is the value that equalizes the area of the annual flow duration curve.

Economic significance of potential water resources of a site. This is influenced by a great number of factors than hydraulic, such as geographical, geological and topographical conditions, energy demand, etc. Ignoring these and comparing relative values of power potential as reflected by hydraulic conditions only, the following four aspects are to be taken into consideration:

a) The absolute quantity of theoretical water power resources, b) The relative share of discharge in the power. Among the hydraulic possibilities representing equal magnitudes, the more advantageous are those where the power in question originates from a smaller flow and a higher head. It is advantageous of highland developments over power stations situated in hilly regions or lowland areas. c) The relative annual fluctuation of available potential power. This can be characterized by the ratio of the values Np50 to the values Np95 (or Np100).

95

50

p

p

NN

=α or, 100

501

p

p

NN

=α

A smaller ratio reflects a more favorable hydraulic possibility. d) The over year or multi-annual variation of potential power. This can be characterized either by a simple diagram showing the annual potential power against time, or by a summation curve of annual values. Power resources can be characterized even by annual values of potential energy in a river, by the quantities of work,

E100 , E95, E50 and Em

All expressed in kilowatt-hours. These values can be computed as areas of the lower parts of the potential power-duration curves. If the head is assumed to be constant, independent of discharge, the computation can be based on the discharge-duration curve. Using the curve,


Figure

( )

( )kWhaFQtQaE

aQtaQE

kWaQHQN

tit

tit

ttp

2424

2424

8.9

365

365

=⎟⎠

⎞⎜⎝

⎛+=

+=

==

∑

∑

Where, t = the duration considered in days, Qt = Selected discharge, Qi = Daily mean of actual discharge at any time, F = Area pertaining to Qt (shaded area). The upper limit of potential energy inherent in the river section is obtained by,

( ) ( )kWhNNE mmmx 876036524 =×=

Where Nm is the annual mean power. The overall coefficient is about 0.75 or 0.80. The equation recommended is,

( )∑−= HQN mmnet 0.84.7

Qm = the arithmetic mean discharge. Net river energy potential,

∑=netnet mNE 8760max

For characterizing the gross potential power of a river basin, the following data should be used,


a) Total annual discharge volume V (m3), b) Medium height of the watershed area H lying upstream above sea level (m), c) Area A of this basin (km2).

Since 1 m3 water weighs 1 ton, the product VH yields the annual gross power pertaining to the selected site in meter-tons,

( )

( )A

VHkmkWhe

VHkWhE

367/

3672 =

=

If the e values are determined along the river basin at several stream sites, then the lines connecting the points of equal e, isopotential lines can be drawn. Flow-Duration Curve If the flows for any unit time are arranged in descending order of time (without regard to chronological sequence), the percentage of time for which any magnitude is equaled or exceeded may be computed. The resulting array is called a flow-duration curve.

Figure. Determination of flow-duration curve (Bayazıt, 2001).

Such curves are useful in determining the relative variability of flow between two points in a river basin or between two basins. For example, if a stream is highly regulated, the curve will approach a horizontal line. The dependable flow is that corresponding to 100 percent of time. The relative variability of two flow records may be compared by converting the discharge scale in terms of a ratio to the mean. Any sub area under the curve represents the volume of annual runoff. Flow-duration curves have been used to approximate the amount of storage needed to increase the dependable flow. For example, the horizontal line AB in below Figure may represent a new dependable flow, and the required storage needed to obtain this flow is indicated by area ABC.


Figure . Duration curve of monthly discharges.

Power production values may be approximated from the duration curve by converting the discharge scale to kilowatts by multiplying by a selected head, efficiency and conversion factors. If the time scale is converted to hours in a year, a unit of are represents kilowatt-hours. The flow-duration curve is particularly useful in combination with a sediment rating curve (river discharge versus the transported sediment load usually expressed in tons per day), to compute total sediment load to be expected in an average year. Flow Mass Curve Total flow volume from a certain initial time t = 0 up to time t can be computed as,

∫=t

QdtH0

In practice, the total volume is computed as,

∑= iitQH

Qi = the average discharge in time interval (month, year) Δti. Flow mass curve is a plot of the cumulative runoff from the hydrograph against time. The time scale is the same as for the hydrograph and may be in days, months or years. The volume ordinate may be in m3-days, m3-months, m3-years, etc. The slope of the mass curve is the derivative of the volume with respect to time or the rate of discharge.


The mass curve usually has a wavy configuration in which the steeper segments represent high flow periods and flatter segments represent low flows. Uniform rates of withdrawal (draft) may be represented as tangent lines drawn from high points to intersect the curve at the next wave. The vertical distance between the draft line and the basic curve represents the cumulative difference between regulated outflow and natural inflow, or the required storage. If the draft line does not intersect the mass curve at the end of a year, it means that the reservoir does not refill with that rate of draft and regulation at the proposed draft rate will extend over two years or more. A typical mass curve is shown in the above Figure. In estimating storage requirements from the mass curve, it is not necessary to assume a constant rate of regulated flow. For example, if the draft rate to meet a demand for irrigation, water supply, or power varies from month to month, the draft line may be a curved or irregular line and the maximum draft may not occur at the low point in the mass curve. An allowance for evaporation should be applied to the mass curve analysis. If the water area does not change significantly during the annual cycle of use, an average correction for each calendar month can be subtracted from the inflow or added to the draft rates. The ordinates of the flow mass curve increase continuously in time. The sum of the differences between the inflow and the yield (average flow) are drawn;


( ) iavei tQQH Δ−=∑0

Reservoir capacity is then vertical distance between the highest and lowest points of the curve.

Figure . Flow mass curve derived using the differences

of the discharges from the yield. Storage-Draft Curve The results of a mass-curve analysis can be plotted as a storage-draft curve. His curve gives the storage needed to sustain various draft rates. Examples of storage-draft curves are shown in the below Figures. Both irrigation requirements and combined irrigation and power requirements are illustrated. These curves were computed from the mass curve. If storage unlimited, the storage-draft curve will approach the available mean flow as asymptote. It is rarely possible to develop mean annual flow of a river basin. For most projects, some spillage will occur in years of runoff. To impound all flood flows will require an extensively large reservoir. Such a reservoir may not fill in many years, and probably could not be justified economically. The selected rate of regulated flow to be developed will depend on; 1. The demand of water users, 2. The available runoff, 3. The physical limits of the storage capacity, 4. The overall economies of the project.

Figure . Storage-draft curves for multipurpose uses.


Selection of Design Flow The hydrologic analyses, combined with economic analyses of costs and benefits for different heights of dam and reservoir capacity will lead to the selection of the reservoir capacity and the corresponding dependable flow that can be justified. The selected design flow may not necessarily be available 100 percent of the time. The propose water use may permit deficiencies at intervals, for example, a 15 percent shortage once in 10 years. Irrigation water supplies may permit greater deficiencies than those for urban and industrial use. Hydroelectric power plants, connected to large systems, may tolerate substantial water supply deficiencies. Final Storage Selection a) Evaporation Losses Detailed evaluation of evaporation losses should be postponed until final operation and routing studies, when the actual variation in water area can be considered as well as the seasonal variation in evaporation. Basic data on water surface evaporation may be obtained from records of pan evaporation. Such records overestimate lake evaporation and must be reduced by a pan coefficient which varies from 0.60 to 0.80 depending on the climate. The collection of evaporation records at a project site should be initiated in the planning stage. Evaporation corrections should be made on a monthly basis using actual past precipitation records at the project site if possible. b) Power Selection of an average flow alone will not permit determination of the benefits from a water resources development project without more detailed studies. Such studies require routing through the reservoir the entire record of flow (corrected for evaporation losses), on a month by month basis, using assumed patterns of use, outlet capacities and, in the case of power, turbine and generator capacities and efficiencies. The reservoir would normally be considered to be full at the start of the operation study, or at least full to normal pool. For power benefits, the energy output will vary in accordance with the inflow, outflow, and change in storage and corresponding head, tailwater elevation, turbine capacity and plant efficiency. If the plant is a part of a system, the output may be subject to varying demands of the system load curve and whether the plant is to be used as a base load plant or a peaking plant. The routing study will indicate the necessary modifications to the head, storage, and even height of the dam to obtain maximum benefit. c) Irrigation Operation studies for irrigation use should be made using seasonal crop demands and selected outlet capacities. Short-term demands may indicate that the storage needed war greater than that required for uniform regulated flow. The proposed annual water use may be greater than that available 100 percent of the time, with the understanding that deficiencies can be tolerated in some years.


d) Water Supply Operation studies for projects providing urban water supplies will be similar to those for irrigation projects in that there may be variations in the seasonal demand, especially where more than one source is available, or where there can be transfers to other regulation reservoirs. However, the degree of dependability of flow must be higher for urban water supply than for irrigation projects. e) Flood Control The storage allocated for flood control in single purpose or multipurpose projects is usually based on a definite design flood the control of which is needed for downstream protection. The required storage capacity is based on routing of the design flood inflow coincident with releases not to exceed downstream channel capacities. Total Storage Requirement The usable storage needed for single purpose projects can be readily determined as described in Sections (a) to (e). The total usable storage needed for multipurpose for multipurpose projects require more complex routing studies and numerous trials to obtain the most economic allocations. In addition to the variable requirement for storage for downstream uses, the total storage may be increased for the following reasons: • Minimum head on power installations. • Allowance for the storage of sediments without loss of usable storage. • Minimum area for recreation use, including seasonal requirements. EXAMPLE : Monthly flow volumes feeding a reservoir are given in the table. Determine the storage capacity required to supply the mean annual flow volume yield. Solution: Cumulative volumes are calculated and given in the table.

Month Volume(106m3)

Cumulative Volume (106m3)

1 296 296 2 386 682 3 504 1186 4 714 1900 5 810 2710 6 1154 3864 7 746 4610 8 1158 5768 9 348 6116 10 150 6266 11 223 6489 12 182 6671


Total volume of flow feeding the reservoir is 6671×106 m3. Annual mean discharge can be calculated as,

smQ 36

21286400365

106671≅

××

=

The reservoir storage capacity required to obtain 212 m3/s yield throughout the year is found by drawing tangents parallel to the average draft line from peak points. The vertical distance is 1800×106 m3 is the required capacity of the reservoir. The reservoir capacity to supply the annual mean discharge can be found out by using the sum of differences method as in table,

Reservoir Capacity

0

1000

2000

3000

4000

5000

6000

7000

8000

Months

Volu

me(

106 m

3 )

V

Month Volume

(106m3) Flow (m3/s)

H0 (m3/s)

ΣH0(m3/s)

1 296 111 -101 -101 2 386 149 -63 -165 3 504 188 -24 -188


4 714 267 55 -134 5 810 335 123 -11 6 1154 431 219 208 7 746 288 76 284 8 1158 432 220 504 9 348 134 -78 426 10 150 56 -156 270 11 223 83 -129 142 12 182 70 -142 0

-300 -200 -100

0 100 200 300

400 500 600

1 2 3 4 5 6 7 8 9 10 11 12

Months

Ho

( )∑ Δ−= iavei tQQH0 Reservoir Capacity = 504 – (-188) = 693 m3/s

Volume = 361018538640031692 m×=××

RESERVOIRS A reservoir is a manmade lake or structure used to store water. A dam reservoir has an uncontrolled inflow but a largely controlled outflow. The water available for storage is totally a function of the natural stream streamflow that empties into it.


Reservoir Capacity Reservoir capacity is the volume of water that can be stored in the particular reservoir. It is the normal maximum pool level behind a dam. This can be calculated by using a topographic map of the region. First, the area inside different elevation contours is measured, and then a curve of area versus elevation can be constructed.

Figure . Area versus elevation for a reservoir

At any given elevation, the increment of storage in the reservoir at that elevation will be Ady, where dy is a differential depth. Then the total storage below the maximum level to any will be given by,

∫=y

AdyV0

Figure. Storage relations for a reservoir with an uncontrolled spillway

Sedimentation in Reservoirs All streams carry sediments that originate from erosion processes in the basins that feed the streams. After a dam is constructed across the team and a reservoir is produced, the velocity in the reservoir will be ne ing into the reservoir will settle down and be trapp ed with enough volume to

sgligible so that virtually all the sediment comed. There be designfore, the reservoir should


hold the sediment and still operate as a water storage reservoir over the project’s design life. For rge projects, the design life is often considered 100 years.

at a delta will be formed. The finer sedim nts will be deposited beyond the delta at the bottom of the reservoir.

la

Sediment carried in a stream is classified as either bed load or suspended load. The bed load consists of the coarsest fractions of the sediment (sands and gravels), and it rolls, slides, and bounces along the bottom of the stream. The finer sediments are suspended by the turbulence of the stream. When the sediment enters the lower velocity zone of the reservoir, the coarser sediments will be deposited first, and it is in this region th

e

The total sediment outflow from a watershed or drainage basin measured in a specified period is the sediment yield. The yield is expressed in terms of tons per square kilometer per year. The engineer designing a reservoir must estimate the average sediment yield for the basin supplying the reservoir to determine at what rate the reservoir will fill with sediment.

Figure. Deposition of sediment in a reservoir For a given reservoir volume, V, the ratio of the reservoir volume decrease due to the deposited sediment can be estimated by this empirical equation,

8.095.061023 ⎟

⎠⎞

⎜⎝⎛×= −

VAGR

ent yield of the basin (kN/km2/year) A = Drainage basin areaV = Reservoir volume (m )

ng R ratio with the design life of the reservoir, T, will yield the percentage of the dead olume in the reservoir. The dead volume can be estimated over the period of design life by,

f the eposit is ρ = 2.65 ton/m3 olume of the reservoir.

Where, G = Sedim

(m2) 3

Multiplyiv

reservoirdead VTRV ××=

EXAMPLE 4.3: The total volume of a reservoir is V = 230×106 m3 with a drainage basin of A = 1200 km2. The design life of the project is T = 100 year and the density (specific mass) od . Calculate the dead v


Solution: The sediment of the river for a river,

( )G 12001421×=

( )volumeyearkmm

A

//280

1421

23

229.0

229.0

=

=−

−

ease every year,

G

GThe ratio of the reservoir volume decr

( )%4.010230

7279000023.0 695.0

⎟⎟⎠

⎜⎜⎝ ×

××=R 101200

000023.0

8.06

8.095.0

=

⎞⎛ ×

⎟⎠⎞

⎜⎝⎛×=

R

VAGR

Reservoir volume decrease due to the sediment deposit every year,

dead

Useful storage is,

museful

Wind-Generated Waves, Setup, and Freeboard

ow over an open stretch of water, waves develop, and the mean level of the ater surface may change. The latter phenomenon, called setup or wind tide, is significant only

in relatively shallow reservoirs. When a dam is designed, the crest of the dam must be made higher than the maximum pool level in the reservoir to prevent overtopping of the dam as the

itional height given to the crest of the dam to ke care of wave action, setup, and possibly settlement of the dam (if it is earthfill) is called

acts on the water surface, and because of this, the surface will tilt, as shown by the roken line in the below Figure.

366 1092.010230004.0 mVdead ×=××=

For the 100 year of design period,

366 10921092.0100 m×=××= V100

( ) 66 101381092230 ×=×−= 3V

henever wind blW

w

wind-generated waves strike the face of it. The addtafreeboard. Setup Consider the basin of water shown in the figure. The solid line depicting the water surface is the case when no wind is blowing; the water surface is horizontal. When the wind is blowing, a shear stressb


Figure. Definition sketch for setup

he amount of setup S is,

T

KDFVS

2

=

D = Aver oi epth (m), V = the wind speed m the surface (km/h)

ind fetch (km) = A constant ≈ 62000

r is oval shaped with a length of 20 km and a width of 10 km. If the lengthwise to the reservoir with a velocity of 130 km/h, what will be

average water depth of the reservoir is 10 m?

W

here,

age reserv r deasured at a height of 10 m from

F = the wKS = Setup (m) EXAMPLE: A reservoiwind blows in a directionthe setup of the

Solution: The setup will be,

m

KDFVS

2

=

55.01062000

=×

Height of Wind Waves and the Run-Up

llowances for wave height and the run-up of wind-generated waves are the most significant omponents of freeboard. The run-up of the waves on the upstream dam face, i.e. the maximum

up a dam face, is equal to H (wave height) for a pical vertical face in deep water, but can attain values over 2H for a smooth slope 1 in 2.

201302 ×

=

Acvertical height attained by a wave runningty A wave height, H (m), (crest to trough) can be estimated by,

426.076.034.0 FFH −+=

here,

W


F = the fetch length (km), H = Wave height (m)

or large values of fetch (F>20 km), the last two terms may be neglected. With the provision ation takes the form of,

Ffor the wind speed, the equ

424.076.0032.0 FVFH −+=

V = Wind velocity (km/hour)

board will be equal to set-up plus run up allowance for settlement of the embankment lus and amount of safety (usually 0.50m).

XAMPLE: Calculate the wind set-up and wave height for a reservoir with 8 km fetch

the minimum freeboard to be given?

Where,

The freep Elength. The average reservoir depth is 15 m. The wind velocity is V = 100 km/h. If the upstream of the dam is vertical, what will be Solution: The wind set-up is,

m

KDFV 2

=S

09.01562000

=×

The wave height,

81002 ×=

mHH 8100032.0

=×=

FVFH

19.1824.076.0

24.076.0032.04

4

−+

−+=

Since the upstream side of the dam is vertical, the run-up height will be taken as the height of the wave. The freeboard,

5.019.119.109.0 +H freeboard +

mmH 00.397.2 ≅=

+=

.5 m is the safety height.

0


LECTURE NOTES – 5





CHAPTER 5

Main Types of High – Head Power Plant Developments

Power plants operating under a head higher than 50 m may be termed as high-head power plants. Three main types of high-head power developments may be discerned.

A) Diversion Canal Type Plant

Figure. General layout and profile of a high-head diversion canal development

The main parts of a high-head diversion canal type plant are:

1) the weir, 2) the canal intake, 3) the head race, 4) the headpond with spillway and gate or valve chamber, 5) the penstock, 6) the powerhouse, 7) the tailrace.


B) Plants Fed by a Pressure Tunnel

Figure. General layout and profile of a pressure tunnel development

1) the dam (sometimes only weir), 2) the intake or headworks, 3) the pressure tunnel, 4) the surge tank, 5) the penstock, 6) the power house, 7) the tailrace.

The tunnel intake should be located close to the bottom of the reservoir to ensure the greatest effective storage volume. Under certain topograhic and geologic conditions, the conveyance of water through a tunnel under the dividing range may, even in case of low dams, be preferable to building a long, meandering power canal. Since here the tunnel is not necessarily a pressure conduit, free-surface flow conditions may prevail therein. Such arrangements should be regarded as diversion canal type developments.

C) Plants with Concentrated Fall Developments, where the powerhouse is located close to, or within, a high dam or high-head river barrage, constitute the third main type of high-head installations. This arrangement, which could be termed plant with concentrated fall, or valley dam station, is essentially similar to that of low-head run-of-river plants. The head for the power station approaches the height of the dam.


Figure. Development with the powerhouse located at the toe of the dam (plant with concentrated fall)

The three main parts of this type of power plants:

1) the water intake (generally built on the upstream face of the dam), 2) the pressure conduit (generally transversing the dam body, sometimes bypassing the

dam adjacent rock), 3) the powerhouse.

The output of diversion canal developments is closely governed by the discharge available in the river, while the small storage capacity created by the low weir is sufficient to meet daily fluctuations in load only. This type may be called as high-head run-of-river plant. The other two types may be referred to as reservoir plants. Pressure tunnel developments are valuable for those fed from a large reservoir under high head. Free Surface Intakes Settling basins and sand traps are very important for high-head water power plants and they should designed on the basis of hydraulic computations. Suspended load, especially sharp edged fine sand transported by mountain streams causes rapid wear of the penstock and steel parts of the turbines. Water flowing at high velocity and carrying heavy sediment load attacks the lining of power canal and power channels. Scratch effects become generally more pronounced with increasing head, therefore, in case of heads higher than 100 m, sand should be carefully settled out and with heads higher than 200 m even the greatest part of silt fraction should be retained. The main parts of an intake are:

1) the inlet section including the sill and coarse rack, 2) the inlet gate and transition section, 3) the settling basin and sand flushing canal.


Figure. General arrangement of intake

Before the inlet section of the intake a bed load deflecting apron should be applied permitting a periodical flushing of bed load hold by the sill. The apron extends to the flushing gate of the weir. Protection against Silting In preventing entrance of bed load, or rather in promoting a desilting effect at the inlet section, in protecting both inlet and canal against sedimentation, the proper choice of intake site is of vital importance.


Changes in the angle of diversion (the angle between the outside wall of the intake structure and the direction of main flow) hardly affect the quantity of sediment entering the canal. More than 90% of transported matter enters the diversion canal branching off at an angle of 300 – 900 from the main course, yet carrying only half of the main discharge. At the same time it can be seen that the angle of diversion has no significant effect upon silting conditions.

The curve shown above illustrates the distribution of discharges and silt quantities for a 300 angle of diversion. A balanced (50 – 50 per cent) distribution of silt is attained only if some 25% of the original discharge is allowed to enter the canal. If the discharge carried by the diversion canal exceeds 60%, the entire sediment load enters to the branch canal. Hydrodynamical considerations will yield a very simple explanation for the above phenomenon, that in case of diversion from a straight stretch, the flow entering the power canal carries extremely great quantities of bed load. The velocity component opposite the inlet section and vertical to it, denoted by v, is due to the transverse head loss Δh, the latter given by the equation,

hgv Δ= 2μ

In one vertical along cross-section (x-x) velocities v due to Δh will be almost uniform. The velocity of direction x-x varies only in direction x within one section and shows an increase from the opposite bank towards the side intake. Since the velocity distribution within the main course above the point of diversion involves a velocity considerably smaller at point B than at the surface, it follows that the deeper the examined subsurface point is situated, the greater the angle formed by the resultant of velocities V and v and by the axis of main flow. Consequently, more water is drawn from deeper layers into the canal than from those nearer to the surface. Thus a lamination according to depth arises in the main flow causing the water entering the power canal to be drawn, for the greater part, from the lower layers that are heavily silted, while water in the upper layers containing considerably less silt overfalls the diversion dam and streams forth in the main course.


Figure. Velocity distribution in the river at the intake

The whole bed load is practically is carried into the power canal for big discharges, and the joint application of high sill, silt-sliding apron and sluiceway gives no substantial relief to the problem. Let us examine the bed load conditions of intakes located in a bend.

Figure. Profile of the water surface in bends

Considering that the water surface as a potential surface normal to the resultant of acting forces, the equation for the sloping water level in the curve may be developed as follows:

mgx

vm

dxdz

2

=


xdxvgdz 2=

Solving the above differential equation,

CLnxvgz += 2

Consequently, at point x = R1, z = 0. Thus,

1

21

2

RxLn

gvz

LnRvC

=

−=

The maximum rise of the surface occurring at the concave bend,

1

22

1

22

log43.02 R

Rg

vRRLn

gvh ××==

Velocity of flows shows a tendency to decrease along the same vertical towards the bottom. Particles of water moving in beds at and near the surface are thus subject to greater centrifugal force than those traveling near the bottom. Consequently, particles at and near the surface are forced towards the concave bank. An equal quantity of water is bound to follow at the bottom in the opposite direction towards the convex band due to the principle of continuity. Particles of water submerging with great velocity cause erosion of the bottom along the concave bank. Part and occasionally the whole of eroded matter is then deposited by the flow slowing down towards the convex bank. So the original rectangular cross-section takes and asymmetrical shown as shown in the Figure.

Figure. Development of spiral flow in bends

Silt of different particle size reaching the bend separates according to size at the peak of the curve. Fine silt settles fairly high on the sand bank formed under water along the convex bank, while coarser grains are carried forth and deposit mostly in or around the holes along the concave bank. It is advisable to have the power canal branch from the concave side in as much as relatively desilted water is required.


Figure. Separation of bed load according to particle size in bends

The intake structure is to be built at a point where the spiral flows is strong and the weir is to be located so that the sluiceway or the lateral opening of the movable gate system also falls within the sphere of spiral flow.

Figure. Spiral flow at the intake

Water discharging into branch does not diverge in a sharp angle but follows a curved route; spiral flow will develop at the upstream end of the canal too, should the diversion be placed either in a straight or in a curved stretch of the water course. Surface flows tend towards the concave side of the curved streamway caused by the diversion, while bottom flow transporting debris is directed towards the canal. With a 50-50 percent ratio of discharges, distribution of bed load is as follows and given in the Figure.

1. Bed load: Canal 100%, main water course 0%, 2. Bed load: Canal 50%, main water course 50%, 3. Bed load: Canal 5%, main water course 95%, 4. Bed load: Canal 100%, main water course 0%, 5. Bed load: Canal 0%, main water course 100%.


Figure. Typical shapes of diversion

No. 5 proved most unfavorable as here silt transportation into canal is intensified by synchronizing spiral flows in both original water course and branch canal. No.1 is also highly unfavorable. No. 2, 3, and 5 may be regarded as favorable. No. 2, diversion is at the upstream end of the curve where spiral flow is not yet fully developed and so the effect of flow in the original water course is largely decreased by spiral flow in the branch canal. No 3. and 5 are the most favorable, as here fully developed spiral flow at the downstream end of the bend cannot be considerably lessened by spiral water movements of opposite direction of the branch. The following basic principles governing selection of the intake site can be suggested:

1. Intakes should be located, whenever possible, on the concave side of a curved stretch, 2. Efficiency of the intake in preventing sedimentation increases with the sharpness of

the bend, 3. The amount of bed load transported into the canal decreases, as the ratio of the total

discharge to the amount increases. 4. Intakes are most favorably located along the downstream reach of the curve, near the

end. 5. The lower the head, the more effective the intake.


6. Conditions in a straight stretch are opposite to those described under No. 3; with diverted flow being constant, any increase in the river discharge will involve more extensive sedimentation in the canal.

7. The silt releasing sluice of the diversion weir, the canal sill and the desilting sluice can only be operated at good efficiency if more or less favorable bend conditions are created through proper design and arrangement in keeping with the above principles.

8. With intakes from straight stretches, but more so along the convex side, the afore- mentioned measures offer no significant contribution to the protection against sedimentation in the canal. In such cases both canal sill and desilting canal give satisfactory results if heads are considerable even during high-head periods.

9. Intakes from straight stretches can be made more favorable by forcing water to follow a curved route with the convex side of stream curve facing the intake structure.

Figure. Intakes from straight stretches with bed contraction, a) with flushing (desilting) canal, b) without flushing canal

This can be achieved by arrangements illustrated in the above Figure, where an inlet section extending crosswise into rive bed, and a weir shorter than the width of flow above the intake make the flow to follow a curved route.

10. The quantity of bed load can be reduced by a longitudinal baffle wall as shown in the below Figure, if Qd < QB where Qd is the diverted discharge and QB the discharge conveyed in the river in width B of the intake. As a result of inequality Qd < QB part of the water is compelled to deviate on a curved path from the bank, thereby bringing about a silt-diverting spiral flow.

B


Protective measures against bed load may be completed by a few remarks.

a) The minimum discharge Q0 capable of inducing bed load movement will be decisive of the choice of the site and the arrangement of the intake. The ratio of the plant discharge capacity Qp and of the above limit discharge Q0 is the one of the factors governing the design of intake.

b) The individual features of an intake may require the discharge diverted into canal, in certain periods, less than available in the river for power generation.

c) Quantitative relations may be established as to the reduction of the discharge diverted. The degree of reduction depends upon the character of the river section.

d) Special care should be taken if a power plant having a relatively great discharge capacity is projected on a mountain river. The necessary reduction of the discharge may in this case permit the diversion of a volume corresponding to the maximum plant discharge capacity at times of flood only.

e) On mountainous rivers carrying a heavy sediment load, the intake works should always be located on the concave side of the bend even if this side is otherwise less favorable.

f) Flow velocity in the inlet section of the intake should preferably be 0.75 m/sec on the average as indicated by experiments conducted between velocity limits of o.50 and 1.10 m/sec.

Care should be taken during the hydraulic designing of the settling basin to ensure the calculated velocity in the structure would range 0.40 to 0.60 m/sec. The hydraulic design in settling basins is broadly outlined in the following paragraph.

1) Exploration of sediment conditions, involving the quantitative and qualitative analysis of sediment carried by the river. In mountain rivers or in steep, upper river sections at the average sediment concentration varies from 2 to 10 kg/m3.

2) Following the investigations of sediment conditions, the necessary degree of load removal, should be determined. Attempts have been made to approximate operating requirements by specifying the diameter of the smallest particles to be settled out (limit particle size). At medium-head plants, the removal of particles larger than 0.2 – 0.5 mm is usually specified. Instead of using the limit particle size, the degree of removal is frequently defined by the removal ratio of concentrations after and before settling expressed in percentages. If the concentration of raw water is C, and that of the clarified water is specified as the permissible value Cp, the required removal ratios is obtained as,

CC p100

By specifying or assuming the limit particle size, the removal ratio may easily be calculated.

3) Having determined the basic data as suggested, design can be done. First the settling velocity of the smallest fraction; i.e., of the limit particle size to be removed should be calculated theoretically or be established by tests.


The so-called horizontal-flow settling system is usually applied at power developments. For this system, the dimensions of the settling basin may in principal be determined by two computation methods. The effect of turbulent flow upon settling velocity is neglected in the simple settling theory. Three basic relations may be written for the determination of the required basin length. Denoting the depth of the basin by h and its width by b, the discharge passing through basin is,

bhVQ = (m3/sec)

Where V is the flow velocity. The second equation expressing the relation between the settling velocity w, the depth of the basin and the settling time t,

wht = (sec)

Finally, the length of the basin will be governed by the consideration saying that water particles entering the basin and sediment particles conveyed by them with equal horizontal velocity should only reach the end of the basin after a period longer than the settling time. Thus even the smallest settling particle may strike the bottom of the basin within the settling zone. The retention period should not be shorter than the settling time. The required length of the basin is,

Vtl = (m)

Eliminating t from the last two relations will be established between the six values governed the hydraulic design:

hVlwbhVQ

==

A solution of the problem is not possible unless four quantities are known. The discharge Q can always be considered given, the settling velocity w is defined by the initially specified degree of removal and can be established by calculation. The highest permissible flow velocity should also be specified in order to prevent particles once settled from picked up again. The actual flow velocity should not exit this limit, whereas excessive dimensions computed by substantially lower velocities would again result in uneconomical design. Velocities higher than the permissible velocity tend to scour the material settling to the bottom, which may even become suspended again. This limit velocity may in fact be considered equal to the theoretical suspending velocity, or to the critical velocity encountered in the theory of sediment transport. The critical velocity,

daV = (cm/sec)

Where d is the diameter of particles in mm and the constant a:


a = 36, for d > 1 mm, a = 44, for 1 mm > d > 0.1 mm

a = 51 for 0.1 mm> d

The fourth dimension that can be assumed in advance is one of the main dimensions of the basin. The depth of the horizontal flow settling basins employed in water power projects is generally between 1.5 and 4.0 m with velocities not higher than from 0.4 to 0.6 m/sec. The water mass conveyed during settling time should equal the capacity of the settling basin. Owing to the retarding effect of turbulent flow on sinking particles, settling is slower in flowing water. By using a lower settling velocity (w – w’), the reduction in settling velocity w’ to be closely related to the flow velocity,

aVw =′ (m/sec)

The coefficient a may be computed from the relation,

ha 132.0=

Where h is the water depth in m. The settling length is therewith,

VhVh

aVwhVl

132.021

23

−=

−= (m)

A negative denominator is an indication of the fact that no settling can be attained under the assume conditions. The computation should be repeated using the modified dimensions.


The most important factors affecting the design of the settling basin are the quality of sediment (specific weight and shape of particles) density of water carrying sediment and water temperature. All estimates involving the direct application of the afore-mentioned data should be regarded as approximate only. The necessary settling length for turbulent flow is computed from the settling velocity in stagnant water w and from the flow velocity. The settling length,

( )2

222

51.72.0

whVl −

=λ (m)

Where λ depends on the removal ratio defined previously. Values of λ defined by the function,

( )wf=λ

W denotes the ratio of settled sediment to the total load entering with the flow and can be computed from the afore-mentioned removal ratio as follow:

CCw P100100−=

The settling velocity pertaining to the limit particle size of the fraction to be settled out without assuming 100% removal. Satisfactory values can be obtained by using coefficients pertaining to a 95 to 98% removal of the limit particle size.


Continuous operation can be ensured by one of the following arrangements:

a) Series of basins, some of which can be flushed while others operating, b) Permanent operation of basins can also be realized by continuously flushing

settled sediment. An inflow exceeding the water demand by 10% should be admitted into the basin and sediment accumulating at the bottom can be flushed continuously by discharging the excess water to waste.

Example: Design a settling basin for a high-head power plant by using the settling theory. The basin should serve to remove particles greater than 0.5 mm diameter from the water conveying mainly sand. The design discharge is 5 m3/sec and assume an initial value of 3.20 m for the depth of the basin. Solution: Determine first the permissible velocity flow velocity. Owing to economical considerations this should equal the critical velocity for which,

sec2.315.04444 cmdV ===

In designing the basin, V = 30 cm/sec flow velocity will be used. The following step is to determine the settling velocity according to the limit particle size of 0.5mm to be removed. From the settling velocity – particle size Figure, w = 6 cm/sec (for γ = 1.064). The required length of the basin is,

mwVhl 16

63020.3 =×==

And the width,

mhVQb 21.5

3.02.35

=×

==

Checking: The settling time is,

sec4.5306.020.3

===wht

The discharge conveyed during this period is,

32674.535 mQtV =×== Should be equal to the capacity of the basin;

32670.1621.520.3 mhblV =××==

Determine the length of the basin using identical basic values by the method of Velikanov’s Figure for a removal ratio of 97% (W=0.97). The Figure yields λ = 1.50 for W = 0.97. The length of the basin is,


( )

( ) ml

whVl

1906.051.7

2.02.33.05.1

51.72.0

2

222

2

222

≅×

−××=

−=λ

Example: Compute for the conditions of the preceding example the settling length by considering the retarding effect of turbulent. Solution: The coefficient governing the reduction of settling velocity is,

0737.020.3

132.0132.0===

ha

And thus the velocity decrement,

sec0221.030.00737.0 maVw =×==′

The settling length,

mww

hVl 30.250221.0060.030.020.3

=−×

=′−

=

The unchanged width of the basin is,

mhVQb 21.5

30.020.30.5

=×

==

And its capacity,

342230.2521.520.3 mhblV =××==

Example: Compute the modified dimensions for a reduced depth of 2.40 m. Solution:

0851.04.2

132.0132.0===

ha

maVw

hVl 90.2030.00851.006.0

30.040.2=

×−×

=−

=

Width of the basin is,

mhVQb 95.6

30.040.25

=×

==


And the reduced capacity, 33489.2095.640.2 mV =××=

Example: A power plant is fed by a river carrying very coarse suspended sediment load. As indicated by the gradation curve obtained for the sediment, 70% are held on the 1 mm screen. In order to protect the turbines the entire over 1 mm diameter should be settled. Solution:

%30100 =CC p

The basin will be designed for a discharge of 12 m3/sec with the retarding effect of turbulent and a depth of 2.80 m will be taken. The critical velocity is,

sec4414444 cmdV =×==

The settling velocity in stagnant water is obtained from the Figure (for γ = 1.064) W = 10 cm/sec. The settling velocity decrement due to the turbulent,

sec0346.044.08.2

132.0132.0 mVh

W =×==′

The actual settling velocity is,

sec0654.00346.0100.0 mWW =−=′−

And the settling length,

mWW

hVl 85.180654.0

44.080.2=

×=′−

=

The required width of the basin is,

mhVQb 74.9

44.080.212

=×

==

A settling basin 20 m long and 10 m wide will have a capacity of,

35608.22010 mV =××=

Compute the length of the basin also by the equation of Velikanov (W = 0.97), λ = 1.50,

( ) ml 50.1210.051.7

2.08.244.050.12

222

=×

−××=



LECTURE NOTES – VI





Chapter 6

The Power Canal

On gentle hill slopes, but especially on steep mountain sides, the canal should closely follow the contour lines of the area. Over sufficiently uniform area, the power canal may be designed with an open cross-section through cuts, overfills and in cut-and-fills as shown in the Figure.

On mountainous slopes it may not be possible to follow the irregular contour lines: deep valleys are to be bridged by aqueducts (such as elevated canals or canal bridges), and high hills crossed by water conveying tunnels. Cross-sections of canals located on steep, hilly mountain slopes are shown in the Figure.



Although a canal located according to these principles involves the construction of relatively more expensive structures, such as bridges and tunnels, the resulting route may still be more economical than that strictly following the contour lines than that strictly following the contour lines of the hilly area, because;

1. The length of the canal will be significantly reduced, 2. The head loss will also be smaller.

It should be kept in mind that geological conditions of the area decisively influence the location of the canal. In order to establish reliable bases for following the alignment and determining the cross-sections, the geologic formations, the dip of layers, the quality of rock should be explored very thoroughly over sections in cut and fill. The proper solution will be governed by,

a) The permissible slope of banks and embankments, b) The depth of the cut, respectively the height of the fill, c) The dimensions and foundations of canal walls, backfills if required, d) The extent and quality of lining.

The construction of open power canals may meet difficulties if,

1) The mountain slope is not stable, 2) The mountain slope is too steep, 3) The mountain slope above the canal is likely to produce much rubble, 4) Snow avalanches are to be expected, 5) During extremely severe and long winter periods the water freezes in the canal.

With the above conditions (occurring separately or simultaneously), the following solutions can be applied in the critical sections;

A) Types of open or closed canals may be constructed of concrete or reinforced concrete, B) The water may be conveyed in tunnels. In order to reduce head losses, such tunnels

should be designed for free surface flow conditions.

Figure. Free-flow tunnels


Thickness and quality of lining in free flow tunnels are governed by rock pressure, hydraulic requirements of water conveyance and water tightness and sometimes by the aggressivity of the waters conveyed. Internal water pressures generally need not be considered. Adequate protection against snow avalanches and freezing can be provided by a closed canal located on the area or even by a simple cover over a canal.

Figure. Closed and covered power conduits


LECTURE NOTES – VII





Chapter 7

The Headpond

The purpose of the headpond (or forebay) is to distribute evenly the water conveyed by the power canal among the penstocks and, at the same time, to regulate the power flow into the latter, as well as to ensure the disposal of excess water. At the headpond the power conduit widens into a basin and thus a part of the suspended sediment still carried by the water settles down. The storage capacity of the headpond tends to drop of water level in case of sudden load increase. Headponds having a storage capacity may even provide daily storage for the plant.

Figure. General arrangement of the headpond

Parts of the headpond are;

1. The basin, 2. The spillway (sometimes of the siphon type), with the overflow weir, 3. The bottom outlet which is generally flushing sluice gate for sediment, 4. The sill equipped with a screen, 5. The gate (valve) chamber, 6. The penstock inlet.


The bottom of the headpond is governed primarily by topographic conditions, the geology of the site should be considered. The site of both the headpond and the powerhouse should be selected simultaneously with a view to ensure the shortest possible penstocks. Parts of the map where the contour lines are close to one other and closely follow the banks of the river course should be investigated as potential sites. The power canal should join the headpond over a gradual transition and the bottom of the basin should have a slope towards the sill. A bottom lining of the basin is indicated only in soils where seepage is to be expected. The spillway is usually an ogee type weir located in the valley side retaining wall of the basin with a sufficient length to discharge the entire full load water supply with a small increase in the basin level. The spillway and the bottom outlet canal should be united immediately at the foot of the basin. Water spilling over the spillway crest and through the bottom outlet can be,

a) Diverted into a suitable river bed in a nearside valley, b) Conveyed by a special chute.

The bottom outlet is controlled by a vertical lift gate. Sediment accumulating before the sill can also be flushed through the bottom outlet. If flushing is not of the desired efficiency and the amount of the remaining in the basin is significant, this should be removed mechanically. The entrance of accumulated sediment into the penstocks is prevented by a sill and a screen. At high head installations the width of the opening is from 15 to 50 mm, depending upon the type of the turbine. Flow velocities related to the overall area of the screen may vary between, V = 0.80 – 0.25 (m/sec) at high-head power plants. When the water is free of sediment load, flow velocities through the gross screen area may have a value as high as 3 to 3.5 m/sec. Screens may either be extended uninterrupted over the entire inflow cross-section, or be arranged between the piers separating the penstock intakes. Flow to the pressure conduits is controlled by vertical lift gates located in the gate chamber. Gates are operated by electric remote control from the switch room of the powerhouse and also directly from the gate house. The gate should close automatically in a case of turbin stop or penstock failure.


LECTURE NOTES – VIII





CHAPTER 8

The Tunnel

Pressure tunnels may be classified according to the head above the arch of the tunnel.

Correspondingly, tunnel may be grouped into,

a) Low-pressure tunnels, with H lower than 5 m, b) Medium-pressure tunnels, with H from 5 to 100 m, c) High-pressure tunnels, with H higher than 100 m.

According to another classification,

a) Unlined, b) Lined, either for structural purposes, or for the purpose water sealing.

Structural linings are called upon to carry the rock pressure and to offer protection against rock splitting from the tunnel roof. Full circular linings, in addition to being capable of resisting external loads, are suitable for;

a) To take internal water pressure, b) To prevent water losses, c) To protect the rock against the aggresivity of conveyed water.

In case of low-pressure tunnels, the trimmed rock may be left unlined except for visible fissures which may be sealed with concrete or cement. In order to reduce hydraulic head losses, rock surfaces should be trimmed smooth or coated with a friction-reducing concrete layer.


A watertight lining is usually required for tunnels operating under medium heads. Seepage is more likely to occur as the head increases. Simple cement-mortar coatings are seldom satisfactory and watertight concrete linings have to be applied in most cases. If the tunnel is unlined, or if the lining serves only for water sealing purposes, i.e., carries no load, the permissible water pressure is determined. h = The depth of overburden over the arch, γ1 = Specific weight of the rock. γ = Specific weight of the water

hpV 11.0 γ= (kg/cm2)

And using a safety factor n, the permissible internal water pressure is,

npp V=

Since,

Hp γ1.0= (kg/cm2)

The permissible head (static and dynamic), with γ = 1 ton/m3,

hn

H

HnHHnnppV

1

1 1.01.01.0

γγγγ

=

===

Practical values for the safety factor n are from 4 to 6. The lower limit should be used for greater depth of overburden and for sound rock on the arch, whereas in case of a shallow cover and poor rock the upper limit is used. Consequently, with the specific weight of rock varying from 2.4 to 3.2 ton/m3, the permissible head in meters related to the depth of overburden above the arch yields,

H = (0.4 to 0.8) h

In pressure tunnels operating under high heads, linings of plain concrete and sometimes even of reinforced concrete are not satisfactory. Steel linings are used.

To reduce construction costs, relatively high flow velocities should be permitted in tunnels, higher than allowed in open channels. Suggested limit velocities,

Very rough rock surfaces → 1.0 – 2.0 m/sec Trimmed rock surfaces → 1.5 – 3.0 m/sec

Concrete surface → 2.0 – 4.0 m/sec Steel lining → 2.5 – 7.0 m/sec


Figure. Steel-lined pressure tunnel

The minimum size of tunnels of circular cross-section is about 1.80 m in diameter. In case of lined tunnels, the computed cross-section should be increased by the thickness of the lining. Friction losses may be calculated by the manning equation,

LRnVhL 34

2

=

L = Length of the tunnel (m), V = Mean flow velocity (m/sec), R = Hydraulic radius (m) n = Manning roughness coefficient.


LECTURE NOTES – IX





Chapter 9

Hydrodynamic Requirements of Intake Design

In designing the general layout of intake structures, the choice of correct intake direction, i.e. of the angle enclosed by the outside wall and the original direction of flow is very important.

Figure Flow pattern at diversion at sharp angle

The angle of inflow varies with the ratio of the diverted discharge to the original. If the ratio of the diverted discharge to the original is small, the angle of diversion will be small, while any increase in the diverted discharge, the angle of diversion will also be wider. The following general rule can be established: in designing intake structures, the inflow angle occurring at times of relatively small discharges should be followed.

⎟⎟⎠

⎞⎜⎜⎝

⎛=

f

c

QQfψ

Figure


Qc = Canal discharge, Qf = Remaining discharge in the watercourse.

=f

c

QQ 0.20 – 0.30 → ψ = 200 - 300

0.1=f

c

QQ → ψ = 450 - 600

The diverted discharge is conveyed by a cross-section of width B in the watercourse having a total width B0.

ψψ

costan bCAB ==

b = Width of the canal, ψ = Angle of diversion, Qc = Discharge in the power canal.

c

f

cfc

qq

Bb

bqBqQ

==

==

ψcos

In case of intake from a reservoir, Vf ≈ 0, qf ≈ 0,

cosψ = 0 → ψ = 900

The greater the velocity Vf and the discharge qf, the smaller the angle of diversion is required. The expression derived by D.Y. Sokolov, corrected angle of diversion,

c

f

f

cf

qq

hhh⎟⎟⎠

⎞⎜⎜⎝

⎛ −−= 1cos εψ

==bbcε Coefficient of contraction at the entrance to the intake canal,

hf = Water depth in the river, hc = Water depth in the power canal. bc = Contracted width of flow in diversion canal. Model tests are recommended. The required area of the inlet section is computed from,


VQ

A p=

Where the mean inflow velocity is assumed as, V = 0.8 – 1.2 m/sec. Intake Headlosses

1. Entrance Loss

Entrance losses are due to two phenomena;

a) Velocity Vf of the flow above the intake site changes to the value V in the inlet section,

b) Sudden contraction of the cross-section causes headloss.

Figure

Maximum entrance loss is,

gV

gVh f

e 223.1

22

ε−=Δ

Factor ε varies from 0.8 to 0.4 depending on the intake angle ψ. The greater values can be assumed in case of intakes under a sharp angle (~300) while the smaller ones are applicable in case of rectangular diversions. In case of intake from a reservoir, Vf ≈ 0,

gVhe 2

3.12

=Δ


In case of a sharp angle diversion and for V ≈ Vf, the minimum value,

gVhe 2

5.02

≅Δ

2) Rack (Screen) Losses For computation of rack losses, O. Kirschmer`s equation may be used.

αϕ sin2

234

gV

bshr ⎟⎠⎞

⎜⎝⎛=Δ

s = Width of rack (screen) bars, in m (or cm), b = Spacing (clearance) between bars, in m (or cm), V = Velocity of flow in front of the screen (m/sec), α = Angle of bars with the horizontal.

Figure

Values of factor φ for different bar cross-sections;

Cross-section φ a 2.42b 1.83c 1.67d 1.03e 0.92f 0.76g 1.79


Example: Calculate the resistance of a screen with an inclination of 750, where the thickness of bars s = 6.2 mm, the spacing between the bars b = 19.2 mm. The bars are of rectangular cross-section. Velocity of flow in front of the bars is V = 1.15 m/sec. Solution:

cmmh

h

gV

bsh

r

r

r

50.30349.0

75sin62.19

15.12.192.642.2

sin2

0234

234

≅=Δ

××⎟⎠⎞

⎜⎝⎛×=Δ

⎟⎠⎞

⎜⎝⎛=Δ αϕ


LECTURE NOTES – X





Chapter 10

Permissible Velocities in Canals

Conditions of Stable Regime

The maximum permissible velocity in open channels will be limited by the resistance of the bed material to erosion or, in case of lined canals, by that of the lining against wear. Maximum Permissible Velocities Some researchers relate permissible bottom velocities to the material of the bed and sides or/and lining, while others suggest values for the permissible mean velocity. The maximum bottom velocity for erosion is given by Sternberg as,

dVb 2ξ= (m/sec) d = Diameter of particles (m) ξ = 4.43 Erosion velocities for various soil particles are given in the Figure after W.P.Craeger and J.D.Justin. The range of maximum permissible mean velocities is given for different grain diameters varying from fine clays to gravel of medium fineness (0.001 – 10mm).


Abscissa represent average particle sizes in millimeters, while mean velocities on the ordinate axes are in m/sec. Functions of both particle size and specific weight of the soil are more accurate.

19.22 194 −= γmdV (m/sec)

dm = Effective size (cm) γ1 = Specific weight of the material (gr/cm3) Maximum permissible mean velocities for loose granular bed material;


In case of depths other than 1 m, velocities will be corrected by,

1VV α=

Maximum permissible mean velocities for solid rocks and α correction factors,

Maximum permissible velocities for cohesive soils are given in the Table. These values can be corrected for the hydraulic radius, R > 3 m by,

1.0

3⎟⎠⎞

⎜⎝⎛=

Rα

If the bed is covered by aquatic growths, mean velocities from 0.8 to 1.8 m/sec can be used.


Lowest Permissible Velocities In order to prevent settling of silt suspended in the water, lowest permissible velocities should also be determined. According to A.Ludin, no sedimentations is likely to occur if the mean velocity,

V > 0.3 m/sec in case of silty water,

V> 0.3 – 0.5 m/sec in case of water carrying fine sand

Silt-Load Carrying Capacity E.A. Zamarin’s empirical equation gives the silt-load capacity as,

wRSV

wVG

00 700= (kg/m3)

G0 = Silt-load carrying capacity of the water (kg/m3), V = Mean flow velocity (m/sec), R = Hydraulic radius (m), w = Mean settling velocity (in still water) (mm/sec), S = Water surface slope w0 = w if w > 2 mm/sec, but w0 = 2 mm/sec, if w ≤ 2 mm/sec. If the silt discharge in the water is smaller than the silt-load carrying capacity of the canal, G < G0, no silting will occur. The above equation applies only to unlined canals free from aquatic growths, to discharges from 0.2 to 150 m3/sec and if V > 0.3 m/sec, w < 10 mm/sec and G < 5 kg/m3. For approximate values of permissible velocities M.M. Grishin suggest the equation,

2.0AQV =


With the following values of coefficient A;

w (mm/sec) < 1.5 1.5 – 3.5 > 3.5A 0.33 0.44 0.55

Defining d0 as the decisive size of screen opening on which 25% of the weight of the natural bed material will be hold. The simple relationship between the diameter d0 (cm) thus defined and the limiting tractive force Tf (kg/m2) is plotted in the Figure.

Curve A represents the limit state of degradation, while curve B should be used for safe design. Considering the two curves more closely, the relationship can be reduced with sufficient accuracy to,

( ) ( )cmdmkgTf 02 ≅

Computing the value of tractive force in (kg/m2) from the equation,

hShSTf 1000== γ (kg/m2)

h = Water depth, S = Slope of water surface. The screen opening parameter d0 in cm concerning to the scaling in the limit state of erosion under the action of the given force. This force Tf is, therefore, referred to as the limiting force for the bed constructed in the material characterized by the parameter d0. Multiplying the decisive parameter considered to the limiting force by 1.25, the curve B suggested for design purposes is obtained by the safety factor η = 1.25.


( ) ( )( ) ( )cmdmkgT

cmdmkgT

f

f

02

02

8.0

25.1

≅

≅

Figure. Flushing of the power canal

Installing a gated spillway of suitable arrangement instead of an overfall weir, a periodical flushing of deposited silt from the power canal and especially from the downstream reaches can be accomplished. In long canals several sluices may be built for this purpose. A water depth exceeding 1.5 – 2.0 meters and a mean velocity of not less than 0.50 m/sec will be sufficient to prevent the growth of plants. Example: A power canal with dimensions h = 2.50 m, R = 1.76 m, and S = 0.0001 has a bed load in the original water course as G = 0.34 kg/m3. The mean settling velocity for a 0.08 mm diameter grain in still water is found to be w = 4.5 mm/sec. The canal is unlined and the average particle size of the bed material is dm = 2 mm. Solution: Using Table I/27, for an average particle size dm = 2 mm, the maximum permissible mean velocity corresponding to a water depth of 1 m, V1 = 0.6 m/sec. The correction coefficient for h = 2.5 m depth is α = 1.20. The maximum permissible velocity is,

( )sec72.060.020.11 mVV =×==α

G0 silt-load carrying capacity is,

wRSV

wVG

00 700=

V = 0.60 m/sec, w = 4.5 cm/sec, w0 = w (because w > 2 cm/sec), R = 1.76 m, S = 0.0001,

30 45.0

5.460.00001.076.1

5.460.0700 mkgG =

××××=


Since,

0.72 m/sec > 0.60 m/sec

0.34 kg/m3 < 0.45 kg/m3

Neither erosion nor silting of the canal is to be feared.


LECTURE NOTES – XI





Prof. Dr. A. Bulu 1

Chapter 11

The Surge Tank

The surge tank is located between the almost horizontal or slightly inclined conduit and steeply sloping penstock and is designed as a chamber excavated in the mountain.

Surge tanks serve as a threefold purpose;

1. Upon the rapid closure of the turbine, water masses moving in the pressure tunnel and in the penstock are suddenly decelerated. Owing to the inertia of moving masses, F = ma, high overpressures develop at the lower end of the penstock, which are propagating upwards in the penstock in the form of pressure wave. The magnitude of the so-called water hammer, caused by the moving masses by closure, will depend upon the dimensions and elastic properties of the conduit. The overpressure due to water hammer travels along the closed conduit and is not relieved until a free water surface is reached.

An important function of the surge tank can be summarized like this. The turbines to the reservoir is practically interrupted by the surge tank to prevent the pressure wave due to the water hammer at the free water surface and to free the pressure tunnel from excessive pressures.

2. The surge provides protection to the penstock against damage of water hammer. The overpressure depends upon the length of the penstock (the closed conduit). The surge tank, by interrupting the closed system of the penstock and of the pressure tunnel, reduces the overpressure due to water hammer.

3. The third purpose of the surge tank is to provide water supply to the turbines in case of starting up. The amount of water required during these changes in operating conditions is supplied by the surge tank installed in the conduit. The capacity thereof should be selected to ensure the required water supply during the most unfavorable increase in demand, until the water mass in the tunnel

Prof. Dr. A. Bulu 2

has attained the necessary velocity. Air should be prevented from entering the penstock even in case of the deepest downsurge in the chamber.

The height of the surge tank is governed by the highest possible water level that can be expected during operation. Variations in demand initiated by a rapid opening or closure of the valve or turbine are followed with a time lag by the water masses moving in the tunnel. Upon the rapid and partial closure of the valve following a sudden load decrease, water masses in the penstock are suddenly decelerated, and one part of the continuous supply from the tunnel fills the surge tank. The water surface in the surge chamber will be raised to above static level. In case of rapid opening, the flow in the tunnel is smaller than the turbine demand to supply water to the turbine. The water surface in the chamber will start to drop to below of the steady-state level. To establish steady-flow conditions, the water surface will again start to rise from the low point, but owing to the inertia of moving water, will again rise over the steady-level. The cycle is repeated all over again with amplitudes reduced by friction, i.e. the oscillation is damped. The phenomenon described is the water surface oscillation. The maximum amplitude of water surface oscillation can be observed when the water demand is suddenly stopped. A wide variety of types has been developed in practice for the surge tank. According to the hydraulic design, the following groups can be distinguished.

1. Simple surge tanks designed as basins, which may be provided with overfall. 2. Special surge tanks:

Surge tanks with expansion chambers, which may be provided with overfall. Surge tank with upper expansion chamber. Surge tank with lower expansion chamber. Double-chamber surge tank.

3. Restricted-orifice type (throttled) surge tanks: Simple restricted-orifice surge tank. Differential (Johnson type) surge tank. Double-chamber, restricted-orifice surge tank.

Prof. Dr. A. Bulu 3

Prof. Dr. A. Bulu 4

Water Surface Oscillations in Simple Surge Tank

The oscillating movement starts as soon as the pressure wave due to a change in the turbine reaches the surge tank after traveling the length penstock.

Placing manometers at the upper and lower end of the penstock, it will be seen that the two react differently to sudden changes in turbine discharges. The lower manometer will be the first to indicate the pressure wave starting from the lower end of the penstock. The upper manometer will indicate the low-frequency oscillations and will also show the water level fluctuations at the same cycle with the surge tank. Waves are damped by roughness conditions.

Figure. Undamped oscillations in the surge tank if frictionless conditions are assumed in the pressure tunnel

Prof. Dr. A. Bulu 5

In the pure theoretical case when no friction is assumed to occur in the pressure tunnel, the water level in the surge tank is on the same elevation as the reservoir whatever the discharge of the system is. Therefore, hydrostatic and hydrodynamic levels are identical. The axis of the undamped oscillation is the hydrostatic (and at the same time hydrodynamic) equilibrium level. The penstock is supplied through a surge tank from the frictionless pressure tunnel. The reservoir level may be considered unchanged. F = Surge tank cross-sectional area, f = Pressure tunnel cross-sectional area, l = Pressure tunnel length. It will be assumed that the time of opening or closure turbine valves is zero (instantaneous). With the above fundamental assumptions, the expressions for the four basic cases are given without derivation. 1. Instantaneous total closure from the maximum discharge of Q0 (so-called total load rejection). It is evident that the total closure at maximum turbine discharge results in the greatest possible surges. This highest value of the ymax surges occurring in the tank upon rejection of different loads will be distinguished by the notation Ymax. The flow velocity in the

pressure tunnel for the discharge Q0 is f

QV 00 = .

The absolute value of the widest amplitude in case of the undamped mass oscillation, i.e. the so-called maximum surge is,

gFlfVY 0max = (m)

The departure of the water level from its initial position at any arbitrary time t (considering the downward branch of the axis y as positive);

tT

SinYy π2max−= (m)

The varying velocity of water flowing in the pressure tunnel at any time t is,

tT

CosVV π20= (m/sec)

At the time t = T/4 (quarter period), the velocity in the tunnel is V = 0, the direction of the flow in the tunnel changes. The velocity of the water level in the surge tank is,

tT

CosVFf

dtdy π2

0−= (m/sec)

Prof. Dr. A. Bulu 6

The time of the total cycle, i.e. the period of the mass oscillation is,

gflFT π2= (sec)

Example: The pressure tunnel length is l = 10 km with a cross-sectional area of f = 10 m2 and steady flow velocity V0 = 2 m/sec at a hydroelectric power plant. Cylindrical surge tank cross-sectional area is F = 100 m2. In case of instantaneous closure, compute the maximum surge height and the period of the oscillation assuming the ideal fluid (frictionless). Solution: Maximum surge height,

mgFlfVY 20.20

10081.9101000020max =

××

×==

The period of mass oscillation,

sec6401081.91001000022 ≅

××

== ππgflFT

Velocities at the maximum surge height in the tunnel and the surge tank are,

01606403602

10010

016064036022

sec1604

6404

0

=⎟⎠⎞

⎜⎝⎛ ×××−==

=⎟⎠⎞

⎜⎝⎛ ××==

===

CosdtdyU

CostT

CosVV

Tt

π

The water will stop at t = T/4 time for the maximum surge case and will begin to drop in the tank.

2. The surge amplitude in case of partial instantaneous closure, from the maximum discharge Q0 to an arbitrary Q1 value is,

( )gFlfVVY 10 −= (m)

Where, V1 = Q1/f is the velocity for the reduced discharge. The position of the water level at any time t is given by the expression,

Prof. Dr. A. Bulu 7

tT

YSiny π2−=

Velocity in the pressure tunnel is,

( ) tT

CosVVVV π2101 −+=

Velocity in the surge tank is,

( ) tT

CosVVFf

dtdyU π2

10 −−==

The period of oscillation is also,

gflFT π2= (sec)

Example: Using the values given in preceding example, compute the maximum surge for the closure from maximum discharge Q0 to 0.5Q0. Solution: The discharge of the full load,

( ) ( ) mgFlfVVY

mV

mQQ

mfVQ

10.1010081.9

101000012

sec11010

sec105.0

sec20102

10

1

301

300

=××

×−=−=

==

==

=×==

The period of oscillation will not change.

3. Oscillations for the instantaneous partial opening from some discharge Q1 to the maximum Q0 (partial load demand) are given by,

( )gFlfVVY 10 −=

The momentary position of the water leveling the surge tank is given by the function,

tT

YSiny π2=

Prof. Dr. A. Bulu 8

Velocities can be computed from the following relations,

( )

( ) tT

CosVVFf

dtdyU

tT

CosVVVV

π

π

2

2

10

100

−==

−−=

The oscillation period equation is the same.

4. The instantaneous total opening from the rest (Q = 0) to the maximum discharging capacity of the turbines Q0 (total load demand) can be characterized by the following relations.

The maximum surge is equal to the value obtained for total closure,

gFlfVY 0max =

The movement of the water surface is,

tT

SinYy π2max=

Velocities are obtained as,

tT

CosVFf

dtdyU

tT

CosVVV

π

π

2

2

0

00

==

−=

Water Surface Oscillations in the Surge Tank by Taking Headloss in the Pressure Tunnel (Damped Oscillations) The frictional resistance developing along the tunnel will be taken into account and its damping effect yielding damped oscillations will be dealt with. The only case of damped mass oscillations for which an exact mathematical solution can be found is the total closure. For other circumstances only approximate mathematical and graphical methods are available. For the examination of instantaneous closure consider the Figure below and notations used therein. The reservoir is connected with a surge tank of cross-sectional area F, by a pressure tunnel of cross-sectional area f, and length l, followed by a penstock conveying a discharge Q0. The hydrodynamic-equilibrium water level in the surge tank for this operating condition is below the hydrostatic level by,

Prof. Dr. A. Bulu 9

Figure

2034

22

00ln VR

Vy β==

Where the static level is equal to that in the reservoir, and V0 = Q0/ f. Hence y0 is the hydraulic resistance of the tunnel at a flow velocity V0. This is the headloss due to the friction in the tunnel computed by the Manning equation as,

213221

032

011

⎟⎠⎞

⎜⎝⎛ Δ==

lhR

nSR

nV

Whence the friction headloss is,

34

220

RlnVh =Δ

The resistance factor of the tunnel is,

34

2

Rnl ⋅

=β

Prof. Dr. A. Bulu 10

In case of instantaneous opening of turbine valves, the discharge for the turbine cannot be supplied by the pressure tunnel because of the velocity differences among the pressure tunnel and penstock. This water volume difference will be supplied by the surge tank initially so that the water level in the surge tank will drop. Air entrance to the penstock should not be permitted in order not to cause bursting of the penstock. There should be minimum water height of 1.50 m over the top of penstock in the surge tank for the minimum water level which is the case of instantaneous opening of turbines for full load. In order to be on the safe side, manning roughness coefficient n should be selected high for concrete lining as n = 0.015 to obtain a higher β resistance factor. Vogt Dimensionless Variables Tables have been prepared to compute surge amplitudes and periods for the surge tanks using dimensionless variables.

( )

0

0

20

20

hyx

VVz

hV

Ff

gl

Δ=

=

Δ⋅⋅=ε

Δh0 = Head loss for the steady flow case (will get negative values since y values are taken positive for upward direction).

a) Instantaneous full closure case Forchhmeir has given for the first maximum surge height for steady flow of Q0 discharge with Δh0 headloss,

εεε212121 maxmax +=⎟

⎠⎞

⎜⎝⎛ +−⎟

⎠⎞

⎜⎝⎛ + xLnx

For m = Damping factor,

20

0

0

22lfV

hgFh

m Δ=

Δ=ε

The equation takes the form of,

( ) ( ) 0maxmax 111 hmyLny Δ+=+−+

m dimensional variables are always negative since ε dimensionless variables are positive and y direction is taken positive for upward direction with Δh negative values. Forchhmeir equation is solved by using the Table.






In order to calculate the other extreme surge values after calculation the first ymax value, Braun equations are used.

( ) ( ) ( ) ( )( ) ( ) ( ) ( )( ) ( ) ( ) ( )( ) ( ) ( ) ( )3344

2233

1122

maxmax11

111111111111

1111

myLnmymyLnmymyLnmymyLnmymyLnmymyLnmy

myLnmymyLnmy

+−+=+−+−−−=−−−+−+=+−+−−−=−−−

The following steps are taken for the solution of the aforementioned equations,

1) ε dimensionless variable is calculated, 2) ymax value is computed by Forchhmeir equation by using the Table for (mΔh0) to

get (mymax) value, 3) After calculation ymax, the other y surge values are calculated by using above

giving equations and the Table.

b) Partial Closure of the Turbine Valve Q0 full load discharge may be reduced to nQ0 for (n < 1). It will be instantaneous full closure if (n = 0). Frank`s Table can be used to calculate the surge values for partial closure. The values in the Table can be defined as,

( )

gflF

tTt

hyx

hgFflV

πτ

ε

2

0

maxmax

20

20

==

Δ=

Δ=

The Table has been prepared for circular simple surge tanks. Example: An hydroelectric power plant with a design discharge Q = 30 m3/sec is fed by a pressure tunnel with a diameter D = 4 m, length l = 5000 m, and Manning coefficient n = 0.014. Compute the extreme surge heights for instantaneous full turbine closure in the surge tank with cross-sectional area F = 150 m2,

a) By using Forchhmeir method, b) By the help of Frank`s table.

Solution:

a) Physical characteristics of the plant are,

222

57.1244

4mDf =

×==ππ


sec39.257.120.30

14

0 mfQV

mDR

===

==

( )78.7

60.515081.939.257.125000

60.51

500057.1239.2

2

2

20

20

34

2

34

220

0

=××××

=Δ

=

=××

==Δ

hgFlfV

mR

lnVh

ε

( )( ) ( ) 257.060.5046.0

046.060.578.7

22

0

0

=−×−=Δ

−=−×

=Δ

=

hmh

mε

From the Forchhmeir Table,

mΔh0 = 0.25 → mymax = -0.551 mΔh0 =0.26 → mymax = -0.559

mΔh0 = 0.257 → mymax = -0.557

my 11.12046.0557.0

max =−−

=

The first minimum level,

( ) ( ) ( ) ( )( ) ( ) ( ) ( )557.01557.0111

1111

11

maxmax11

+−+=−−−−−−=−−−

LnmyLnmymyLnmymyLnmy

( ) ( ) 114.1443.0557.111 11 =−=−−− myLnmy

114.0114.11

0

0

=Δ=Δ+

hmhm

From the Frank`s Table,

mΔh0 = 0.11 → mymax = -0.399 mΔh0 = 0.12 → mymax = -0.413

mΔh0 = 0.114 → mymax = -0.405

my 80.8046.0405.0

1 −=−

=


Second maximum level,

( ) ( ) ( ) ( )( ) ( )

114.0114.1519.0595.01

405.01405.0111111

0

0

0

1122

=Δ=+=Δ+

−−−=Δ++−+=+−+

hmhm

LnhmmyLnmymyLnmy

The same mΔh0 value has been obtained coincidentally.

mΔh0 = 0.114 → my2 = -0.405

my 80.8046.0405.0

2 ==

Second minimum level,

( ) ( ) ( ) ( )( ) ( )

065.11405.01405.011

1111

0

0

2233

=Δ++−+=Δ+

−−−=−−−

hmLnhm

myLnmymyLnmy

mΔh0 = 0.065 → my3 = -0.318

my 91.6046.0318.0

3 −=−

=

Third maximum level,

( ) ( ) ( ) ( )065.1383.0682.01

318.01318.0111

0

44

=+=Δ+−−−=+−+

hmLnmyLnmy

mΔh0 = 0.065 → my4 = -0.318

my 91.6046.0318.0

4 ==

b) Franks Table will be used for surge calculations for instantaneous closure, n = 0.

36.078.7

1178.7 ==→=ε

ε

The first maximum level for n = 0,

35.01=

ε → x = -2.24 , τ = 0.293

40.01=

ε → x = 1.88 , τ = 0.300

36.01=

ε → x = -2.17 , τ = 0.294


Oscillation period,

sec49057.1281.9

150500022 =××

== ππgflFT

Δh0 = -5.60m

( ) ( )

sec144490294.015.1260.517.2

max

0maxmax

=×=×==−×−=Δ×=

Ttmhxy

τ

First inflection point,

35.01=

ε → x = -0.45 , τ = 0.520

40.01=

ε → x = -0.41 , τ = 0.525

36.01=

ε → x = 0.44 , τ = 0.521

( ) ( )

sec255521.0490

46.260.544.0

1

1

inf

inf

=×=

=−×−=

t

my

First minimum level,

35.01=

ε → x = +1.63 , τ = 0.797

40.01=

ε → x = +1.34 , τ = 0.805

36.01=

ε → x = +1.57 , τ = 0.799

( )

sec392799.0490

79.860.557.1

1

1

min

min

=×=

−=−×=

t

my

Second inflection point,

35.01=

ε → x = 0.256 , τ = 1.030

40.01=

ε → x = 0.218 , τ = 1.037

36.01=

ε → x = 0.248 , τ = 1.031


( )sec505031.1490

43.160.5256.0

2

2

inf

inf

=×=

=−×=

t

my

Second maximum level,

35.01=

ε → x = -1.274 , τ = 1.300

40.01=

ε → x = -1.025 , τ = 1.309

36.01=

ε → x = -1.224 , τ = 1.302

( ) ( )

sec638490302.1

13.760.5274.1

2

2

max

max

=×=

=−×−=

t

my

Placing the surge values to the table,

y Forchheimer Frank ymax 12.11 12.15 y1 -8.80 -8.79

The values are close for the both methods.


Table 18. Discharge increase from nQ0 to Q0


c) Instantaneous Opening of the Turbines The discharge increase to the turbines by instantaneous opening is done from nQ0 to Q0 (n < 1). If the turbines are not running, there will be no discharge feeding the penstock which is (n =0) case. Instantaneous partial opening case may be computed to find out the surge heights by using Frank`s Table 18.

The column with 01=

ε value corresponds to ε→∞. Since l, f, V0

2, F are physical

magnitudes, this ε→∞ corresponds to Δh0 = S0l → 0 which is the ideal fluid case. The x and τ values of this column can only be used for ideal fluids which is no friction losses would occur in the plant. Example: An hydroelectric power plant with a pressure tunnel of the length l = 5000 m, diameter D = 4 m, and Manning coefficient n = 0.014 is feeding the turbines. The cross-sectional area of the cylindrical surge tank is F = 150 m2. Calculate the extreme surge levels by using Frank tables for,

a) Instantaneous discharge increase from 0 m3/sec to 10 m3/sec, b) Instantaneous discharge increase from 10 m3/sec to 30 m3/sec.

Solution:

a) Q0 = 10 m3/sec,

sec80.057.12

10

144

4

57.1244

4

0

222

mfQV

mDR

mDf

===

===

=×

==ππ

mR

lnVh 63.01

5000014.080.034

22

34

220

0 =××

==Δ

( )

12.087.68

11

87.6863.015081.9

80.057.1250002

2

20

20

==

=××××

=Δ

⋅⋅=

ε

εh

VFf

gl

sec49057.1281.9

150500022 =××

== ππgflFT

First minimum surge tank level for n = 0 by using Table 18,


10.01=

ε → x = 10.10 , τ = 0.255

15.01=

ε → x = 6.75 , τ = 0.258

12.01=

ε → x = 8.76 , τ = 0.256

( )

sec125256.0490

52.563.076.81min

=×=

−=−×=

τ

my

First maximum level,

10.01=

ε → x = -5.00 , τ = 0.760

15.01=

ε → x = -2.13 , τ = 0.768

12.01=

ε → x = -3.85 , τ = 0.763

( ) ( )sec374763.0490

43.263.085.31max

=×=

=−×−=

τ

my

Second minimum level,

10.01=

ε → x = 5.80 , τ = 1.261

15.01=

ε → x = 3.15 , τ = 1.272

12.01=

ε → x = 4.74 , τ = 1.265

( )sec620265.1490

00.363.074.42min

=×=

−≅−×=

τ

my

b) Instantaneous discharge increase from Q = 10 m3/sec to Q0 = 30 m3/sec.

Q0 = 30 m3/sec , f = 12.57 m2 , R = 1 m.

( )m

hV

Ff

gl

mR

lnVh

mV

78.760.515081.9

39.257.125000

60.51

5000014.039.2

sec39.257.12

30

2

2

20

20

34

22

34

220

0

0

=××××

=Δ

⋅⋅=

=××

==Δ

==

ε


36.078.7

11==

ε , 333.0

3010

==n

Interpolation will be done for the required ε1 , and n values using Table 18.

Minimum surge tank level for n = 0.333,

n = 0 → 35.01=

ε → x = 2.96 , τ = 0.272

n = 0 → 40.01=

ε → x = 2.61 , → τ = 0.276

n = 0 → 36.01=

ε → x = 2.89 , τ = 0.273

n = 0.5 → 35.01=

ε → x = 1.83 , τ = 0.297

n = 0.5 → 40.01=

ε → x = 1.06 , τ = 0.306

n = 0.5 → 36.01=

ε → x = 1.80 , τ = 0.299

n = 0.333 → 36.01=

ε → x = 2.17 , τ = 0.290

( )

sec14249029.0

15.1260.517.2

1

1

min

min

=×=

−=−×=

t

my

First maximum surge tank level,

n = 0 → 35.01=

ε → x = 0.52 , τ = 0.817

n = 0 → 40.01=

ε → x = 0.68 , τ = 0.833

n = 0 → 36.01=

ε → x = 0.55 , τ = 0.820

n = 0.5 → 35.01=

ε → x = 0.78 , τ = 0.834

n = 0.5 → 40.01=

ε → x = 0.85 , τ = 0.853

n = 0.5 → 36.01=

ε → x = 0.79 , τ = 0.838

n = 0.333 → 36.01=

ε → x = 0.71 , τ = 0.832


( )sec408832.0490

98.360.571.01max

=×=

−=−×=

t

my

Second minimum surge tank level,

n = 0 → 35.01=

ε → x = 1.16 , τ = 1.345

n = 0 → 40.01=

ε → x = 1.09 , τ = 1.370

n = 0 → 36.01=

ε→ x = 1.15 , τ = 1.35

n = 0.5 → 35.01=

ε → x = 1.08, τ = 1.364

n = 0.5 → 40.01=

ε → x = 1.04 , τ = 1.393

n = 0.333 → 36.01=

ε → x = 1.10 , τ = 1.36

( )sec66636.1490

16.660.510.12min

=×=

−=−×=

t

my

Stability Conditions of the Surge Tanks Stability conditions of the surge tanks were first established by D. Thoma and F. Vogt. They stated that in order to prevent the development of unstable oscillations the cross-section of the surge tank should exceed a critical value. According to the Thoma equation suggested in small oscillations, the limit cross-sectional area of the surge tank is,

02 HglfkFF thm β

=> (m2)

k = Factor of safety, V0 = Pressure tunnel velocity for the new dynamic equilibrium level, i.e. to the power output to be succeeded after opening (m/sec), β = Resistance factor of the pressure tunnel (sec2/m), l = Length of the tunnel (m), H0 = H – βV0

2 = H – Δh0 = net head (by neglecting the headloss in the penstock) (m). Substituting the damping factor m,

lfgFm β2

=

The minimum value of head succeeding surge stability in case of a given cross-sectional area F of the surge tank is,


mkH

gmlf

HgklfF

=

==

0

0 22 ββ

Assuming local headlosses can be neglected with respect to friction losses, and with the substitution,

20

34

20

34

34

2

22 ngHfRk

lngHlfRkF

Rln

thm ==

=β

Is obtained, which can be simplified in case of a circular pressure tunnel cross-section, R = D/4 as hydraulic radius, f = πD2/4 cross sectional are, to the form of,

20

34

234

62.1944 nHDDkF×××

=π

20

310

160 nHDkF =

A safety factor k of 1.5 to 1.8 may be adopted. As can be seen from the equation, the lower the friction factor β, the larger the cross-sectional area of the surge tank. Limit values of F are thus obtained by simultaneous assumption of the highest safety factor k and lowest Manning coefficient n. Substituting the pairs of values k = 1.5, n = 0.014 as well as k = 1.8, n = 0.0106, we obtain,

0

310

0

310

22

0

310

0

310

21

1000106.01608.1

50014.0160

5.1

HD

HDF

HD

HDF

≅××

=

≅××

=

21

2 =FF

In case of a concrete lined pressure tunnel, the deviation depending on the choice of the friction coefficient n, as well as on the safety factor k, is considerable between extreme F2/F1 = 2. For a lining carried out with steel, the mean value n = 0.0143 – 0.0133 may be applied.


For great amplitudes the Thoma equation was modified by Ch. Jaeger, demonstrating that the safety factor can n0 longer be considered constant. According to Jaeger, the cross-sectional area necessary for stability should not be less than,

02

34

0 22 HgnfRk

HglfkF ∗∗ ==β

For a circular cross-section,

02

310

160 HnDkF ∗=

The safety factor is,

0

max482.01Hyk +=∗

ymax is the amplitude of the undamped (frictionless) surge.

References Çeçen, K. (1974). Su Kuvvetleri, İ.T.Ü., İnşaat Fakültesi. Mosonyi, E. (1963). Water Power Development, Publishing House of the Hungarian Academy of Sciences, Budapest, hungary. Öziş, Ü. (1991). Su Kuvveti Tesislerinin Planlama Esasları, Dokuz Eylül Üniversitesi Mühendislik – Mimarlık Fakültesi Yayınları, No.197, İzmir. Scheleiermacher, E. (1967). Su Kuvveti Tesisleri İnşaat ve Proje Esasları, Teknik Üniversite Matbaası, İstanbul. Ünsal, İ. (1977). Su Kuvvetleri, Elazığ Devlet Mühendislik ve Mimarlık Akademisi. Ünsal, İ. (1978). Değişken Akımların Hidroliği, İ.T.Ü., İnşaat Fakültesi.

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HYDROELECTRIC POWER PLANTS - … · LECTURE NOTES – I « HYDROELECTRIC POWER PLANTS » Prof. Dr. Atıl BULU Istanbul Technical University College of Civil Engineering Civil Engineering