�Surge Tanks� Surge tanks may be considered essentially as a forebay close

to the machine.� They are essential part of the conveyance pressure conduit

whenever such systems are long.� Their primary purpose is the protection of the long pressure

tunnel in medium- and high-head plants against high waterhammer pressures caused by sudden rejection or acceptanceof load.of load.

� The surge tank converts these fast (water hammer) pressureoscillations into much slower – and lower – pressurefluctuations due to mass oscillation in the surge chamber.

� The threefold purposes served by surge tanks may besummarized as

(1) flow stabilization to the turbine,(2) water hammer relief or pressure regulation, and(3) storage function.

� The surge chamber (see Fig. ) dividing the pressure tunnel into a

short high-pressure penstock downstream and a long low-pressure

tunnel upstream thus functions as a reservoir for the absorption or

delivery of water to meet the requirements of load changes.

� It quickly establishes the equilibrium of the flow conditions, which

greatly assists the speed regulation of the turbine.

� The surge tank protects the low-pressure conduit/tunnel system from� The surge tank protects the low-pressure conduit/tunnel system from

high internal pressures.

� Due to the surge tank, the entire pressure conduit on the upstream

side of the surge tank can be designed as a low pressure system,

while the penstocks between the surge tank and the power house

will be designed as conduits which can resist high water hammer

pressures.

� Surges in surge chambers

� Sudden changes in load conditions of the turbine cause massoscillations in surge tanks which are eventually damped out bythe hydraulic friction losses of the conveyance.

� The amplitude of these oscillations is inversely proportional tothe area of the surge tank, and if the area provided were verylarge dead beats would be set in the tank.

� Although these conditions would be favorable to achieving the� Although these conditions would be favorable to achieving thenew equilibrium state very quickly, the design would beuneconomical.

� On the other hand, if too small an area is provided theoscillations (surge tank) become unstable; this isunacceptable.

� It is therefore essential to choose a section in which theoscillations become stable within a short period of time.

� The critical section for stability is given by the equation,known as Thoma criteria;

� where At and Lt are the tunnel cross-sectional area andlength respectively, and the suffix ‘0’ defines the steadystate conditions prior to the load variation (see Fig.).state conditions prior to the load variation (see Fig.).

� A stable tank area is usually chosen with a safety factorof about 1.5.

Hydraulic Design of Simple Surge Tanks� Consider the simple surge tank installation shown below

where the initial flow through the control gate is cut offrapidly

� It is desired to find the maximum upsurge in thesurge tank and the time at which this upsurgeoccurs.

� The equations are derived based on theassumptions that hydraulic losses are negligible inthe simple surge tank, the velocity head in the pipecan be neglected, and the rigid water column theorycan be neglected, and the rigid water column theoryof water hammer is sufficient.

� This is justified if the pressure rise is small and thereis neither appreciable stretching of the pipe norcompressing of the water.

� Prior to the gate closure, the mass of water which is

moving in the tunnel is Lt At γγγγ/g. Upon gate closure, the

unbalanced force acting on this water column is

(i)

� From Newton’s 2nd Law of motion, the deceleration of the

water column in the tunnel is

(ii)

� From the condition of continuity of flow, following

complete gate closure, the flow of water into the surge

tank is the same as that out of the tunnel, that is;

(iii)

� The simultaneous solution of equations (ii) and (iii) is

performed with the following boundary conditions:

o when t = 0, Z = 0, and

o dZ/dt = Q0/As.

� Then,

(iv)

� From which

(v)

� And the time required to reach the maximum upsurge is;

(vi)

Where

Z = upsurge in surge tank above the static water level, m;

Q0 = water discharge in the pipe before gate closure, m3/s;

As = cross-sectional area of surge tank, m2;

Lt =length of pipe (low-pressure tunnel) from surge tank tot

the reservoir (open water surface, m;

At = cross sectional area of pipe (tunnel), m2;

t = instantaneous valve closure time, sec;

T = time required to reach the maximum upsurge in the

tank, sec.

� The last two equations can be useful for preliminaryanalysis and in cases where the magnitude of hydrauliclosses is small in the pipe extending upstream from thesurge tank.

� In actual installations, frictional effects are not negligible.

� Jaeger has recommended the use of the following� Jaeger has recommended the use of the followingapproximate formula for the calculation of upsurge incase where friction is taken into account.

� Hence, for sudden 100% load rejection, maximumupsurge will be;

(vii)

WhereZ = maximum upsurge friction taken into account, m;

Zmax = maximum upsurge with negligible friction, m;and � maximum downsurge,

(viii)(viii)

� For 100% load demand, the maximum downsurge;

(ix)

� where P0 is given by the following relation

(x)

= the friction head loss in the steady state = the friction head loss in the steady state condition, m.

� The maximum upsurge and down surge should be

contained within the surge chamber.

� The range of surge levels (amplitudes) must not be too

large to minimize the governing difficulties.

� The maximum upsurge and downsurge are computed for

extreme conditions, i.e.

� the top level of the surge chamber is governed by the� the top level of the surge chamber is governed by the

maximum upsurge level when the reservoir level is at its

maximum, and

� the bottom level of the chamber is controlled by the

maximum downsurge level when the reservoir is at its

lowest drawdown level.

� Instantaneous 100% demand conditions result in toolarge a maximum downsurge, as the normal practiceis to allow for 0–10% of full load demand quickly but,thereafter, the unit is brought to full load onlygradually.

� Maximum downsurges are normally calculatedagainst 75–100% of full load and, once again, theagainst 75–100% of full load and, once again, thebottom level of the chamber is controlled by thereservoir at its lowest drawdown level condition.

� This condition is invariably more critical than the onegoverned by the maximum downsurge after a loadrejection.

� Excessive surges may occur if several quick load variationsare imposed on the unit (overlapping surges).

� These may create additional governing difficulties and the topand bottom levels of the surge chamber may have to bemodified to accommodate these excessive surges.

� In order to achieve conservative designs of the surge chamberit is usual to assumeit is usual to assume

o a lower conduit friction factor than average for calculating themaximum upsurge, and

o a higher friction factor for the maximum downsurge.

� In all cases due consideration of the effect of ageing of thetunnel must be given.

� Head losses of all types are assumed to be proportional to V2.

� Types of surge tanks

1. Simple surge tanks.

� The simple surge tank (Fig. (a)) is of uniform cross-section and

is open to the atmosphere, acting as a reservoir.

� It is directly connected to the penstock so that water flows in

and out with small head losses when load variations occur.and out with small head losses when load variations occur.

� It is usually large in size with expensive proportions and

sluggish in responding to damping surges and unstable.

� These are very rarely used in modern practice except in

installations where load changes are either small or very

gradual.

2. Throttled tank.

� In the throttled tank (restricted orifice type tank) the restrictedentry (Fig. (b)) to the surge tank creates retardation andacceleration conditions of flow in the tunnel upstream of it,thus reducing the storage requirement and minimizing themaximum up and downsurges.

� This improves the stability damping quality of the surge tank� This improves the stability damping quality of the surge tankoscillations.

� Although this type of surge tank is economical (because of itssmaller size) compared with the simple tank section, the rapidcreation of retarding and accelerating heads complicates thegoverning mechanism, requiring additional inertia in the turbo-generator units

3. Surge tank with expansion chambers.

� This type of surge tank (Fig. (c)) consists of a narrow riser

(main surge shaft); attached to it at either end are large

expansion chambers.

� The narrow riser reacts quickly, creating accelerating or

decelerating heads, and at the same time the expansiondecelerating heads, and at the same time the expansion

chambers minimize the maximum up- and downsurge levels,

thus limiting the range of surge levels (i.e. easier governing).

� In order to reduce the costs of the structure, spilling

arrangements may sometimes be provided either to wastage

(if water is not scarce) or back to the penstock.

4. Differential surge tank.

� This type (also known as Johnson’s differential tank – Fig. (d))consists of an internal narrow riser shaft with an orifice entry tothe larger outer shaft at the bottom.

� As the central riser is narrow it responds instantaneously duringthe upward phase; at the same time the maximum amplitude isrestricted to its top level, any excess water spilling back into therestricted to its top level, any excess water spilling back into theouter chamber.

� Similarly, during the downward phase water spills into the narrowriser while the riser itself responds quickly to maintaining thedesired level.

� The differential tank with an extended penstock, which acts as acentral riser, is shown in Fig. (e).

5. Surge tanks with venturi mounting.� Considering the velocity energy under the surge tank (V2/2g = E0),

Thoma’s critical section can be written as

� More economical sections may result by providing a venturicontraction (Fig. (f)) under the surge tank (thus increasing thevelocity head, E0.

ExamplesExample 1. Given: H0 = 152 m, As = 29 m2, Dt = 3.0 m,

Lt = 915 m, and Q0 = 23.9 m3/sRequired: Zmax and T

Solution:At=p/4 x32 = 9p/4 m2At=p/4 x3 = 9p/4 m

Example 2.

In a hydropower project, water is delivered from an

impounding reservoir through a low-pressure tunnel and

four high-pressure penstocks to the four turbine units. The

elevation of the reservoir water level is 1500 m a.m.s.l, andelevation of the reservoir water level is 1500 m a.m.s.l, and

the elevation of the tailwater is 1200 m a.m.s.l. The

maximum reservoir storage which can be utilized

continuously for a period of 48 hours is 15 x 106 m3.

The low-pressure tunnel is constructed as follows:

� Length, Lt = 4 km

� Diameter, Dt = 8 m

� Friction factor, f =0.028

The high-pressure penstocks (4 in no.) are constructed asfollows:follows:

� Length of each penstock, Lp = 500 m

� Diameter of each penstock, Dp = 2.0 m

� Friction factor, f = 0.016

� Turbine efficiency, ηt = 90%

� Generator efficiency, ηg =90%

i. Determine the maximum power output from the

installation;

ii. If a simple surge tank 6 m in diameter is provided at the

end of the low-pressure tunnel, estimate

a) The maximum upsurge and downsurge in the surge a) The maximum upsurge and downsurge in the surge

tank for a sudden rejection of one unit, and

b) The maximum downsurge for a sudden demand of

one unit

Solution� The discharge available, Q = 15 x 106/48 x 60 x 60 =

86.8 m3/s

i. Power output� Velocity in tunnel,

,

� Therefore, head loss in tunnel,

� Discharge per penstock, Qp = 86.8/4 = 21.7 m3/s� Velocity in penstock, Vp = 21.7x4/4π = 6.91 m/s

� Head loss in penstock,

� Gross head at the turbine = 1500 – 1200 = 300 m� Hence, net head, H = 300 – 2.13 – 9.73 = 288.14 m� Power output per turbine,

� Total power output,Ptot = 4 x 55.20 = 220.80 MW

� The net output of the generator Pnet= 0.90 x 220.80 = 198.72 MW

ii. Surge Tank� Area of surge tank, As = π x 62/4 = 28.27 m2

� Area of tunnel, At = p x 82/4 = 50.27 m2

� Length of tunnel, Lt = 4000 ma) Sudden rejection of one unit

� Therefore, maximum upsurge, Z, will be,

� Hence, Z = Zmax x 0.8817 = 11.62 x 0.8817 = 10.25 m� Maximum downsurge, Z,� Maximum downsurge, Z,

Z(min.) = 11.62 x (- 0.70) = - 8.14 m

b) Maximum downsurge for sudden demand of one unit,

Z(min.) = 11.62 x (-1.023) = -11.9 m

Exercise 1. A power station is fed through a 10,000m long concreteExercise 1. A power station is fed through a 10,000m long concretelined tunnel of 5.0 m diameter operating under a gross head of 200m. The discharge through the tunnel is 30m3/s. A surge tank of 300m2area has been provided at the end of the tunnel. Calculate:a) The maximum upsurge in the tank,b) The minimum downsurge in the tank.

Assume a friction factor f for the concrete lined tunnel as 0.016.

Penstocks� The penstocks are pipes of large diameter, usually of

steel or concrete, used for conveying water from thesource (reservoir or forebay) to the power house.

� They are usually high-pressure pipelines designed towithstand stresses developed because of static andwater hammer pressures created by sudden changes inpower demands (i.e. valve closures and openingspower demands (i.e. valve closures and openingsaccording to power rejection and demand).

� The provision of such a high-pressure line is veryuneconomical if it is too long, in which case it can bedivided into two parts,� a long low-pressure conveyance (tunnel) followed by short high-

pressure pipeline (penstock) close to the turbine unit, separatedby a surge chamber which absorbs the water hammer pressurerises and converts them into mass oscillations.

Figure Components of a penstock

� Classification of penstockClassification may be based on:

1) The material of fabrication/construction

2) Method of support.

3) Rigidity of connection and support

4) Number of penstocks

1. Material of fabrication/constructionFactors for the choice of material are:

� the head to which the penstock is subjected,� the topography of the terrain, and � the discharge to be handled.

� Various materials used are steel, R.C., PVC, wood stave

pipes, banded steel, etc.

� Steel penstock has become the most common type of

installation due to simplicity in fabrication, strength and

assurance that they can perform in a wide variety of

circumstances.circumstances.

� The penstocks required to withstand high pressures because

of very high heads, are fabricated usually as banded steel

pipes.

� Banded steel pipes are thin-walled but strengthened with the

help of hoops of high-strength steel slipped over them.

� Very large diameters of precast or cat-in-place R.C.

penstocks are impractical.

� They are usually limited to heads of less than 30 m

and working pressures of the order of 5 to 15 kg/cm2

(500 to 1500 kPa).(500 to 1500 kPa).

� Diameters may vary from 0.3 m to 3.0 m, and in 3 m

to 7 m sections.

� The following factors have to be considered whendeciding which material to use for a particular project:

o Required operating pressure,

o Diameter and friction loss,

o Weight and ease of installation,

o Accessibility of site,

o Cost of the penstock,

o Design life,

o Availability,

o Weather conditions.

2. Method of support

� A penstock may be either buried or embedded

underground (or inside dams) or exposed above ground

surface & supported on piers.

� Buried penstocks: are supported on the soil in a trench at� Buried penstocks: are supported on the soil in a trench at

a depth of 1 to 1.5m and backfilled.

� For buried penstocks the general topography of the land

should be gently sloping and of loose material.

Burried Penstocks

Advantages Disadvantages

Continuity of support given by thesoil provides better structuralstability

Difficulty of inspection

Protection of the pipe againsttemperature fluctuations with thehelp of small overburden

Difficulty of maintenance

Conservation of natural landscape Possibility of sliding on steepslopes

Protection from slides, storms &sabotage

Expensive for large diameter inrocky soils

Advantages Disadvantages

Ease of inspection of faults Direct exposure to weather effect

Economy in rocky terrain & large diameters

Development of longitudinal stress due to support and

Exposed penstocks: Exposed to view and supported on piers or saddles.

diameters stress due to support and anchorage, thus requiring expansion joints

Stability is ensured with proper anchorage

When the situation warrants, partly buried system, may be adopted thatcombines the advantages of both system.

3. Rigidity of connection & Support

� There are three possible methods of support,

a) Rigid pipe support: Here every support is an anchorage so

that any movement is checked completely. This type is

suitable when the temperature condition is moderate.

b) Semi-rigid pipes: Here each member of the pipe line is fixedb) Semi-rigid pipes: Here each member of the pipe line is fixed

at one and leaving the possibility of movement over the other

support.

c) Flexible support (Flexible or loose-coupled pipes): Here

expansion joint are introduced between each adjacent

section

4) Number of Penstocks

� The number of penstocks used at any particular installation can be

single or multiple.

� The general trend at older power stations was to use as many

penstocks between the forebay/surge tank and the powerhouse as the

number of units installed.

� The recent trend is to use a single penstock, unless the size or thickness� The recent trend is to use a single penstock, unless the size or thickness

of the penstock involves manufacturing difficulties.

� When a single penstock feeds a number of turbines, special sections

called manifolds are used at the lower end of the penstock to direct flow

to individual units.

� The design of such sections is an intricate job and has to be analyzed

carefully.

� The advantages of using a single penstock over the use

of multiple penstocks are:

o The amount of material required to manufacture is less,

making it economical.

o The cost of civil engineering components such as penstock

supports and anchors is less.supports and anchors is less.

� On the other hand, the use of a single penstock means

reduced safety of operation and complete shutdown will

become necessary in case of repair.

� Furthermore, significant losses are usually experienced

at the manifolds.

� In general, the use of multiple penstocks is preferablyemployed for low-head plants with short penstocks;whereas for high-head plants requiring long penstocks,provision of a single penstock with manifold at the endusually proves economical.

�Design Criteria for Penstocks

� For the purpose of engineering feasibility and preliminarydesign, there are three major considerations that needengineering attention:

1. The head loss through the penstock,

2. The safe thickness of the penstock shell (wall), and

3. The economical size of the penstock.

� Head Losses� The head losses consist of

o Trash rack losses,o Entrance losses,o Friction losses in the pipe,o Bend losses, ando Bend losses, ando Stop log, gate slot, and transition losses

� Thickness of Penstock� According to ASME wall thickness may be computed

using the following formula

Where t = wall thickness, cm

P = pressure, kg/cm2

R = internal radius, cm

S = design stress, kg/cm2

η = joint efficiency factor and 0.15 cm is allowance for corrosion.

� Under normal flow, the penstock is subjected to only� Under normal flow, the penstock is subjected to onlyinternal hydrostatic pressure.

� But when turbine gates at the end of penstock are closedsuddenly, there is a sudden pressure rise, i.e. waterhammer pressure.

� In this case the following steps are followed to computethe wall thickness.

1. Calculate the pressure wave velocity, a, as;

Wherea = pressure wave velocity, m/sa = pressure wave velocity, m/sρ =mass density of water, kg/m3

k = bulk modulus of water ≈ 2 x 109 N/m2

D = internal diameter of penstock, mE = modulus of elasticity of penstock material = 200 x 109

N/m2

t = thickness of the penstock wall, m

2. Calculate velocity, surge head (∆h), and total head (htot)

1. For steel penstock, if it is subjected to corrosion and welding

and rolling defects, the effective thickness is less than the

nominal thickness quoted by the manufacturer.

� Therefore, to find the effective thickness (teff),� Welding divide t by 1.1� Flat-rolled divide by 1,2� Corrosion subtract 1mm for 10 years life

Subtract 2mm for 20 years life4. Calculate factor of safety (F.S.) as follows:

Where teff = effective wall thickness, mS = ultimate tensile strength of penstock material, N/m2

D = diameter of penstock, m5. If safety factor is less than 3.5, reject the penstock option and

repeat the above steps for stronger wall option

� Economic Diameter of a Penstock

� For successful operation, the size of the pipe for a given

discharge may vary between wide limits, but usually there is

one size that will make for the greater economy in design.

� The diameter of the penstock is determined from economic

consideration and then checked to see that acceptableconsideration and then checked to see that acceptable

velocities are not exceeded.

� The following two methods are used to determine the

size/diameter of a penstock

o Empirical equations

o Graphical (economic analysis)

� Empirical Relations� These are over-simplified rule of the thumb relations

based on available experience.1) USBR Formula

V = optimum velocityV = optimum velocityThe formula generally applies up to middle range of heads.2) Sarkaria developed an empirical approach for

determining steel penstock diameter by using data fromlarge hydro projects with heads varying from 57 m to 313m and power capacities ranging from 154 MW to 730MW.

� He reported that the economical diameter of the penstock is given by the equation

D = Penstock diameter (m)P =rated hp (metric) of the turbines,P =rated hp (metric) of the turbines,H = Net head at the end of the penstock, m.

3. Donald’s formula

4. Gordon and Penman: for steel penstocks used insmall hydropower installations;

5. Fahlbusch reformulated the objective of the economicanalysis in terms of the amount of the invested capitaland the capitalized value of the lost energy, and arrivedand the capitalized value of the lost energy, and arrivedat the conclusion that the most economical diametercan be computed within an accuracy of about ±10%from

where P is the rated capacity of the plant (kW), H is therated head (m), and D is the diameter (m).

� Graphical Method

� There are various combinations of V and D which would

give the same discharge.

� The frictional loss in the penstock, also

depends on diameter and in turn determine efficiency on

the penstock pipe.

� Thus, the larger the diameter for a given discharge, the

smaller will be the head losses and greater will be the net

head available to the turbines, resulting in a greater

power development.

� On the other hand, greater size would mean less velocity

and greater capital investment.

� We should choose, therefore, a size which would give the

least annual costs.

� If the total annual cost of the penstock corresponding to� If the total annual cost of the penstock corresponding to

any given diameter is plotted, installation cost results.

� On the other hand, if the cost (which is the price of

energy lost in friction at the prevalent rate of sale of

energy) due to the horse power lost is plotted in the same

figure, power lost would result.

� The total cost would be the sum of the ordinates of thetwo curves corresponding to any diameter.

� Then, the total cost curve is plotted in the same figure.� The least cost corresponds to the optimum diameter of

the penstock.

� Penstock Joints

� Penstock pipes are generally supplied in standard

lengths, and have to be joined together on site.

� There are many ways of doing this, and the following

factors should be considered when choosing the best

jointing system for a particular scheme.

o Relative costs

o Ease of installation

o Suitability for chosen pipe material

o Degree of joint flexibility

� Methods of pipe jointing fall roughly into four

categories:

o Flanged joints

o Spigot and socket joints

o Mechanical jointso Mechanical joints

o Welded joints

� Flanged Joints

� Flanges are fitted to each end of individual pipes during

manufacture, and each flange is then bolted to the next

during installation as shown in Figure below

� A gasket or other packing material is necessary between eachflange of a pair.

� Flange jointed pipesare easy to install,but flanges can add tothe cost of the pipe.� Flanged joints do not allow any flexibility.� Flanged joints do not allow any flexibility.� They are generally used to join steel pipes, and occasionally

ductile iron pipes.

� Spigot and Socket Joints: are made by either fitting acollar to, or increasing the diameter during manufacture of,one end of each pipe such that the internal diameter of thecollar or increased internal diameter of the pipe is the same asthe external diameter of the pipe.

� The plain end of each pipe can thus be pushed into the collar

or ‘socket’ in the next as shown in Figure below.

� A good seal is

required between

each pipe section,each pipe section,

and this is achieved by either providing a rubber seal or special

glue called solvent cement, depending up on the material of

which the pipes are made.

� Spigot and socket joints are generally used to join ductile

steel, PVC, and concrete pipes.

� Mechanical Joints

� Mechanical joints are rarely used on penstocks because

of their cost.

� One important application of it is for joining pipes of

different material or where a slight deflection in the

penstock is required that does not warrant installing a

bend.

� Welded Joints

� Welded joints are used on penstocks made of steel.

� Steel pipes are brought to the site in standard lengths,

and then welded together on site.

� One advantage of welding on site is that changes in the� One advantage of welding on site is that changes in the

direction of the pipe can be accommodated without

preparation of a special bend section.

� It is relatively cheap method, but has the drawback of

needing skilled site personnel.

� Expansion Joints� A penstock, specially exposed ones, will change in length

depending on temperature fluctuations.� If it is fixed the thermal expansion forces are substantial.

It is possible to relieve these forces by incorporatingspecial joints called expansion joints, which allow thepipe to expand and contract freely.pipe to expand and contract freely.

� For short penstocks, provision of a single expansion jointmay be sufficient, but for long penstocks with a multipleanchor blocks expansion joints should be placed beloweach anchor block.

� Another alternative to take care of thermal expansion isto take in to account the forces that result from it indesigning anchors.

� Penstock Supports and Anchors� Slide blocks, anchors, and thrust blocks all serve the

same basic function: to constrain movement of thepenstock.

� Different terms are used with these structures simply toindicate which specific function they serve (see figure)

� Slide Blocks� Slide Blocks� A slide block, also called supporting pier, carries the

weight of pipe and water, and restrains the pipe fromupward and sideway movements, but allows it to movelongitudinally. In most cases the spacing between slideblocks are assumed equal to the length of each pipe.

� If the penstock is buried, slide blocks are unnecessary;rather instead the pipe is laid in a trench on a bed of sandor gravel of consistent quality, with no big stones whichcould cut into the pipe or cause stress concentrations onthe pipe wall.

� Forces that act on slide blocks� Weight of the pipe and enclosed water: As slide blocks� Weight of the pipe and enclosed water: As slide blocks

do not resist longitudinal forces, only the component ofthe weight perpendicular to the pipe will be considered.

� Friction forces on the blocks: This is due to thelongitudinal movement of the pipe over the blocks causedby thermal expansion and contraction.

� Weight of the block

Figure Forces on slide block

� Anchor Blocks

� An anchor block consists of a mass of reinforced

concrete keyed to the penstock so that the penstock

cannot move in any way relative to the block.

� It is designed to withstand any load the penstock may� It is designed to withstand any load the penstock may

exert on it.

� Anchors are often used at bends (horizontal and vertical)

and before entrance to the powerhouse.

� They can also be used along long straight sections of

penstock, each one next to expansion joint.

� Because an anchor is keyed to the penstock pipe and isalso frequently located at a bend in the pipe, more forcesact on an anchor than on a slide block.

� The major forces which act on anchor blocks are thefollowing:

� Weight of the pipe and enclosed water

� Hydrostatic force on a bend� Hydrostatic force on a bend

� Friction forces on slide blocks located between the anchor

and expansion joint

� Thermally induced stresses, when expansion joints are not

incorporated

� The weight of the anchor block itself

� Thrust Blocks

� These are a special form of anchor whose sole purpose

is to transmit forces primarily caused by hydrostatic

pressures at horizontal bends along a buried penstock to

undisturbed soil which provides the reaction force.undisturbed soil which provides the reaction force.

� However, if the bend is vertical, an anchor block is still

used if the back filled soil is not able to resist this force.

� Penstock Valves� Valves are usually installed at two places in a penstock.� One valve is provided at the upstream end of the

penstock, i.e., at the forebay or immediately after thesurge tank, and is called penstock inlet valve, while thesecond is provided at the downstream end of the conduit,immediately ahead of the turbine, and is named asturbine inlet valve.turbine inlet valve.

� The upper valve is sometimes replaced by a gate.� The main purpose of penstock inlet valve is for

dewatering of the penstock in case maintenance of thepenstock is required.

� But, it can be omitted for short penstocks where theclosure of the power canal or power tunnel is possiblefrom the intake.

� The main purpose of turbine inlet valve is to close the

penstock while the turbine is inoperative.

� It can also act as an emergency shut-off device.

� This valve cannot be omitted except under special case where

the penstock supplies a single unit having installed the

penstock inlet valve.

� The number of turbine inlet valves required at a power station

is governed by the number of turbine units installed, but not by

the number of penstocks, as a single penstock can serve a

number of units through a manifold at the end.

� There are various types of valves for use in hydropower

installations. The most frequently applied include:

o Butterfly valves

o Spherical valves

o Needle valveso Needle valves

o etc

� The type to be applied should be determined individually

for each case after considering the various factors

involved.

� Power House� The power house structure can be divided in two

sections,o a substructure supporting the hydraulic and electrical

equipment, ando a superstructure housing the equipment.

� The substructure is usually a concrete block with all the� The substructure is usually a concrete block with all thenecessary waterways formed within it.

� The scroll case and draft tube are usually cast integrally(especially in large low-head plants) with the substructure withsteel linings.

� The superstructure usually houses the generating units andexciters, the switch board and operating room.

� Vertical-axis units (whose turbines are placed just below the

floor level) generally require less floor space than those

mounted on horizontal axes.

� The cost of the superstructure can be reduced considerably by

housing individual generators only (outdoor power house),

although it has the disadvantage that maintenance works havealthough it has the disadvantage that maintenance works have

to be restricted to good weather conditions only.

� Under certain topographic conditions, particularly when the

power plant is situated in narrow canyons with no convenient

site for a conventional type of power house, this may be

located underground.

� It is essential to equip a power house with a crane to lift

and move equipment for installation and maintenance

purposes.

� Travelling cranes spanning the width of the building and

capable of traversing its entire length is normally used.capable of traversing its entire length is normally used.

� The crane rail elevation depends on the maximum

clearance required when the crane is in operation which,

in turn, determines the overall height of the

superstructure.

Section through a power station

� Tail Race� The tail race is the waterway into which the water from

the turbine units is discharged (through draft tubes ifreaction-type units are used).

� It may be very short and if the power house is close tothe stream the outflow may be discharged directly into it.

� On the other hand, if the power house is situated at a� On the other hand, if the power house is situated at adistance from the stream the tail race may be ofconsiderable length.

� Proper tail race design ensures, especially in low-headplants, that more of the plant gross head is available forpower development.

� The tail race in the vicinity of the draft tube exit (head of tail

race) must be properly lined, as it may otherwise degrade and

cause lowering of the tailwater elevation of scouring of the

channel bottom.

� Should this be allowed to progress the designed turbine

setting level would alter, thus causing reduced efficiency of the

turbine (cavitation in the turbine runner), and remedial

measures (artificial raising of the water level) would have to be

taken.

� The tail race channel may sometimes aggrade, in which case

the gross head at the plant decreases, with a resulting

reduction in power output.

� This situation may arise if the main spillway outflow is

close to the tail race without an adequate separating wall.

� Gates, with an appropriate hoisting mechanism, must be

provided at the draft tube outlet (between the piers and

tail race) to isolate the draft tube for maintenance works.tail race) to isolate the draft tube for maintenance works.

� The tail race of the underground power house is

invariably a horizontal tunnel into which the turbine units

discharge the water.

� Such tunnel flow could sometimes take place under

pressure