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HYDRAULIC 1.Introduction: Hydraulic (water) turbines are the machines which convert the water (Hydro power) into Mechanical energy. The water energy may be either in the form of as we find in , reservoirs, or in the form of in flowing water. The shaft of the directly coupled to the electric generator which converts mechanical energy in to electrical energy. This is known as " Hydro-Electric power". 2.Classification of Hydraulic Turbines: Water turbines are classified into various kinds according to i) the action of water on blades, ii) based on the direction of fluid flow through the runner and iii) the specific of the machine. (i) Based on the action of Water on Blades : Thesemaybeclassifiedinto:1)typeand2)Reactiontype In impulse turbine, the pressure of the flowing fluid over the runner is constant and generally equal to an atmospheric pressure. All the available potential energy at inlet will be completely converted intokinetic energy using nozzles, which in turn utilized through a purely impulse effect to produce . Therefore, in impulse turbine, the available energy at the inlet of a turbine is only the kinetic energy. In reaction turbine, the turbine casing is filled with water and the water pressure changes during flow through the rotor in addition to kinetic energy from nozzle (fixed blades).As a whole, both the pressureenergy and kinetic energy are available at the inlet of reaction turbines for producing power. (ii) Based on the direction of Flow of Fluid through Runner : Hydraulic machines are classified into : a) Tangentialorperipheralflow b) Radial inwardor outwardflow c) Mixedordiagonalflow d) Axialflow types. a) TangentialFlow Machines : In tangential flow turbines, the water flows along the tangent to the path of rotation of the runner. Example: b) Radial Flow Machines : In radial flow machine, the water flows along the radial direction and flow remains normal to the axis of rotation as it passes through the runner. It may be inward flow or outward flow. In Inward flow turbines, the water enters at the outer periphery and passes through the runner inwardly towards the axis of rotation and finally leaves at inner periphery. Example: . In outward flow machines the flow direction is opposite to the inward flow machines.

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c) Mixed or Diagonal Flow : In this type of turbine, the flow of fluid may enter at the outer periphery, passes over the runner inwardly and leaves axially or parallel to the axis of rotation and vice-versa. Examples: Modern Francis turbine, Deriaz turbine. d)Axial Flow Devices : In this type of turbine, the waterflows along the direction parallel to the axis of rotation.Examples: , propeller turbine etc., (iii) Based on : Hydraulic turbines are classified into : (a) Low Specific Speed: Which employs high head in the range of 200m up to 1700 m. These machines requires low discharge.Examples: Pelton wheel.

NS = 10 to 30 single jet and 30 to 50 for double jet Pelton wheel. (b) Medium Specific Speed : Which employs moderate heads in the range of 50m to

200 m.Example: Francis turbine, NS = 60 to 400. (c) High Specific Speed : Which employs very low heads in the range of 2.5 m to 50

m. These requires high discharge.Examples: Kaplan, Propeller etc., NS = 300 to 1000. 3 Pelton Wheel : This is a impulse type of tangential flow hydraulic turbine. It mainly posses:(i) Nozzle (ii) runner and buckets (iii) casing (iv) brake nozzle. Fig. 1 shows general layout of hydro-electric power plant with pelton wheel. The water from the is made to flow Headrace h through the penstock. At Dam the end of the penstock, Casing nozzle is fitted which Penstock Vanes convert the potential Nethead (H) energy into high kinetic energy. The speed of the jet Nozzle Grosshead issuing from the nozzle can (H) be regulated by operating g Jetofwater the spear head by varying Spear Tailrace the flow area. The high velocity of jet impinging over the buckets due to which the runner starts Fig.1LayoutofHydro-electricpowerplant rotating because of the impulse effects and thereby hydraulic energy is converted into mechanical energy.After the runner, the water falls into tail race. Casing will provide the housing for runner and is open to atmosphere. Brake nozzles are used to bring the runner from high speed to rest condition whenever it is to be stopped. In order to achieve this water is made to flow in opposite directionthrough brake nozzle to that of runner .

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4.0 Heads and Efficiencies of Hydraulic Turbines: 4.1 Hydraulic Heads : (a) Gross head: It is the difference between the head race and tail race level when there is no flow.As such it is termed as static head and is denoted as Hs or H g . (b) Effective head: It is the head available at the inlet of the turbine. It is obtained by considering all head losses in penstock. If hf is the total loss, then the effective headabove the turbine is H = Hg - h f (1) 4.2 Efficiencies : Variousefficiencies of hydraulic turbines are:

i) Hydraulicefficiency (hH ) ii) Volumetric efficiency ( hvol ) iii) Mechanicalefficiency (hmech ) iv) Overall efficiency ( h 0 )

i) Hydraulic efficiency (hH ): It is the ratio of power developed by the runner to the water power available at the inlet of the turbine. . (m /g ) (U V± U V )r Q(U V ± U V ) H ______c. 1 u1 2 u2______1 u1 2 u2 ___ e hH = = = (2) (m gH ) / gc r Q g H H H - h h = ______f . Where H is effective head at the inlet of turbine. H H ii) Volumetric efficiency (hvol ): It is the ratio of the quantity of water actually striking the runner to the quantity of water supplied to the runner.

QQ-Qa __D _ hvol = = DQ = amount of water that slips directly to the tail race. QQth =loss. (3)

iii) Mechanicalefficiency (hmech ): It is the ratio of shaft power output by the turbine to the power developed by the runner. Shaft output power h = ______mech Power developed by the runner ______SP hmech = (4) rQ (U1 V u1± U 2 V u2 )/g c

iv) Overall efficiency (ho ) : It is the ratio of shaft output power by the turbine to the water power available at inlet of the turbine. Inlet Shaft output power ______shaftoutput ho = Water power at inlet Turbine Generator ______SP ho = rQ gH h h Ifh =1 H ï mech ï ho =( hH hvol ) h mech h ï ho = hH,act hmech (5) Fig.7.2 Variousefficienciesofpowerplant

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5.0 WorkDonebythePelton Wheel:

V1=Vu1 U Vr1

b2 Nozzle q

0 b2=180 - q

Vr2

Vf V2

b2

ïU ïVu2 ï Fig3Shapeofbucket& Velocitydiagram

Whereq is the angle through which the jet is deflected by the bucket.b2 is the runner tip angle = 180 -q . Fig. 3 shows the inlet and outlet velocitytriangles . Since the angle of entrance of jet is zero, the inlet velocity triangle collapses to a straight line. The tangential component of absolute velocity at inlet Vu1 = V1 and the relative velocity at the inlet is V r1 = V 1 - U. From the outlet velocityDle .

Vu2 = V r2 cosb 2 – U ... =CVC)b V r1 cosb 2 – U ( r2 = b V r1 ... Vu2 = (V1 – U)C b cosb 2 -U ( V r1 = V 1 – U) Work done / kg of water by the runner

W = U (Vu1 + V u2 ) / g c ( + ve sign for opposite direction of V u1 and V u2 ) = U [Vu1 + (V1 – U)C b c osb 2 – U] / g c

= U [(V1–U)C + (V 1 – U) b cosb 2 ] / gc

W = U [(V1 – U) (1+Cb cosb 2 )] / g c (6) The energy supplied to the wheel is in the form of kinetic energy of the jet which is equal to 2 V1 /2gc . W U [ (V – U) ( 1+C cosb )] / g Hydraulic efficiency,h =______= ______1 b 2 c H 2 2 V1 / 2gc V1 / 2gc 2U (V – U) ( 1+C cosb ) h =______1 b 2 (7) H 2 V 1

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dh For maximum hydraulic efficiency,____H = 0 dU 2(1+ C cosb ) as______b 2 0 , V – 2U = 0 2 1 V 1

\V1 = 2U or f = U/V 1 = 0.5

U =½ V1 = 0.5 V 1 (7.8) This shows that the tangential velocity of bucket should be half of the velocity of jet for maximum efficiency. Then, h 2U (2U – U) ( 1+C cosb ) = ______b 2 H,max 2 (2U) h 1+C cosb =______b 2 (9) H,max 2 0 If Cb = 1, then the above equation gives the maximum efficiency forb2 = 0 5.1 WorkingProportionsofPelton Wheel:

(i) Ideal velocity of jet from the Nozzle, Vth =Ö2gH (10)

&Actual velocity of jet , V1 = C v Ö2gH

Cv = coefficient of velocity for nozzle is in the range of 0.97 to 0.99 ( ii) Tangential velocity of buckets, U=f Ö2gHÞ f =U/(Ö 2gH ), (11) Wheref =Speedratioandisintherangeof0.43to0.48 (iii) Least diameter of the jet, (d) 2 2 pd pd Totaldischarge,Q =nCV =n Ö2gH (12) T 4 1 4 v Where n = number of jets (nozzles) (iv) Mean Diameter or Pitch diameter of Buckets or Runner : (D) pDNUx 60 Tangentialvelocity,U=______orD= ______(13) 60p N (v)Number ofBucketsrequired(Z): The ratio of mean diameter of buckets to the diameter of jet is known as "Jet Ratio". i.e., m = D/d. m D \ Z=___ +15= ____ +15wheremrangesfrom6to35. (14) 2 2d

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Example1 : Pelton wheel has to be designed for the following data : power to be developed = 5880 kW, Net head available = 300m, Speed = 550 RPM, ratio of jet diameter to wheel diameter = 1/10 and overall efficiency = 85%. Find the number of jets, diameter of jet,

diameter of the wheel and the quantity of water required. Assume CV = 0.98,f = 0.46.

Solution :Given: P= 5880 kW, H = 300m, N = 550 rpm, d/D= 1/10,ho = 0.85, C V = 0.98, f = 0.46 Totaldischarge Q : ______Px 1000 ______Px 1000 3 hO= or Q = = 2.35 m /s rQg HhO r gH

V1 =C V ÖÖ2gH = 0.98 2 x 9.81 x 300 =75.19 m/s Tangential velocity of wheel, U=f Ö2gH = 0.46 xÖ 2 x 9.81 x 300 = 35.29 m/s Diameter of wheel, D U =p DN/60 35.29 =p x Dx 550/60 D = 1.225 m. Diameter of jet d, d/D = 1/10\ d =D/10 = 1.225/10 = 0.1225 m 2 Totaldischarge,QT =n(p d /4)V1 2 2.35=np x0.1225 x75.19/4 Number ofJets, n=2.65» 3. Example 2 : A single jet impulse turbine of 10 MW capacity is to work under a head of 500m. If the specific speed = 10, over all efficiency = 0.8 and the coefficient of velocity = 0.98, find the diameter of the jet and bucket wheel.Assumef = 0.46.

Solution:Given: P =10,000kW,H=500m,NSOV =10,h =0.8,C =0.98,d=?,D=?, f =0.46 1/2 ______N(P) ______N Ö10,000 Ns = = =10 H5/4 (500)5/4 N=236.4rpm

Velocityofjet, V1 =C V Ö2gH =97.06m/s Tangentialvelocityofbucket,U=f ÖÖ2gH =0.46 2gH U=45.56m/s U=p DN/60= p Dx236.4/60=45.56m/s Diameter ofawheel ,D=3.68m

hO =Px1000/(r QgH) 0.8=10,000/(9.81x500xQ) 2 2 3 Q=2.548m /s=p d /4 V1 = p d /4x97.06 Diameter ofjet, d=0.183mor18.3cm

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Example 3 : A double jet pelton -wheel is required to generate 7500 kW when the available head at the base of the nozzle is 400m. The jet is deflected through 165O and the relative velocity of the jet is reduced by 15% in passing over the buckets. Determine (i) the diameter of each jet (ii) Total flow (iii) Force exerted by

the jets in the tangential direction. Assume generator efficiency is 95%,hO = 80%, speed ratio = 0.47 Solution: OO Given: n=2,Pg =7500kW,H=400,b 2 =15 =(180-165 ), V r2 =0.85 V r1 , h g =0.95 ho=0.8, f =0.47=U/Ö2gH

ïV1 = V u1 ï ï b2 o ïU ï Vr1 ï q=165 P Turbine G O/L

Vr2 V2 Vf

b2 ïU ï ïVu2 ï

Now,hg =(O/P)/(I/P)=P g /P orP=P g /h g =7500/0.95=7894.74kW OutputbytheturbineP =7894.74kW Px1000P h =______= _____ 0 rQgHQgH P 7894.74 Totalflow, Q=______= ______=2.515m3 /s gHhO 9.81x400x0.8

Velocityofjetfromnozzle, V1 = Ö2gH

=Ö2x9.81x400 =88.6m/s= Vu1 2 2 ____pd 2x ______p d Alsoflow,Q =2.515=nV1 = 88.6 T 4 4 Diameter ofthejet, d=0.1344m 0.47=U/88.6Þ U=41.64m/s Vr1 =(V1 -U)=88.6-41.64=46.96m/s Vr2 =0.85x46.96=39.92m/s 0 Vf =V r2 sin15Þ V f =10.33m/s Totaltangentialforce, 0 (U - Vu2 )= V r2 cosb 2 =39.92cos15 =38.56m/s FT =r Q(Vu1 - V u2 )/g c V =41.64-38.56 =1000x2.515(88.6-3.08)/1000 u2 Vu2 =3.08m/s FT =215.08kN

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6.0 Reaction Turbine: In reaction turbines, only part of total head of water at inlet is converted into velocity head before it enters the runner and the remaining part of total head is converted in the runner as the water flows over it. In these machines, the water is completely filled in all the passages of runner. Thus, the pressure of water gradually changes as it passes through the runner. Hence, for this kind of machines both pressure energy and kinetic energy are available at inlet. e.g., Francis turbine, Kaplan turbines, Deriaz turbine.

6.1 Francis turbine Shaft

Scroll casing

Guidevane Regulatingrod Runner Drafttube Tailrace

Scroll casing Runner Link

Guidevane

Guidewheel

Fig..4Francisturbine

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Francis turbine which is of mixed flow type is as shown in Fig. 4. It is of inward flow type of turbine in which the water enters the runner radially at the outer periphery and leaves axially at its center. Thisturbinemainly consistsof:(i)Scrollcasing(ii)Stayring(iii)Guidevanes (iv)Runner(v)Drafttube. (i) Scroll Casing : The water from penstock enters the scroll casing (called spiral casing) which completely surrounds the runner. The main function of spiral casing is to provide an uniform distribution of water around the runner and hence to provide constant velocity. In order to provide constant velocity, the cross sectional area of the casing gradually decreses as the water reaching runner (ii) Stay ring :The water from scroll casing enters the speed vane or stay ring. These are fixed blades and usually half in number of the guide vanes. Their function is to (a) direct the water over the guide vanes, (b) resist the load on turbine due to internal pressure of water and these load is transmitted to the foundation. (iii) Guide Vanes: Water after the stay ring passes over to the series of guide vanes or fixed vanes. They surrounds completely around the turbine runner. Guide vanes functions are to (a) regulate the quantity of water entering the runner and (b) direct the water on to the runner. (iv) Runner : The main purpose of the other components is to lead the water to the runner with minimum loss of energy. The runner of turbine is consists of series of curved blades (16 to 24) evenly arranged around the circumference in the space between the two plate. The vanes are so shaped that water enters the runner radially at outer periphery and leaves it axially at its center. The change in direction of flow from radial to axial when passes over the runner causes the appreciable change in circumferential force which in turn responsible to develop power. (v) Draft Tube : The water from the runner flows to the tail race through the draft tube. A draft is a pipe or passage of gradually increasing area which connect the exit of the runner to the tail race. It may be made of cast or plate steel or concrete. The exit end of the draft tube is always submerged below the level of water in the tail race and must be air-tight. The draft tube has two purposes : (a) It permits a negative or suction head established at the runner exit, thus making it possible to install the turbine above the tail race level without loss of head. (b) It converts large proportion of velocity energy rejected from the runner into useful pressure energy. 7.0 Draft Tubes: There are different types of draft tubes which are employed to serve the purpose in the installation of turbine are as shown in Fig.5. It has been observed that for straight divergent type draft tube, the central cone angle should not be more than 8O . This is because, if this angle is more than 8O the water flowing through the draft tube without contacting its inner surface which result in eddies and hence the efficiency of the draft tube is reduced.

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Hs Hs Tailrace Tailrace

(a)Conicaldraft (b)Simpleelbow (c)MoodySpreading tube tube tube

Hs Tailrace

(d)Drafttubewithcircularinletandrectangularoutlet Fig.5. Typesofdrafttubes (a) Straight divergent conical tube (b) Moody spreading tube (or Hydracone) (c) Simple elbow tube (d) Elbow tube having circular cross section at inlet and rectangular cross section at outlet. 8.0 Work Done and Efficiencies of Francis Turbine: Inlet InletInlet V1 V Vr1 f1 V1 V a1 b1 1 Vr1 Vf1 V ï Vu1 ï V f1 Runner r1 b blade 1 a1 b1 a1

ï U1 ï ï U1 ï ï Vu1 ï ï Vu1 ï V =V Vr2 w 2 f2 (a) When V >U (b)If WhenU > V a u1 1 1 u1 2 b2 O O Outlet orb1 <90 orb1 >90 U ïï 2 Outletvelocitytriangle D2 Radialdischarge D1 ie., V2 = V f2 U a D Vr2 O 1 1 V2=Vf2 &a2 =90 U2 a D2 or Vu2 =0 b2 V =0 Runner u2 U2 ïï Fig.6 Velocitytrianglesfordifferentconditions

The absolute velocity at exit leaves the runner such that there is no whirl at exit ie., Vu2 = 0. The Inlet velocity triangle are drawn for different conditions as shown in Fig 6 (a) & (b).

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(i) Work done / kg of water , W = (U1 Vu1 - U 2 V u2 )/g c As Vu2 = 0 in francis turbine (axial outlet velocity) W =U1 Vu1 /g c (15)

(ii) Hydraulic efficiencyhH : Workdonebytherunner h = ______H Availableenergyatinlet UV rQ[ ______1 u1 [ g He h = ______c = H ______rQgH H gc 2 2 Also gH - (V / 2g ) H - (V /2g) h =______4 c = ______4 (16) H gH H

Where H is the head at inlet, V4 = velocity of water at exit of the draft tube

(iii) Volumetric efficiency (hvol ): It is the ratio of quantity of water through the runner (Q2 ) to the quantity of water suppled (Q 1 ) ______(Q1 -D Q) hvol= Q 2 / Q 1 = Q1 DQ = loss = Q1 - Q 2

(iv) Mechanical efficiency (hmech ): Power output at the shaft end h = ______mech Power developed by the runner

(v) Overall efficiency (ho ): Power output at the shaft Shaft power h =______= ______o Power available at inlet of the turbine Water power at inlet

ho=( h H x h vol ) h mech = h H,act x h mech (17) 8.1 Working Proportions of Francis Turbine:

(1) Tangential velocity of runner (U1 ): Speed ratio,f = U1 /Ö2gH , wheref ranges from 0.6 to 0.9 (18)

(2) The ratio of flow velocity Vf1 at inlet tip of the vane to the spouting velocity (Ö2gH ) is known as the flow ratio

\ y = Vf1 /Ö2gH ,wherey rangesfrom0.15to0.30(19) (3) If n is the number of vanes in the runner, 't' is the thickness the vane at inlet

and B1 is the width of the wheel at inlet then, thearea of flow at inlet,

Af1 = (p D - n t)B1 = C p D 1 B 1 (20)

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where C = Contraction factor which represents the area occupied by the vane thickness in the runner and it is a few percentage ofp D1 . Discharge, Q =Af Vf = Cp D 1 B 1 V f 1 = C p D 2 B 2 V f 2 Normally it is assumed that D1 = 2D 2 & Vf 1 = V f 2 , B 2 = 2B 1 2 ______H-(V4 /2g) Note : Hydraulic efficiency is given byhH = if the velocity V the exit of H 4 2 draft tube is given since V4 /2g represents the head loss at the exit of draft tube.

Efficiency of the draft tube : V1 2 ______Actual regain of pressure head h = V = V d Velocityhead at entrance to draft tube. 2 3

2 2 2 2 [][](V - V )/2g -hf (V - V )/2g -hf h =______2 4 = ______3 4 (21) d 2 2 3 V2 /2g V 3 /2g

Where V3 = Velocityof water at exit of runner = V2 V4 V4 = Velocityof water at exit of draft tube. 4 hf = head lose in draft tube. .Example 4 : Show that the utilization factor for an inward flow reaction turbine with relative velocity component at inlet perpendicular to the tangent of the wheel and the 2 2 absolute velocity at the exit is radial is given byÎ = 2 cos a1 / (1+ cos a 1 ) Wherea1 is the angle made by the entering fluid with tangent of the whee. Solution : CombinedInlet&outletvelocitytri nglea

V V b = a V V a b a b Utilization, ïU = V ïU ï 2 E UV U V =0 Î=______= ______1 u1 = ______1 2 E+ V /2g U V + V2 /2g U2 + V2 /2 2 c 1 u12 c 1 2 2U2 2U2 ______1 ______1 ... 2 2 2 = = ( V2 = V 1 - U 1 ) 2U2 + V2 -U 2 U2 + V2 1 1 1 1 1 2 2 2 2V cos a 2cos a1 Î =______1 1 = ______22 2 2 V cosa + V 1+cos a1 1 1 1

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Example 5 : In a Francis turbine, the discharge is radial. The blade speed at inlet = 25 m/s. At the inlet tangential component of velocity = 18 m/s. The radial velocity of flow is constant and equal to 2.5 m/s. Water flows at the rate of 0.8 m3 /s. The utilization factor is 0.82. Find: i) Euler's head ii) Power developed iii) Inlet blade angle iv) Degree of reaction (R). Draw the velocity triangles. 3 Solution : Given : U1 = 25 m/s, Vu1 =18 m/s, V f = 2.5 m/s, Q = 0.8m /s, (i) Euler's head : gHe = E = (U 1 Vu1 ) /g c as V u2 = 0, E = 25 x 1 8/1 = 450 J/kg = gHe \ He = 450/9.81 = 45.87 m. (ii) Power developed (P) :

P=r QE = r Q (gHe )= (1000 x 0.8 x 450) / 1000 P= 360 kW

(iii) Inlet blade angle (b1 ): Inlet Outlet

V1 Vr2 Vr1 V2 =Vf2 Vf

a1 b1 a2 b2 ï V ï ï U ï ï U ï V =0

2 2 1/2 V1 = (Vf +V u 1 ) = 18.17 m/s Vf1 =V f 2 = V f = V 2 = 2.5 m/s

tanb1 = V f /(U 1 - V u1 ) = 2.5 / (25 - 18) 0 b1 = 19.65 (iv) Degree of reaction (R) 2 2 2 2 He - [ (V - V )]/2g 45.87 - [(18.17 - 2.5 )/2 x 9.81] R =______1 2 = ______He 45.87 R = 0.64

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9.0 Propeller & Kaplan Turbines The propeller turbine consists of an axial flow runner with 4 to 6 blades of aerofoil shape. The spiral casing and guide blades are similar to that of the Francis turbines. In the propeller turbines, the blades mounted on the runner are fixed and non-adjustable. But in Kaplan turbine the blades can be adjusted and can rotate about the pivots fixed to the boss of the runner. This is only the modification in propeller turbine. The blades are adjusted automatically by a servomechanism so that at all loads the flow enters without shock. Shaft Scrollcasing Guidevane Guidevane

Inletofrunnervanes Runnervanes Outletofrunnervanes Boss Tailrace d D

Drafttube

Fig.7.7MaincomponentsofKaplanturbine.

The Kaplan turbine is an axial flow reaction turbine in which the flow is parallel to the axis of the shaft as shown in the Fig.7.7. This is mainly used for large quantity of water and for very low heads (4-70 m) for which the specific speed is high. The runner of the Kaplan turbine looks like a propeller of a ship. Therefore sometimes it is also called as propeller turbine. At the exit of the Kaplan turbine the draft tube is connected to discharge water to the tail race. Inlet Outlet 1) At inlet, the velocityDle is as V1 V r2 V =V shown in Fig.8. f 2 Vr1 Vf b 2) At the outlet, the discharge is a b1 always axial with no whirl ï U ï ï U ï

velocity component i.e., outlet ï Vu1 ï Vu2=0 velocity triangle just a right angle triangle as shown Fig.8. Fig.8 VelocitytrianglesforKaplan Turbine.

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9.1 Working Proportions of Kaplan Turbine

i) Tangentialvelocity of blades based on tip diameter (UD )

UD = f Ö2gH (22) where, D = Tip diameter or Outer diameter of the runner d = Hub diameter or Boss diameter of the runner \ d/D = 0.35 to 0.60 ii) Flow velocity (Vf )

y = Vf /Ö2gH Usually, 0.35

Solution: P =80,000HP =58,800kW,H=25m,hO =0.9, f =1.6, y =0.5, d/D=0.35,Tofind :D,N.

h0 =P/( r QgH)

Q = P/(h0 r gH) = 58,800 / (0.9 x 9.81 x 1 x 25) Q = 266.4 m3 /s p2 2 p 2 2 Again, Q =___ (D - d )y 2gH = ____ D 1- (d/D) x y 2gH 4Ö 4 [] Ö ____p 2 2 266.4 = D (1 - 0.35 )x (0.5) Ö2 x 9.81 x 25 4 D = 5.91 m

Also,f = UDD /ÖÖ2gH = U /(2 x 9.81 x 25) = 1.6

UD = 35.44 m/s =p DN/60 = p x 5.91 x N / 60 \ N = 114.5 rpm.

Example 7:Determine the efficiency of a Kaplan turbine developing 2940 kW under a head of 5m. It is provided with a draft tube with its inlet diameter 3m set at 1.6m above the tail race level.Avacuum pressure gauge connected to the draft tube inlet indicates a reading of 5m of water.Assume that draft tube efficiency is 78%.

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Solution:Given: P =2940kW,H=5m,Hs =1.6m,p 2 /r g=-5m, h draft =0.78, h o =? ApplytheBernoulli'sequationbetween(2)&(3)weget 2 2 p2 /r g = -[] H s + (V2 - V 3 )/2g 2 d =38m 2 2 2 - 5 = - 1.6 -[] (V2 - V 3 )/2g 2 2 2 H (V2 - V 3 )/2g = + 3.4 m Z1 s p2 /r g=5 2 2 H =1.6m (V2 - V 3 ) = + 66.708 (1) s Actualpressuregainindrafttube Drafttubeefficiencyh = ______3 d Pressureatinlettothetube. Z2 Datumline (V2 -V 2 )/2g(V 2 - V 2 ) =______2 3 = ______2 3 =0.78 2 2 V2 /2g V 2 2 \ 1- (V3 /V 2 ) = 0.78 2 2 V3 = 0.22 V 2 (2) 2 \In equation (1), V2 (1 - 0.22) = 66.708 ... V2 = 9.25 m/s = Vf ( Axial discharge in Kaplan turbine) p 2 Now discharge, Q =A V =____ x 3 x 9.25 = 65.38 m3 /s f f 4 Overall efficiency of a Kaplan turbine, Power output P h =______= ______0 Hydropoweratinletr QgH 2940x1000 h =______=91.68% 0 1000x9.81x65.38x5

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