HYDRAULIC TURBINES 1.Introduction : Hydraulic (water) turbines are the machines which convert the water energy (Hydro power) into Mechanical energy. The water energy may be either in the form of potential energy as we find in dams, reservoirs, or in the form of kinetic energy in flowing water. The shaft of the turbine directly coupled to the electric generator which converts mechani mec hanical cal ener energy gy in to ele electr ctrica icall ene energ rgy y. Thi Thiss is know known n as " Hyd Hydroro-Ele Electr ctric ic powe power". r".
2.Classif 2.Cl assificati ication on of Hydr Hydrauli aulicc Turbi urbines nes : Water turbines are classified 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 speed spe ed of the mac machin hine. e. (i)) Ba (i Base sed d onthe a cti ction on of Wat ater er on Bla Blades: des:
These may be classified into:1) Impulse type and 2) Reaction type In im impul pulse se tu turb rbin ine, e, th thee pr pres essu sure re of th thee fl flow owin ing g fl flui uid d ov over er th thee ru runn nner er is co cons nsta tant nt an and d generally equal to an atmospheric pressure. All the available potential energy at inlet will will be completely converted into kin kineti eticc ener energy gy usi using ng nozz nozzles les,, which in turn utilized through a purely impulse impulse effect effect to produce work. Therefore, in impulse turbine, turbine, the available energy at th thee in inle lett of a tu turb rbin inee is onl only y th thee ki kine neti ticc ene energ rgy y. 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 pressure en ener ergy gy and ki kine neti ticc en ener ergy gy are available at the inlet of reaction reacti on turbi turbines nes for for produc producing ing power power.. (ii)) Ba (ii Basedon sedon thedire thedirecti ction on of Fl Flow ow of Fl Fluidthro uidthroug ugh h Ru Runn nner er : Hydrau Hyd raulic lic mac machin hines es are cla classi ssifie fied d int into o: a) Tangen angentialor tialor peri peripher pheral al flow b) Ra Radi dial al in inwa ward rd or ou outw twar ard d fl flow ow c) Mi Mixedor xedor dia diagon gonalflow alflow d) Axi Axialflow alflow typ types es.. a) Tange angentia ntiall Flow FlowMach Machines ines : In ta tange ngent ntia iall fl flow ow tu turb rbin ines es,, th thee wa wate terr fl flow owss al alon ong g th thee ta tang ngent ent to th thee pa path th of ro rota tati tion on Pelton ton whe wheel el of the run runner ner.. Exam Example ple:: Pel b) Ra Radia diall Fl Flow ow Ma Machi chines: nes: In ra radi dial al fl flow ow ma machi chine, ne, th thee wa wate terr fl flow owss al along ong th thee ra radi dial al di dire rect ctio ion n an and d fl flow ow re rema main inss normal to the axis of rotation as it passes through the runner. It may be inward flow or outward outwar d flow flow.. In In Inwa ward rd fl flow ow tu turb rbin ines es,, th thee wat water er ent enter erss at th thee out outer er pe peri riph pher ery y an and d pa pass sses es th thro roug ugh h the run runner ner inw inward ardly ly tow toward ardss the axi axiss of rot rotati ation on and fin finall ally y lea leaves ves at inn inner er per periph iphery ery.. Example: Fran Francis cis turb turbine ine. In outward flow machines the flow direction is opposite to the inward inwar d flow machin machines. es. Dr.M.S.Govinde Dr .M.S.Govinde Gowda, Principal, Principal , Alva’s Inst. Of Engg & Tech, Tech, Moodbidri
2
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 water flows along the direction parallel to the axis of rotation. Examples: Kaplanturbine, propellerturbineetc., (iii) Basedon Specific Speed: 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. N = 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, N = 6 0 to 4 00. (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., N = 300 to 1000.
3 PeltonWheel : 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 da m is m ad e to flo w Head race h through the penstock. At Dam Casing the end of the penstock, Penstock Vanes nozzle is fitted which Net head c o n ve r t t h e p o te n ti a l (H) energy into high kinetic energy. The speed of the jet Nozzle Gross head issuing from the nozzle can (Hg) be regulated by operating Jet of water the spear head by varying Spear Tail race the flow area. The high velocity of jet impinging over the buckets due to Fig.1 Layout of Hydro-electric power plant which the runner starts 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 direction through brake nozzle to that of runner . Dr.M.S.Govinde Gowda, Principal, Alva’s Inst. Of Engg & Tech, Moodbidri
3
4.0 Heads andEfficiencies of HydraulicTurbines : 4.1 Hydraulic Heads : (a) Gross head: It is the difference between the head race and tail race level when there isno flow.As such itis termed asstatic headand isdenoted asHs orHg . (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 headabovetheturbine is H= Hg -hf (1) 4.2 Efficiencies : Various efficiencies of hydraulic turbines are:
i) Hydraulic efficiency ( )
ii) Volumetric efficiency (vol)
iii) Mechanical efficiency (mech)
iv) Overall efficiency (0)
i) Hydraulic efficiency ( ): It is the ratio of power developed by the runner to the water power available at the inlet of the turbine.
Q(U1Vu1U2Vu2) ___ (m /gc ) (U1Vu1 U2Vu2) ____________ He ___________________ H = = = (2) Q g H (m gH ) / gc H H - hf . H = ______ Where H is effective head at the inlet of turbine. H ii) Volumetric efficiency ( vol): It is the ratio of the quantity of water actually striking the runner to the quantity of water supplied to the runner. Q Q - Q vol = = __ _ Q = amount of water that slips directly to the tail race. Q Q = loss. (3) .
.
iii) Mechanical efficiency ( mech): It is the ratio of shaft power output by the turbine to the power developed by the runner. Shaft output power mech = ___________________________ Power developed by the runner SP mech = ____________________ (4) Q (U1Vu1 U2Vu2)/gc iv) Overall efficiency ( o ) : It is the ratio of shaft output power by the turbine to the water
power available at inlet of the turbine. Shaft output power o = __________________ Water power at inlet
Inlet shaft output Turbine
Generator
SP
o = _______ Q gH o = ( vol) mech o = mech
H (5)
mech
If
=1
Fig. 7.2 Various efficiencies of power plant
Dr.M.S.Govinde Gowda, Principal, Alva’s Inst. Of Engg & Tech, Moodbidri
4
5.0 Work Done by the Pelton Wheel : V1 =Vu1 Vr1
U
2
Nozzle
2= 180 - 0
V V
V
2
U
V
Fig 3 Shape of bucket & Velocity diagram Where is the angle through which the jet is deflected by the bucket. 2 is the runnertip angle = 180 - . Fig.3 shows theinletand outlet velocity triangles . Sincethe angle of entrance of jet is zero, the inlet velocity triangle collapses to a straight line. The tangential component of absolutevelocity atinlet Vu1 = V1 and the relative velocityat the inlet isVr1 = V1 - U. Fromthe outletvelocity
le.
Vu2 = Vr2 cos 2 – U = C bVr1 cos
2 – U
(
.
Vu2 = (V1 – U) Cb cos 2 -U
(
.
.
.
.
.
Vr2 = CbVr1 ) Vr1 = V1 – U)
Work done /kg of water by the runner W =U (Vu1 +V u2)/ gc (+ ve signforopposite directionof Vu1 andVu2 ) = U [Vu1 + (V1 – U) Cb c os 2 – U] / gc = U [(V1 – U )+ (V1 – U) Cb cos 2] /gc W = U [(V1– U) (1+C b cos 2 )] / gc
(6)
The energy supplied to the wheel is in the form of kinetic energy of the jet which isequal to 2
V1 /2gc. Hydraulic efficiency,
W U [ (V1 – U) ( 1+Cb cos2 )] / gc _________________________ = _______ = 2 2 V 1 / 2gc
V 1 / 2gc
2U(V1 –U)( 1+Cb cos2 ) = ______________________ 2
(7)
V1
Dr.M.S.Govinde Gowda, Principal, Alva’s Inst. Of Engg & Tech, Moodbidri
5
For maximum hydraulic efficiency, 2(1+ C b cos2) __________
as
2 V1
V1 = 2U or
dH ____ =0 dU
0, V1 – 2U = 0
= U /V1 = 0.5
U = ½ V1 = 0.5 V1
(7.8)
This shows that the tangential velocity of bucket should be half of the velocity of jet for maximum efficiency. Then, 2U (2U – U)( 1+C b cos2) ________________________ = H, max 2 (2U)
H, max
1+C b cos2 ___________ = 2
(9)
If C b = 1, then the above equation gives the maximum efficiency for 2 = 0 5.1 Working Proportions of Pelton Wheel : (i)
Ideal velocity of jet from the Nozzle, Vth = 2gH
(10)
&Actualvelocityof jet,V1 =Cv 2gH Cv = coefficient of velocity for nozzle is in the range of 0.97 to 0.99 ( ii) Tangential velocityof buckets,
U = 2gH
= U/(2gH ),
(11)
Where = Speed ratio and is in the range of 0.43 to 0.48 (iii) Leastdiameterof the jet, (d)2 2 d d Total discharge, QT = n Cv 2gH V1 = n 4 4 Where n = number of jets (nozzles) (iv) Mean Diameter or Pitch diameter of Buckets or Runner: (D)
Tangential velocity, U =
DN _______ 60
or D =
U 60 _______ N
(12)
(13)
(v) Number of Buckets required (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 = ____ + 15 where m ranges from 6 to 35. (14) 2 2d Dr.M.S.Govinde Gowda, Principal, Alva’s Inst. Of Engg & Tech, Moodbidri
6
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, = 0.46. Solution :Given: P= 5880 kW, H= 300m, N= 550 rpm, d/D= 1/10, o = 0.85, CV = 0.98, =0.46 Totaldischarge Q : Px 1000 Px 1000 = _________ or Q = _________ = 2.35 m3 /s Qg H gH
2gH = 0.98 2 x 9.81 x 300 =75.19 m/s Tangential velocity of wheel, U= 2gH = 0.46 x 2 x 9.81 x 300 = 35.29 m/s V1 =CV
Diameter of wheel, D U = DN/60
35.29 = x Dx 550/60 D=1.225 m.
d =D/10 = 1.225/10 = 0.1225 m
Diameterof jetd, d/D = 1/10 2
Total discharge, Q T = n ( d /4)V1 2
2.35 = n x0.1225 x75.19 /4 Number of Jets, 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. Assume = 0.46. Solution :Given: P =10,000 kW, H =500m, N =10, = 0.8,C =0.98, d = ?, D = ?, 1/2 = 0.46 N (P) 10,000 _________ N ____________ Ns= = = 10 H5/4 (500)5/4 N = 236.4 rpm
Velocity of jet,
V1 = CV 2gH = 97.06 m/s
Tangential velocity of bucket, U =
2gH = 0.46 2gH
U = 45.56 m/s U = DN/60 = Dx236.4/60 = 45.56 m/s Diameter of a wheel , D = 3.68 m
= Px1000 / ( QgH) 0.8 = 10,000/(9.81 x 500 x Q) Q = 2.548 m3/s =
2
2
d /4 V1 = d /4 x 97.06
Diameter of jet, d= 0.183 m or 18.3 cm Dr.M.S.Govinde Gowda, Principal, Alva’s Inst. Of Engg & Tech, Moodbidri
7
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 165 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%, = 80%, speed ratio=0.47 Solution:
2 =15 = (180-165 ), Vr2 =0.85 Vr1 , g =0.95
Given : n =2, Pg =7500 kW, H =400,
o=0.8, =0.47=U/2gH
V =V
U
2
V
165
P
o
Turbine
O/L V
G
V
V
2
Now,
U
V
g = (O/P)/(I/P) = Pg /P or P=Pg/g = 7500/0.95 = 7894.74 kW
Out put by the turbine P = 7894.74 kW Px1000
P
0 = ________ = _____ QgH QgH Total flow,
Q=
P 7894.74 ________ _____________ = = 2.515 m3/s gH 9.81x400x0.8
Velocity of jet from nozzle, V1 = 2gH =
2 x 9.81 x 400 = 88.6 m/s = V 2
2
d V 2_______ x d ____ Also flow, QT = 2.515 = n = 88.6 4 0.47=U/88.6 U= 41.64 m/s Diameter of the jet, d = 0.1344 m Vr1= (V - U) = 88.6 - 41.64 = 46.96 m/s Vr2= 0.85 x 46.96 = 39.92 m/s Total tangential force, FT = Q (Vu1- Vu2) /gc
Vf =10.33 m/s (U - Vu2 ) = Vr2 cos 2 = 39.92 cos 15 = 38.56 m/s Vf =Vr2 sin 15
Vu2 = 41.64 - 38.56 = 1000 x 2.515 (88.6 - 3.08)/1000 Vu2 = 3.08 m/s FT= 215.08 kN Dr.M.S.Govinde Gowda, Principal, Alva’s Inst. Of Engg & Tech, Moodbidri
8
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
Guide vane
Regulating rod Runner Draft tube
Tail race
Scroll casing
Runner
Link
Guide vane
Guide wheel
Fig. .4 Francis turbine Dr.M.S.Govinde Gowda, Principal, Alva’s Inst. Of Engg & Tech, Moodbidri
9
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 itscenter. This turbine mainly consists of: (i)Scroll casing (ii)Stay ring (iii)Guide vanes (iv) Runner (v) Draft tube. (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) Stayring : The water from scroll casing enters the spee d 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 8 . This is because, if this angle is more than 8 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. Dr.M.S.Govinde Gowda, Principal, Alva’s Inst. Of Engg & Tech, Moodbidri
10
Hs
Hs Tail race
(a) Conical draft tube
Tail race
(b) Simple elbow tube
(c) Moody Spreading tube Hs
Tail race
(d) Draft tube with circular inlet and rectangular outlet
Fig.5. Types of draft tubes (a) Straight divergentconical 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 : V1
1
1
Runner blade
Inlet
Inlet Vf1
Vr1 Vu 1
V1
V2=Vf2
Outlet
2
U2
U1
Vf1
Vr1
Vu 1
U1 D1 U2 D2
Vu 1
1
O
or 1 < 90
D1
U1
(b) If When U1 > Vu
O
Vr1
Vf1
(a) When Vu 1> U1
2
D2
Runner
V1
Vr2
Inlet
or 1 > 90
Outlet velocity triangle Radial discharge ie., V2 = Vf2 Vr2
V2=Vf2
O
& 2 = 90
or Vu2 = 0
2 Vu 2=0
U2
Fig.6 Velocity triangles for different conditions 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 Fig6(a)&(b). Dr.M.S.Govinde Gowda, Principal, Alva’s Inst. Of Engg & Tech, Moodbidri
11
(i) Workdone/kgofwater, W = (U1Vu1 - U 2Vu2)/gc AsV u2 = 0 in francis turbine (axial outlet velocity)
W = U1 Vu1/gc (ii) Hydraulic efficiency
(15)
H : Work done by the runner
H = ________________________ Available energy at inlet U V
1 u1 Q [ ______ [ g
H
c = H = _____________ QgH H ______
gc
Also
2
2
gH - (V / 2gc) __________ H - (V4 /2g) H = ___________ = gH H
(16)
Where H isthe head atinlet, V4 = velocity ofwaterat exitof the draft tube (iii) Volumetric efficiency ( vol): It is the ratio of quantity of water through the runner (Q2 ) to the quantity of water suppled (Q1 )
(Q1 - Q) vol = Q2 / Q1= __________
Q=loss=Q1 -Q 2
Q1
(iv) Mechanicalefficiency ( mech):
Power output at the shaft end Power developed by the runner
mech = ____________________________ (v) Overall efficiency ( o): Power output at the shaft Shaft power ________________ o = _________________________ = Power available at inlet of the turbine Water power at inlet
o = (H x vol) mech = H,act x mech
(17)
8.1 Working Proportions of FrancisTurbine : (1) Tangential velocity of runner (U1 ):
Speed ratio, = U1 /2gH, where ranges from 0.6 to 0.9
(18)
(2) The ratio of flow velocity Vf1 at inlet tip of the vaneto the spoutingvelocity ( 2gH ) isknownasthe flow ratio
= Vf1 / 2gH, where ranges from 0.15 to 0.30
(19)
(3) If n isthe number ofvanes in the runner, 't' isthe thickness the vane at inlet and B1 is thewidthofthewheelatinlet then,the area of flow atinlet, Af1 = ( D - n t)B1 = C D1 B1 Dr.M.S.Govinde Gowda, Principal, Alva’s Inst. Of Engg & Tech, Moodbidri
(20)
12
where C = Contraction factor which represents the area occupied by the vane thickness in the runner and it is a few percentage of D1 . Discharge, Q = Af Vf = C D1 B1 Vf1= C D2 B2 Vf 2 Normally it is assumed that D1 = 2D2 & Vf1= Vf 2, B2 = 2B1 H- (V42 /2g) __________ = if the velocity V4 the exit of H draft tube is given since V42 /2g represents the head loss at the exit of draft tube.
Note : Hydraulic efficiency is given by
Efficiencyof thedraft tube :
2
Actual regain of pressure head Velocity head at entrance to draft tube.
d = _______________________________ 2
2
2
V
V=V
2
(V2 - V4 )/2g-hf ________________ (V3 - V4 )/2g-hf ________________ d = = 2 2 V2 /2g V3 /2g
(21)
3
Where V3 = Velocity ofwaterat exitof runner= V2
V
V4 = Velocity of water at exit of draft tube.
4
hf = head losein 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
absolute velocity at the exit is radial is givenby = 2 cos
1 / (1+ cos21 )
Where 1 isthe angle made by the enteringfluid withtangentof the whee. Solution : Combined Inlet & outlet velocity tria ngle V
V V
Utilization, 2 U V U E 1 u1 ______________ _______________ 1 = ____________ = = 2 2 2 2 E + V /2gc U1 Vu1 + V2 /2gc U + V /2 2
1
=0
U
(
.
.
.
2
2
2
V2 = V1 - U1 )
1
2
2
2
2V cos 1 ____________ 1 2 V 1
2
2U 2U ____________ _________ 1 1 = = 2 2 2 2 2 2U + V 1 -U 1 U + V1
=
V
2
1
2
U =V
=
V
2
cos
1 +
2 V 1
=
2 cos 1 ____________
1
2
1+ cos
Dr.M.S.Govinde Gowda, Principal, Alva’s Inst. Of Engg & Tech, Moodbidri
13
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 m /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, Vf = 2.5 m/s, Q = 0.8m /s, (i)
Euler's head : gHe = E = (U1Vu1) /gc
as Vu2 = 0,
E=25x18/1=450J/kg=gHe
He= 450/9.81 = 45.87m.
(ii) Power developed (P) :
P= QE =
Q (gHe)= (1000 x 0.8 x 450)/ 1000
P=360kW (iii) Inlet blade angle ( 1):
Inlet
Outlet
V V
2
U
2
V1 = (Vf +Vu 1 )1/2 = 18.17 m/s
V
V
=V
V
V
U
V
=0
Vf1 =Vf 2 = Vf = V2 = 2.5 m/s
tan 1 =Vf /(U1 -Vu1)= 2.5/ (25- 18) 0
1 =19.65
(iv) Degree of reaction (R) 2
2
H - [ (V 2 -V 2 )]/2g ___________________________ 45.87 - [(18.17 - 2.5 )/2 x 9.81] _________________ 1 2 R= = H 45.87 R=0.64
Dr.M.S.Govinde Gowda, Principal, Alva’s Inst. Of Engg & Tech, Moodbidri
14
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 Scroll casing Guide vane
Guide vane
Runner vanes
Boss
d D
Inlet of runner vanes Outlet of runner vanes
Tail race
Draft tube
Fig.7.7 Main components of Kaplan turbine.
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 1) At inlet, the velocity shown in Fig.8.
Outlet
is as le
2) At the outlet, the discharge is always axial with no whirl velocity component i.e., outlet velocity triangle just a right angle triangle as shown Fig.8.
V1
Vr2 Vf
1
U Vu 1
Vf =V2
Vr1
U
Vu 2=0
Fig.8 Velocity triangles for Kaplan Turbine.
Dr.M.S.Govinde Gowda, Principal, Alva’s Inst. Of Engg & Tech, Moodbidri
15
9.1 Working Proportions of Kaplan Turbine i) Tangential velocityof blades based on tip diameter (U )
U =
2gH
(22)
where, D = Tip diameter or Outer diameter of the runner d = Hub diameter or Boss diameter of the runner d/D = 0.35 to0.60 ii) Flow velocity(Vf )
= Vf / 2gH
Usually, 0.35 <
< 0.75
(23)
iii) The discharge (Q)flowingthrough the runner is given by 2 2 2 2 ____ ____ Q= (D - d )Vf = (D - d ) 4 4
2gH
(24)
Note: Example 6 : A kaplan turbine produces80,000 HP(58,800 kW) under a head of 25m which has an overall efficiency of 90%.Taking the value of speed ratio = 1.6, flow ratio = 0.5 and the hub diameter = 0.35 times the outer diameter. Find the diameter and the speed of the turbine. Solution : P = 80,000 HP = 58,800 kW, H = 25m, = 0.9, = 1.6, = 0.5, d/D =0.35, To find :D, N.
= P/(QgH) Q = P/( gH)= 58,800 / (0.9 x 9.81 x 1 x 25) 3
Q = 266.4 m /s 2 2 2 2 ___ ____ Again, Q = (D - d ) 2gH = D 1- (d/D) x 2gH 4 4
[
]
2 2 ____ D (1 - 0.35 )x (0.5) 2x9.81x25 4 D=5.91m
266.4 =
Also,
= UD 2gH = UD / (2x9.81x25) = 1.6 UD = 35.44 m/s = DN/60 = 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.A vacuum pressure gauge connected to the draft tube inlet indicates a reading of 5m of water. Assume that draft tube efficiency is 78%.
Dr.M.S.Govinde Gowda, Principal, Alva’s Inst. Of Engg & Tech, Moodbidri
16
Solution :Given : P = 2940 kW, H = 5 m, Hs =1.6 m, p2 / g = - 5 m, draft = 0.78,
o = ?
Apply the Bernoulli's equation between (2) & (3) we get p2/g = - [Hs + (V 22 - V 32 )/2g ]
d2 = 38 m
- 5 = - 1.6 - [ (V 22 - V 32 )/2g] 2 (V 2
2 -V 3 )/2g = 2 2 (V 2 - V 3 ) =
Draft tube efficiency
2
+3.4 m
Z1
+ 66.708
Hs
Actual pressure gain in draft tube
d = ___________________________ Pressure at inlet to the tube.
=
V 22
=
(V 22 - V 32 ) __________ V 22
/2g
p2/ g =5 Hs =1.6 m
(1)
(V 22 -V 32 )/2g ____________
2
3
Z2
Datum line
= 0.78
1-(V 3/V 2) 2 = 0.78 V 32 = 0.22 V 22
(2)
n equation (1), V22 (1 - 0.22) = 66.708 V 2 = 9 .25 m /s = Vf Now discharge, Q =Af Vf =
(
.
.
.
Axial d ischarge i n Kaplan t urbine)
2 3 ____ x 3 x 9.25 = 65.38 m /s 4
Overall efficiency of a Kaplan turbine, Power output P 0 = __________________ = ________ QgH Hydro power at inlet 2940 x 1000 1000 x 9.81 x 65.38 x 5
0 = _____________________ = 91.68 %
Dr.M.S.Govinde Gowda, Principal, Alva’s Inst. Of Engg & Tech, Moodbidri
