1

The external diameter of an inward flow reaction turbine is 0.5 m. The

width of the wheel at inlet is 150 mm and the velocit y of flow at inlet is

1.5 m/s. Find the rate of flow passing through the turbine.

Solution:

D1 = 0.5 m, B1 = 0.15 m, V f 1 = 1.5 m/s, Q = ?

Discharge through the turbine = Q = D1 B1 V f 1 = x 0.5 x 0.15 x 1.5

Q = 0.353 m 3 /s (Ans)

The external and internal diameters of an inward flow reaction turbine are

600 mm and 200 mm respectivel y and the breadth at inlet is 150 mm. If

the velocit y of flow through the runner is constant at 1.35 m 3 /s, find the

discharge through turbine and the width of wheel at outlet.

Solution:

D1 = 0.6 m, D2 = 0.2 m, B1 = 0.15 m, V f 1 = V f 2 = 1.35 m/s, Q = ? , B2 = ?

Discharge through the turbine = Q = D1 B1 V f 1 = x 0.6 x 0.15 x 1.35

Q = 0.382 m 3 /s (Ans)

Also discharge is given b y Q = D2 B2 V f 2 = x 0.2 x B2 x 1.35 0.382

B2 = 0.45 m/s (Ans)

An inward flow reaction turbine running at 500 rpm has an external

diameter is 700 mm and a width of 180 mm. If the gu ide vanes are at 20

to the wheel tangent and the absolute velocit y of water at inlet is 25 m/s,

find (a) discharge through the turbine (b) inlet vane angle.

Solution:

N = 500 rpm, D1 = 0.7 m, B1 = 0.18 m, a = 20, V1 = 25 m/s, Q = ? , = ?

We know that th e peripheral velocit y is given b y

smND

u /33.1860

5007.0

601

1

From inlet velocit y triangle, we have

V f 1 = V1 Sin x Sin 20 = 8.55 m/s

2

Vw1 = V1 Cos = x Cos 20 = 23.49 m/s

657.133.1849.23

55.8Tan

11

1

uV

V

w

f

= 58.89 (Ans)

Q = D1 B1 V f 1 = x 0.7 x 0.18 x 8.55 = 3.384 m 3 /s (Ans)

A reaction turbine works at 450 rpm under a head of 120 m. Its diameter

at inlet is 1.2 m and the flow area is 0.4 m 2 . The angle made b y the

absolute and relative velocities at inlet are 20 and 60 respectivel y with

the tangential velocit y. Determine (i) the discharge through the turbine

(ii) power developed (iii) efficiency. Assume radi al discharge at outlet .

Solution:

N = 450 rpm, H = 120 m, D1 = 1.2 m, a1 = 0.4 m2 , = 20 and = 60

Q = ?, = ?, Vw2 = 0

We know that the peripheral velocit y is given b y

smND

u /27.2860

4502.1

601

1

u1

V1

Vr1

Vw1

Vf1

3

27.2860Tan

Tan

1

1

11

1

w

f

w

f

V

V

uV

V

Hence V f 1 = (Vw 1 28.27) Tan 60 (01)

Further 20TanTan1

1 w

f

V

V

Hence V f 1 = (Vw 1) Tan 20 (02)

From equa tions 1 and 2, we get

(Vw 1 28.27) Tan 60 = Vw 1 Tan 20

Hence Vw 1 = 35.79 m/s

V f 1 = 35.79 x Tan 20 = 13.03 m/s

Discharge Q = D1 B1 V f 1 = a1 V f 1 = 0.4 x 13.03 = 5.212 m 3 /s (Ans)

Work done per unit weight of water =

NmkNuVg w

/178.10127.2879.3510

1111

Water Power or input per unit weight = H = 120 kN-m/N

Hydraulic efficiency = %31.84120

178.101

The peripheral velocit y at inlet of an outward flow reaction turbine is 12

m/s. The internal diameter is 0.8 times the external diameter. The vanes

are radial at entran ce and the vane angle at outlet is 20. The velocit y of

flow through the runner at inlet is 4 m/s. If the final discharge is radial

and the turbine is situated 1 m below tail water level, determine:

1. The guide blade angle

2. The absolute velocity of water leav ing the guides

3. The head on the turbine

4. The h ydraulic efficiency

Solution:

u1 = 12 m/s, D1 = 0.8 D2 , = 90, = 20 , V f 1 = 4 m/s, Vw2 = 0, Pressure

head at outlet = 1m, = ?, V1 = ?, H = ?, h = ?

4

From inlet velocit y triangle,12

4Tan

1

1 u

V f , Hence = 18.44

Absolute velocit y of water leaving guide vanes is

m/s65.12412 2221211 fVuV

60and

602

21

1

NDu

NDu

Comparing the above 2 equations, we have2

2

1

1

2

2

1

1 henceand6060

D

u

D

u

D

u

D

u

Hence m/s158.0

12

1 12

2 uDD

u

From outlet velocit y triangle, V2 = V f 2 = u2 tan 20 = 15 tan 20 = 5.46 m/s

As Vw2 = 0

Work done per unit weight of water = m/NkN4.1410

121211

g

uVw

Head on turbine H

Energy Head at outlet = WD per unit weight + losses

u2

V2=Vf 2

Vr1= Vf1V1

u1

Vr2

5

g

uVw

g

VH 11

22

21 and hence

m89.164.142

46.51

2

gH

Hydraulic efficiency = %26.8510089.1610

121211

Hg

uVwh

Jan/Feb 2006

An inward flow water turbine has blades the inner and outer radii of

which are 300 mm and 50 mm respectively. Water enters the blades at the

outer periphery with a velocit y of 45 m/s making an angle of 25 with the

tangent to the whe el at the inlet tip. Water leaves the blade with a flow

velocit y of 8 m/s. If the blade angles at inlet and outlet are 35 and 25

respectivel y, determine

(i) Speed of the turbine wheel

(ii) Work done per N of water (08)

Solution:

D1 = 0.6 m; D2 = 0.1 m, V1 = 45 m/s, 25, V2 = 8 m/s, 35, 25,

N = ?, WD/N = ?

423.0251

1 SinV

VSin f

Hence V f 1 = 0.423 x 45 = 19.035 m/s

466.025tan1

1 w

f

V

VTan

Hence Vw1 = 40.848 m/s

111

1

848.40

035.107.035tan

uuV

VTan

w

f

u1 = 13.655 m/s

601

1

NDu

and hence RPM434.65

6.0

655.136060

1

1

D

uN (Ans)

6

m/s552.460

3.8691.0

602

2

ND

u

Ignoring shock losses, V r 2 = V r 1 = m/s187.3335sin

035.19

sin1

fV

Vw2 = V r 2 cos - u2 = 33.187 cos 25 4.552 = 25.526 m/s

Work done per unit weight of water = 22111

uVuVg ww

m/s4.67552.4526.25655.13848.401/ g

NWD (Ans)

July/Aug 2005

A reaction turbine 0.5 m dia develops 200 kW while running at 650 rpm

and requires a discharge of 2700 m 3 /hour; The pressure head at entrance

to the turbine is 28 m, the elevation of the turbine casing above the tail

V2Vr2

u2

Vf2

Vw2F

GH

E

V1 Vr1

u1

Vf1

Vw1

B

D

CA

Tangent

Tangent

7

water level is 1.8 m an d the water enters the turbine with a velocit y of 3.5

m/s. Calculate (a) The effective head and efficiency, (b) The speed,

discharge and power if the same machine is made to operate under a head

of 65 m

Solution :

D = 0.5 m, P = 200 kW, N = 650 rpm, Q = 2700/602 = 0.75 m 3 /s,

V1 = 3.5 m/s, mg

p281

The effective head =H=Head at entry to runner Kinetic energy in tai l race

+ elevation of turbine above tailrace

m29.18758.1102

5.328

2

2221

g

V

g

pH

(Ans)

Hydraulic efficiency = %36.911001875.2975.0101000

10200 30

HQg

P

Further unit quantit ies are given b y

Unit speed =2

2

1

1

H

N

H

NN u

Unit Discharge =2

2

1

1

H

Q

H

QQu

Unit Power =2

3

2

2

23

1

1

H

P

H

PPu

31.120651875.29

650 2 N

Nu

N2 = 969.97 rpm (Ans)

1388.0651875.29

75.0 2 Q

Qu

Q2 = 1.119 m3 /s (Ans)

268.1651875.29

200

232

23

P

Pu

P2 = 664.49 kW (Ans)

8

A Francis turbine has inlet wheel diameter of 2 m and outlet diameter of

1.2 m. The runner runs at 250 rpm and water flows at 8 c umecs. The

blades have a constant width of 200 mm. If the vanes are radial at inlet

and the dis charge is radiall y outwards at exit, make calculations for the

angle of guide vane at inlet and blade angle at outlet (10)

Solution:

D1 = 2 m, D2 = 1.2 m, N = 250 rpm, Q = 8 m3 /s, b = 0.2 m, Vw1 = u1 ,

Vw2 = 0, = ?, = ?

601

1

NDu

m/s18.26

60

2502

m/s71.1560

2502.1

602

2

ND

u

Q = D1b V f 1 = D2b V f 2

8 = x 2 x 0.2 x V f 1

Hence V f 1 = 6.366 m/s

Similarl y 8 = x 1.2 x 0.2 x V f 2

V2= Vf 2

u2

Vr1= Vf 1

u1= Vw 1

V1

Vr2

9

V f 2 = 10.61 m/s

18.26

366.6tan

1

1 u

V f

= 13.67 (Ans)

71.15

61.10tan

2

2 u

V f

= 34.03 (Ans)

Determine the ov erall and h ydraulic efficiencies of an inward flow

reaction turbine using the following data. Output Power = 2500 kW,

effective head = 45 m, diameter of runner = 1.5 m, width of runner = 200

mm, guide vane angle = 20 , runner vane angle at inlet = 60 and specific

speed = 100.

Solution:

P = 2500 kW, H = 45 m, D1 = 1.5 m, b1 = 0.2 m, = 20 , = 60 ,

N s = 110, o= ? , h = ?

We know that specific speed is given b y

u2

Vr2Vf2=V2

u1

V1

Vr1 Vf1

Vw1

10

45

H

PNN s and hence rpm

P

HNN s 233

2500

45100 45

45

smND

u /3.1860

2335.1

601

1

But from inlet velocity triangle, we have

tantan11

1ff VVu

60tan20tan3.18 11 ff

VV and hence V f 1 = 8.43 m/s

m/s16.2302tan

43.8

tan1

1 f

w

VV

Vw 2 = 0 and hence

%18.941004510

3.1816.2311

Hg

uVwh (Ans)

Q = D1 b1 V f 1 = x 1.5 x 0.2 x 8.43 = 7.945 m 3 /s

%93.6910045945.7101000

102500 3

HQg

Po

(Ans)

Determine the output Power, speed, specific speed and vane angle at exit

of a Francis runner using the following data. Head = 75 m, Hydraulic

efficiency = 92%, overall efficiency = 86 %, runner diameters = 1 m and

0.5 m, width = 150 mm and guide blade angle = 18 . Assume that the

runner vanes are set normal to the periphery at inlet.

Solution:

Data: H = 75 m, h = 0.92, o = 0.86, D1 = 1 m, D2 = 0.5 m, = 18 ,

Vw1 = u1 , P = ?, N = ?, = ?

Hg

u

Hg

uVwh

2111

u12 = 0.92 x 10 x 75 = 690

u1 = 26.27 m/s

11

u1=Vw1

Vr1=Vf1V1

V2=Vf2Vr2

u2

smNND

u /27.2660

0.1

601

1

N = 501.7 RPM

Vf1 = u1 tan 26.27 x tan 18 = 8.54 m/s

Q = D1 b1 V f 1 =

= x 1.0 x 0.15 x 8.54 = 4.02 m 3 /s

2

2

1

1

D

u

D

u and hence u2 = 0.5 x u1 = 13.135 m/s

Assuming V f 1 = V f 2

From outlet velocit y triangle, we have

65.0135.13

54.8tan

2

2 u

V f

Hence = 33

86.07502.4101000

P

HQg

Po

Hence P = 2592.9 kW (Ans)

Specific speed = RPM75.11575

9.25927.501

45

45

H

PNN s

The following data is given for a Francis turbine. Net Head = 60 m; speed

N = 700 rpm; Shaft power = 294.3 kW; o = 84%; h = 93%; flow rat io =

0.2; breadth ratio n = 0.1; Outer diameter of the runner = 2 x inner

diamete r of the runner . The thickness of the vanes occupies 5%

circumferential area of the runner, velocity of flow is constant at inlet and

outlet and discharge is radial at outlet. Determine:

(i) Guide blade angle

(ii) Runner vane angles at inlet and outlet

(iii) Diameters o f runner at inlet and outlet

(iv) Width of wheel at inlet

Solution

H = 60 m; N = 700 rpm; P = 294.3 kW; o = 84%; h = 93%;

12

flow ratio = 2.02

1 Hg

V f

m/s928.6601022.01 fV

Breadth ratio 1.01

1 D

B

D1= 2 x D2

V f 1 = V f 2 = 6.928 m/s

Thickness of vanes =

5% of circumferential area of runner

Actual area of flow = 0.95 D1B1Discharge at outlet = Radial and hence

Vw 2 = 0 and V f 2 = V2

We know that the overall efficiency is

given b y

60101000

103.29484.0;

3

0

QHQg

P

Q = 0.584 m 3 /s

Q = 0.95 D1B1V f 1=0.95 D1 x (0.1 D1) x 6.928 = 0.584

Hence D1 = 0.531 m (Ans)

1.01

1 D

Band B1 = 53.1 mm (Ans)

m/s46.1960

700531.0

601

1

ND

u

Hydraulic efficiency6010

46.1993.0; 111

wwhV

Hg

uV

Vw 1 = 28.67 m/s

From Inlet velocit y triangle 242.067.28

928.6tan

1

1 w

f

V

V

Hence Guide bla de angle = = 13.58 (Ans)

u2

Vr2Vf2=V2

u1

V1

Vr1 Vf1

Vw1

13

752.046.1967.28

928.6tan

11

1

uV

V

w

f

Vane angle at inlet = = 37 (Ans)

m/s73.9

60

7002531.0

602

2

ND

u

From outlet velocit y triangle, we have

712.073.9

928.6tan

2

2 u

V f

= 35.45 (Ans)

Diameters at inlet and outlet are D1 = 0.531m and D2 = 0.2655 m

A Kaplan turbine develops 9000 kW under a net head of 7.5 m. Overall

efficiency of the wheel is 86% The speed ratio based on outer diameter is

2.2 and the flow ratio is 0.66. Diameter of the boss is 0.35 times the

external diameter of the wheel. Determine the diameter of the runner and

the specific speed of the runner.

Solution:

P = 9000 kW; H = 7.5 m; o = 0.86; Speed ratio = 2.2; flow ratio = 0.66;

Db = 0.35 Do ;

2.22

1 Hg

u

m/s94.265.71022.21 u

66.02

1 Hg

V f

m/s08.85.710266.01

fV

5.7101000

10900086.0;

3

0

QHQg

P

Q = 139.5 m 3 /s

14

5.13908.835.044

221

22 oofbo DDVDDQ

Do = 5.005 m (Ans)

m/s94.2660

005.5

60

NNDu o

N=102.8 rpm (Ans)

(Ans)rpm76.7855.7

90008.102

45

45

H

PNN s

A Kaplan turbine working under a head of 25 m develops 16,000 kW shaft

power. The outer diameter of the runner is 4 m and hub diameter is 2 m.

The guide blade angle is 35 . The h ydraulic and overall efficiency are

90% and 85% respectivel y. If the velocit y of whirl is zero at outlet,

determine runner vane angles at inlet and outlet and speed of turbine.

Solution

H = 25 m; P = 16,000 kW; Db = 2 m; Do = 4 m; = 35; h = 0.9;

o= 0.85; Vw 2 = 0; = ?; = ?; N = ?

25101000

101600085.0;

3

0

QHQg

P

Q = 75.29 m 3 /s

29.752444 1

221

22 ffbo VVDDQ

V f 1 = 7.99 m/s

From inlet velocit y triangle,

1

1tanw

f

V

V

m/s41.1135tan

99.71 wV

From Hydraul ic efficiency

Hg

uVwh

11

15

2510

41.119.0 1

u

u1 = 19.72 m/s

9614.041.1172.19

99.7tan

11

1

w

f

Vu

V

= 43.88 (Ans)

For Kaplan turbine, u1 = u2 = 19.72 m/s and V f 1 = V f 2 = 7.99 m/s

From outlet velocit y triangle

4052.072.19

99.7tan

2

2 u

V f

= 22.06 (Ans)

m/s72.1960

4

6021

NNDuu o

N = 94.16 rpm (Ans)

Vr2

u2

V2=Vr2

u1

V1

Vf1

Vw1

Vr1

16

A Kaplan turbine works under a head of 22 m and runs at 150 rpm. The

diameters of the runner and the boss are 4.5 m and 12 m respectivel y. The

flow ratio is 0.43. The inlet vane angle at the extreme edge of the runner

is 16319. If the turbine discharges radiall y at outlet, determine the

discharge, the h ydraulic efficiency, the guide blade angle at the extreme

edge of the runner and the outlet vane angle at the extreme edge of the

manner.

Solution:

H = 22 m; N = 150 rpm; Do = 4.5 m; Db = 2 m; 16319 V

V V f 2 = V f 1 ; Q = ?; h = ?; 43.0

21 Hg

V f

,

m/s34.3560

1505.4

6021

ND

uu o

Vr2

u2

V2=Vr2

u1

V1

Vf1

Vw1

Vr1

17

m/s02.92210243.01 fV

111180tan

w

f

Vu

V

2997.034.3502.9

19'163180tan1

wV

Vw 1 = 5.24 m/s

Hydraulic efficiency is gi ven b y

%17.842210

34.3524.511

Hg

uVwh

72.124.5

02.9tan

1

1 w

f

V

V

= 59.85 (Ans)

2552.034.35

02.9tan

2

2 u

V f

= 14.32 (Ans)

A kaplan turbine is to be designed to develop 7,350 kW. The net available

head is 5.5 m. Assume that the speed ratio as 0.68 and the overall

efficiency as 60%. The diameter of the boss is r d of the diameter of the

runner. Find the diameter of the runner, its speed and i ts specific speed.

Solution:

P = 7350 kW, H = 5.5 m

68.02

1 Hg

V f

and hencem/s13.75.510268.01 fV

09.22

1 Hg

u

and hencem/s07.235.51022.21 u

5.5101000

1073506.0;

3

0

QHQg

P

18

Q = 222.72 m 3 /s

72.22213.7344

22

122

oofbo

DDVDDQ

Do = 6.69 m (Ans)

m/s07.2360

69.6

60

NNDu o

N=65.86 rpm (Ans)

(Ans)rpm37.6705.5

735086.65

45

45

H

PNN s