1Unit QuantitiesIn the studies of comparison of the performances of turbines of differentoutput, speeds and different heads, it is convenient to determine theoutput, the speed and the discharge, when the head on the turbine isreduced to unit y, i .e. 1 m. The conditions of the turbine under unit headare such that the efficiency of the turbine remains unaffected. Thus thevelocit y triangles under working conditions and under unit head aregeometricall y similar.

Given a turbine every velocit y vector ( V1 , U1 , Vw 1 , V f 1 ) is afunction of H where H is the head on the turbine. With this basic concept,we can determine the speed, discharge and power under unit head.

Unit Speed (Nu):This is the speed of a turbine working under a unit headLet N be the speed of tu rbine, H be the head on the turbine and u be theperipheral velocit y.

We know that the peripheral velocit y u is given b y

60ND

u where D is the mean diameter of the runner which is treated as

constant and N is the speed of the runner,

Hence u NBut HKu u 2 and hence HuHence HN

i .e. HKN 1 , where K1 is the proportionalit y constant.From definition of Unit speed, it is the speed of a turbine when workingunder unit head. Hence at H=1 , N=N u . Substituting, we get

11KN u

HNuN orH

NNu -------------------------------- ------------------------------ (01)

Unit Discharge (Qu):This is the discharge through the turbine working under a unit head

2Consider the Q as the discharge through a turbine. From dischargecontinuit y equation, Q = a x V , where a is the cross -sectional area of flowand V is the mean flow velocit y.

For a given turbine, the cross -sectional area is constant and hence Q VBut HgCV v 2 and hence HV and hence HQi .e., HKQ 2 where K2 is the proportionalit y constant.From definition of Unit discharge, it is the discharge through the turbinewhen working under unit head. Hence at H=1 , Q=Qu . Substituti ng, we get

12KQu

HQuQ orH

QQu -------------------------------- ------------------------------- (02)

Unit Power (Pu):This is the Power developed b y the turbine working under a unit headConsider the P as the power developed by the turbine.We know that the efficie ncy of turbine is given b y

HQP

Where is the weight densit y of the fluid/water passing through the

turbine, Q is the discharge through the turbine and H is the head underwhich the turbine is working. But efficiency of a turbine and we ightdensit y of water are constants and hence, we can write

P Q HFrom discharge continuit y equation, Q = a x V , where a is the cross -sectional area of flow and V is the mean flow velocit y.

For a given turbine, the cross -sectional area is constant and he nce Q VBut HgCV v 2 and hence HV and hence HQSubstituting, we get

HHP or HHKP 3where K3 is the proportionalit y constant.

3From definition of Unit Power, it is the power developed b y the turbinewhen working under unit head. Hence at H=1 , P=Pu . Substituting, we get

113KPu

HHPuP or2

3H

PHH

PPu -------------------------------- ------------------- (03)

Unit Speed, Unit discharge and Unit Power is definite characteristics of aturbine.If for a given turbine under heads H1 , H2 , H3 ,. the corresponding speedsare N1 , N2 , N3 ,, the corresponding discharges are Q1 , Q2 , Q3 ,. and thepowers developed are P1 , P2 , P3 ,. Then

Unit speed =3

3

2

2

1

1

HN

HN

HNN u

Unit Discharge =3

3

2

2

1

1

HQ

HQ

HQQu

Unit Power =2

3

3

3

23

2

2

23

1

1

3

3

2

2

1

1 orH

P

H

P

H

PP

HHP

HHP

HHP

P uu

Thus if speed, discharge and power developed b y a turbine under a certainhead are known, the corresponding quantities for an y other head can bedetermined.

Specific Speed of a Turbine (N s)The specific speed of a turbine is the speed at which the turbine will runwhen developing unit power under a unit head. This is the t ypecharacteristics of a turbine. For a set of geometricall y similar turbines thespecific speed will have the same value.Consider the P as the power developed by the turbine.We know that the efficiency of turbine is given b y

HQP

Where is the weight densit y of the fluid/water passing through the

turbine, Q is the discharge through the turbine and H is the head under

4which the turbine is working. But efficiency of a turbine and weightdensit y of water are constants and hence, we can write

P Q HDischarge is given by the product of cross sectional area of flow and the

flow velocit y. Cross sectional area is d b and hence

Q = D b V fBut HgCV v 2 and hence HV And HKu u 2 and hence HuFurther, the peripheral velocit y is given by

60ND

u where D is the mean diameter of t he runner.

From the above two equations of u , we can write that

b

D

b

D

5HDN

orNHD

But in turbines the width of flow area b is proportional to the diameter D .

Hence D b, with which

NHb

Hence HNH

NHP

or 2

25

NHP

orP

HKP

HN2

52

52

Simplifying further

PHKN

45

But from the definition of specific speed, it is the speed of a turbine whenit is working under unit head developing unit power. Hence when H = 1and P = 1. N = N s . Hence

ss NKKN and11 4

5

Substituting we get

PHNN s

45

or4

5H

PNN s

Examination questionsDec/Jan 07Define unit power, unit speed, unit discharge and specific speed with

reference to h ydraulic turbines . Derive expressions for these terms. (06)

6July 06Derive expressions for specific speed of a turbine (06)A turbine is to operate under a head of 25 m at 200 rpm. The discharge is9 m3 /s. If the efficiency is 90 %, determine :i) Power generated ii) Speed and Power at a head of 20 m. (07)

Jan 06What is specific speed of a turbine and ex plain its significance (04)Jan 05Define and derive an expression for specific speed of a turbine, indicatingits significance. (08)Define the terms unit po wer, unit speed and unit discharge with referenceto a h ydraulic turbine. Also give the expressions for these terms (06)July 04

Define specific speed of a h ydraulic turbine. Derive an equation forspecific speed in terms of operating speed, power and head (08)

7Dec/Jan 07Suggest a suitable type of a turbine to develop 7000 kW power under ahead of 20 m, while operating at 220 rpm. What are the considerations fo ryour suggestion? (04)Solution:P = 7000 kW; H = 20 m; N = 220 rpm.

We know that speci fic speed of a turbine is given b y

45

H

PNN s

Substituting , we get

rpm43520

70002204

5 sN

As the specific speed is between 300 and 1000, Kaplan turbine can besuggested .Jan 06A turbine is to operate under a head of 25 m at 200 rpm. The disc harge is9 m3 /s. If the efficiency is 90%, determine the performance of the turbin eunder a head of 20 m (08)Solution:

H1 = 25 m; N1 = 200 rpm; Q = 9 m 3 /s; = 0.90, H2 = 20 m; N2 = ?; Q2 = ?P1 = ?, P2 = ?

We know that

HQP

2591000109.0 P

P1 = 2025 kWFurther unit quantit ies are given b y

Unit speed =2

2

1

1

HN

HNN u

8Unit Discharge =2

2

1

1

HQ

HQQu

Unit Power =2

3

2

2

23

1

1

H

P

H

PPu

402025

200 2 NN uN2 = 178.9 rpm(Ans)

8.12025

9 2 QQuQ2 = 8.05 m 3 /s (Ans)

2.162025

20252

32

23

PPu

P2 = 1449 kW (Ans)July 04Suggest a suitable type of turbine to develop 7500 kW of power under ahead of 25 m, while operating at 220 rpm. If the same turbine has to workunder a head of 10 m, what power would devel op? What would be the newspeed? (12)Solution:P1 = 7500 kW; H1 = 25 m; N1 = 220 rpm; H2 = 10 ; P2 = ? ; N2 = ?We know that specific speed of a turbine is given b y

45

H

PNN s

Substituting , we get

rpm8.34025

75002204

5 sN

As the speci fic speed is between 300 and 1000, Kaplan turbine can besuggested .Further unit quantit ies are given b y

9Unit speed =2

2

1

1

HN

HNN u

Unit Power =2

3

2

2

23

1

1

H

P

H

PPu

441025

220 2 NN uN2 = 139.14 (Ans)

601025

75002

32

23

PPu

P2 = 1897.4 kW (Ans)