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1. Specific Speed
Its Significance
Numerical
Satish G. Taji
Assistant Professor
Civil Engineering Department
SRES’s Sanjivani College of Engineering, Kopargaon1
Hydraulic Turbines
1

2. SPECIFIC SPEED (Ns)
 The specific speed of a turbine is defined as the speed of
a turbine which is identical in shape, geometrical
dimensions, blade angles, gate opening, etc. which
would develop unit power when working under a unit
head.
 P = 1 kW and H = 1 m, the speed N = specific speed Ns
Prepared By Prof. S. G. Taji

3. SPECIFIC SPEED (Ns) – Expression
 In the expression, we need to find relationship of
Specific Speed (Ns) with Speed (N) , Power (P) and Head
(H)
 Let, P = Shaft power,
 Q = Discharge through turbine,
 H = Head under which turbine is working,
 w = Weight density of water = ρ x g
 D = Diameter of actual turbine,
 N = Speed of actual turbine,
 u = Tangential velocity of the turbine,
 Ns = Specific speed of the turbine, and
 V = Absolute velocity of water.
Prepared By Prof. S. G. Taji

4. SPECIFIC SPEED (Ns) – Expression
 The overall efficiency (η0) of any turbine is given by,
Now, select one variable from this eq (say Power, P)
P = ηo × w Q H
In above eq, w, and ηo are constant.
∴ P ∝ Q × H -------------(1)
We need to keep only Ns, N, P and H in the relationship
∴ find value of Q in terms of N, P or H
Prepared By Prof. S. G. Taji

5. SPECIFIC SPEED (Ns) – Expression
Now,
Q ∝ Area x Velocity -------------(2)
Velocity = Cv√2gH  ∴ V ∝ √H
Area = πd2 /4 OR B x D  ∴ A ∝ D2
but, u = π D N / 60  ∴ u ∝ D N ----(3)
also, u = Ku √(2gH)  ∴ u ∝ √H
Eq. (3) √H ∝ D N  ∴ D ∝ √H / N
Now, A ∝ D2  A ∝ (√H / N)2
Put values of V and A in eq (2)
Q ∝ (√H / N)2 x √H  Q ∝ H3/2 / N2
put this value in Eq (1)
Prepared By Prof. S. G. Taji

6. SPECIFIC SPEED (Ns) – Expression
Q ∝ H3/2 / N2
∴ P ∝ Q × H  P ∝ [H3/2 / N2 ] x H
P ∝ H5/2 / N2
P = K H5/2 / N2  K = Constant of Proportionality
Now, according to definition,
when, H= 1m and P=1KW , then N=Ns
1=K (1)5/2 / Ns2  K = Ns2
put this value in above eq.
P = Ns2 H5/2 / N2
Ns = N √ P / H5/4
Prepared By Prof. S. G. Taji

7. SPECIFIC SPEED (Ns) – Turbine Selection
Prepared By Prof. S. G. Taji
Turbine Specific Speed Remark
Pelton Wheel
With single jet
8.5 to 30
Upto 43 with
double jet
Francis
Turbine
50 to 340 --
Kaplan
Turbine
255 to 860 --

8. SPECIFIC SPEED (Ns) – Significance
 Specific speed plays an important role in the selection of
the type of turbine.
 By knowing the specific speed of turbine the
performance of the turbine can also be predicted.
 If a runner of high specific speed is used for a given head
and power output, the overall cost of installation is
lower.
 The runner of too high specific speed with high available
head increases the cost of turbine on account of high
mechanical strength required.
 The runner of too low specific speed with low available
head increases the cost of generator due to the low
turbine speed.
Prepared By Prof. S. G. Taji

9. SPECIFIC SPEED (Ns) - Numerical
 A turbine is to operate under a head of 25 m at 200 r.p.m
The discharge is 9 m3/s. If the overall efficiency is 90 per
cent, determine : (i) Power generated; (ii) Specific speed
of the turbine;(iii) Type of turbine.
 Solution. Given data, Head, H = 25 m; Speed, N = 200
r.p.m.; Discharge, Q = 9 m3/s; Overall efficiency, η0 =
90%.
(i) Power generated, P :
P = ηo × ρgQH = 0.9 ×1000x9.81 × 9 × 25 = 1986.5 kW
(ii) Specific speed of the turbine, Ns :
Ns = N √ P / H5/4 = 159.4 r pm
(iii) Type of Turbine : As the specific speed lies between
50 and 340, the turbine is a Francis turbine.
Prepared By Prof. S. G. Taji

10. SPECIFIC SPEED (Ns) - Numerical
 In a hydroelectric station, water is available at the rate of
175 m3/s under a head of 18 m. The turbines run at a
speed of 150 r.p.m. with overall efficiency of 82%. Find
the number of turbines required if they have the
maximum specific speed of 460. [GATE]
 Solution. Given data, Q = 175 m3/s; H = 18 m; N = 150
r.p.m.; ηo = 82%; Ns = 460.
Number of turbines required = Max Power that can be
generated / Power that can be generated under specific
conditions
Max Power that can be generated:
Pm = ηo x ρ x g x Q x H = 1000 x 9.81 x 175 x 18
= 25339.23 kW
Prepared By Prof. S. G. Taji

11. SPECIFIC SPEED (Ns) - Numerical
 Power Generated under specific conditionsS
Ns = N √ P / H5/4
460 = 150 √ P / (18)5/4
P = 12927.5 kW
Number of turbines required = Pm / P
= 25339.23 / 12927.5
= 1.96 say ~ 2
Prepared By Prof. S. G. Taji

12. SPECIFIC SPEED (Ns) - Numerical
Prepared By Prof. S. G. Taji