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1. Unit Quantities
Unit Speed
Unit Discharge
Unit Power
Numerical
Satish G. Taji
Assistant Professor
Civil Engineering Department
SRES’s Sanjivani College of Engineering, Kopargaon1
Hydraulic Turbines
1

2. Performance of Turbine – Unit Quantities
 Turbines are often required to work under varying
conditions of head, speed, output and gate opening.
 As such, in order to predict their behavior, it is essential
to study the performance of the turbines under the
varying conditions.
 Head, Speed , Power and Gate Opening
 These variable conditions can be developed only for
laboratory turbines or those in the test plant and are
otherwise uncommon
 To facilitate comparison between the performances of
the turbines of the same type but having different
outputs and speeds and working under different heads,
it is often convenient to express the test results in terms
of certain unit quantities.
Prepared By Prof. S. G. Taji

3. Performance of Turbine – Unit Quantities
 These quantities are known as unit quantities.
 The conditions of the turbine working under unit head
are such that the efficiency of the turbine is remained
unaffected.
 In short, unit quantity (say speed, discharge or power)
is that quantity which is obtained when turbine is
working under unit head.
Condition – Efficiency should be constant or unaffected
Prepared By Prof. S. G. Taji

4. Performance of Turbine – Unit Quantities
1. Unit Speed (Nu):
 Speed of turbine working under unit head
 Let, N=speed of turbine under head, H
H=head under which turbine is working
u = tangential velocity
Now, u = π D N / 60  ∴ u ∝ N (given turbine, D const)
also, u = Ku √(2gH)  ∴ u ∝ √H
∴ N ∝ √H  N = K1 √H ----- (1)
When, H= 1 m (unit head) N = Nu (Unit Speed)
Nu = K1 √1  Nu = K1 - ---- Put in eq (1)
N = Nu √H  Nu = N / √H
Prepared By Prof. S. G. Taji

5. Performance of Turbine – Unit Quantities
2. Unit Discharge(Qu):
 Discharge passing through turbine working under unit
head
 Let, Q = Discharge passing through turbine under H
= Area of flow x Velocity
Now, Velocity = Cv√2gH  ∴ V ∝ √H
For given turbine, D or Area of flow is constant)
∴ Q ∝ √H  Q = K2 √H ----- (1)
When, H= 1 m (unit head) Q = Qu (Unit Discharge)
Qu = K2 √1  Qu = K2 - ---- Put in eq (1)
Q = Qu √H
Qu = Q / √H
Prepared By Prof. S. G. Taji

6. Performance of Turbine – Unit Quantities
3. Unit Power (Pu):
 Power output of turbine working under unit head
 Let, P = power developed by turbine under head H
Now, ηo = P/ ρ g Q H  ∴ P = ηo × ρg Q H
also, P ∝ Q x H ∴ P ∝ √H x H
∴ P ∝ H3/2  P = K3 H3/2 ----- (1)
When, H= 1 m (unit head) P = Pu (Unit Speed)
Pu = K3√1  Pu = K3 - ---- Put in eq (1)
P = Pu H3/2
Pu = P / H3/2
Prepared By Prof. S. G. Taji

7. Unit Quantities – Use
 If a turbine is working under different heads the
behaviour of the turbine can be easily known from the
values of the unit quantities as follows :
Let, H1, H2,…… = Heads under which a turbine works,
N1, N2,…… = Corresponding speeds,
Q1, Q2,…… = Corresponding discharges, and
P1, P2,……. = Corresponding powers developed.
 Then using unit quantities equations, we obtain
Prepared By Prof. S. G. Taji

8. Unit Quantities – 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 efficiency is 90%
determine the performance of the turbine under a head
of 20 m.
 Solution. Head under which turbine works, H1 = 25 m
Speed of the turbine, N1 = 200 r.p.m.
Discharge through the turbine, Q1 = 9 m3/s
Efficiency (overall), ηo = 90%
Performance of turbine under H2= 20 m; (N2 ? Q2? P2 ?):
P1 = ηo × ρgQH = 0.9 ×1000x9.81 ×9× 25 = 1986.5 kW
Prepared By Prof. S. G. Taji

9. Unit Quantities – Numerical
 A
Prepared By Prof. S. G. Taji

10. SPECIFIC SPEED (Ns) – Expression
Prepared By Prof. S. G. Taji