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1. Pelton Wheel Turbine
Design Aspects
Numerical
Satish G. Taji
Assistant Professor
Civil Engineering Department
SRES’s Sanjivani College of Engineering, Kopargaon1
Hydraulic Turbines
1

2. Pelton Wheel Turbine – Design Aspects
1. Velocity of Jet
 V1= Cv √2gH  Cv =Coeff. Of Vel. ,H = Net Head
2. Mean Diameter of the Wheel (Pitch dia of pitch
circle): can be calculate using relation
3. Velocity of Wheel (if not given)
 u = Ku√(2gH)
 Where, Ku = speed ratio = u / V1
 it’s value varies from 0.43 to 0.48
4. Area of Jet
 a = Q / V1
 If Q is not given, then compute Q from overall efficiency (ήo) or Power
Prepared By Prof. S. G. Taji

3. Pelton Wheel Turbine – Design Aspects
5. Angle of Deflection
 Varies from 160o to 170o  =180o - ϕ
6. Jet Ratio (m): it is ratio of pitch dia of wheel to dia of jet
 m = D / d  varies between 11 to 14  can take m=12
7. Number of Buckets on Runner
 Z = 15 + 0.5 m or Z = 15 + 0.5 (D/d)
8. Number of Jets
 Nj= Total Discharge(Q) / Discharge from one jet (q)
9. Bucket Dimensions
 L = Length or height of bowl = 2.4 d to 3.2 d
 B = Width of bucket = 4d to 5d
 T = Depth of bucket = 1.2 d to 1.9 d
 Angle ϕ = 10o to 20o Angle β = 5o to 8o
Prepared By Prof. S. G. Taji

4. Pelton Wheel Turbine – Summary
Work Done /sec =(ρ a V1) x [Vw1 ± Vw2] x u N-m/s or W
(when, β=90O , Put Vw2 = 0)
ήh = [Vw1 ± Vw2] x u / g H OR 2[Vw1 ± Vw2] x u / V12
ήo = P / ρ g Q H
 If only numerical mentions only ‘efficiency’ is mentioned
then consider it as hydraulic efficiency
 If ‘Q’ is not given, then calculate ‘Q’ using overall
efficiency eq.
 If ‘u’ is not given, then calculate it as u = Ku√(2gH) and
then use eq. u1 = u2 = u = π D N / 60 to calculate “D”
 Once calculated no. of jets and rounded to next full
integer, then re-calculate diameter of jet for modified
discharge.
Prepared By Prof. S. G. Taji

5. Pelton Wheel Turbine – Numericals
Pelton wheel is to be designed for the following specifications :
Power (brake or shaft) =9560 kW; Head =350 metres;
Speed = 750 r.p.m. ; Overall efficiency = 85%
Jet diameter = not to exceed 1/6 th of the wheel diameter
Determine : (i) The wheel diameter, (ii) Diameter of the jet, and
(iii) The number of jets required. Cv = 0·985, Speed ratio = 0.45.
Solution.
Shaft or brake power = 9560 kW
Head, H = 350 m
Speed, N = 750 r.p.m.
Overall efficiency, η0 = 85%
Ratio of jet diameter to wheel, d/D = 1/6  d = (1/6) x D
Co-efficient of velocity, Cv = 0.985; Speed ratio, Ku = 0.45
Prepared By Prof. S. G. Taji

6. Pelton Wheel Turbine – Numericals
(i) The wheel diameter, D :
Velocity of jet, V1= Cv √2gH = 0.985 √ 2 × 9.81 × 350 = 81.62 m/s
The velocity of wheel, u = u1 = u2 = Ku × √ 2gH
= 0.45 √ 2 × 9.81 × 350 = 37.3 m/s
But, u = πDN/60
∴ 37.3 = πD x 750 /60
∴ D = 0.95 m
(ii) Diameter of the jet, d :
d = (1/6) D = (1/6) x 0.95 = 0.158 m
(iii) The number of jets required :
Nj= Total Discharge from turbine(Q) / Discharge from one jet (q)
Discharge from one jet, q = Area of jet × velocity of jet
= (πd2 /4 ) x 81.62 = 1.6 m3 /s
Prepared By Prof. S. G. Taji

7. Pelton Wheel Turbine – Numericals
(iii) The number of jets required :
Nj= Total Discharge from turbine(Q) / Discharge from one jet (q)
Total Discharge from turbine, (use overall efficiency eq)
ή0 = (Shaft Power) / (Water Power)
Shaft Power = 9560 KW = 9560 x 103 W
Water Power = KE/Sec = ρ g Q H
ή0 = 9560 x 103 / (1000 x 9.81 x Q x 350)
Q = 9560 x 103 / (1000 x 9.81 x 0.85 x 350) = 3.27 m3 /s
Nj= Total Discharge from turbine(Q) / Discharge from one jet (q)
= 3.27 / 1.6 = 2 jets
Prepared By Prof. S. G. Taji

8. Pelton Wheel Turbine – Numericals
A Pelton wheel is receiving water from a penstock with a
gross head of 510 m. One-third of gross head is lost in
friction in the penstock. The rate of flow through the nozzle
fitted at the end of the penstock is 2·2 m3/s. The angle of
deflection of the jet is 165°. Determine : (i) The power given
by water to the runner, and (ii) Hydraulic efficiency of the
Pelton wheel.
Take Cv (co-efficient of velocity) = 1·0 and speed ratio = 0·45.
Solution. Gross head, Hg = 510 m;
Head lost in friction, hf =510 / 3 = 170 m
∴ Net head, H = Hg – hf = 510 – 170 = 340 m
Discharge, Q = 2·2 m3/s; Angle of deflection = 165°
Power = (ρ a V1) x [Vw1 ± Vw2] x u = ??? Find V1, Vw1 , Vw2 & u
V1=Cv√2gH, u=Ku ×√2gH, Vw1=V1,& Vw2(use outlet vel triangle)
ήh = RP / WP = (ρ a V1) x [Vw1 ± Vw2] x u / ρ g Q H
Prepared By Prof. S. G. Taji

9. Pelton Wheel Turbine – Numericals
∴ Angle, φ = 180° – 165° = 15°; Cv = 1.0, Ku = 0.45
1. The power given by water to the runner :
Power = (ρ a V1) x [Vw1 ± Vw2] x u = ??? Find V1, Vw1 , Vw2 & u
V1=Cv√2gH = 1.0 √ 2 × 9.81 × 340 = 81.67 m/s
u=Ku ×√2gH = 0.45 √ 2 × 9.81 × 340 = 36.75 m/s
From inlet velocity triangle Vr1=V1 – u = 44.92 m/s (u1 = u2 = u)
Also, Vw1=V1= 81.67 m/s
Now, for Vw2, consider outlet
velocity triangle
Prepared By Prof. S. G. Taji
V1
u1 Vr1
Vw1
u1 = u2 = u
Vr2
u2 Vw2
V2
Vf2
βϕ
ϕ
Dirn of Motion

10. Pelton Wheel Turbine – Numericals
1. The power given by water to the runner :
From outlet velocity triangle, we have: Vr2 = Vr1 = 44.92 m/s
Also, cos φ = u2 + Vw2 / Vr2  Vr2 cos φ = u2 + Vw2 = u + Vw2
or, Vw2 = Vr2 cos φ – u = 44.92 cos 15° – 36.75 = 6.64 m/s
Work done / per second = ρQ (Vw1 + Vw2) × u
= 1000 × 2.2 (81.67 + 6.64) × 36.75
Power given to the runner=7139863 Nm/s = 7139.8 kW
2. Hydraulic Efficiency of the Pelton Wheel:
ήh = RP / WP = (ρ a V1) x [Vw1 + Vw2] x u / ρ g Q H
Power = 7139.8 kW
ρ g Q H = 1000 x 9.81 x 2.2 x 340
ήh = 97.3 %
Prepared By Prof. S. G. Taji
Vr2
u2 Vw2
V2
Vf2
βϕ

11. Pelton Wheel Turbine – Numericals
Prepared By Prof. S. G. Taji

12. Pelton Wheel Turbine – Numericals
Prepared By Prof. S. G. Taji

13. Pelton Wheel Turbine – Numericals
Prepared By Prof. S. G. Taji

14. Thank You….!!!