Pelton Turbines
By Dr. Laith Ayad Salman
Pelton turbines
Hydraulic Power plants
Advantages of hydraulic power plants:
higher efficiency
• opemtional tlexibility
• ease of maintenance
• long wear and tear
• potentially inexhaustible source of energy
• no atmospheric pollution
• an attraction for tourism
When can Pelton Turbines be choosen
Pelton turbine is chosen when operating head is more than 300 m.
One of the largest single unit installed at Newcolgate Power Station,
California, USA has rating of 170 MW
Description of pelton turbines
Pelton turbine is impulse turbine
No pressure drop across the buckets
Flow is axial
No change in the peripheral velocity
Water inter and leave the bucket at the same radius
How does pelton turbines work?
Water is supplied through penstock
Water accelerated in the nozzle
Head converted into speed in the form of jet at the atmospheric pressure
Jets strikes deflected buckets attached to a rotating runner
Jet losses its kinetic energy and water is discharged at low speed to
reservoir or tail race
The tail race is used to prevent submerging the wheel during flood
When a lot of water is available :
Two wheels is connected to single shaft.
Two or more jets are arranged to a single wheel.
Aspear moving inside the nozzle controls water to the turbine
Pelton wheel Installation
• Water is supplied through a reservoir
• Surge tank is used to dump out pressure fluctuations
• At the end of the penstock a nozzle is installed to convert head to velocity as
jet
H1 = total head available
hf = head losses due to friction
hn = head losses in the nozzle
Net head available = H = H1- hf-hn
Hf = (flv/2gd)
l = length of the penstock
V= velocity in the penstock
d = diameter of the penstock
f = coefficient of friction
In practice the net head is 85-95 % of the total head
Analysis of Pelton wheel
Jet velocity (v1) = (2gh)0.5
Total transferred energy (E) = (U1Vu1-U2Vu2)/g
Since the turbine is axial U1 =U2 = U
So E = U(Vu1-Vu2)/g
As the jet strike the bucket horizontally ,
Vu1 = V1
Vw1 = V1- U
Pelton Wheel efficiencies
Hydraulic efficiency= (γ Q E)/ (γ Q H)
Mechanical efficiency = power developed by turbine shaft (p)/ (γ Q E)
Total efficiency = p / (γ Q H) = η t = ηh x ηm
Pelton Wheel Design
V1 = (2gH)0.5 where H is the net head
Va = actual velocity of the jet = Cv (2gH)0.5 where Cv is between 0.98 to 0.99
Power available is (p) = γ Q H
where γ (gama ) is the specific weight of water, in N/m3, Q is the flow rate in m3/S, H
head in meters
Diameter of the jet (d) = π/4 d2 x V1 = π/4 d2 x Cv (2gH)0.5
Speed ratio = U/V1
U = (πdN)/ 60 so mean diameter of the wheel is D = (60 x U)/ πN
Jet ratio (m) = D/d
Number of buckets (z) = D/2d + 15 or Z = 0.5m +15
Example
A Pelton wheel develops 67.5 kw under a head of 60
m of water. It rotates at 400 rev/min. The diameter of
penstock is 200 mm. The ratio of bucket speed of jet
velocity is 0.46 and overall efficiency of the
installation is 83%. Calculate: (a) Volumetric flow
rate (b) Diameter of the jet (c) Wheel diameter
Solution :
P = 67.5 Kw , H= 60 m , ω = 400 rev/min , U/V1 = 0.46, over all
efficiency = 0.83
η o = P / (Y Q H ) = 67.5 x1000/0.83x 9800x60 = 0.138 m3/s = Q
V1 = (2gH)0.5 = (2x9.81x60)0.5 = 34.2 m/s
Q = π/4 d2 x V1
d2= 0.138x4 /(πx 34.2) = 71.1mm
U/V1= 0.46 so U = 0.46 x 34.2 = 15.7 m/s
D = 60 x U / π x N = 60 x 15.7 / π x 400 = 0.75 m
Turbine specific speed Wt = ω (P/ρ)0.5 / (gH) 5/4
ω = 2 π x N /60 = 41.8 rad/s, so Wt = 0.11
