POWER WORKING OF PELTON WHEEL

2. Name Roll
 Waqas 10-MCT-19
 Shaheryar 10-MCT-57
 Usama 10-MCT-63
 Ishtiaq 10-MCT-64
 M.Ehsan 10-MCT-66

3. Pelton Wheel is the only
impulse turbine, named in
honour of L.A. Pelton (1829-
1908) of California, USA.
It was invented in 1878.
It is an efficient turbine
particularly suited to high
heads with efficiencies often
more than 90%.
Single nozzle impulse
turbines have a flat efficiency
curve and may be operated
down to loads of 20% of rated

4. The water flows along the
tangent to the path of the
runner.
When the water-jet contacts
the bucket, the water exerts
pressure on the bucket and
the water is decelerated as it
does a "u-turn" and flows out
the other side of the bucket at
low velocity.
A very small percentage of
the water's original kinetic
energy will still remain in the
water; however, this allows the
bucket to be emptied at the
same rate it is filled

7. *Work done is a parameter of physical
performance.
Work done by a pelton wheel per second
is determined,using torque(T) revolving
the wheel and the angular
displacement(d) in radian per second and
is given by:
Work=T*d
d is the angle between the whirl flow
direction and the vane.Torque is
calculated by angular velocity measured
by TachoMeter in RPM

9. Power of the pelton
wheel does not depend
upon diameter of the
wheel.
It depends upon the
head and amount of water
applied to it.
We can increae our
power by using double
nozzles in pelton wheel.
Power of pelton is also
depends upon head.
Head of 1200 to 2000 is
common occurence head
in pelton wheel.
Potential
Energy
Kinetic
Energy
Electrical
Energy
Mechanical
Energy
Electricity

10. The power P = Fu = Tω, where ω is
the angular velocity of the wheel.
 Substituting for F, we have P = 2ρQ(Vi − u)u.
 To find the runner speed at maximum
power, take the derivative of P with respect
to u and set it equal to zero, [dP/du = 2ρQ(Vi −
2u)].
 Maximum power occurs
whenu = Vi /2. Pmax = ρQVi
2/2.
Substituting the initial jet power Vi = √(2gh),
this simplifies to Pmax = ρghQ.
This quantity exactly equals the kinetic
power of the jet, so in this ideal case, the
efficiency is 100%, since all the energy in the
jet is converted to shaft output.

11. A wheel power divided by the
initial jet power, is the turbine
efficiency.
 η = 4u(Vi − u)/Vi
2.
It is zero for u = 0 and for u = Vi.
 As the equations indicate, when
a real Pelton wheel is working close
to maximum efficiency, the fluid
flows off the wheel with very little
residual velocity.
 This basic theory
does not suggest that efficiency
will vary with hydraulic head.

15. The work done /s by unit weight of water striking/s is given
by:
Work=1/g[(Vw1+Vw2)u]
Where
g is gravitational acceleration;
Vw1 is velocity of whirl at inlet;
Vw2 is velocity of whirl at outlet and
u is tangential velocity of vane or bucket.