PELTON WHEEL

POWER WORKING OF PELTON WHEEL

1. Group Members Name Roll Waqas 10-M CT- 19 Shaheryar 10-M CT- 57 Usama 10-M CT- 63 Ishtiaq 10-M CT- 64 M.Ehsan 10-M CT- 66

2. 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 capacity. Pelton Wheel

3. 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 Working features

4. Working of Pelton Wheel

5. Working pelton wheel

6. * 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 Workdone by Pelton W heel

7. W orkdone system design

8. 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. Power of pelton wheel Potential Energy Kinetic Energy Electrical Energy Mechanical Energy Electricity

9. The power  P  =  Fu  =  T ω , where  ω   is the angular velocity of the wheel. Substituting for  F , we have  P  = 2 ρ Q ( V i  −  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  = 2 ρ Q ( V i  − 2 u )]. Maximum power occurs when u  =  V i  /2.  P max  =  ρ QV i 2 /2. Substituting the initial jet power  V i  = √(2 gh ), this simplifies to  P max  =  ρ ghQ . 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. Equations of power

10. A wheel power divided by the initial jet power, is the turbine efficiency.   η  = 4 u ( V i  −  u )/ V i 2 . It is zero for  u  = 0 and for  u  =  V i . 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. Efficiency of pelton wheel

11. How to calculate efficiency?

12. Graph (efficiency v/s speed)

13. 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. Another Relation For 'W'