Correct Answer - Option 2 : shaft power/ runner power
Explanation:
The efficiency of any mechanical device is defined as the ratio of power output to energy input.
Hydraulic efficiency : \({\eta _h} = \frac{{Power\;delivered\;to\;runner}}{{Power\;supplied\;at\;inlet}} = \frac{{Runner\;Power}}{{Water\;Power}}\)
Mechanical efficiency : \({\eta _m} = \frac{{power\;delivered\;to\;shaft}}{{power\;delivered\;to\;runner}} = \frac{{Shaft\;Power}}{{Runner\;Power}}\)
Overall efficiency : \({\eta _o} = {\eta _h} \times {\eta _m} = \frac{{Runner\;Power}}{{Water\;Power}} \times \frac{{Shaft\;Power}}{{Runner\;Power}}\)
∴ \({\eta _o} = \frac{{Shaft\;Power}}{{Water\;Power}}\)
The overall efficiency of a turbine is the measure of the performance of the turbine.
The overall efficiency of Pelton wheel is approximately 0.85.
Because there are losses like mechanical friction loss, aerodynamic drag on the buckets or windage loss, friction along the inside walls of the buckets, nonalignment of the jet and bucket when bucket turns, back splashing and nozzle losses.
Also, the mechanical friction and aerodynamic drag losses increase with the increase in speed.
- Theoretically, a Pelton turbine can have a 100% efficiency if the speed of the incoming water jet is two times the speed of the bucket (i.e. V = 2u) or bucket speed u = 0.5 V.
- In this case, water will attain a zero velocity with respect to earth (stationary reference frame) and transfer all its energy to the turbine.
- But for practical turbines, the bucket speed is 0.44 - 0.46 of the nozzle velocity.
