Editing Pelton wheel (section)

=== Final jet velocity ===

Assuming that the jet velocity is higher than the runner velocity, if the water is not to become backed-up in runner, then due to conservation of mass, the mass entering the runner must equal the mass leaving the runner.  The fluid is assumed to be incompressible (an accurate assumption for most liquids).  Also, it is assumed that the cross-sectional area of the jet is constant. The fluid impacts the runner, slowing down and transferring momentum from the jet to the wheel. Assuming no losses, the fluid leaving the runner's velocity is reduced by the velocity difference between the jet and the runner. The jet ''[[speed]]'' remains constant relative to the runner.  So as the jet recedes from the runner, the jet velocity relative to the runner is: − (''V''<sub>''i''</sub>&nbsp;− ''u'')&nbsp;= −''V''<sub>''i''</sub>&nbsp;+ ''u''. Note the minus sign indicates the reduction in stream velocity from impacting the runner. With the assumption that jet velocity is higher than runner velocity, the result is a "rebound" resulting in the jet flowing away from the runner. In the standard reference frame (relative to the earth), the final velocity is then: ''V''<sub>''f''</sub>&nbsp;= (−''V''<sub>''i''</sub>&nbsp;+ u)&nbsp;+ ''u''&nbsp;= −''V''<sub>''i''</sub>&nbsp;+ 2''u''.

The value ''V''<sub>''f''</sub> is bounded by two cases: a stationary runner and a runner moving at the velocity of the stream, ''V''<sub>''i''</sub>. For the stationary runner case, the stream velocity is -Vf, indicating the fluid fully reversed direction. In this case the force on the wheel is the highest (due to the largest possible velocity change), but power delivered is zero, since there is no movement. For the case where the runner is moving at the speed of the stream, the velocity of the wheel is the highest, but power delivered is also zero, since there is no torque or force imparted on the wheel (due to no change is stream velocity). 
<!-- Note: reference is believed to have two sign errors, that are believed to be corrected here.  The two sign errors cancel each other, so the end result is the same.-->