the jet is constant. 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: − (Vi − u) = −Vi + u. In the standard reference frame (relative to the earth), the final velocity is then: Vf = (−Vi + u) + u = −Vi + 2u. Optimal wheel speed The ideal runner speed will cause all of the kinetic energy in the jet to be transferred to the wheel. In this case the final jet velocity must be zero. If −Vi + 2u = 0, then the optimal runner speed will be u = Vi /2, or half the initial jet velocity. Torque By Newton's second and third laws, the force F imposed by the jet on the runner is equal but opposite to the rate of momentum change of the fluid, so F = −m(Vf − Vi)/t = −ρQ[(−Vi + 2u) − Vi] = −ρQ(−2Vi + 2u) = 2ρQ(Vi − u), where ρ is the density, and Q is the volume rate of flow of fluid. If D is the wheel diameter, the torque on the runner is. T = F(D/2) = ρQD(Vi − u). The torque is maximal when the runner is stopped (i.e. when u = 0, T = ρQDVi). When the speed of the runner is equal to the initial jet velocity, the torque is zero (i.e., when u = Vi, then T = 0). On a plot of torque versus runner speed, the torque curve is straight between these two points: (0, pQDVi) and (Vi, 0). Nozzle efficiency is the ratio of the jet power to the waterpower at the base of the nozzle. Power 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 when u = Vi /2. Pmax = ρQVi2/2. Substituting the initial jet power Vi = , 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. Efficiency A wheel power