Simple impulse turbine optimum speed ratio:\nFor a simple impulse stage with nozzle exit speed V at angle α to the wheel tangent, the rotor (diagram) efficiency is maximum when the blade speed Vb equals ________.

Introduction / Context:
In a simple impulse stage, the pressure drop occurs in the nozzle and the rotor primarily turns the flow to extract work. The rotor (diagram) efficiency depends on the blade speed ratio and the nozzle exit angle. There is an optimal blade speed that maximizes this efficiency for given V and α.

Given Data / Assumptions:

• Impulse stage, negligible losses in ideal analysis.
• Nozzle exit absolute velocity magnitude V at angle α to the tangent.
• Constant axial velocity through the rotor, symmetric turning for the ideal case.

Concept / Approach:
Maximizing rotor efficiency amounts to maximizing the work extracted per unit kinetic energy at rotor inlet. The classical result for the ideal impulse stage gives the optimum blade speed ratio φ_opt = Vb/V = (cos α)/2. Therefore Vb = 0.5 * V * cos α at the condition for maximum diagram efficiency.

Step-by-Step Solution:

Verification / Alternative check:
Substituting φ_opt back into the expression yields the well-known maximum ideal rotor efficiency η_r,max = cos^2 α, consistent with impulse stage theory when inlet and outlet relative flow angles are equal and friction is neglected.

Why Other Options Are Wrong:

• Vb = V cos α: Too high; would not maximize rotor efficiency.
• Terms with V^2: Dimensionally incorrect for a speed; these combine velocity and velocity squared improperly.

Common Pitfalls:
Confusing the optimum for rotor efficiency with overall stage efficiency, which also depends on nozzle and mechanical losses.

Final Answer:
Vb = 0.5 V cos α