Parsons (50% reaction) turbine — minimum leaving kinetic energy A Parsons stage has blade peripheral speed U = 320 m/s at the mean radius, and the rotor-blade exit (relative) angle β2 = 30°. For minimum kinetic energy of steam leaving the stage (i.e., zero whirl at exit), determine the absolute steam velocity at the rotor exit V2.

Introduction / Context:
This problem tests steam-turbine velocity triangles for a Parsons (50% reaction) stage and the design condition for minimum leaving kinetic energy (LKE). Minimizing the absolute exit velocity reduces exhaust kinetic-energy losses and improves stage efficiency. The key geometric link is between blade speed, relative flow angle at rotor exit, and the absolute exit velocity.

Given Data / Assumptions:

• Blade (peripheral) speed at mean radius U = 320 m/s.
• Rotor exit relative flow angle β2 = 30° (measured from the direction of wheel motion).
• Design for minimum LKE implies zero whirl component at exit, Vw2 = 0.
• Axial velocity is the absolute exit velocity when Vw2 = 0.

Concept / Approach:
For the rotor exit triangle: the relative velocity Vr2 leaves at angle β2 to wheel direction. Its tangential component is Vr2 * cosβ2 and axial component is Vr2 * sinβ2. The absolute tangential (whirl) component at exit is Vw2 = Vr2 * cosβ2 − U. Setting Vw2 = 0 gives Vr2 = U / cosβ2. Then the absolute exit speed equals its axial component Va2 = Vr2 * sinβ2 = U * tanβ2. Therefore, V2 = U * tanβ2 at the minimum LKE condition.

Step-by-Step Solution:
Write the exit-whirl condition: Vw2 = Vr2 * cosβ2 − U = 0.Solve for Vr2: Vr2 = U / cosβ2.Compute absolute exit velocity: V2 = Va2 = Vr2 * sinβ2 = (U / cosβ2) * sinβ2 = U * tanβ2.Substitute U = 320 m/s and β2 = 30°: tan30° ≈ 0.577 ⇒ V2 ≈ 320 * 0.577 ≈ 184.6 m/s ≈ 185 m/s.

Verification / Alternative check:
With Vw2 = 0, the leaving velocity is purely axial and minimized for the given U and β2. Any nonzero Vw2 would increase the magnitude of V2, raising exhaust losses (V2^2/2), confirming 185 m/s as the minimum-LKE result.

Why Other Options Are Wrong:
213 m/s and 107 m/s correspond to other arbitrary combinations (e.g., 640/3 or 320/3) not supported by the zero-whirl condition. 640 m/s implies an unrealistically high exit speed. 355 m/s would require tanβ2 ≈ 1.11, not 0.577.

Common Pitfalls:
Mixing up β2 with an inlet angle, or using φ = U/V1 relationships mistakenly at exit. Also, confusing the “minimum LKE” condition with “maximum work” generally—here both align with Vw2 = 0 for the single stage.

Final Answer:
185 m/s