A hydraulic turbine develops a discharge of 5 m^3/s under a head of 20 m at a rotational speed of 500 rpm. If the turbine must operate under a reduced head of 15 m for the same discharge (geometrically similar operation), what will be the approximate.

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Solve this multiple-choice question and choose the correct option.

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Introduction / Context:Turbine similarity (unit quantities) relates speed, discharge, and head. For dynamically similar operation, rotational speed varies with the square root of head. This relation is widely used to predict operating points when head fluctuates in hydropower schemes. Given Data / Assumptions: Initial head H1 = 20 m, speed N1 = 500 rpm, discharge Q is unchanged. New head H2 = 15 m, geometric similarity assumed. Affinity law: N ∝ √H for similar operation. Concept / Approach:For turbines, the specific speed and similarity conditions yield N2 = N1 * √(H2/H1), when comparing points of similar efficiency profiles and geometry. Since only head changes, speed scales by √H. Step-by-Step Solution: Compute head ratio: H2/H1 = 15/20 = 0.75.Take square root: √0.75 ≈ 0.866.Calculate N2: 500 * 0.866 ≈ 433 rpm. Verification / Alternative check: Check unit speed n11 = N * √D / √H (for fixed geometry D constant, n11 roughly constant), confirming the proportionality. Why Other Options Are Wrong: 403 and 388 rpm underestimate the √H scaling; 627 rpm contradicts the reduction in head; 500 rpm implies no scaling with head. Common Pitfalls: Confusing pump and turbine affinity relations or neglecting the square-root dependency on head. Final Answer: 433

A hydraulic turbine develops a discharge of 5 m^3/s under a head of 20 m at a rotational speed of 500 rpm. If the turbine must operate under a reduced head of 15 m for the same discharge (geometrically similar operation), what will be the approximate new speed (rpm)?

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