Draft tube performance definition If V1 and V2 are the absolute water velocities at the inlet and outlet of a draft tube, respectively, the efficiency of the draft tube is:

Introduction / Context:
A draft tube is fitted at the outlet of reaction turbines (e.g., Francis, Kaplan) to convert kinetic energy at runner exit into static pressure, allowing recovery of part of the velocity head and enabling the turbine to be installed above tail water level without excessive loss of head.

Given Data / Assumptions:

• Inlet absolute velocity to draft tube = V1; outlet absolute velocity = V2.
• Same datum for assessing kinetic head; density and gravity constant.
• Neglecting elevation changes inside the short tube for the definition (focus on kinetic head recovery).

Concept / Approach:
The draft tube efficiency quantifies the fraction of the inlet kinetic head recovered as useful static head. Kinetic head is proportional to V^2. Therefore, recovered fraction equals (V1^2 − V2^2)/V1^2, a non-dimensional number between 0 and 1 in ideal cases (losses make it less than 1).

Step-by-Step Solution:
Kinetic head at inlet = V1^2 / (2 g).Kinetic head at outlet = V2^2 / (2 g).Recovered head = (V1^2 − V2^2) / (2 g).Efficiency η_d = (Recovered head) / (Inlet kinetic head) = (V1^2 − V2^2)/V1^2.

Verification / Alternative check:
If V2 ≈ 0 (excellent diffusion), η_d → 1. If no recovery (V2 = V1), η_d = 0. These limits match physical expectations.

Why Other Options Are Wrong:
Reversing numerator sign makes η_d negative. Velocity ratios (c) and (d) ignore the squared relation of kinetic energy. Option (e) uses the wrong reference in the denominator.

Common Pitfalls:
Mixing velocity head with velocity; forgetting that energy terms scale with the square of velocity.

Final Answer:
η_d = (V1^2 - V2^2) / V1^2