What is the approximate efficiency of a Kaplan turbine developing 3000 kW under a net head of 5 m? It is provided with a draft tube with its inlet (diameter 3 m) set 1·6 m above the tailrace level. A vacuum gauge connected to the draft tube indicates a reading of 5 m of water. Assume draft tube efficiency as 68% and acceleration due to gravity as 10 m/s². Neglect head losses in draft tube.
- A85%Correct
- B90%
- C80%
- D95%
Solution
Concept
Draft tube analysis and overall efficiency of a turbine
Formula
ηd = (V₁²/2g − V₂²/2g − hf) / (V₁²/2g); ηo = P / (ρ g Q H)
Step-by-step solution
1. Draft tube efficiency 0.68 with no friction loss: V₂²/2g = 0.32 V₁²/2g. 2. Bernoulli from the inlet to the tailrace: −5 + V₁²/2g + 1.6 = V₂²/2g, so 0.68 V₁²/2g = 3.4 and V₁²/2g = 5 m, V₁ = 10 m/s. 3. Q = (π/4) × 3² × 10 = 70.7 m³/s. 4. Water power = 1000 × 10 × 70.7 × 5 = 3534 kW. 5. Efficiency = 3000 / 3534 = 0.85 = 85%.
Explanation
The vacuum reading at the draft-tube inlet and its height above the tailrace fix the velocity there. From the velocity the discharge and the power in the water follow, and the turbine efficiency is the output divided by this power.
Why the other options are wrong
90%, 80% and 95% would need water powers of 3333, 3750 and 3158 kW, which do not match the discharge found.
Shortcut
V₁²/2g = 3.4/0.68 = 5 m → V₁ = 10 m/s.
Common mistake
Forgetting the draft tube efficiency while applying Bernoulli.
Correct answer: A) 85%