Day 12 Hydraulic Turbines – ECET 2026 Mechanical

Why This Topic is Important for ECET

Hydraulic turbines are a core subject in Hydraulics & Fluid Mechanics, directly applied in hydropower plants. In ECET exams, questions come from types of turbines, efficiencies, governing, specific speed, and velocity triangles. Mastering this ensures easy scoring + practical knowledge for future engineering applications.

📘 Concept Notes

🔹 What is a Hydraulic Turbine?

A hydraulic turbine is a rotating machine that converts water energy (potential/kinetic) into mechanical shaft work. This shaft work is later converted into electrical energy by generators.

🔹 Classification of Hydraulic Turbines

Based on Energy Conversion:
- Impulse Turbine: Uses only kinetic energy of water.
  Example → Pelton Wheel.
- Reaction Turbine: Uses both pressure & kinetic energy.
  Examples → Francis Turbine, Kaplan Turbine.
Based on Flow Direction:
- Radial flow (inward/outward)
- Axial flow
- Mixed flow
Based on Head of Water Supply:
- High head (>250 m) → Pelton turbine
- Medium head (60–250 m) → Francis turbine
- Low head (<60 m) → Kaplan turbine

🔹 Main Components of a Turbine

Nozzle: Directs water jet (Pelton).
Runner: Wheel/blades that rotate.
Shaft: Transmits power.
Casing: Prevents water splash & directs flow.
Draft Tube: Helps recover pressure at exit (reaction turbines).

🔹 Important Terms

Hydraulic Efficiency: Ratio of power delivered to runner to water power.
Mechanical Efficiency: Ratio of shaft power to runner power.
Overall Efficiency: Ratio of shaft power to water power.
Specific Speed (Ns): Turbine speed when producing unit power under unit head.

🔹 Examples

Pelton Wheel: High head → mountain regions.
Francis Turbine: Medium head → multipurpose projects (e.g., Nagarjuna Sagar Dam).
Kaplan Turbine: Low head → river-based stations.

⚙️ Formulas

Water Power:

$P_w = \rho g Q H$

Hydraulic Efficiency:

$\eta_h = \frac{P_{runner}}{P_w}$

Mechanical Efficiency:

$\eta_m = \frac{P_{shaft}}{P_{runner}}$

Overall Efficiency:

$\eta_o = \frac{P_{shaft}}{P_w}$

Specific Speed:

$N_s = N \sqrt{P} / H^{5/4}$

Unit Speed:

$N_u = N / \sqrt{H}$

Unit Power:

$P_u = P / H^{3/2}$

Unit Discharge:

$Q_u = Q / \sqrt{H}$

🔟 10 MCQs

Q1. Which of the following is an impulse turbine?
a) Francis
b) Kaplan
c) Pelton
d) Propeller

Q2. A Kaplan turbine is best suited for:
a) High head
b) Medium head
c) Low head
d) Ultra-high head

Q3. The function of a draft tube in a reaction turbine is:
a) Increase velocity
b) Decrease pressure
c) Recover pressure head
d) Prevent cavitation

Q4. The specific speed of a Pelton wheel is generally:
a) 8–30
b) 50–250
c) 250–500
d) 500–1000

Q5. If a turbine develops 2000 kW under 100 m head at 300 rpm, the specific speed is approximately:
a) 20
b) 30
c) 40
d) 60

Q6. The efficiency ratio of shaft power to water power is called:
a) Hydraulic efficiency
b) Mechanical efficiency
c) Overall efficiency
d) Volumetric efficiency

Q7. Francis turbine is a:
a) Axial flow reaction turbine
b) Radial flow impulse turbine
c) Mixed flow reaction turbine
d) Radial flow impulse turbine

Q8. In Pelton wheel, the bucket deflects the water jet by:
a) 60°–90°
b) 90°–120°
c) 120°–160°
d) 160°–170°

Q9. For maximum efficiency of Pelton wheel, the speed ratio (runner speed/jet speed) should be around:
a) 0.25
b) 0.46
c) 0.75
d) 1.0

Q10. A turbine works under 80 m head and produces 1000 kW power. Water flow rate = 2 m³/s. The overall efficiency is:
a) 50%
b) 60%
c) 70%
d) 80%

✅ Answer Key

Q	Ans
1	c
2	c
3	c
4	a
5	b
6	c
7	c
8	d
9	b
10	c

🧠 Explanations

Q1: Pelton is impulse turbine (only kinetic energy). → (c)
Q2: Kaplan = axial flow, low head (<60 m). → (c)
Q3: Draft tube recovers pressure head & prevents cavitation. → (c)
Q4: Pelton wheel specific speed very low (8–30). → (a)
Q5: $N_s = N \sqrt{P} / H^{5/4}$ → $N_s = 300 \times \sqrt{2000} / (100^{1.25}) = 300 \times 44.7 / 316 \approx 42.5$ ≈ 40 → (b)
Q6: Overall efficiency = shaft/water power. → (c)
Q7: Francis = mixed flow reaction turbine. → (c)
Q8: Pelton buckets deflect jet by 160°–170°. → (d)
Q9: Optimum speed ratio ~0.46. → (b)
Q10: Water power = $\rho g Q H = 1000 \times 9.81 \times 2 \times 80 = 1569.6 kW$ . Efficiency = 1000/1569.6 = 0.637 = 63.7% ≈ 70% → (c)

🎯 Motivation / Why Practice Matters

In ECET 2026, Hydraulic turbines questions are scoring because formulas are direct and concepts are repetitive. With practice, you can solve numerical problems in <1 minute. Mastering this boosts your accuracy, speed, and confidence—all essential to secure a top rank in ECET Mechanical.

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Day 12 Hydraulic Turbines – ECET 2026 Mechanical

Why This Topic is Important for ECET