⚡ Hydraulic Turbine Design
Design Pelton, Francis, and Kaplan turbines. Compute specific speed, runner diameter, efficiency envelopes, shaft power, and cavitation safety.
⚡ Fortran 90 Engine
Double Precision (IEEE 754)
✓ ISO / ASME Validated
Fluid Mechanics
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📦 Fortran Code
4.5 KB
📅 Released
Jun 2026
⏱️ Latency
< 1 ms
🔄 Turbine Schematic
📝 Configuration
Key Equations:
Phydro = ρgQH
Ns = N√PkW/H5/4
Nq = N√Q/H3/4
U = ϕ√(2gH); D = 2U/ω
Pshaft = η Phydro
Phydro = ρgQH
Ns = N√PkW/H5/4
Nq = N√Q/H3/4
U = ϕ√(2gH); D = 2U/ω
Pshaft = η Phydro
📊 Results
Configure inputs and click Design to view results.
📘 Methodology
Specific Speed
The metric specific speed Ns = N√P/H5/4 determines the optimal turbine type: Pelton for low Ns (<60), Francis for medium (60–300), Kaplan for high (>300). Nq uses flow instead of power.
Euler Equation
The peripheral speed U relates to runner diameter through U = ωD/2. A flow coefficient ϕ scales U relative to the spouting velocity √(2gH). Runner diameter follows from the chosen speed.
Assumptions
- Empirical Gaussian efficiency envelopes vs Ns.
- Simplified cavitation sigma from Nq correlation.
- Single-jet Pelton; no multi-jet correction.
- Constant head and flow (design point).
📘 Calculation Methodology: Hydraulic Turbines (Pelton, Francis, Kaplan) Performance
Mathematical Model & Theory
Hydraulic turbines convert potential and kinetic energy of water flow into mechanical shaft power with efficiency $\eta = P_{shaft} / (\rho g Q H_{net})$:
$$P_{available} = \rho g Q H_{net}, \quad P_{shaft} = \eta \rho g Q H_{net}$$
$$N_{st} = \frac{N \sqrt{P_{shaft}}}{H_{net}^{5/4}} \quad (\text{Specific Speed})$$
Assumptions
- Incompressible, steady water flow through runner.
- Net head $H_{net} = H_{gross} - h_{losses}$.
Academic References
- Dixon, S. L., & Hall, C. A.: Turbomachinery, Butterworth-Heinemann.
- Munson, B. R. et al.: Fluid Mechanics, Ch. 12.
Worked Engineering Example
Problem Statement:
A hydroelectric turbine with $H_{net} = 120\text{ m}$, $Q = 15\text{ m}^3/\text{s}$, and $\eta = 91\%$ drives a generator ($\eta_{gen} = 97\%$). Calculate electrical output.
Step-by-step Solution:
1. $P_{hyd} = 1000 \times 9.81 \times 15 \times 120 = 17.658\text{ MW}$.
2. $P_{elec} = 17.658 \times 0.91 \times 0.97 \approx 15.59\text{ MW}$.
Final Result:
Electrical generation output is $\mathbf{15.59\text{ MW}}$.
A hydroelectric turbine with $H_{net} = 120\text{ m}$, $Q = 15\text{ m}^3/\text{s}$, and $\eta = 91\%$ drives a generator ($\eta_{gen} = 97\%$). Calculate electrical output.
Step-by-step Solution:
1. $P_{hyd} = 1000 \times 9.81 \times 15 \times 120 = 17.658\text{ MW}$.
2. $P_{elec} = 17.658 \times 0.91 \times 0.97 \approx 15.59\text{ MW}$.
Final Result:
Electrical generation output is $\mathbf{15.59\text{ MW}}$.