Hydropower

Overview

Engineering reference data for Hydropower in pumps.

Key Formulas

Pump Power

$P = \frac{Q \cdot H \cdot \rho \cdot g}{\eta}$

Hydraulic power / efficiency.

NPSH Available

$NPSH_a = \frac{P_s}{\rho g} + \frac{v_s^2}{2g} - \frac{P_v}{\rho g}$

Net Positive Suction Head available.

Affinity Laws

$Q \propto n, \quad H \propto n^2, \quad P \propto n^3$

Flow, head, power vs speed.

Variables

Practical Examples

Example 1: Hydroelectric Power Output

Calculate the theoretical power available from a flow of 1 m³/s water with a head (fall height) of 100 m:

$P_{th} = \rho \cdot q \cdot g \cdot h = (1000 \, \text{kg/m}^3)(1 \, \text{m}^3/\text{s})(9.81 \, \text{m/s}^2)(100 \, \text{m})$

$P_{th} = 981{,}000 \, \text{W} = 981 \, \text{kW}$

Example 2: Potential Energy in Elevated Water

Calculate the energy stored in 10 m³ of water elevated 10 m above the turbine:

$W = \rho \cdot V \cdot g \cdot h = (1000 \, \text{kg/m}^3)(10 \, \text{m}^3)(9.81 \, \text{m/s}^2)(10 \, \text{m})$

$W = 981{,}000 \, \text{J} = 981 \, \text{kJ} = 0.27 \, \text{kWh}$

---

Efficiency Considerations

Due to energy losses (friction, turbine inefficiency, electrical losses), the practically available power is always less than theoretical power. Typical hydroelectric system efficiencies:

The practical available power is calculated by applying the efficiency factor to the theoretical power formula:

$P_a = \mu \cdot \rho \cdot q \cdot g \cdot h$

where $\mu$ is the system efficiency (dimensionless, between 0 and 1).

---

Reservoir Energy Estimation

For natural reservoirs where the surface area varies with elevation, total potential energy is estimated by integrating the potential energy contributions from horizontal water segments across the full depth. This accounts for the varying cross-sectional area at different heights, providing a more accurate energy estimate than using a single average volume.

References

• Original Source