1. What is Specific Energy in Hydraulics?

Definition: Specific energy represents the total energy per unit weight of fluid at a particular point in a hydraulic system.

Purpose: It helps engineers analyze and design hydraulic systems, channels, and pipelines by accounting for velocity head, elevation head, and pressure head.

2. How Does the Calculator Work?

The calculator uses the formula:

Where:

• \( E \) — Specific energy (ft)
• \( V \) — Velocity of flow (ft/s)
• \( g \) — Gravitational acceleration (32.174 ft/s²)
• \( h \) — Elevation height (ft)
• \( P \) — Pressure (psi)
• \( \gamma \) — Specific weight of fluid (lb/ft³)

Explanation: The formula combines kinetic energy (velocity head), potential energy (elevation head), and pressure energy (pressure head) into a single energy value.

3. Importance of Specific Energy Calculation

Details: Specific energy is crucial for determining flow characteristics, identifying critical flow conditions, and designing efficient hydraulic structures.

4. Using the Calculator

Tips: Enter the flow velocity, elevation height, pressure (default water γ = 62.4 lb/ft³). All values must be ≥ 0, with γ > 0.

5. Frequently Asked Questions (FAQ)

Q1: Why is gravitational acceleration constant?
A: 32.174 ft/s² is standard gravity at Earth's surface, used for most engineering calculations.

Q2: What's a typical specific weight for water?
A: Fresh water at 68°F is 62.4 lb/ft³. Saltwater is about 64 lb/ft³.

Q3: How does pressure conversion work?
A: The calculator automatically converts psi to lb/ft² (1 psi = 144 lb/ft²) for consistent units.

Q4: When would specific energy be negative?
A: Specific energy is always positive in physical systems. Negative values indicate input errors.

Q5: How is this different from total head?
A: Specific energy is similar to total head but expressed in feet of fluid rather than energy units.