Physik
Verifizierter Rechner mit transparenter Formel
Centrifugal Pump Power Calculator.
Calculates the hydraulic power required to pump a fluid based on flow rate, head, and fluid density.
Ihre Eingaben
So funktioniert es
- 1
Convert flow rate from m³/h to m³/s by dividing by 3600.
- 2
Multiply by gravitational acceleration (9.81 m/s²).
- 3
Multiply by head and fluid density.
- 4
Divide by 1000 to get power in kilowatts.
(flow_rate / 3600) * 9.81 * head * density / 1000Häufig gestellte Fragen
What is hydraulic power?
It is the theoretical power needed to move fluid at a given flow rate and head, ignoring pump and motor inefficiencies.
How do I account for pump efficiency?
Divide the hydraulic power by the pump efficiency (as a decimal) to get the shaft power required.
Why is density important?
Denser fluids require more energy to lift, so power increases with density.
Diese Rechner-Kategorie erkunden
Ergebnisse
Formel geprüftHydraulic power
2,725kW
Schätzung nur zur allgemeinen Orientierung – wichtige Entscheidungen mit einem geeigneten Fachmann verifizieren.
So funktioniert es
Calculates the hydraulic power required to pump a fluid based on flow rate, head, and fluid density.
- Convert flow rate from m³/h to m³/s by dividing by 3600.
- Multiply by gravitational acceleration (9.81 m/s²).
- Multiply by head and fluid density.
- Divide by 1000 to get power in kilowatts.
Formeln
Die Mathematik hinter diesem Rechner, aufgeschrieben, damit Sie das Ergebnis überprüfen können.
Hydraulic Power
Power in kW equals flow rate (m³/s) times density (kg/m³) times gravity (9.81 m/s²) times head (m), divided by 1000.
Example:
Input: Q = 50 m³/h, H = 20 m, ρ = 1000 kg/m³
Calculation: (50/3600) × 1000 × 9.81 × 20 / 1000
Result: ≈ 2.73 kW
Anwendungsfälle im Alltag
Wo diese Berechnung im täglichen Leben vorkommt.
Sizing a pump motor
Estimate the minimum motor power needed for a given duty.
Example: Selecting a motor for a 50 m³/h, 20 m head water pump.
Comparing pump options
Evaluate energy requirements for different flow and head conditions.
Example: Choosing between two pumps with different efficiencies.
Energy cost estimation
Calculate power consumption to estimate operating costs.
Example: Estimating annual electricity cost for a pumping system.
Tipps und häufige Fehler
Tips
- Use actual fluid density for accurate results (e.g., 998 kg/m³ for water at 20°C).
- Remember that hydraulic power is theoretical; actual power draw is higher due to losses.
- Check units: flow rate must be in m³/h, head in meters, density in kg/m³.
- For pump selection, divide by efficiency (e.g., 0.7 for 70%) to get shaft power.
Common Mistakes to Avoid
- Forgetting to convert flow rate from m³/h to m³/s.
- Using density of water for other fluids without checking.
- Ignoring pump efficiency when sizing the motor.
Annahmen und Einschränkungen
- Use the stated inputs and units.
- Results are estimates for planning and education.
- Check measurements and source data before making an important decision.