Result
Preliminary calculation: This tool is for quick sizing and information. Electrical installation designs and cable and protection selection must be done according to the relevant standards and by an licensed electrical engineer. Results are approximate.
Formulas
Q_c = P × ( tan φ₁ − tan φ₂ )S = P / cos φ, I = S / (√3 × V)Inductive loads such as motors, transformers and compressors draw reactive power. Compensation generates part of this reactive power on site with capacitors, which reduces line current, losses and voltage drop. Over-compensation is also undesirable.
- For variable loads, an automatic stepped panel is used
- In systems with inverters, harmonic filters and reactors should be considered
- Capacitors require discharge resistors and protection
Frequently asked questions
Why is reactive power charged for?
The distribution grid also carries infrastructure for reactive power; reactive consumption above certain ratios can incur an extra charge. See the tariff regulations for current limits.
What should the target cosφ be?
It depends on the facility and regulations; a range of 0.95-0.99 is often targeted.
Should compensation be done per motor or at the panel?
For large motors it is done on site; for general load a central automatic panel is preferred. It is determined for each project.