Marekebisho ya sababu ya nguvu ni nini?

Power factor correction (PFC) is the process of adding capacitors to an electrical system to reduce reactive power demand and bring the power factor closer to unity (1.0). This reduces current draw, lowers losses, and avoids utility penalties.

Ninawezaje kuhesabu ukubwa wa capacitor unaohitajika?

Qc = P × (tan φ1 − tan φ2), where P is active power in kW, φ1 = arccos(PF_current), φ2 = arccos(PF_target). The result Qc is in kVAR. For capacitance in μF: C = Qc × 10⁹ / (2π × f × V²).

Ni lengo gani la sababu ya nguvu ninastahili kukusudia?

Huduma nyingi za umeme zinahitaji PF ≥ 0.90 ili kuepuka faini. Lengo la 0.95–0.98 linashauriwa. Epuka kusahihisha kupita kiasi juu ya 1.0 (PF inayoongoza) kwani hii inaweza kusababisha ongezeko la voltage na matatizo ya resonance.

Faida za marekebisho ya sababu ya nguvu ni zipi?

Benefits include: reduced electricity bills (avoiding reactive power charges), lower current in cables (freeing capacity), reduced I²R losses, improved voltage regulation, and compliance with utility requirements.

Je, ninapaswa kutumia benki za capacitor za kudumu au za kiotomatiki?

Benki za kudumu zinafaa kwa mizigo thabiti (mfano, motor moja kubwa). Benki za kiotomatiki zenye contactors na kidhibiti cha PF zinafaa zaidi kwa mizigo inayobadilika — zinawasha/kuzima hatua za capacitor ili kudumisha lengo la PF.

Zana ya Uhandisi Isiyolipisha

Kalkuleta ya Urekebishaji wa Sababu ya Nguvu

Calculate required capacitor bank kVAR and μF to improve power factor. Qc = P × (tan φ₁ − tan φ₂).

kVAR Sizing Capacitor μF Power Triangle

Active Power P (kW)

Sababu ya Nguvu ya Sasa (cos φ₁)

Sababu ya Nguvu Inayolengwa (cos φ₂)

System Voltage (V, mstari hadi mstari)

Frequency (Hz)

Muunganiko wa Capacitor

Quick presets

Results

Benki ya Capacitor Inayohitajika

—

Capacitance kwa Kila Awamu

—

Current Before (S₁ / I₁)

—

Mkondo Baada (S₂ / I₂)

—

Upunguzaji wa Mkondo

—

Reactive Power Compensation

Uwezo katika μF

Mfano — mzigo wa kW 100, PF 0.75 → 0.95

tan φ₁ = tan(acos 0.75) = 0.8819, tan φ₂ = tan(acos 0.95) = 0.3287

Qc = 100 × (0.8819 − 0.3287) = 55.3 kVAR

⚠️ Note: Usiofanya marekebisho kupita kiasi zaidi ya umoja wa PF (inayoongoza). Hii inaweza kusababisha kupanda kwa voltaji, mwangwi na transfoma, na uharibifu wa vifaa.

Kikokotoo cha Marekebisho ya Sababu ya Nguvu | Zana ya Bure ya kVAR & μF

Published by Nikolai Shelkovenko on Febuari 15, 2026 Aprili 5, 2026

Results

Reactive Power Compensation

Uwezo katika μF

Duka

Support

Contact