Kev paub

Dab tsi yog qhov zoo tshaj Q, thiab nws txiav txim siab ua haujlwm li cas?

May 26, 2026 Tso lus

Q: Q txhais li cas, xam, thiab ua kom zoo rau kev sim teb?

 

A: Q=X_L / R=lub zog khaws cia / lub zog dissipated × 2π. Nws txiav txim siab voltage amplification thiab input zog txuag.

Cheebtsam ntawm tag nrho cov kuj R:

• Reactor tooj liab: 40–60%. Core poob: 15-25%. Lead tsis kam: 5-10%.

• Load dielectric (tanδ): 10–25%. Corona/PD: 0-5%.

 

Yam Q tus nqi los ntawm kev thauj khoom:

Load Hom

Capacitance

Hom Q

Fais fab transformer

5-20 nF

30–60

GIS/substation

1–50 nF

40–100

MV kab (< 1 km)

0.1–0.5 μF

30–50

HV cable (>5 km)

1–5 μF

15–30

Lub tshuab hluav taws xob stator

0.5–5 μF

20–50

Capacitor bank

10–100 μF

10–20

Q vs. Input Power (rau 500 kVA tso zis):

Q=10 → 50 kW (loj diesel)|Q=30 → 16.7 kW (nruab nrab gen)

Q=50 → 10 kW (small gen)|Q=80 → 6.25 kW (mains)|Q=100 → 5 kW (mains)

Factors cuam tshuam Q:

• Reactor: loj dua qhov sib txawv → Q. Siv nplej-oriented steel. Litz hlau> 200 Hz.

• Ntau zaus: siab dua f → qis Q (cov nyhuv ntawm daim tawv nqaij).

• Load: siab dua C → qis Q. Voltage: siab dua V → qis Q (corona poob).

Teb kwv yees: Q_est ≈ 1/(tanδ_specimen + tanδ_reactor).

Yog tias tanδ_specimen=0.005 thiab tanδ_reactor=0.02 → Q ≈ 40.

⚠ Ib txwm xav tias Q 20% qis dua nominal rau lub tshuab hluav taws xob loj.

Xa kev nug