功率因数补偿计算器
求出提高功率因数所需的电容器 kvar。
Reads AS/NZS 3000:2018.
尝试一个工程 10
每项工作都会预填示例数值。请替换为您自己的数值。
- Correct a factory power factorThe kvar and bank steps to lift a 250 kW load from 0.78 to 0.95.
- Free supply capacity with power factor correctionThe kVA a 400 kVA site gets back when its power factor rises to 0.98.
- Correct a motor boardThe kvar to lift a 75 kW motor load from 0.82 to 0.95.
- Choose the capacitor stepsThe bank in 25 kvar steps for a 300 kVA load.
- Check a bank for resonanceThe bank against the fault level and 20 % harmonic load.
- Correct a single‑phase loadThe kvar for a 15 kW single‑phase load at 0.7.
- Size a capacitor at a motorThe largest capacitor a pump motor can take, and its new line current.
- Correct a small motor at the motorWhy a small motor capacitor stops short of the target power factor.
- Cut the line current of a large motorThe current a capacitor at a large motor takes off its cable and starter.
- Check a motor capacitor for self-excitationThe no-load limit on a capacitor at a small conveyor motor.
输入
使用默认输入值的示例计算。
- 正确
- 负载或配电板
- 相
- 三相
- 电压
- 400 V
- 负荷单元
- kW
- 负荷
- 100
- 当前功率因数
- 0.75
- 目标功率因数
- 0.95
- 电容组级
- 25 kvar
- 电流谐波畸变率（THDi）
- 0 %
- 电池组处的短路电流水平
- 0 MVA
需要修正
55.32kvar
- 已装电容组
- 3 × 25kvar
- 电容器组的功率因数
- 0.991
- 之前的电流
- 192.45A
- 之后的电流
- 145.59A
- 释放容量
- 32.47kVA
- 电容组导体，至少
- 146.14A
此结果仅供设计参考。请对照现行标准进行核验。
计算步骤
- 有功功率100kW
- 现有功率因数下的无功功率88.19kvar
- 目标功率因数下的无功功率32.87kvar
- 需要修正结果55.32kvar
- 电容器组级数
3 × 25 kvar
- 带电容器组的功率因数0.99
- 之前的电流192.45A
- 之后的电流145.59A
- 释放容量32.47kVA
- 电容器组的额定电流108.25A
- 电容器组导体的最小载流量，135 %146.14A
AS/NZS 3000:2018 Cl 4.15.2.3
关于此计算器
How this is calculated
The tool finds the capacitor bank that lifts a load to a target power factor. It then gives the current before and after.
- Real power. You enter kW, or kVA with a power factor. The tool finds the real power.
- Reactive power. It finds the reactive power at the present power factor. It does the same at the target.
- Correction required. The difference is the kvar the bank must supply.
- Bank steps. It rounds up to whole steps of the size you enter. It then recalculates the power factor with that bank.
- Currents. It finds the current before and after. The difference is the capacity the bank releases.
- Bank conductors. It sizes the bank conductors at 135 % of the bank rated current. This follows AS/NZS 3000 Cl 4.15.2.3.
AS/NZS 3000 Cl 4.15.2.2 covers switching, and Cl 4.15.3.1 covers discharge.
Worked example
- 100 kW three‑phase load at 400 V, 0.75 to 0.95, 25 kvar steps. The correction required is 55.32 kvar. The bank is 3 × 25 kvar. The power factor with the bank is 0.991. Current falls from 192.45 A to 145.59 A. The released capacity is 32.47 kVA. The bank conductors need at least 146.14 A.
One motor
Set Correct to one motor to size a capacitor fitted at a motor and switched with it. The capacitor must not supply more than the motor's own magnetising kvar. If it does, the motor, still turning after it is switched off, can excite itself from the capacitor and raise its voltage. The tool takes the no-load current as all magnetising and keeps the capacitor to 90 % of the no-load kvar, a rule of practice. It fits the smaller of that and the kvar to reach the target at full load, then gives the line current with it from the motor tool's full load current.
- A 30 kW motor at 0.85 with 17 A no-load current can take 10.6 kvar at most. The 9.7 kvar to reach 0.95 fits, and the line current falls from 56.6 A to 50.64 A.
- An 11 kW motor at 0.8 with 7.5 A no-load current can take 4.68 kvar at most, which lifts it to 0.939 and cuts the line current to 18.79 A. Correct the rest at the board.
- A 55 kW motor at 0.75 with 30 A no-load current can take 18.71 kvar at most, which lifts it to 0.867 and cuts the line current to 101.78 A.
Limits
The tool does not assess harmonics unless you enter the THDi. Switch each step with a device rated for capacitors. The tool does not model resonance with the supply.
See also: Power factor and reactive power, Power factor correction for a 250 kW factory and Three phase power.
问题
结果仅供设计参考。请依据现行标准核对。
电动机处最大可安装多大的电容器？
不超过电动机空载无功功率（kvar）的约 90%，这是实践中的经验规则。若超过此值，电动机在断开后可能发生自励磁。请将工具设为“单台电动机”，并输入空载电流。
kvar 如何求出？
工具根据有功功率，分别求出当前功率因数和目标功率因数下的无功功率。两者之差即为电容器组需提供的 kvar。然后向上取整至电容器组的整档。
为何修正后电流会下降？
有功功率保持不变，但电容器组在现场提供无功功率。因此供电系统承载的视在功率减少。电流按比例下降，所释放的千伏安即为释放出的容量。
何时需要抗谐波电容器组？
变频器、整流器和 LED 驱动器会产生谐波电流。普通电容器可能与电源发生谐振并放大谐波。经验法则是：THDi 约超过 10 % 时，应加装带串联电抗器的失谐电容器组。超过 20 % 时，应进行谐波分析。
谐振次数是什么意思？
电容器组与电源构成调谐电路。其谐振次数等于电源短路容量与电容器组 kvar 之比的平方根。接近第 5、7、11 或 13 次谐波时，本工具会发出警告。
计算示例
背景
本工具读取的条款和表格
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