华为UPS不间断电源UPS2000-G-10KRTL塔式机10KV9000W

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如何提高ups电源功率因数有哪些方法，对设备有哪些意义？企业所用交流设备多数为感性负载。

　　在感性电路中，感性负载的功率因数

　　也就是说，电路中还有一部分能量并没有消耗在负载上，而是与电源之间反复进行交换，这就是无功功率，它占用了电源的部分容量。

　　一、提高功率因数的意义

　　1.充分利用电源设备的容量

　　每个供电设备都有额定的容量，即视在功率。供电设备输出的总功率s中，一部分为有功功率，另一部分为无功功率越小，电路中的有功功率就越小，提高的值，可使同等容量的供电设备向用户提供更多的功率。因此，提高供电设备的能量的利用率。

　　设电源容量为sn=40kva，则带40w（=0.4）的荧光灯，可带400盏；带40w（=1）的白炽灯可带1000盏。

　　可见，功率因数从0.4提高到1，发电机正常供电的用电器的个数即从400个提高到1000个，使同样的供电设备为更多的用电器供电，大大提高供电设备的能量利用率。

　　在ups电源电压一定的情况下，对于相同功率的负载，功率因数越低，电流越大，供电线路上的电压降和功率损耗也越大。

　　我们知道，故用电器的功率因数越低，则用电器从电源吸取的电流就越大，输电线路上的电压降和功率损耗就越大；用电器的功率因数越高，则用电器从电源吸取的电流就越小，输电线路上的电压降和功率损耗就越小。故提高功率因数，能减少供电线路上的电压降能量损耗。

　　如，220v/40w的白炽灯电流为0.18a；而220v/40w的荧光灯，因其cosφ=0.4，所以电流为0.455a，比前者大得多，显然，经过线路电阻带来的电压降和功率损耗也要大得多。

　　如果供电线路上的电压降过大，就会造成电网末端的用电设备长期处于低压运行状态，影响其正常工作。为了减少电能损耗，改善供电质量，就必须提高功率因数。

how can huawei ups improve the power factor of ups power supply?

huawei ups

what are the methods and significance of improving the power factor of ups power supply for e? most of the communication devices used by enterprises are inductive loads.

in inductive circuits, the power factor of inductive loads

that is to say, there is still a portion of energy in the circuit that is not consumed by the load, but is repeatedly exchanged with the power supply, which is reactive power, which occupies a portion of the power supply's capacity.

1、 the significance of improving power factor

1. fully utilize the capacity of power e

each power supply device has a rated capacity, which is the apparent power. in the total power output s of power supply e, one part is active power, and the other part is reactive power. the smaller the reactive power, the smaller the active power in the circuit. the increased value can enable power supply e of the same capacity to provide more power to users. therefore, improving the energy utilization rate of power supply e.

if the power capacity is sn=40kva, a 40w (=0.4) fluorescent lamp can be e with 400 units; a 40w (=1) incandescent lamp can carry 1000 units.

it can be seen that when the power factor is increased from 0.4 to 1, the number of electrical appliances normally powered by the generator increases from 400 to 1000, allowing the same power supply e to supply more electrical appliances, greatly improving the energy utilization rate of the power supply e.

when the ups power supply voltage is constant, for loads of the same power, the lower the power factor, the greater the current, and the greater the voltage drop and power loss on the power supply line.

we know that the lower the power factor of electrical appliances, the greater the current they draw from the power source, and the greater the voltage drop and power loss on the transmission line; the higher the power factor of the electrical appliance, the smaller the current drawn by the appliance from the power source, and the smaller the voltage drop and power loss on the transmission line. therefore, improving the power factor can reduce the voltage drop and energy loss on the power supply line.

for example, the current of a 220v/40w incandescent lamp is 0.18a; and the 220v/40w fluorescent lamp, due to its cos φ= 0.4, so the current is 0.455a, which is much larger than the former. obviously, the voltage drop and power loss caused by the line resistance are also much greater.

if the voltage drop on the power supply line is too large, it will cause the electrical e at the end of the power grid to operate at low voltage for a long time, affecting its normal operation. in order to reduce power loss and improve power supply , it is necessary to increase the power factor.