華為UPS-如何提高UPS電源功率因數(shù)？

       華為UPS
如何提高UPS電源功率因數(shù)有哪些方法，對(duì)設(shè)備有哪些意義？企業(yè)所用交流設(shè)備多數(shù)為感性負(fù)載。

　　在感性電路中，感性負(fù)載的功率因數(shù)

　　也就是說(shuō)，電路中還有一部分能量并沒有消耗在負(fù)載上，而是與電源之間反復(fù)進(jìn)行交換，這就是無(wú)功功率，它占用了電源的部分容量。

　　一、提高功率因數(shù)的意義

　　1.充分利用電源設(shè)備的容量

　　每個(gè)供電設(shè)備都有額定的容量，即視在功率。供電設(shè)備輸出的總功率S中，一部分為有功功率，另一部分為無(wú)功功率越小，電路中的有功功率就越小，提高的值，可使同等容量的供電設(shè)備向用戶提供更多的功率。因此，提高供電設(shè)備的能量的利用率。

　　設(shè)電源容量為SN=40kVA，則帶40W（=0.4）的熒光燈，可帶400盞；帶40W（=1）的白熾燈可帶1000盞。

　　可見，功率因數(shù)從0.4提高到1，發(fā)電機(jī)正常供電的用電器的個(gè)數(shù)即從400個(gè)提高到1000個(gè)，使同樣的供電設(shè)備為更多的用電器供電，大大提高供電設(shè)備的能量利用率。

　　在UPS電源電壓一定的情況下，對(duì)于相同功率的負(fù)載，功率因數(shù)越低，電流越大，供電線路上的電壓降和功率損耗也越大。

　　我們知道，故用電器的功率因數(shù)越低，則用電器從電源吸取的電流就越大，輸電線路上的電壓降和功率損耗就越大；用電器的功率因數(shù)越高，則用電器從電源吸取的電流就越小，輸電線路上的電壓降和功率損耗就越小。故提高功率因數(shù)，能減少供電線路上的電壓降能量損耗。

　　如，220V/40W的白熾燈電流為0.18A；而220V/40W的熒光燈，因其cosφ=0.4，所以電流為0.455A，比前者大得多，顯然，經(jīng)過(guò)線路電阻帶來(lái)的電壓降和功率損耗也要大得多。

　　如果供電線路上的電壓降過(guò)大，就會(huì)造成電網(wǎng)末端的用電設(shè)備長(zhǎng)期處于低壓運(yùn)行狀態(tài)，影響其正常工作。為了減少電能損耗，改善供電質(zhì)量，就必須提高功率因數(shù)。

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.