Home › Electrical Engineering Forum › General Discussion › How to minimize the phase difference to installation of equipments › Re: How to minimize the phase difference to installation of equipments

**Ally Kanyondo said: **

ABKHAN said:Dear Alls,

can you tell me the right way to minimize the phase difference to install the single phase equipments, i.e. our company is a software house, they have 350 computers,each compoter has almost 350~500 watt supply and installed 20 nos. 2 ton air conditioners, these are all single phase supply 220V, 17 nos 4ton 3 phase air conditioners, and our transformer is 200kVA, is there is any formula to apply there i minimize the phase difference, and tell me it the 200kVA transformer is sufficient or not and how much capacity we of tranformer we required.

Regards,

AB KHAN

Dear Mr AB KHAN

Firstly Can you elaborate your quastion about minimizing phase difference.secondly, to know whether your transformer is sufficient or not, you should calculate the total powe consumed by computers and air conditoners,

To calculate the total power do the following:

1. For computers, total active power(Pp)=350*500= 175kW2.FOR AIR CONDIOTIONERS1 TON=3.504kW, soTotal active Power for air condioners(Pa)= 3.504*(20*2*1.73+17*4)=480.7488kW,NOTING THAT THE HIGHLTED NUMBER 1.73 USED TO CONVERT SINGLE PHASE POWER TO THREEPHASE.Then the total active power(P)=Pp+pa=480.7488kW+175kW=655.7488kWAssume the diversity factor is 0.7Then the diversified power=0.7*P=0.7*655.7488=460kWAssuming also the power factor is 0.85The total Apparent Power(S)=P/0.85=540kVA.From there you can know know whether your chosen tr is right or not.In your calculations you should not forget to calculate the total power for lighting then add to Pp and Pa.Regards.

Dear Mr. Ally ;

Refer to your above calculation I have the following points :

– You applied the Diversity Fcator ” Kc “, but you forgeten to apply the ” Utilisation Factor – Ku “, as there's certainly some computers that are switched-off.

– The Electric Power for ” AC 1 ton ” is not in all case ” 3.504 kW “, because it depends on the Compressor type ” Piston or Rotary ” ( Exp. 1 ton / Rotary type → ≈ 6 Amp. ).

– In this case, it will be better to do al follow :

– First, prepare the ” Loads Distribution Table “.

– Calculate the Current for each Phase ” IL1, IL2, IL3 “.

– Calculate the Active Power ” P ” accordignly to the bigest current's value.

– Multiply the ” P ” by ” Ku & Kc ” to have the real need Active Power.

– Assume ( if not knowing ) the Power Factor value, and then caculate the need Apparent Power.

Regards.