# How to minimize the phase difference to installation of equipments

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

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• #10560
Keymaster

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

#12344
Ally Kanyondo
Participant

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= 175kW

2.FOR AIR CONDIOTIONERS

1 TON=3.504kW, so

Total 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=655.7488kW

Assume the diversity factor is 0.7

Then the diversified power=0.7*P=0.7*655.7488=460kW

Assuming also the power factor is 0.85

The 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.

#12358
erickench
Participant

If you mean load balamcing then you would have to make a chart. See below.

A          B          C          N

etc.

Total

You then add up the full load amps for each phase. The idea is to get the phase amps as close to being equal as possible. For eample:

If a phase has too many amps(alot more than the other phases) then you must move the load from that phase to another phase until the total amps for each phase is closely equal to each other. This could be done by moving the circuit breakers around in a paneboard/switchboard.

Remember: certain loads do not have a neutral connection and therefore the neutral amps for that load is zero.

#12398
Ally Kanyondo
Participant

Hi,

Erickench is right, you should arrange you loads in the distribution panel or distribution board diagonal in order to ensure load balancing.

See the table below.

 R B Y N LOAD 1 X LOAD 2 X LOAD 3 X LOAD 4 X LOAD 5 X LOAD 6 X LOAD 7 X

Where R,B and Y are phases and X is the total load of each load.

From the that you can play with you loads in order to balance the phase differences.

Regards.

#12390
Spir Georges GHALI
Participant

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= 175kW

2.FOR AIR CONDIOTIONERS

1 TON=3.504kW, so

Total 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=655.7488kW

Assume the diversity factor is 0.7

Then the diversified power=0.7*P=0.7*655.7488=460kW

Assuming also the power factor is 0.85

The 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.

#12389
Ally Kanyondo
Participant

Spir Georges GHALI said:

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= 175kW

2.FOR AIR CONDIOTIONERS

1 TON=3.504kW, so

Total 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=655.7488kW

Assume the diversity factor is 0.7

Then the diversified power=0.7*P=0.7*655.7488=460kW

Assuming also the power factor is 0.85

The 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.

Spir Georges GHALI said:

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= 175kW

2.FOR AIR CONDIOTIONERS

1 TON=3.504kW, so

Total 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=655.7488kW

Assume the diversity factor is 0.7

Then the diversified power=0.7*P=0.7*655.7488=460kW

Assuming also the power factor is 0.85

The 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.

Dear Mr Ghali

Thanks for your above comments. i didnt know if the rotary type have the power of 6kW for 1 ton.

I am taking it as a note. I always thankfull for your advice. Please don forget to work on my previous post about the formula for calculating max area of lightning protection.

Regards

#12403
Spir Georges GHALI
Participant

Dear Mr. Ally ;

For ” 1 Ton – rotary type ” the nominal current is around ” 6 A “ ( and not  6 kW ).

Regards.

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