Calculation of Unbalanced 3 Phase Power?

Dear Sir,

There are meters available to measure power(Kw) and energy ( Kwh).You

can use them. Since you have taken the Voltage & current and the pf ,a simple

( not technically correct ) way is 1.732 x Line Voltage x Line Current X pf ( take

average value for Volts and current). This shoul give fairly good value for power.

For your reading -1.732 x 419 x 22.33 x .944 = 15,297 W ( 15.297 Kw)

Manroop Chennai.

Please understand, this had been covered many times earlier too.

The power is instantaneous reading (the moment you have done the measurement)

Most likely (may be wrong) you are interested in Energy consumption.

That can not be done by this method, since the currents will keep on variying as a light is switched off, an AC goes into sleep, Fridge compressor trips,...

To know energy consumption, you have to integrate the ∫VICosΦdt

All this effort can be done by a simple KW meter as mentioned in previous post.

Do you want to know the power consumed for a particular period?...Get the help of reading from energymeter. If you want to know the total load of each office, then calculate each circuit load and add . Is it the current reading you mentioned is full load condition? I beleive the readings are given for one phase of 3 offices.( or 3 phases of 1 office?)

Dont you have any load schedule for the distribution board?... What is the utilization factor and demand load of each distribution board?

You have to collect maximum demands of all offices in your commercial building. You must have a clear idea about the max. load of all offices when all loads are ON. If you have three offices which are being fed by three different phases then you can get the max. loads on each phase. Same time you collect the energy meter readings of each offices in Kwh and divide the value by 24x30 hours to get the average load per day in each phase.

In basic load calculation during the erection of the building first get the maximum demand load to determine your main MDB size. Distribute your circuits closely balance to each phases or per leg relevant to the rating of your PB main circuit breaker. It is not the supply voltage to be balanced but the current load of each phase. In voltage meter reading you can notice 1 or 2 volt or more difference between the phases... why? because the current load in any particular phase varries... as the current load increases the voltage drop will generates within the circuit especially when the wires or cables is not properly size relevant to the max. full load amps and length of wires to the farthest point of the circuit. Notice that in your given data 24A 29A 14. Phase L1L3 = 41.67%; L1L2 = 17.24% , L2L3 = 51.72% phase current load difference. At least not more than at max. of 33% per phase.

Get a class 1 power meter that connects with clamp on CTs and get a real reading over a weeks time - then move to the next office. Most of the suggestions are only going to give you a swag. The name plate power requirements for many pieces of equipment are very different from actual - could be either way, higher or lower.

Dear Sir,

The calculation of unbalanced power supply may be done by summarizing of every phase power, so:

P1 = 419V/sqrt(3)*24A*0.944 = 5481 (W)

P2 = 418V/sqrt(3)*29A*0.944 = 6607 (W)

P3 = 420V/sqrt(3)*14A*0.944 = 3205 (W)

Ptot = P1+P2+P3 = 15,293 W

Thats all

Your honour,

what is this formula sir, how did you get P1,P2 and P3 using this formula, sir?....

I am not the original replyer, however the answer makes sense. When have 3 phase balanced load, you can use (sqr.3 x IL x V LL x Pf). If you do not have a balanced load you use: (V Phase X IL x pf), you do not multiply by sqr 3. However assuming a Y connected load to get the Vphase you will divide VL by Sqr 3.

Eg. If you have a balance load of 3 phase, 10A, VL = 208

Then P=sqr3 x 10 x 208 = 3600W

If you have a balance unbalanced load of 3 phase 10A, 8A, 6A, VL = 208

Then Vp = 208/(sqr 3) = 120V

P1=P Aphase = 10A x 120V = 1200 W

P2=P Bphase = 8A x 120V = 960 W

P3=P Cphase = 6A x 120V = 720 W

To find the total Power = P1 + P2 +P3 = 1200+960+720 = 2880W

This is the same as using the average of 10A, 8A, and 6A = 8A

If you use 8A then P total = (sqr 3) x 8 x 208 = 2880W

The reply by Guest in #7 will help you and is correct for each office circuit. But this is the load at the time of measurement and should be monitored for some time or for several intervals and then averaged for a period of time covering the office hours...You must monitor to find the peak load for each...

For a better measurement is to use an energy meterto record the total energy consumed for a period of time and deduct an average. But you should also consider the peak value for each circuit (This was suggested before).

Thank you all !!!!

I have taken your experienced comments and will work accordingly!!!

Hope to continue the fruitful discussion further.

Thank You !!!

Harmeet Singh

Kolkata

[Ptotal= √3(Vline)(Lline)(pf.)]

P1= √3(419V)(19A)(0.944)=16442.12W

P2= √3(418V)(29A)(0.944)=19820.15W

P3= √3(420V)(14A)(0.944)=9614.13W

Way solve three phase power in balanced system but this can not use to solve unbalanced system .

Is there a simple formula to solve current per Leg in an unbalanced 3ph delta if only the L-L current is known?

Ex.

L1:L2 - 91amps

L2:L3 - 78amps

L3:L1 - 91amps

How do I determine the line current of each leg?