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how to calculate total Load using Current 3

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Hafzkk

Electrical
Aug 21, 2024
4
When I measured the current of a SMDB incomers using clamp meter, I got 32.24 A for red wire, for Yellow wire I got 37.5 A and for blue wire I got 30.23 A. How can I calculate the total load at that time using these information.
The nominal electricity supply voltage is 415 for three phase and 240 for single phase
 
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Sounds like your meter is measuring one phase at a time on a three-phase system.
Unless you can measure all three phases (voltage AND current) simultaneously, you will not know the total electrical power - which means not knowing the electrical load.

Side note - don't assume everyone on here understands the abbreviations and/or acronyms you use frequently. For example, I don't think the term SMDB refers to a "Standardized Material Data Base".

Converting energy to motion for more than half a century
 
Hello,

Thanks for the reply. Sorry for the abbreviations, by SMDB I meant Sub Main Distribution Board.

so is there a way to calculate the load of Sub Main Distribution Boards or Final Distribution Boards using clamp meter?
 
If the load is a single phase load, then you can get a reasonable approximation by measuring the voltage and current in that phase. If the load is multi-phase, then no - you need a measurement from all applicable phases simultaneously.

Converting energy to motion for more than half a century
 
You may calculate the loads in VA or KVA with Amp readings and Volt readings.
For a wye connected board, you may use A phase Amps times A phase Volts to neutral plus B phase Amps times B phase Volts to neutral plus C phase Amps time C phase Volts to neutral = demand in VA.
For power (Watts) you must know the power factor or the angular displacement between the Amps and the Volts.
A Watt Meter will read Watts or Kilo-Watts directly or with the use of a multiplier factor.
For the main service power, I generally take several spot readings on the revenue meter, each the consumption averaged over a minute or so.
How to take a spot reading on the revenue meter?
That is a chapter on its own.
There is a factor on the meter which indicates the Watt-Hours per revolution of the disk or cycle of the display on an electronic meter.
Note; Watt hours per revolution, not Kilo-Watt-Hours.
The time is typically in the range of less than 20 seconds per revolution, but it depends.
I generally time about 10 revolutions and then calculate the average consumption for that time.
It depends on the actual load, I may vary my criteria to account for heavier or lighter loads.

--------------------
Ohm's law
Not just a good idea;
It's the LAW!
 
'..You may calculate the loads in VA or KVA with ... For a wye connected board, .. use A phase Amps times A phase Volts to neutral plus B phase Amps times B phase Volts to neutral plus C phase Amps time C phase Volts to neutral = demand in VA. For power (Watts) you must know the power factor or the angular displacement between the Amps and the Volts'[/.
1. I fully agreed with Mr Waross's learned advice. It is based on the first principle. It does NOT matters what is "SMDB".
2. BTW: the units used KVA, KW, KWh.. should be kVA, kW, kWh...In SI, k=kilo=10[sup]3[/sup]=1000. K=kelvin, symbol K. e.g. temperature rise from 40[sup]o[/sup] C to 100 [sup]o[/sup]C is 60K. Attention: 60 K, not 60 [sup]o[/sup] K.
Che Kuan Yau (Singapore)
 
3 phase power formula
P(kW)= (I * V * pf * 1.732)/1000

You can not directly read power factor using a volt meter or an amp meter, so you cant calculate power this way, unless you know from other means what the power factor is. This stipulation has been stated above.

 
It's surprising how many electrical professionals don't know this:

Screenshot_2024-08-21_at_17-01-21_Fundamentals_of_Electricity_fve9jn.png


Link to full article and full instructions.

--------------------
Ohm's law
Not just a good idea;
It's the LAW!
 
This is the equivalent digital electronic meter.
There is a series of dashes across the bottom of the display.
The dashes either highlight or dim in sequence across the display.
One cycle or pass of the dashes is equivalent to one revolution.
The Kh factor on this meter is 1.0, that is; One Watt hour per pass.

V2S-2S_DIGITAL_KWH_METER_FORM_2S_240V_200A_mzd1tr.png


--------------------
Ohm's law
Not just a good idea;
It's the LAW!
 
Thanks Guys,

My purpose of asking this question was that one of our client want to install 20 Kilowatt machine for them without doing any load upgradation and other paper works that comes with it. So our idea is to calculate the existing Maximum demand and then comes to a decision.
Thus we turned ON all equipment simultaneously and measured the current using clamp meter. That's how we got the current values of 3 phase wires in my first question.

When I asked chatGPT about this it gave me the equation for total current I = SquareRoot(Ir[sup]2[/sup]+Iy[sup]2[/sup]+Ib[sup]2[/sup])
Then multiplied this current with supply voltage 415V.
But I don't think that is the correct way to calculate.
 
The easy way is to use the phase to neutral voltage. For a 415 Volt system, the phase to neutral voltage is 240 Volts. so (Ia+Ib+Ic) x 240 Volts = KVA demand. The calculation is valid even for a delta system where there no actual connection for 240 Volts.
But
In Canada, the code allows demand factors to be applied to loads when calculating the minimum size of the main service.
When a substantial load is to be added to an existing service the load calculations must be redone to verify that the main service is adequately sized.
For new loads added to an existing service there is an exception;
"If it can be shown that the existing demand is less than the existing service capacity then additional loads may be added."

Thus we turned ON all equipment simultaneously and measured the current using clamp meter. That's how we got the current values of 3 phase wires in my first question.
There are a couple of issues with this method:
1. If the equipment is not loaded, the currents may be lower than when working.
2 Demand factors assume that not all equipment will be loaded 100% at any time. The measured currents may be higher than when the plant is working.
One accepted method in Canada is to connect recording meters to the service for a time acceptable to The Authority Having Jurisdiction. (The inspection department.)
The last time that I saw this done, the AHJ demanded several months of recordings.
But there is an easier, faster, cheaper, more accurate, documented method.
If your plant is charged for demand, then each months power bill will show the maximum demand for that month.
For both new equipment and for power factor correction, I visit the accounts payable department and request copies of the last 12 or 24 months of the power bills.
The information that you need is all there, documented and measured to revenue metering accuracy.
Typically, a reasonable AHJ will allow one peak demand to be ignored as an outlier, not liable to be repeated, and as a reading that may not have been captured with an alternate, acceptable determination of demand.
Disclaimer:
Good engineering and due diligence suggest that the system should be evaluated for the presence of any large single phase loads that may push the current in one or two phases too high.
Hope this helps
Good luck.

--------------------
Ohm's law
Not just a good idea;
It's the LAW!
 
An additional note to your method;
You should be concerned that no one phase is overloaded.
To that end, it may be safer to work in Amperes.
That is; add 28 Amps for the new machine to the highest phase current and compare the sum with the main service capacity.
That said, I would discount the measured 38.5 Amps for the reasons given and go with the power bills.

--------------------
Ohm's law
Not just a good idea;
It's the LAW!
 
I think before that you prepare the load list of all the equipments then calculate the total load of the property .Check the infra if this is ok to addtional load and whether you have to increase the demand load of your customer .Hope it helps.

 
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