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Total Building Load in KVA

Michael Brennan
Hi All,


I have been asked to provide the total electrical load of a building.  The max load per phase is 426.3A, on a 3-phase supply.  The majority of this is made up of 1-phase (230V) loads, with a few 3-phase loads mixed in there.  I have designed the electrical distribution such that the loads are split as equally as possible, but I have a question on how to calculate the total KVA value for this supply.  I'm getting a little confused with the information provided online and thought I would check this with a reliable source (you guys!).  Can someone confirm which of the following is the correct calculation, please?


1) I = 426.3A (per phase).  V = 400V (per phase).  Therefore: (426.3 * 400 * 1.732) / 1000 = (426.3 * 692.8) / 1000 = 295.34KVA


2) I = 426.3A (per phase).  V = 230V (per phase).  Therefore: (426.3 * 230 * 1.732) / 1000 = (426.3 * 398.36) / 1000 = 169.82KVA


Many thanks in advance,


MJB
  • 0
    Multiply the phase to neutral voltage by the current on each phase.

    230/400 volt supply with 100 amps on each phase is 69KVA.

    In practice it would be often be prudent to allow a bit more to allow for the loading being unbalanced.

    As a simplified example, a total load of 300 amps spread "equally" on a 3 phase supply could be in practice have at least 110 amps on one phase.

    Allow a margin also for supply voltage variations, and possibly for load growth.
  • 0
    Excellent, thank you for your quick reply!  So from your answer, the first calculation I proposed was correct...  I thought this was the case but I was being very wary to not accidentally oversize the supply due to a calculation error.


    Anyways, thanks again for your help.


    MJB
  • 0
    Michael Brennan:

    ... the first calculation I proposed was correct... 


    Not quite! It should be 426.3 (which seems very precise) x 230 x 3 ÷ 1000 = 294 kVA.


  • Former Community Member
    0 Former Community Member
    Equation 1


    Equation 2 is not correct on a 400V 3 phase system


    As a quick sanity check, and using 240V, we know 1kw is 4A at 240V


    So, as you have 425A/phase and reasonable balance, then divide by 4 = 106, then multiply by 3 (for the 3 phases) gives you 320kVA - which is close enough to your equation 1 answer.


    Regards


    OMS
  • 0
    The precision was more to check some figures in my spreadsheet straight off the values I had rather than rounding up later on but thank you for the clarity with your answer.
  • 0
    As a sanity check in these cases you can ditch the 1.732 which is just the 400V/230V ratio and not very exactly , and assume all loads are just the sum of the single phases


    (426.3Amps  *230volts ) * 3phases = 294147 watts = 294kW.


    The rounding errors will be dwarfed by the reality of the supply being anything between 240V near the intake, and more like 220 at the outbuildings.

    How are you handling load diversity ? A lot of the time the max demands end up oversized on bigger projects.


    To be unbalanced by 20 to 30% between currents on the phases is not uncommon, agonizing to the hundreds of milliamps is not realistic.

  • 0
    Okay thank you, that's a very fair point and makes sense.  I feel like the total load is larger than required by gut feeling but perhaps it is about right.


    I have 7 sub DB's feeding from one main DB.  I have taken diversity into account separately on each DB but not between the DB's.  I was not sure if I should take that into account additionally or not.  I have assumed the following rules for diversity:
    • Socket circuits: 100% largest circuit, 50% remaining circuits.

    • Lighting (LED): 90% of all lighting circuits.

    • Heating: 100% heating control panel circuit

    • Ventilation: 100% first circuit, 75% remaining circuits

    • Chillers: 100% first chiller, 75% second chiller

    • Cookers: 100% largest circuit, 80% second-largest circuit, 60% remaining circuits

    • Dishwasher: 75% circuit

    • Walk-in Refrigerators: 100% largest, 75% second largest


    I have assumed these rules for each individual DB.  Each DB serves different areas, so I'm not sure if I can apply diversity between the DB's additionally to the diversity already applied separately?

  • 0
    Michael Brennan:

    Okay thank you, that's a very fair point and makes sense.  I feel like the total load is larger than required by gut feeling but perhaps it is about right.


    I have 7 sub DB's feeding from one main DB.  I have taken diversity into account separately on each DB but not between the DB's.  I was not sure if I should take that into account additionally or not.  I have assumed the following rules for diversity:
    • Socket circuits: 100% largest circuit, 50% remaining circuits.

    • Lighting (LED): 90% of all lighting circuits.

    • Heating: 100% heating control panel circuit

    • Ventilation: 100% first circuit, 75% remaining circuits

    • Chillers: 100% first chiller, 75% second chiller

    • Cookers: 100% largest circuit, 80% second-largest circuit, 60% remaining circuits

    • Dishwasher: 75% circuit

    • Walk-in Refrigerators: 100% largest, 75% second largest


    I have assumed these rules for each individual DB.  Each DB serves different areas, so I'm not sure if I can apply diversity between the DB's additionally to the diversity already applied separately?

     


    Probably OK but we need to know more about the loads to ascertain if this is reasonable.

    Socket outlets, probably fine.

    Lighting., 90% of the installed load might not be enough, it is reasonable to assume that it will all be in use in working hours. I would allow 110% of the installed load to take account of manufacturing tolerances and mains voltage variations. if 90% of the circuit ratings has been allowed, then that is probably generous.

    Heating controls, probably fine.

    Ventilation, it depends on how controlled, it might all run in working hours, in which case 110% of the connected load.

    Chillers, might all run in hot weather, unless duty/standby.

    Cookers probably fine, unless unusually intensive use is expected.

    Dishwasher, probably fine.

    Walk in fridges, probably fine.


    In general it is the installed load that should be counted, not the size of the sub circuits.

    As a simplified example, consider 200 LED lights each with an input current of 0.25 A, connected to ten sub circuits, each 16 amps. I would take the maximum demand as being 200 multiplied by 0.25 to give a load of 50 amps, plus say 10% for mains voltage variations and manufacturing tolerances. 55 amps.

    I would not take it as being 160 amps, the sum of the sub circuit ratings, nor as 144 amps, 90% of the the total sub circuit ratings.


    Likewise a 66 amp chiller I would count as being 66 amps, plus again a 10% margin for voltage variations and manufacturing  tolerances. The chiller might well be on a 100 amp circuit, but I would not count it as 100 amps when assessing total load.


    Socket outlets are the big unknown as we dont really know what will be plugged into them. A fixed allowance per square meter of floor area can be a better approach for large areas. in shops and offices.

    This however sounds like a catering operation, in which case one should be very generous WRT to socket outlets and the assumed loading thereof. IT only takes a few tabletop fryers each with two 3 Kw elements for it all to go orribly wrong.


  • 0
    Thank you Broadgage, I really appreciate your comments.  The loads I have considered are total current of the socket circuits (20A radials mostly), and everything else I have looked at installed equipment data and taken the FLC's of the actual equipment (plus a little more for tolerance) as you mention above.  Your comment about the kitchen sockets is very appreciated, I will make sure to take this into consideration.


    Ultimately, the building has many different uses as the spaces are diverse within it, and as such it is hard to define the exact times of use etc.  I.e. offices, meeting rooms, commercial kitchen & cafe, auditorium, dance studio, etc.


    Many thanks for your comments though, I appreciate everyone on these forums who is able to offer support!!
  • Former Community Member
    0 Former Community Member
    Hello Michael,
    I would advise you to use suitable calculators. On the net you can find many free ones.  For example the MeteorSPEC:
    d650a4c56dbd95178b6e2afe8cac649b-original-power_demand_calc.png
    You must to define a summarized power factor for the load. It is usually in the limits 0.92- 0.97 for buildings. Please, see the History list on the window above.

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