# Installed apparent power (kVA)

The installed apparent power is commonly assumed to be the arithmetical sum of the kVA of individual loads. The maximum estimated kVA to be supplied however is not equal to the total installed kVA.

The installed apparent power is commonly assumed to be the arithmetical sum of the kVA of individual loads. The maximum estimated kVA to be supplied however is not equal to the total installed kVA.

The apparent-power demand of a load (which might be a single appliance) is obtained from its nominal power rating (corrected if necessary, as noted above for motors, etc.) and the application of the following coefficients:

η = the per-unit efficiency = output kW / input kW

The apparent-power kVA demand of the load

${\displaystyle Pa={\frac {Pn}{\eta \cos \varphi }}}$

From this value, the full-load current Ia(A)[1] taken by the load will be:

• ${\displaystyle {\mbox{Ia}}={\frac {{\mbox{Pa}}\times {10^{3}}}{\mbox{V}}}}$

• ${\displaystyle {\mbox{Ia}}={\frac {{\mbox{Pa}}\times {10^{3}}}{{\sqrt {3}}\times {\mbox{U}}}}}$

V = phase-to-neutral voltage (volts)
U = phase-to-phase voltage (volts)

It may be noted that, strictly speaking, the total kVA of apparent power is not the arithmetical sum of the calculated kVA ratings of individual loads (unless all loads are at the same power factor).

It is common practice however, to make a simple arithmetical summation, the result of which will give a kVA value that exceeds the true value by an acceptable “design margin”.

When some or all of the load characteristics are not known, the values shown in Figure A10 may be used to give a very approximate estimate of VA demands (individual loads are generally too small to be expressed in kVA or kW). The estimates for lighting loads are based on floor areas of 500 m2.

Fig. A10 – Estimation of installed apparent power
Fluorescent lighting (corrected to cosφ = 0.86)
Type of application Estimated (VA/m2) fluorescent tube with industrial reflector[a] Average lighting level (lux =lm/m2)
Roads and highways storage areas, intermittent work 7 150
Heavy-duty works: fabrication and assembly of very large work pieces 14 300
Day-to-day work: office work 24 500
Fine work: drawing offices high-precision assembly workshops 41 800
Power circuits
Type of application Estimated (VA/m2)
Pumping station compressed air 3 to 6
Ventilation of premises 23
Electrical convection heaters: private houses
flats and apartments
115 to 146
90
Offices 25
Dispatching workshop 50
Assembly workshop 70
Machine shop 300
Painting workshop 350
Heat-treatment plant 700
1. ^ example: 65 W tube (ballast not included), flux 5,100 lumens (Im), luminous efficiency of the tube = 78.5 Im / W.

## Notes

1. ^ For greater precision, account must be taken of the factor of maximum utilization as explained in Estimation of actual maximum kVA demand