Installed apparent power (kVA): Difference between revisions
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{{def | {{def | ||
|η|the per-unit efficiency {{=}} output kW / input kW | |η|the per-unit efficiency {{=}} output kW / input kW | ||
| | |cos φ|the power factor {{=}} kW / kVA}} | ||
The apparent-power kVA demand of the load | The apparent-power kVA demand of the load | ||
| Line 101: | Line 101: | ||
{{fn-detail|1|For greater precision, account must be taken of the factor of maximum utilization as explained in [[Estimation of actual maximum kVA demand]]}} | {{fn-detail|1|For greater precision, account must be taken of the factor of maximum utilization as explained in [[Estimation of actual maximum kVA demand]]}} | ||
</references> | </references> | ||
Latest revision as of 09:48, 22 June 2022
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
cos φ = the power factor = kW / kVA
The apparent-power kVA demand of the load
[math]\displaystyle{ Pa=\frac{Pn}{\eta\cos\varphi} }[/math]
From this value, the full-load current Ia(A)[1] taken by the load will be:
- [math]\displaystyle{ \mbox{Ia}=\frac{\mbox{Pa}\times {10^3} }{\mbox{V} } }[/math]
for single phase-to-neutral connected load
- [math]\displaystyle{ \mbox{Ia}=\frac{\mbox{Pa}\times {10^3} }{\sqrt3\times\mbox{U} } }[/math]
for three-phase balanced load where:
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.
| 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 | |
- ^ example: 65 W tube (ballast not included), flux 5,100 lumens (Im), luminous efficiency of the tube = 78.5 Im / W.
Notes
- ^ For greater precision, account must be taken of the factor of maximum utilization as explained in Estimation of actual maximum kVA demand
