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| {{Menu_General_rules_of_electrical_installation_design}}
| | #REDIRECT [[Estimation of actual maximum kVA demand#Diversity factor - Coincidence factor (ks)]] |
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| The term diversity factor, as defined in IEC standards, is identical to the factor of simultaneity (ks) used in this guide.In some English-speaking countries however (at the time of writing) diversity factor is the inverse of ks i.e. it is always ≥ 1. <br>
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| [[Image:Fig-A14.jpg]] | |
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| '''''Fig A14:'''''<i> An example in estimating the maximum predicted loading of an installation (the factor values used are for demonstration purposes only)</i> <br>
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| ==== <br>Choice of transformer rating<br> ====
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| When an installation is to be supplied directly from a MV/LV transformer and the maximum apparent-power loading of the installation has been determined, a suitable rating for the transformer can be decided, taking into account the following considerations '''(see Fig. A15):'''
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| *The possibility of improving the power factor of the installation (see chapter L)
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| *Anticipated extensions to the installation
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| *Installation constraints (e.g. temperature)
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| *Standard transformer ratings
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| {| cellspacing="1" cellpadding="1" width="396" align="left" border="1"
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| | bgcolor="#0099cc" rowspan="2" |
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| '''Apparent power kVA'''
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| | bgcolor="#0099cc" colspan="2" | '''In (A)'''
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| |-
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| | bgcolor="#0099cc" | '''237 V'''
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| | bgcolor="#0099cc" | '''410 V'''
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| |-
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| | 100
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| | 244
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| | 141
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| |-
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| | 160
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| | 390
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| | 225
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| |-
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| | 250
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| | 609
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| | 352
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| |-
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| | 315
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| | 767
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| | 444
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| |-
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| | 400
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| | 974
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| | 563
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| |-
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| | 500
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| | 1218
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| | 704
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| |-
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| | 630
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| | 1535
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| | 887
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| |-
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| | 800
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| | 1939
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| | 1127
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| |-
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| | 1000
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| | 2436
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| | 1408
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| |-
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| | 1250
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| | 3045
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| | 1760
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| |-
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| | 1600
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| | 3898
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| | 2253
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| |-
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| | 2000
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| | 4872
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| | 2816
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| |-
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| | 2500
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| | 6090
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| | 3520
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| |-
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| | 3150
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| | 7673
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| | 4436
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| |}
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| '''''Fig. A15'''''<i>: Standard apparent powers for MV/LV transformers and related nominal output currents</i>
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| The nominal full-load current In on the LV side of a 3-phase transformer is given by:
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| <math>\mbox{In}=\frac{\mbox{Pa}\times10^3}{\mbox{U}\sqrt3}</math>
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| *Pa = kVA rating of the transformer
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| *U = phase-to-phase voltage at no-load in volts (237 V or 410 V)
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| *In is in amperes.
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| For a single-phase transformer:
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| <math>\mbox{In}=\frac{\mbox{Pa}\times10^3}{\mbox{V}}</math> <br>where
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| *V = voltage between LV terminals at no-load (in volts)
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| Simplified equation for 400 V (3-phase load)
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| *In = kVA x 1.4
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| *The IEC standard for power transformers is IEC 60076.
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