will_87;First off, let me welcome you to ECN!

Now to address your queries...

Terms Used:

- TRUE POWER: Wattage "W" and Kilowatts "KW",
- REACTIVE POWER: Volt-Amp Reactive "VAR" and Kilo Volt-Amp Reactive "KVAR",
- APPARENT POWER: Volt-Amp "VA" and Kilo Volt-Amp "KVA".

The Maximum Demand Volt-Amp Load is used to determine the capacity (KVA Rating) of the Transformer.

Per the loads, the total Apparent Power (VA / KVA), includes both the True Power (Wattage / Kilowatts) and the Reactive Power (VARs / KVARs).

The Power Factor is the ratio between the drawn True Power (W) and the complete Apparent Power "Package" (VA).

Power Factor is not relevant if the "Actual E × I" (Volts times Full Load Amperes) of the connected Loads is known.

Where an Equipment's rating is listed only in Kilo Watts, knowing the Power Factor for the Equipment would be necessary to obtain the Load's KVA value.

This would be a situation where the Nameplate Rating only contains the Maximum True Power Draw (Wattage or H.P.), and does not include the Design Voltage, Full Load Amperes, and Phase.

***EXCEPTION***Where the Equipment / Load is a "Pure Resistance" Load - such as an Incandescent Lamp, Quartz-Halogen Lamp, or Resistance Element Heater, the listed KW rating will equal the Load KVA.

These Loads will be 1.0 P.F.

The total of KW and KVARs drawn from the source equals the KVA value.

TRIVIAL INFORMATION FUN-FACTS:These values are obtained, as described below:

Example #1:

Figuring VA from known W and VAR

Use the Pythagorean Theorem (Right-Triangle Trigonometry

) to find the unknown value from two known values.

--known values--

Reactive Power = 3 VARs,

True Power = 4 Watts.

--known values--

Apparent Power = 5 VA.

The formula is:

C²=A²+B²

Where:

"A" = True Power (Wattage), or Sine;

"B" = Reactive Power (VARs), or Cosine,

"C" = Apparent Power (VA), or Tangent.

The Square Root of "C²" will be the unknown value.

(VA = Square Root of W² + VAR²)

........................

Example #2:

The Power Factor is found from known VA and Wattage.

Using the same values as above (3 VARs, 4 Watts, 5 VA), we may find the Power Factor:

4 Watts ÷ 5 VA = 0.8 Power Factor.

(Phase Angle is somewhere around 37º and 39º)

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So, as you have stated in the latest post, KVA/KW = P.F. is correct.

All these Power values: "KW" and "KVAR", are various levels of Energy Consumed by the connected loads - more correctly described as "Energy drawn from the Power Source".

The Wattage drawn produces heat and does actual work, whereas Magnetizing and / or Capacitive Charges drawn and stored by Transformers, Induction Motors, and other Reactive Loads, is the Reactive Power (VAR) component.

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< end of trivial information section >

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And is this the system load I would need to consider to correctly select the KVA size?

Get the actual E×I values from the Equipment to be driven by the new Transformer, and use this for determining the KVA rating needed (barring Load Diversity, and etc.).

For 3 Phase Loads, multiply the Equipment's Full-Load Amperes (FLA) by 719 to obtain the 3 Phase KVA value.

BTW: "719" is 415 Volts multiplied by 1.732 (Square Root of 3).

For L-L Single Phase Loads, multiply the FLA by 415.

Use a Panel Schedule Spreadsheet to calculate the Loads and arrive at the KVA requirement.

Let me know if you need a Panel Schedule Spreadsheet.

-- Scott