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Transformer replacement #199813
03/09/11 01:36 PM
03/09/11 01:36 PM
W
will_87  Offline OP
New Member
Joined: Mar 2011
Posts: 5
england
hello all,
I am new to the site so please bear with me,
I'am have some trouble calculating KVA for a replacment 11KV/415V transformer that I suspect to be under rated.

Would I be right in thinking that the fundamental formula (assuming a balanced load)would be V(L-L)*1.732*I(Line current)/1000 ? or will this need to include P.F
where (V*1.732*I*.85)/1000
And is this the system load I would need to consider to correctly select the KVA size?


many thanks in advance!

Last edited by will_87; 03/09/11 01:38 PM.
Tools for Electricians:
Re: Transformer replacement [Re: will_87] #199838
03/10/11 12:21 PM
03/10/11 12:21 PM
J
JBD  Offline
Member
Joined: Jul 2001
Posts: 599
WI, USA
You do not include PF when calculating kVA.

Re: Transformer replacement [Re: will_87] #199852
03/11/11 11:41 AM
03/11/11 11:41 AM
W
will_87  Offline OP
New Member
Joined: Mar 2011
Posts: 5
england
thanks JBD.

So the PF should only be used to calculate KW with KVA?

KVA/PF = KW?

in power terms and not energy consumed

Re: Transformer replacement [Re: will_87] #199861
03/12/11 05:58 AM
03/12/11 05:58 AM
Scott35  Offline

Broom Pusher and
Member
Joined: Oct 2000
Posts: 2,708
Anaheim, CA. USA
will_87;

First off, let me welcome you to ECN! thumbs

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 wink ) 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:

C=A+B
Where:
"A" = True Power (Wattage), or Sine;
"B" = Reactive Power (VARs), or Cosine,
"C" = Apparent Power (VA), or Tangent.
The Square Root of "C" will be the unknown value.
(VA = Square Root of W + VAR)

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

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 37 and 39)

---------------------------------------------------------

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.

_____________________________________________________
< end of trivial information section >
_____________________________________________________

Quote


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



Get the actual EI 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


Scott " 35 " Thompson
Just Say NO To Green Eggs And Ham!
Re: Transformer replacement [Re: will_87] #199908
03/14/11 08:44 AM
03/14/11 08:44 AM
W
will_87  Offline OP
New Member
Joined: Mar 2011
Posts: 5
england
Wow, many thanks for that Scott couldn't ask for anymore. Very clear explanation. I find it sometimes difficult to find an answer using many peoples opinions.
And a Panel Schedule would be grand!

Sorry to ask another question to you're answer but would I also be right in thinking that the KVA requirement would be selected on the highest KVA value of load in the schedule and not overall system load as the transformer KVA rating will allow a tolerance over that value as not to over rate the install?


Re: Transformer replacement [Re: will_87] #200103
03/21/11 02:24 AM
03/21/11 02:24 AM
Scott35  Offline

Broom Pusher and
Member
Joined: Oct 2000
Posts: 2,708
Anaheim, CA. USA
Will,

Please check your Private Messaging Mailbox for my reply.

-- Scott


Scott " 35 " Thompson
Just Say NO To Green Eggs And Ham!

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