ECN Electrical Forum - Discussion Forums for Electricians, Inspectors and Related Professionals
ECN Shout Chat
ShoutChat
Recent Posts
Increasing demand factors in residential
by gfretwell - 03/28/24 12:43 AM
Portable generator question
by Steve Miller - 03/19/24 08:50 PM
Do we need grounding?
by NORCAL - 03/19/24 05:11 PM
240V only in a home and NEC?
by dsk - 03/19/24 06:33 AM
Cordless Tools: The Obvious Question
by renosteinke - 03/14/24 08:05 PM
New in the Gallery:
This is a new one
This is a new one
by timmp, September 24
Few pics I found
Few pics I found
by timmp, August 15
Who's Online Now
1 members (CoolWill), 250 guests, and 13 robots.
Key: Admin, Global Mod, Mod
Previous Thread
Next Thread
Print Thread
Rate Thread
#128540 05/22/03 06:16 PM
Joined: May 2003
Posts: 18
D
ds247 Offline OP
Member
I am confused about kva ratings on transformers. For example, suppose I have a transformer that has a 1 kva rating. I think this means that if I have a strictly resistive load then the load can be up to 1 kilowatt or 1 kva. But if I have an inductive load with a power factor of 80% then the load can not be more than .8 kva. Is this correct? Also does anyone know why va is used instead of watts? Thanks.

Stay up to Code with the Latest NEC:


>> 2023 NEC & Related Reference & Exam Prep
2023 NEC & Related Reference & Study Guides

Pass Your Exam the FIRST TIME with the Latest NEC & Exam Prep

>> 2020 NEC & Related Reference & Study Guides
 

#128541 05/23/03 03:39 PM
Joined: Apr 2002
Posts: 2,527
B
Moderator
Generally, the transformer can be loaded to 1000 voltamperes {1kVA} at it’s rated voltage and frequency, regardless of power factor. The transformer could be seen as “transparent” to power factor, but limited by a maximum voltampere load.

It’s a little confusing to discuss only watts and power factor. Voltamperes and power factor is easier to understand. As described, the 80%PF load could not be greater than 800 watts...that is still 1000 voltamperes, so the transformer is not yet overloaded.

In basic terms, because PF represents sort of an electrical efficiency, the closer to 100% or 1.0, the lower the circuit losses.




[This message has been edited by Bjarney (edited 05-23-2003).]

#128542 05/25/03 10:20 AM
Joined: May 2003
Posts: 18
D
ds247 Offline OP
Member
thanks, Bjarney

#128543 05/27/03 03:03 PM
Joined: May 2003
Posts: 107
J
Member
transformers power ratings are expressed as VA due to the much required prevention of overload.
If transformers were to be rated as kw ignoring power factor,well theoretically only resistive components could be used as you rightly said.

eg
If i had a heating element rated at 1000w 240v ,the load current would be around 4amps.


Now if I had a pump also rated at 1000w 240v but with a power factor as bad as say 0.5

then the load current would be

1000 w/240 v/0.5pf
8.3amps a diffrence of 4.3amps, over double the resistive load with the same power rating
but with the added pf of 0.5.

Hence it is quite important to take note between kw and kva.


Another example is to disconnect power factor correction capacitors from any highly inducdive circuit( under controled conditions of course!! )what you should find is an increase in current due to the absence of the power factor correction from the capacitors,forcing the circuit current to lag the circuit voltage.

hope this helps you!!




[This message has been edited by james S (edited 05-27-2003).]

[This message has been edited by james S (edited 09-19-2003).]

#128544 05/29/03 06:10 PM
Joined: May 2003
Posts: 18
D
ds247 Offline OP
Member
Thanks James.

#128545 09/05/03 02:59 PM
Joined: Jul 2003
Posts: 36
J
Member
Transformers are AC devices and it is possible for the current and the voltage to be out of sync with each other. In dc circuits, the voltage and current are always in sync.

Power is measured in units of watts (1 W = 1 J/s) and is defined as the rate at which energy is changed. Energy can be changed from an electrical form (the product of voltage and current) to another form, such as mechanical work or heat. In DC systems or resistive AC systems where the voltage peaks and valleys are in sync, the power is simply the product of the voltage and the current.

In AC systems, where the current and voltage peaks and valleys are not in sync, the actual power is not a simple product of the rms voltage and current, but the product of the voltage, current and the cosine of the angle between them. Inductive devices tend to cause the voltage to "lead" the current and capacitive devices tend to cause the voltage to "lag".

The VA (volt-ampere) is the product of the current and voltage not taking into account of the leading or lagging, but it is not equal to the real power of the system because no "work" is being done by this power, which is called apparant power. Thus it can not be measured in watts, as watts is a unit of real power.

A transformer is rated in volt amps because it doesn't actually "consume" power, except for a small about of losses due to magnetization. The transformer passes the power on to other devices. The volt-ampere rating is how much power the transformer can actually pass onto the load connected to it. A transformer has a peak voltage at which it can function or else the voltage wave will be clipped, and harmonmics and other bad effects will be induced in the system causing the core to overheat (effects of eddy currents). The transformer also has a maximum allowable current that can pass through its windings. If the limit is exceeded, the windings will over heat (I^2R losses). It makes no difference whether the current or voltage are in sync or not as to whether the trnasformr will be damaged if either or both are exceeded. A low voltage during a high current portion of the cycle can damage the transformer just as much as a high voltage during the low current portion of the cycle can do the same.

Thus, the importance of an absolute volt-ampere rating and not a actual power rating for a transformer.

#128546 09/08/03 12:59 AM
Joined: Oct 2000
Posts: 2,723
Likes: 1
Broom Pusher and
Member
Quote
A transformer is rated in volt amps because it doesn't actually "consume"
power, except for a small about of losses due to magnetization.

Good point made here.

Transformers will only "Reflect" an Impedance to the Generating Device, which will result in the correct level of True Power being supplied, then transfered to the load device.
They need to be able to carry the load's complete Volt-Amp level, hence the KVA rating.

If the load's KVA rating is equal to the KW rating (100% / 1.0 P.F., or completely True Power load), then the Kilo Watts (KW) will equal Kilo Volt-Amperes (KVA) - and the Transformer's rating will be reflected this way.

If the load is not 100% True Power (1.0 P.F.), then the total Apparent Power (Volt-Amps) must be figured.
The VA figure would be True Power (Watts) and Reactive Power (Volt-Amps Reactive, or VARs).
There is a thread somewhere which explains how to calculate Apparent Power / Power Factor. It may be in this forum area (Electrical Tech. Area), or in another area.
If elsewhere, try a subject search, or maybe someone can paste a link in this thread.

Power Generating devices have True Power ratings.

Most Load devices will have True Power ratings - unless they are Reactive devices, which will either list FLA or KVA (or both).

Thanks to all for the contributions in this thread!

Scott35


Scott " 35 " Thompson
Just Say NO To Green Eggs And Ham!
#128547 12/07/03 01:25 PM
Joined: Oct 2003
Posts: 10
T
Member
A good way to represent this whole concept of reactive power as opposed to treu power is to plot it on a graph.( The circle known as the KVA arc used in the 1st and 4th quadrants. If you follow the arc from the X-axis as if the reactive power were increasong ie pos or neg, you will see the real power decrease in proportion to the reactive power while the KVA stays the same.O
using the X and Y axis as the center of the circle, the circumference of the circle will be the limit of the transformer say 25kva, the X axis will always be used as real power and the Y axis will be used to denote the reactive power.
By observation when reactive power is equal to zero that is when the Y component is zero the KVA=KW, and as soon as there is a reactive component present the KVA will change in accordance to the square-root sum of the squares of the real compnent and the reactive component.
As an example if Real Power is 300 watts and the reactive Q is 400 vars the KVA is " the square root of 300(squared) + 400(squared) which would give 500 Volt-Amps and an angle of 53.13 degrees
the angle or power factor is ArcTan (X/R) or in this case reactive power/real power.
Once the power factor is known the whole problem falls apart.
Tom


Thomas-F

Link Copied to Clipboard
Powered by UBB.threads™ PHP Forum Software 7.7.5