primary and secondary KVA on plate iron

[Opcom]

From time to time the KVA is stated instead of Amperes on a transformer.

I understand KVA is not always watts, so how is this example explained?

AMERTRAN
PRI 115/230V 50/60 hZ 1.66kva
SEC 4730/2365V 2.34KVA
RMS TEST 12KV DC OPERATING 5KV
TYPE W
RISE 45 DEG C
SPEC 27614
CAT 30715
SER 640994
So, the secondary is rated 1.4 times the KVA of the primary. This magical transformer actually makes some of its own electricity.

@230V primary, 1.66KVA is 7.22 Amperes.
I can't assume that is in phase with the voltage because it is KVA not watts.

@4730V secondary, 2.34KVA is 495mA. Do I assume that this is the DC average current that can be drawn? The DC voltage will generally be less with a choke input filter. Maybe 3800VDC. -1881 watts of DC.

That is close to the 1.66KVA primary winding rating.


Now this one:
MOLONEY ELECTRIC CO (ST LOUIS MO)
INPUT 6.7KVA
PRI 120/230V 60 HZ
SEC 4430-0-4430
RISE 55 DEG C ABOVE 40 DEG AMBIENT
SERVICE SINGLE PHASE FW RECTIFIER
APPROX IMP. 34%
TOTAL WT. 323 LBS
SER 920462
There is no secondary KVA rating. But maybe it is like the first transformer and the secondary rating is 1.4 times the primary KVA rating or 9.5KVA. This assumption is totally groundless.. but it would mean:

@230V primary, 6.7KVA is 29 Amperes.
I can't assume that is in phase with the voltage because it is KVA not watts.

KVAsec=1.4*KVApri..

@4430V secondary, 9.5KVA is 2.1 Amperes.

Do I assume that this is the DC average current that can be drawn? The DC voltage will generally be less with a choke input filter. Maybe 3540VDC. -7500 watts of DC.

The goal is not to try to get more juice from the already huge transformers but to understand the relationship between primary and secondary KVA and why they are different. Also, how to determine KVA and DC Amperes when it is not stated, or if that is even important.

I had proceeded from these premises:
1.) The current in the primary and secondary would be in phase
2.) The voltage across the primary and secondary would be in phase.
3.) The phase relationship between current and voltage is not known.
4.) Primary to Secondary KVA relationship is assumed similar among transformers of similar size, but is not known unless stated.

[KA2DZT]

I'm not understanding the Amertran ratings.  Not sure why the primary would have a lower KVA rating than the secondary.  I would think it would be the opposite with loses and then the difference wouldn't be that much.

Are you sure you read the ratings correctly??

Maybe the xfmr has a poor power factor.  With a choke input filter the current in the secondary would be lagging the voltage.  So multiplying them gives the KVA load on the secondary.  Because the current is not in phase with the voltage it puts less of a load on the primary.  Hence the lower KVA rating for the primary.  The RMS TEST voltages are correct.  1/2 the TEST voltage minus 1KV is 5KV operating.

Not sure if any of this is correct.

Maybe you'll get more input from others.

Fred

[Opcom]

Here's the nameplate

I read as "KVA  PRI  1.66  SEC  2.34".

also revisited the PSUDII simulation with 2340VA in mind.
That's more than the primary VA but the object was to use close to the secondary VA rating. Unfortunately it really does not calculate this, or anything on the primary. I guess the RMS values are the ones to use?

[KA2DZT]

I looked at the nameplate and it does look like that's what the KVA ratings are.  I've read a number of explanations for KVA ratings for xfmrs.  Seems that not everyone is working from the same page.  Hard to tell exactly what Amertran was calculating when they came up with those ratings.

Fred

[Opcom]

not knowing what the application was.. Perhaps it was intended for even more reactive loads.

[Tom WA3KLR]

Patrick,

I’m NOT a transformer designer.  60 Hertz Power transformers are designed to a Volt Amp spec.  From what I have seen in the past, there is one VA rating, based on the secondary output.  But the designer starts with a primary VA rating for the primary winding and core designs.  The transformer should be around 90 % efficiency.

For the Moloney transformer, if the primary VA is 6700, then the secondary VA is around 6030 VA.  For the 4430 volts secondary this would imply 1.36 amps out, I would think.

I only have worked with transformers here at home in the few hundred watts category and found that the power to weight factor is 15 – 30 watts per pound for 60 Hertz transformers.  The Moloney comes out to 19 Watts per pound.  

For a cross-check what does the Amertran puppy weigh?

I was looking for one of those simplified transformer and rectifier configurations/ratings schematics custom transformer catalogs supply and I have seen a number of times over the years but cannot find right now.  

However, I did find a similar reference in the trusty Reference Data for Radio Engineers, Fifth edition, posted in tabular style, (for engineers !).  (PDF attached.) It’s a little more confusing, BUT does have the transformer primary and secondary VA rating listed, which wouldn’t be done in the basic transformer/rectifier/filter design schematic guides.  I think the whole thing is normalized to 1 Volt and 1 Amp.  

SURPRISE - If you look near the bottom of the second page under the full-wave center tap column you will see that the secondary VA rating is 1.57 and the primary VA is 1.11; a 1.41:1 ratio.

[Opcom]

I don't know the weight of the Amertran. It's in Gainesville right now. 80 miles away. I think it is  about 100 LBs or so from last time I picked it up from a table and carried it but I'm not a good scale. I always underestimate when picked up from a table and overestimate when picked up from the floor! It's a gravity thing!

Very interesting the 1.4 value appears in the volume. The 1.11 value seems familiar from the Handbook.

[Tom WA3KLR]

Here's another table from a couple pages later in the RDRE chapter, specifically for single phase rectifier/filter configurations  (Many numbers seem counter-intuitive, but remember the law of conservation of energy (and power) applies and the TRUE RMS value of currents stated can be counter-intuitive also.

[Tom WA3KLR]

O.k.  Here is the typical technical data I was hoping to post originally.  This pdf is from Amveco, a toroidal transformer manufacturer.  But the numbers should be the same:

[Opcom]

It looks like the KVA relationships are special and could relate to a specific application. Those document clarify how to best use the transformers. Thanks!

[WA1GFZ]

A good transformer will be about 90% efficient so the secondary will deliver less power than the primary. Usually a balance between primary resistive losses, core loss and secondary resistive loss. Maybe in some cases the secondary is over designed to handle higher currents due to reactive loads to further reduce losses. Maybe since it is step up the wire size may be increased to reduce resistive losses.

[The Slab Bacon]

I have always been a big follower of the T.I.T. (Timtron institute of Technology) for coming up with comfortable continuous ratings for big iron.
Timmy always said to allow 10w of continuous carrier outpoot of the transmitter per pound of plate iron weight. I have found this "seat of the pants" rule of thumb to work pretty well for mod iron as well.

Good example: the plate iron for MOST 1kW broadcach rigs weigh right around 100lbs.

this formula is for AM service, in light duty SSB service, you can push them a lot harder than that, but for AM, that will give you a comfortable starting point
     

[KM1H]

The BC-1T mod iron weighs 54# and the plate iron is 89# and the swinger is 73#.
These are all original Gates parts that have original schematic symbols stenciled on, and same manufacturer, so Im guessing they are the originals and worked fine in that rig until late 2010.

And as mentioned in another thread the mod reactor is 72# and that looks to be a replacement from Gates.

Carl

[The Slab Bacon]

The plate iron in my 4X1 rig came out of a gates (I dont know what model)
It has Gates Radio stenciled on it. I assume by the size that it is out of a 1kW model. It weighs 106 lb. That is what the ups sticker on the shipping box said and styrofoam doesn't weigh that much. So without the cardboard and foam packing it was prolly right at 100 lb. IIRC, it was somewhere around 40 or 45 bux to ship it ground, but it's been a few years ago.

[Opcom]

10W carrier per LB.- that would be only as a plate modulated system right?
it has to take into account modulator idling power too.

This older Moloney transformer (early 1950's?) might not do 10W per LB, maybe 8.5W per lb. by the figures below. Its construction could be weightier.
In a linear amp, the carrier wattage would be only about 37% of the watts per LB as in a plate modulated stage because even with a goodly modulator dissipation assumed for a plate modulated transmitter, the linear amplifier is so inefficient at carrier levels.

The earlier posted history of Moloney was not quite complete regarding the Canadian Moloney parent and does not say when U.S. production began.
http://www.moloney-electric.com/aboutus.php

The transformers you guys have could be a lot newer, better more high tech cores, higher temperature ratings maybe.

BC-1T plate iron 89# = 890W = 11.2 lbs per watt.
gates 1KW: 106# = 1060W = 10.6# per watt


The Moloney:
For a linear  amplifier, just saying, @carrier it is a lower figure.
323# = 3230W potential carrier by the 10W/LB rule.

In a plate modulated transmitter:
Starting with a 10W/LB rating and assuming an AM TX input power ratio of 1.5:1 (signal) from 0 to 100% modulated:
3230W RF is 4000W to final (80% eff.)
plus 650W to resting modulator
- it's a total of 4650W DC from the iron @carrier if modulator and final are considered.

- but when 100% modulation is applied and Final input goes to 1.5x, the 7000W DC requirement (4K RF + 3K mod) would overload the transformer.
(3KW DC to the modulator at 2KW audio output=100% mod., max. 67% eff.)

-- derating all proportionally to 6000W DC @ 100%, makes the carrier 2768W (11KW PEP).
 - That is 8.5W per LB no big deal but when there are 300+ LBs it makes a little difference.

 - used for an RF amplifier that is 67% efficient at max power, the setup will make less carrier.
 - for an amp. with eff. = 30% @ carrier, 6KW DC input:
CCS Single tone would be only 6000*.67 = 4020W and therefore a carrier of 1050W.

[The Slab Bacon]

Geeeezee, Pat, dont go gettin all fancy and high tech on us. 

It's just a simple rule of thumb for guestimating what a transformer is good for while "flying by the seat of your pants"

Oh, yea, and you forgot to include hypersil iron cores       

[KA2DZT]

Just use the xfmr for whatever you're building.  Run the xmtr at full tilt for a while and feel how warm the xfmr gets.  All those type xfmrs are rated at CCS.  You'll be using it at ICAS, the xfmr won't even work up a sweat.

Don't over think the whole thing.  You don't need to worry about every bit of the fine details of the ratings.  And forget about all the fancy computer simulations.  All that is just nonsense.  Except for the Slab Bacon, hams have been building HV PS for decades with whatever xfmrs they had on hand and everything worked out just fine.

As a side note;  if you see smoke, you made a mistake.

Fred

[WA1GFZ]

Computer simulation is not nonsense. It saves many hours of guessing by swapping parts around. A lot easier to understand what is going on before you drill, blast and melt solder.

[KA2DZT]

Computer simulation is not nonsense. It saves many hours of guessing by swapping parts around. A lot easier to understand what is going on before you drill, blast and melt solder.

Maybe,  I've been building PS for nearly 50 yrs, haven't needed any computers.

Fred

[Opcom]

I never used to use computers. Maybe that I do now is part of the problem and has caused a distraction. The pile of numbers and stuff is not at all high tech, just an attempt at understanding.

One thing the simulation showed me is that I have to use a soft start or there will be a 6KV+ surge when the supply is turned on. The current builds up through the chokes as the cap charges, then once it is charged, the stored energy of the choke gives it an extra kick above the rectified level. But few equipments that size don't have some sort of soft start.

[N4LTA]

Power transformers are rated in KVA(or VA) and not power mainly because it is current that  saturates the core.

At a 1.0 power factor KVA and KW are equal but at very low power factors the current soars and saturates the core long before the actual power (at unity power factor) is reached. With a power factor of zero, you have zero power transferred and the transformer core could be fully saturated (zero power factor would be when voltage and current are exactly 180 out of phase.

A 10 KVA transformer could be fully loaded at 1000 volts at 10 amps yet if the current and voltage were 180 degrees out of phase - no power would be transferred and the same transformer would transfer 10KW if the voltage and current were in phase - actually 10KW x efficiency which is 90% in smaller transformers and 98% or better in large transformers.

Most of the losses are magnetizing and eddy current losses in the core plus winding IR losses.

Pat
N4LTA

[KA2DZT]

Power transformers are rated in KVA(or VA) and not power mainly because it is current that  saturates the core.

At a 1.0 power factor KVA and KW are equal but at very low power factors the current soars and saturates the core long before the actual power (at unity power factor) is reached. With a power factor of zero, you have zero power transferred and the transformer core could be fully saturated (zero power factor would be when voltage and current are exactly 180 out of phase.

A 10 KVA transformer could be fully loaded at 1000 volts at 10 amps yet if the current and voltage were 180 degrees out of phase - no power would be transferred and the same transformer would transfer 10KW if the voltage and current were in phase - actually 10KW x efficiency which is 90% in smaller transformers and 98% or better in large transformers.

Most of the losses are magnetizing and eddy current losses in the core plus winding IR losses.

Pat
N4LTA

Zero power factor is when the current and voltage are 90degs out of phase.  Current lags the voltage in an inductive circuit.  PF is equal to the cosine of the angle between the current and voltage.  Cosine 90degs is zero.

It's not possible for the current to lag the voltage more than 90degs.  If it did the circuit would be capacitive where the current is leading the voltage.  The PF would then be leading.

Fred

[N4LTA]

Correct - I had a senile moment -  but you see what I meant  - 90% out of phase would be zero power factor.

Power factor is normally denoted at leading or lagging. In the power inductry lagging is almost always the case as most power loads are inductive except in the case of correction capacitors and synchronous generators (which can be set to lead or lag by adjusting the field)

In any case you can have a transformer loaded fully and transfer little or no power.

Pat
N4LTA

[KA2DZT]

This stuff gets a little tricky at times.  I just corrected my post to remove the % sign that I used for degs.  Also corrected negative PF to read PF leading.

You're right most loads in industry are inductive.  In a power supply with an input choke the filter caps help to correct the PF back or close to resonance.  So the xfmr will end up always supplying some level of average power.  A resonant circuit has a PF of 1.

It would be almost impossible for a circuit to be 100% inductive, since in the real world any circuit element has some properties of other circuit elements.  Any inductor has some capacitance along with DC resistance.

When I studied this stuff in college my professor had a PHD in EE.  I still had to help with a few of the real world questions asked by other students  The professor did end up asking me not to take any more courses he taught. After that I just kept quiet.  He did know the math forwards and backwards.

Fred

[N4LTA]

Yep, A pf of zero is not a real world thing - I guess you could get close with a superconductor and at least get rid of the R but there is always stray C.

Probably a good capacitor would be the cheapest way to get close but never perfact.

My initial EE electric circuits professor was a retired Naval Academy prof - he knew the practical stuff pretty well and he could bust your chops in a heartbeat. He was famous (infamous) up and down the east coast. He turned out to be my advisor and I liked him a lot end the end.

WP Seagraves (papa as he was call behind his back)  - not any like him left.  He was the Dean of undergraduate EE at NC State for a long time.


Pat
N4LTA