Reality check on transformer power dissipation

[Shackled Designer]

I see that there has been a previous post similar to what I am about to ask. However, in that case, one or two of the responders asked why the question mattered, and the OP apparently never offered an answer.

So, I have a buck transformer bumping 277VAC (single-phase) down to 230VAC. I am interested in calculating what the power dissipation would be at full load. My assumption is that power dissipation would be greatest at full load, but if this is wrong, please feel free to better inform me. So far, I have not found a manufacturer that publishes either full load transformer power dissipation or a dissipation curve vs. load %.

Before continuing with the numerical data, why is this important? Well in this case, the transformer is being installed in an enclosure with a number of other devices which also dissipate heat, and so I am attempting to calculate watt dissipation and thereby how much enclosure cooling I need.

Now, according to the transformer labeling, this is a 1.5 kVA transformer. However, in its buck configuration, it is acting as an 8.6 kVA transformer. According to the manufacturer's literature, in this arrangement and at full load, the transformer will convert 276VAC to 230VAC with respective primary and secondary currents of 31.3 and 37.5 amps. Calculating the power on both sides of the transformer, I have . . .

Primary: 276V * 31.3A = 8,638.8 VA
Secondary: 230V * 37.5A = 8,625 VA

Using only significant digits,

Primary: 8,640 VA
Secondary: 8,630 VA

The difference, then is 10 VA, and if the numbers are accurate, then I suppose this is my power loss. However, I am surprised that the number is this low. In fact, the difference might be a result of nothing more than decimal rounding. Is a mere 10 VA realistic for this case? Seems a bit low to me.

I reckon that the most reliable way to determine power loss is to measure primary/secondary voltage/current in the final system. However, that does not really help with a project in the planning stages.

So if the value I have calculated is not reality, what is a better way to estimate watt dissipation for this case?

Many thanks,
The Shackled Designer

 

[gar]

140822-1207 EDT

You are taking the difference between two large numbers and that can introduce large errors.

I will make a guess that the power dissipation in the buck/boost at its full load is less than 10% of its kVA rating and more than 1%.

Measure the unloaded power input to a 1.5 kVA transformer and double this for full load on the assumption that the design is for equal dissipation in core and copper at full load. Then double this again for a safety margin in the estimate.

I have a 175 VA transformer that has a 5 W no load input. So 10 W out out 175 W is 5.7 %. On your 1500 VA this would imply 86 W. Efficieny is a function of transformer design and size.

.

 

[Shackled Designer]

Thank you, Gar.

That's a pretty interesting estimation method you have suggested since it relies on the actual kVA of the transformer instead of the buck-rated kVA, which I guess makes sense.

I initially had figured an educated guess efficiency of maybe 97% -- maybe a tad optimistic, but hopefully not off by an order of magnitude -- but for an 8.6 kVA, that still made for > 250 watts. It does look a good deal nicer, even if I figure 5% to 10% loss, for a 1.5 kVA. Does the connection of a transformer as an autotransformer, i.e. partially common primary and secondary, bypass some of the efficiency losses?

Since this is for a cooling calculation, and because the transformer is one of numerous heat generators in the cabinet, I might skip the unloaded power measurement in favor of a 10% worst-case (hopefully) scenario on the losses.

Further comments are welcome.

Kind regards,
Shak

 

[gar]

140822-1316 EDT

Power dissipation in a transformer will monotomically increase with load.

We have a 2 kVA transformer on a chop saw to simply power a light and solenoid valves. Thus, it is virtually unloaded. The surface temperature of it unloaded is too hot to touch, if it has been on for a while. A disconnect preceeds the transformer so it is not left on un-necessarily.

An autotransformer will be slightly more efficient than a two coil buck transformer.

You can get better transformer efficiency with a tape wound core using special materials and no air gap.

.

 

[Jraef]

Thank you, Gar.

That's a pretty interesting estimation method you have suggested since it relies on the actual kVA of the transformer instead of the buck-rated kVA, which I guess makes sense.

I initially had figured an educated guess efficiency of maybe 97% -- maybe a tad optimistic, but hopefully not off by an order of magnitude -- but for an 8.6 kVA, that still made for > 250 watts. It does look a good deal nicer, even if I figure 5% to 10% loss, for a 1.5 kVA. Does the connection of a transformer as an autotransformer, i.e. partially common primary and secondary, bypass some of the efficiency losses?

Since this is for a cooling calculation, and because the transformer is one of numerous heat generators in the cabinet, I might skip the unloaded power measurement in favor of a 10% worst-case (hopefully) scenario on the losses.

Further comments are welcome.

Kind regards,
Shak

For the purposes of determining heat losses inside of an enclosure, I was taught, oh so many years ago by Hoffman for determining air conditioner sizes, to assume 2% total losses based on load on the transformer. With newer TP-1 based transformer standards that is now supposed to be less than that, but you can't assume a BB transformer will be designed to TP-1 standards because its use is outside of the requirement.

So in the older designs the core losses are fixed, usually assumed to be 0.5%; they do not change regardless of loading. But the copper losses, which change with load, are assumed to be 1 to 1-1/2%. So even though a BB transformer is used as an autotransformer, the windings are all still in the circuit so I don't see that changing. But you are correct in that the 2% losses would be based on the actual kVA that the BB transformer is being used for.

 

[Shackled Designer]

So, Jraef, it sounds like you are saying that since I am using a 1.5 kVA xformer as an 8.6 kVA (equivalent BB transformer), that my 2% loss should be based on the 8.6 kVA value, right?

10% of 1,500 VA = 150 VA
2% of 8,600 VA = 172 VA

Following Gar's post, I did use 10% on my 1.5 kVA-based estimate. Fortuitously, the 22 VA difference is easily covered by the size of the AC unit I'm likely to spec.

Of course, if my loss is much worse than 2%, then I may have to reconsider, but from what you suggest about estimated efficiency, I really should not be miles away from where I need to be.

Many thanks,
Shak

 

[GoldDigger]

No. Jraef was saying that if you use your 1.5 kVA transformer to handle only .5kVA of power the losses will be lower than 2% of 1.5.
The fact that the buck/boost autotransformer connection takes the place of a 7.5kVA isolation transformer does not increase the losses in any way.
BB is more efficient since the majority of the load power does not pass through the primary.

 

[gar]

140826-1632 EDT

Some experiments on a Sq-D 500 SV1B transformer, about 20 years old.

Measured unloaded power input and turns ratio. Applied 125.2 V to get 480 V on the primary. This is a ratio of 0.260 .

Measured unloaded input power was 14.7 W. This is 2.94 % of the transformer rating.
Measured primary resistance = 10 ohms (480 connection), and secondary = 0.8 ohms (120 connection). Using the measured ratio 10 ohms reflects to 0.26^2 * 10 = 0.68 ohms as viewed at the secondary. A reasonable balance of primary and secondary copper. Measured the equivalent input excitation current at 120 and reflected to the primary it is 0.44 * 0.26^2 = 0.03 A.

The following are calculated values for an assumed full load. Assume 480 input and 125.2 as the open circuit output voltage. Primary internal resistance as reflected to the secondary is 0.68 ohms, and equivalent secondary resistance including the reflected primary is 0.68 + 0.8 = 1.48 ohms. Secondary current is 500 / 120 = 4.17 A. And 1.48 * 4.17 = 6.17 V. Fairly close to 5.2 V. So the assumptions are somewhat reasonable.

Resistive power loss in the secondary at full load is approximately 4.17^2 * 0.8 = 13.9 W. Assume the total input current from the secondary load as reflected to the primary is 4.17 * 0.26 = 1.08 A. But total input current has to include the current for the core excitation. I will simply add the two input current components for an estimate. This will over estimate the RMS input current slightly. On this assumption the total input current is 1.08 + 0.03 = 1.11 A. Thus, primary power dissipation at full load is 1.11^2 * 10 = 12.3 W.

All together total input power at full load is estimated at 14.7 + 12.3 + 13.9 = 40.9 W or 100 * 40.9 / 500 = 8.2 % of the transformer rating.

See if my calculations are about correct.

If we assume your autotransformer efficiency is 5%, then 1500 *0.05 = 75 W might be an estimate of the added power to your cabinet.

.

 

[Sahib]

If all the output power from the transformer is utilized by the devices within the cabinet, it may be assumed conservatively for air-conditioning calculation purpose that the total heat dissipation is equal to the power input of the transformer without any need for individual device power loss.

 

[Phil Corso]

Efficiency of BB Xfmr Connected as an AT!

Efficiency of BB Xfmr Connected as an AT!

Shak,

For the 95% value you cited for the BB transformer, connected as 2-ckt Xfmr, then its lossses at full-load and unity PF, are:

0.05 * 1,500 / 0.95 or 78.95 W

 

[Phil Corso]

Efficiency of BB Xfmr Connected as an AT!

Efficiency of BB Xfmr Connected as an AT!

Shak,

For the 95% value you cited for the 1,500VA BB transformer, connected as 2-ckt Xfmr, then its losses at full-load and unity PF are,

(0.05 * 1,500 / 0.95 or 78.95 W

Similarly, when connected as an AT, the losses are still 78.95, but only [ 78.95 / (8,630+78.95) ] or 0.00906 of the input, yielding an efficiency of 99.1%!

Regards, Phil Corso ( Cepsicon@ao.com )