Applying Power Adjustment Factor

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Wilkes

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A 240 volt single phase circuit that has a load of 2616 va has to have a power adjustment factor of 1.16 applied to it due to the voltage of the circuit being changed to 277 volts. When applying this is the paf multiplied by the va or is the va divided by the paf?
 
131106-2102 EST

Can you more clearly present your question? I can guess, but your words don't compute. What is the load and what are it characteristics?

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Wilkes said:
A 240 volt single phase circuit that has a load of 2616 va has to have a power adjustment factor of 1.16 applied to it due to the voltage of the circuit being changed to 277 volts. When applying this is the paf multiplied by the va or is the va divided by the paf?
Click to expand...
First, I get 277 divided by 240 to be 1.154 which rounds to 1.15 not 1.16.
Assuming that the load can tolerate the extra voltage AND that it is a resistive load, then the power dissipated in the load will be 1.16 x 1.16 (~ 1.33) times the power at 240 volts.
If it is a constant current load, then the power will be the power at 240 times 1.16.
If it is a motor load, then the current may go down as the voltage changes leaving the power almost unchanged.

You have not provided enough information about the load for us to say which of the three analyses (or none) are right for your situation.
 
131106-2150 EST

Wilkes:

You have a resistive load.

If it is on off control for regulation, then on current is 277/240 times higher, and on power is (277/240)^2 times greater when on. Average power is the same. So on time is shorter.

If phase shift controlled, and regulated to temperature then RMS current is constant. Input volt-amperes are 277/240 times greater than at 240.

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131106-2344 EST

Wilkes:

You also must determine if your heat strip is rated for 277 V operation.

Note:

With phase shift control the RMS current is the same at either 240 or 277 applied voltage. The volt-amperes at 277 is 277/240 times the volt-amperes at 240 because the load is a constant average RMS current over a half cycle, and only the voltage was scaled up.

In a sense we can argue the same is true for the on-off control if we average over a longer time period. However, during the on time both the heat strip and the source's internal resistance are stressed more greatly. I don't really know if this matters if the heat strip is rated for 277. Transformers are somewhat tolerant of short time overloads, and it is unlikely you will overload the transformer. Furthermore a heat strip is probably only active in cold weather.

The lifetime of most products is reduced as temperature rises.

.
 
gar said:
131106-2344 EST

Wilkes:

You also must determine if your heat strip is rated for 277 V operation.

Note:

With phase shift control the RMS current is the same at either 240 or 277 applied voltage. The volt-amperes at 277 is 277/240 times the volt-amperes at 240 because the load is a constant average RMS current over a half cycle, and only the voltage was scaled up.

In a sense we can argue the same is true for the on-off control if we average over a longer time period. However, during the on time both the heat strip and the source's internal resistance are stressed more greatly. I don't really know if this matters if the heat strip is rated for 277. Transformers are somewhat tolerant of short time overloads, and it is unlikely you will overload the transformer. Furthermore a heat strip is probably only active in cold weather.

The lifetime of most products is reduced as temperature rises.

.
Click to expand...
The heat trace is rated for 277v and is energized at 44 degrees farenheit. The heat trace is self regulating.
 
Wilkes said:
A 240 volt single phase circuit that has a load of 2616 va has to have a power adjustment factor of 1.16 applied to it due to the voltage of the circuit being changed to 277 volts. When applying this is the paf multiplied by the va or is the va divided by the paf?
Click to expand...

Wilkes said:
The heat trace is rated for 277v and is energized at 44 degrees farenheit. The heat trace is self regulating.
Click to expand...

Self regulating heat trace doesn't have a constant power load. The number you list for 240V of 2616VA is for a specific temperature. Generally, heat trace has a rated watts/foot - which again doesn't match except for a specific temp.

A lot of heat trace is negative temp coefficient semiconductor - it is not linear. The watts output (at a specific temp) roughly changes as the square of the voltage (as gar and gold said) - but not real linearly.

Qs:
Are you looking for the CB size to feed the trace when it is changed to 277V? And the concern is the loading when energized at 44F?

If the concern is overheating - it won't, it's self regulating. As it warms up, it will taper off. And I'm thinking you knew this.

Did you not have quite enough heat at 240V and are looking to get a bit more at 277V and you are wondering how much more? If it is this, I did one of these several years ago. Trace was rated for 240V, we were feeding it with 208V. Buck-boosted to 240V. At 33F we were getting about 25% more watts. That was enough.

So where am I headed? Go look up the mfg spec. They will have pretty good charts showing voltage/watts/temp.

ice
 
I'm trying to calculate the maximum allowable footage allowed for 5 watt heat trace fed with a 20 amp cb at 277v. I can't figure out how the power adjustment factor needs to be applied to this because the manufacturers spec sheet only shows the maximum length at 240 volts but gives a paf of 1.16 if supplied by 277v.
 
Wilkes said:
I'm trying to calculate the maximum allowable footage allowed for 5 watt heat trace fed with a 20 amp cb at 277v. I can't figure out how the power adjustment factor needs to be applied to this because the manufacturers spec sheet only shows the maximum length at 240 volts but gives a paf of 1.16 if supplied by 277v.
Click to expand...

There are two (closely related) components in the maximum length, IMHO:
1. The maximum length fed from one end based on voltage drop and still getting the rated output at the far end, and
2. The maximum length based on the total current pulled, in terms of the connections at one end and the longitudinal conductors inside the heat trace.
The third factor is just how much current it will draw on a 20a circuit. (I assume the manufacturer specifies a 20A max OCPD?)

Basically though, you will get the 5 watt per foot output at a lower average current when you feed the stuff from 277V. And with a higher temperature in the active element. If 1 and 2 are not factors, then you could go 1.16 times as far. (That is, put 1.16 times as much length on the fixed current breaker.)

But the one thing that I think you are assured of is that the maximum heat output of the trace will be (5 x 1.2) watts per foot when driven from the higher voltage. That means that if you have a high enough thermal load on it, it could end up using 1.2 times as much current for the same length. That would force you to cut the length by 1.2. Fortunately, it is unlikely that the design will be requiring maximum output from the tape. Unfortunately you cannot count on that without some engineering. :happysad:
 
Wilkes said:
I'm trying to calculate the maximum allowable footage allowed for 5 watt heat trace fed with a 20 amp cb at 277v. I can't figure out how the power adjustment factor needs to be applied to this because the manufacturers spec sheet only shows the maximum length at 240 volts but gives a paf of 1.16 if supplied by 277v.
Click to expand...

Okay. This should help:

Here is an example of a cable I'd have used.
http://www.thermon.com/catalog/us_pdf_files/TEP0067.pdf

And here is their method of dealing with the different voltages.
http://www.thermon.com/paf/tt131.pdf

They include an example.

Edit to add:
Two things I'd like to point out:
Thermon says to get their program.

Even though the 5W/ft cable has 22% more power at 277V over 240V, Thermon says you can increase the circuit length 1%. I don't see where you can use the paf to calculate the increase/decrease in circuit length for a change in supply voltage.

ice
 
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