Breaker Size: Calculating RMS current from datasheets impossible?

[pyro214]

I'm trying to spec breaker sizes for many non-linear devices however the datasheets typically only provides operating voltage and real power (W). To properly spec a breaker size I'd like to use the RMS current; for a single-phase load I'd solve for current using either one of these equations

• P(Watts) = pf x I(Amp) x V(Volts)
• I = VA/V

In addition to voltage and real power I'd require either current, apparent power, or expected power factor to solve for the RMS current.

What do you do in this situation? Is this common? Thank you in advance.

 

[kingpb]

I think you will find that for your non-linear loads the PW = PVA.

 

[charlie b]

I think you will find that for your non-linear loads the PW = PVA.

You lost me here. Are you saying that power factor is always unity for non-linear loads? I think not. So what are you saying? :?

 

[charlie b]

In addition to voltage and real power I'd require either current, apparent power, or expected power factor to solve for the RMS current.What do you do in this situation? Is this common?

It's common. I usually select a reasonable value for power factor, based on what I know about the loads.

 

[Besoeker]

You lost me here. Are you saying that power factor is always unity for non-linear loads? I think not.

I agree. Power factor as in kW/kVA comes in different flavours. Conventional displacement PF (inductive loads typically) and distortion power factor with non-linear loads like rectifiers, VSDs, and pretty much all electronic gizmos.

 

[kingpb]

You lost me here. Are you saying that power factor is always unity for non-linear loads? I think not. So what are you saying? :?

Give me a load list and I'll tell you, but when someone makes a blanket statement that their single phase non-linear loads are X number of watts, I can reasonably make the assumption they are talking about non-industrial single phase loads and therefore switched-mode power supplies are involved.

In today's world, these power supplies are driven by standards that require the use of power factor corrected power supplies. The PFC is to make the power factor as close to unity as possible, and the reactive power consumption is theoretically zero (or nearly). Therefore, for all intensive purposes PW=PVA.

Harmonics, now that's another issue.

 

[gar]

131025-0935 EDT

pyro214:

As "charlie b" stated it is not clear what is your question.

If I assume from your name you are working with ovens or heaters, then these are likely resistive loads. To adjust temperature it is possible you are using phase shift controlled SCRs or Triacs to ajust power and therefore temperature.

I also assume that you are concerned with steady state current, and not a turn on transient inrush current of some sort.

The RMS current to a constant resistance load from a constant voltage sine wave will never be greater than the value obtained when full sine wave excitation is applied to the resistance. As you chop part of the sine wave away by a phase shift delay to turn on of the electronic switch the heating of the resistance will drop and that means the RMS current drops. You can write the equation for RMS current using the square root of the integral from T1 to T2 of a sine wave squared divided by time per cycle. Do this for a half cycle.
The integral of sin^2 x dx = x/2 - (sin 2x) / 4.

Assume turn off occurs at Pi and the time duration is Pi, and turn on is at K*Pi, where K ranges from 0 to 1, then
Irms^2 = (Pi / (2*Pi) ) - PI*K / (2*Pi) + sin (2*Pi*K) / 4*Pi
Irms^2 = 1/2 - K/2 + (sin (2*Pi*K) ) / 4*Pi.

For K = 0 (a full half cycle)
Irms^2 = 1/2, or the sq-root is 0.707 .

Turn on the sine wave at 90 deg, K=0.5, and the result is
Irms^2 = 1/2 -0.5/2
Irms^2 = 0.25, or the sq-root is 0.500 .

Thus, the RMS current for a sine wave on from Pi/2 to Pi, and 3*Pi/2 to 2*Pi is
0.5/0.707 = 0.707 times the RMS current of a full on sine wave as applied to a fixed resistance. If you design for full excitation current and the load resistance is constant, then reduced on time will not cause greater heating in the supply side than full excitation. A circuit breaker for a full sine wave will work with a partial sine wave.

Ckeck my math.

.

 

[pyro214]

I agree. Power factor as in kW/kVA comes in different flavours. Conventional displacement PF (inductive loads typically) and distortion power factor with non-linear loads like rectifiers, VSDs, and pretty much all electronic gizmos.

Thanks.

I'm dealing with communication equipment (power factors being as low as 0.6 so far). If I start assuming low power factors for some equipment I'll ensure the cable sizes are also increased accordingly.

 

[gar]

131027-1118 EDT

pyro214:

You really need to provide more information for the type of question you asked. And you need to make some measurements.

What is communication equipment? Is this radio transmitters, computer modems, or what?

An example load with a switching power supply, my HP 7955 computer. Nameplate at 120 V is 4 A. As I use this machine 123 V, 1.25 A, 93 W, 154 VA, and 0.60 PF. Nameplate VA = 480.

How would you design the supply for many of these computers? What is the probability of ever producing a load of 4 A?

.