Confusing Surge Rating Description in my Inverter Manual

st4rgut

I just purchased a 1500 Watt 24V Reliable Pure Sine Wave Inverter to run a SuperFlo Vs pool pump that uses around 1550 Watts starting up. The inverter has a surge rating of 3000 watts so I thought I would be safe. However, the manual doesn't mention how long the inverter can supply surge power. Instead there seems to be an odd disclaimer under the Surge Power Rating section: 

During start up, certain loads require considerably higher surge of power for short duration (lasting from tens of milliseconds to few seconds) as compared to their Maximum Continous Running Power Rating. The inverter continuous power should be higher than the surge power rating of these appliances. "

I don't understand what the point of having a surge rating is if it's not to power loads using surge power. What should I make of the above?

mcgivor

Induction motors are particularly hard on inverters because they draw  very high current on startup. There are two stages, inrush which lasts approximately 150 milliseconds and surge which is typically around 3 seconds. The inrush or lockded rotor amperage  (LRA) can reach as much as 500% if the rated run current, the surge current and duration will be dependent on application, for example a pump with high head would be different to the same pump with low head.

There is also power factor to be considered when sizing an inverter to run an indication motor, one with a 1500W rating (real power) with a power factor of 0.7  would mean the inverter would need supply 1950W (apparent power) , a good rule of thumb is to have a rated capacity double the nameplate wattage of the motor. "Simplified explanation" 

The AC current on inrush will be reflected in the DC side, this is dependent on the conversion ratio, a 24 to 120 V inverter would have a 5:1 value, therefore a 10A load at 120VAC would be 50A at 24VDC. The high inrush of an induction motor along with subsequent surge can place very high current demands on the battery along with the feed conductors resulting in voltage sag which further increases current.

The disclaimer is likely published because there is no way the manufacturer can predict the end users application or load power factor, usually a cheap high frequency inverter cannot support its surge rating of 200% for much more than a fraction of a second, whereas a low frequency type can support 300% for several seconds, the Reliable inverter is a high frequency type.

Could have just replied with the previous paragraph but thought a little background would be helpful.  

RCinFLA

Depends on the startup mechanism of motor.  There are commonly three different single phase inductiion motor startup mechanisms.

Air conditioners have just run capacitor that provides startup phase shift,  refrig compressor have positive temp coeffficient thermistors,  water pumps and air compressors have a large value electrolytic startup cap temporarily connected which are disengaged by a centrifugal switch after shaft spins up enough.

For pumps there can be a significant time variation depending on centrifugal switch variability.  Pumps also have a significant initial mechanical load on motor during startup making spin up tougher to accomplish. Range is typically between 400 msecs and 700 msecs.  

An inverter surge specification at 1 msec is just a marketing spec play as 1 msec is not going to be long enough for motor, or much of anything.  You need surge current out to a few hundred milliseconds.

High frequency inverters are limited by their DC to DC converter electrolytic filter capacitor value and their ferrite transformer core abrupt saturation characteristics.  They just don't store enough energy to handle much of a surge.  The DC to DC converter MOSFET's will shoot up to very high current and can blow out if not shut down quickly when ferrite transformer core goes into magnetic flux saturation at high load surge.

Low frequency 'heavy iron' transformer inverters have much better surge capability because the large silicon-iron core low freq transformer has soft saturation characteristics, compared to high freq ferrite transformer, and inverter doesn't rely on a filter capacitor to provide the surge energy.  You still have to make sure your battery and battery cables do not have too much series resistance or the voltage at inverter input will slump too much. Typical surge capability for low freq heavy iron transformer inverters are 2x to 3x rated VA at 100 milliseconds.

BB.

I think that you have a bit of translation and cover their behind wording...

The inverter can probably output 2x rated Watts (VA) for a few seconds... And it should (normally) shutdown relatively safely (and no damage) if somewhat overloaded (a bit too large of induction motor or such).

An issue, probably, is the whole stuff that makes AC Math "hard"... One of the useful equations for power:

• Power = Voltage * Current -- The "standard power equation"--Works pretty well with DC circuits.

For AC power systems, the more complete equation(s) is(are):

• Power = Voltage * Current * Cosine phase angel between Voltage and Current
• Power = Voltage * Current * Power Factor (PF)
• PF ~ Cosine of phase angle

What you here is the "difference" between Watts and VA (volt amps).

PF (and Cosine) can easily run from 0.50 to 1.0 ... And what this means is that "ideal" AC loads (resistors, PF corrected power supply inputs, filament lamps, etc.) are near PF/Cos ~0.95-1.0

And from the equation, you can see that means Watts ~ VA with "good" PF.

And with "bad" PF/Cos, you can be seeing 0.5 to 0.65 or so... 1/0.5 = 2x ... Where VA = 2 x Watts... Or specifically with 120 VAC circuits. that means that to Get (say 1,200 Watts):

• Power = V*I*PF
• Power = 1,200 Watts = 120 Volts * 10 Amps * 1.0 PF (with "perfect" power factor)
• Power = 1,200 Watts = 120 Volts * 20 Amps * 0.5 PF (with "bad" power factor)
• VA (with "bad" power factor) = 120 Volts * 20 Amps = 2,400 VA

So, the wiring, transformers, internals of AC inverter, genset alternator, etc. needs to send about 2x more current than is needed for 1,200 Watts because of bad power factor (poorly designed induction motor, starting induction motor, "cheap" lamp ballast, etc.).

With most AC inverters, they are spec'ed at (for example) 1,200 Watts and 1,200 VA... If you have a poor power factor load, then the AC current is used "less efficiently" and you need heavier wire/AC inverter/Genset to run the specific loads (typically induction motors, some lighting, older computers, etc.).

More or less the battery bank (the DC side of things) will still be supplying ~1,200 Watts of power (even with 2,400 VA "bad" PF) or the genset still sucking the same amount of fuel (1,200 Watt load's worth of fuel flow).

Why is high VA and bad "PF" bad? One reason is resistance. More current, more heating--Needing larger wiring/transformer/alternator windings:

• Power = V*I
• Power = Volts^2 / R (heating)
  
• Power = Amps^2 * R (heating)

And you notice that if you 2x more current, you get 4x more Watts (Heating)... So, you need larger wire/more heat sinking, etc...

There are other issues too (inductors and transformers have maximum current ratings or the metal cores "saturate" magnetically, etc.).

-Bill