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Kw & KV question?


solarman
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On most inverters it seems that a 5kva is equal to a 4KW. I've seen this on Aperts, Victrons etc. 

I have now found an Axpert and and MLT inverter where it says the 4KW is equal to 4 KVA (see attached). I understand it has to do with power factor?

Can someone please explain how this works in layman's terms?

Inverter.png

Edited by solarman
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Layman's terms? Okay, I will give it a shot. I think I'll number my thoughts.

  1. First, I assume that you remember high school physics and that P = VI, that is watts = volts multiplied by amps. For DC, this remains exactly the same.
  2. For AC, things are not that straight forward. The voltage is not constant, it cycles between -325V and +325V 50 times a second. Without getting into the math so much, as long as this is a sine wave, then the average value (or the RMS value) is at 0.707 of the peak, which works out around 230V. So that is how we get that value.
  3. Now, if the load is a simple load without any funny characteristics, something like a kettle, then the current will also be a sine wave. Because I=V/R, and R is more or less constant for such a load, the current draws a sine wave that looks exactly like the one for the voltage. When the voltage goes up, so does the current. If you show the two waveforms next to each other, they will exactly line up, their peaks will be in the same place, and they will cross the zero line in the same place. We say that the voltage and current is exactly in phase.
  4. Referring to point 2, this sine wave for the current will also have an RMS/average value that is 0.707 of the peak.
  5. For such simple loads as in point 3, you can multiply the RMS voltage with the RMS current, and you will get the real power. No funny business.
  6. VA, as you might have guessed, is the RMS voltage multiplied by the RMS current. For the simple load in point 3, the VA will exactly equal the power. We say that the real power (W) and the apparent power (VA) is the same. We also say that this kind of load has a unity power factor.
  7. However... not all loads are such simple loads. And this is where things go a bit squivy.
  8. Imagine a game where you and a friend are throwing a ball at each other. The ball goes back and forth (much like the voltage on an AC system). Each time you catch the ball, it takes some energy to stop the ball and reverse its momentum before you can throw it back. For a ping-pong ball this will be negligible. For a bowling ball it won't be.
  9. When reversing the direction of the light ball, the energy you exert with your muscles will more or less be in phase with the movement of the ball.
  10. When reversing the direction of the heavy ball, the energy you exert around the turning point is hugely out of phase with the movement of the ball.
  11. It is the same with some loads on an AC system. The current and voltage waveforms stop lining up. Their peaks don't line up, and their zero-crossings don't line up.
  12. When this happens, you can no longer multiply the RMS voltage with the RMS current to get power.
  13. What you have to do is pick 100 (or however many you want) points on each chart, take the voltage at THAT point and multiply it by the current at THAT point, and then add all these little results together. To get more accurate, you use more sample points, ideally infinitely many. More perceptive readers may recognise that this is Calculus, it is integration.
  14. If the voltage and current waveforms are still perfect sine waves that are merely out of phase, then we can present the relationship between W and VA as follows: P = V * I * cos(theta), where V and I are the RMS values referred to earlier. Theta is a greek letter that represents the "phase angle", essentially how far apart the waveforms are shifted.
  15. Because the maximum value a cosine will yield is 1, it means P is always strictly less or equal to VI (in other words, W is equal or less than VA). Because V is constant (230VRMS), this necessarily means that higher VA values correspond to higher average current (we will use this result later).
  16. cos(theta) is also called the power factor, so sometimes you might simply say that "this has a power factor of 0.5" instead of saying "the phase angle is 60 degrees", because the former is easier to work with.
  17. To make matters even worse, the current waveform isn't always a perfect sine wave. We call these non-linear loads. There is no point in discussing them here, so you may relay :-)

Okay, now we get to the second half of the explanation, which will hopefully be a bit less unwieldy.

Modern inverters consist of two stages.

First there is a boost stage, that takes your 48VDC and boosts it to 325VDC (usually a bit higher). If you paid attention, you recognise that voltage from the prior discussion.

Secondly there is an inverter/final stage that neatly carves out a sine wave out of the 325VDC to make 230VAC RMS.

Now, the boost stage doesn't care about the apparent power. It's DC so it is the same thing. The average current is just the current.

For the final stage, however, has to care about the average current, because the components (transistors, IGBTs, FETs) are rated for their current capabilities.

The boost stage works hard in an inverter. This is usually the more stressed part, makes the most heat, and uses the most expensive switching components.

The final stage doesn't work as hard, so it is easier to design one that is a bit oversized without incurring additional costs.

If you are still here, you know where this is going. As an inverter designer you design the boost stage for the real power, and the final stage for the apparent power. Because it is easy and cheap to oversize the latter, this is usually done. This means many inverters can handle the higher peaks of loads that have a lower power factor as long as their real power is within the capacity of the boost stage.

As it goes with these things, marketing wants to print the higher number on the brochure. So they will rather market it as a 5kva than a 4kw.

What they don't tell you, is that if you use loads with a really poor power factor, that 5kva inverter might only be good for 2kw.

To be fair though, most induction motors are not far off a power factor of 0.8. So this design choice is not completely without merit.

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I suppose another analogy is to think of a steady stream running into a holding dam (aka DC), and then someone sitting at the outlet opening and closing a valve really quickly to create a pulsed flow (aka AC). If the water level in your holding dam remains on the same level, then it means the average flow in and out is the same. The pipe running out of the dam has to be able to take higher peak pressures though because of that monkey sitting there opening and closing the valve.

The boost stage is a bit like the inlet pipe. It's DC, so steady flow. The final stage is like that monkey flipping the valve. Same amount of water, but higher pressure spikes. Got to rate the pipe according to the peak pressure :-)

It's not completely accurate, but it might help :-)

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1 hour ago, Youda said:

if an inverter is rated for 5kVA/4kW - what is the maximum resistive load it can safely handle?

The maximum resistive load it can handle is 4kw, which will be the same as 4kva (because the power factor is 1).

Let's consider another real life example. In my home I replaced 50W MR16 downlights (that's the 12V kind with the old iron transformers) with 5W LED lamps. This creates a particularly bad power factor, which I will get into again below (because some days I like rambling :-) ).

The power factor in this case is around 0.2. It is non-linear to boot. This means that each lamp uses 5W, but 25VA.

So if i were to put 200 of them on a 5000VA inverter, the final stage will be maxed out. The real power drawn on the battery side (and going through the boost converter) will be only 1kw.

So why does this load have such a poor power factor? Old-school iron transformer. When this thing is properly sized (in other words, the load on the output matches the transformer size), the magnetic field is charged up via the primary winding, and then discharged via the secondary winding, so that at the end of the half-cycle there is little residual magnetism left in the core. in my application, the 50W lamp was replaced with a 5W lamp, so the secondary winding does not properly consume the magnetic flux in the core. The only place for this to go is back out via the primary, in reverse. This causes a high out of phase current at the start of the next half-cycle. Possible solutions: Put more lamps on each transformer, or use smaller transformers.

Edited by plonkster
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Okay...is there a situation where one could pull 5kVA from a 5kVA/4kW rated unit? I don't think so...

So, in this particular case the 5kVA label has no information value for any user. It's same like saying that there's 8 or 12 mosfets in the inverter - it might be true, but for me as an user/installer, it does not mean that 12 is better than 8, because it says nothing about the quality or rating of the components.

Contrary, another unit, rated as 5kVA/5kW sounds logical - with a resistive load, I can draw 5kVA/kW. With capacitive or inductive load, it depends on actual PF of that load.

Am I right, please?

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27 minutes ago, Youda said:

Okay...is there a situation where one could pull 5kVA from a 5kVA/4kW rated unit? I don't think so...

Yes, if your power factor is 0.8. Then both the boost stage and the final stage are maxed out.

28 minutes ago, Youda said:

rated as 5kVA/5kW sounds logical - with a resistive load, I can draw 5kVA/kW

Correct. And that is essentially the difference between the 4048 and the 5048 models of that particular inverter.

33 minutes ago, Youda said:

With capacitive or inductive load, it depends on actual PF of that load.

Correct. Both numbers are important of course. Especially on smaller units. If you want to run a couple of low-wattage LED lamps from a 350VA unit for example, watch out :-)

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4 minutes ago, plonkster said:

Yes, if your power factor is 0.8. Then both the boost stage and the final stage are maxed out.

Okay, now I see that even that strange unit rated as 5kVA/4kW has a meaningfull existence. As with all the modern gadgets, switched power suplies and LED lights, a typical home clearly has an overall PF of 0.8 or similar.

Thanks :)

 

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On 2018/07/24 at 9:35 PM, Youda said:

 

Okay, now I see that even that strange unit rated as 5kVA/4kW has a meaningfull existence. As with all the modern gadgets, switched power suplies and LED lights, a typical home clearly has an overall PF of 0.8 or similar.

Thanks :)

 

And, that, brings you to an interesting system design question: What do you sell the client? From the same brand, do you sell the right one for his application, or the one he can afford?

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30 minutes ago, SilverNodashi said:

What do you sell the client?

Something with some margin, so that this question essentially becomes moot. Or that is what I would do. Reason: Have a customer who complained that his Inverter/Charger only charges at 63A when the box said 70A. Then you have to explain that the 70A is at a specific temperature, the thing derates when it runs at peak values for long periods. Moral of the story: If you need the full 100% capacity for your largest load... then the inverter is too small.

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