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 > Converter Left On Powered by Inverter--Measurements

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BFL13

Victoria, BC

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Posted: 02/11/13 07:25am Link  |  Print  |  Notify Moderator

I suppose if all were right, then the converter and inverter would cancel. The point is they don't and you lose more than you get.

So 56 in from the converter should balance 56 out? But there is 100 out and 56 in. 44 more out than 56 in is 44/56 =78% ?


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HiTech

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Posted: 02/11/13 07:49am Link  |  Print  |  Notify Moderator

Yes I believe the total watts produced at the inverter is as reasonable a place to start as any, since the converter + inverter creates a feedback loop circuit. The negative amps out of the battery represent the total loss of the system vs 100% efficiency. The total watts divided by the voltage at the DC amp meter should be very close equivalent total DC amps in+out.

I'm thinking more like 100 Amps out of the battery (as you said, depending what part of the 12v system you use for voltage), 56 back in, 44 net out, so 56% overall efficiency, and 44% loss. If it were a magically 100% efficient perpetual motion machine there would be 0 amps out of the battery after startup.

I assume the bulk of that loss is in the DC-AC-DC conversion of the inverter and converter, plus a little for wire heating, and possibly small Peukert (on edit, I gues you would not see the Peukert losses since they are before the meter, except in the form of the voltage dropping faster, requiring more amps for constant wattage) or other discharge losses from the battery.

It's not unreasonable, just very eye opening. In thinking about it, I never really stacked up the efficiencies one after another to consider what the real world loss would be. Imagine what would happen with undersized wiring or a 100 amp converter running.

If you had a big enough converter relative to your inverter you would have a failsafe to prevent it happening, but I'm not sure how hard that would be to constantly trigger the inverter over wattage protection.

Thanks for measuring this!

Jim

* This post was last edited 02/11/13 08:19am by HiTech *   View edit history

pianotuna

Regina, SK, Canada

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Posted: 02/11/13 08:23am Link  |  Print  |  Notify Moderator

Hi Jim,

Always measure at the battery. Failing that always measure at the same point in the system.


Regards, Don
My ride is a 28 foot Class C, 256 watts solar, 556 amp hours of AGM in two battery banks 12 volt batteries, 3000 watt Magnum hybrid inverter, Sola Basic Autoformer, Microair Easy Start.

BFL13

Victoria, BC

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Posted: 02/11/13 08:31am Link  |  Print  |  Notify Moderator

I have a 2000w inverter and a 400w inverter I could also try this out on. The batteries are nearly charged up now (left them on the 7355 overnight), but I guess it doesn't matter what SOC they are at for this exercise.

The inverter's readout for voltage and watts doing this, was 11.6v ISTR and 1003w so that makes amps at 86a, if that means anything at all.

Also the converter doing 56amps at current limit, would have its own voltage sagging from 13.8 no load to somewhere in the low 13s.

EDIT--that 56amps is an ASSumption too. Not measured. Reasonable assumption though IMO.

I have no idea what to try to match up with what to explain the "missing 44." I didn't think it would be that big of a difference in and out though. I thought it didn't matter a lot if you forgot the converter was on for a while, because it would lose at a slower sort of rate. Nope--it is big. [emoticon]

* This post was last edited 02/11/13 09:03am by BFL13 *   View edit history

12thgenusa

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Posted: 02/11/13 08:38am Link  |  Print  |  Notify Moderator

100 amps out * .78 efficiency = 78 amps to the converter

78 amps * .7 PF = 54.6 amps output from converter

100 - 54.6 = 45.4 amps net loss in the conversion

plus wire losses at high amperage

Peukert would only affect the apparent quantity in the "tank" and would not show in the readings. However, it would actually add to the net loss in the conversion by also reducing the remaining amount available.

On edit: The current between the inverter and converter is AC so the amperage figure is not correct. However, if the conversion is made to watts, the final answer is still the same.

100 amps out * 12.5 = 1250 watts out of battery
1250 * .78 = 975 watts out of inverter
957 * .7 = 682.5 watts out of converter
682.5 / 12.5 = 54.6 amps returned to battery

* This post was edited 02/11/13 09:29am by 12thgenusa *


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HiTech

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Posted: 02/11/13 09:34am Link  |  Print  |  Notify Moderator

12thgenusa wrote:



78 amps * .7 PF = 54.6 amps output from converter



Correct me if I am wrong but I believe that power factor is for sizing and does not imply any efficiency loss at all. It is a phase change in the loading (ELI the ICE man), not any loss of energy from the system.

Can any EE's confirm or correct this?

70% efficiency for the converter is not an unreasonable number but I do not think that is the same as the power factor, is it? Converter watts are still watts, they just take that extra .3 of energy in *and return it* when there is a non 0 reactance. When there is no reactance with a pure resistive load, PF is 1.

Jim

BFL13

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Posted: 02/11/13 10:09am Link  |  Print  |  Notify Moderator

I agree that PF is not "efficiency." We had had this confusion before.

EG, a PD9280 is rated to do 80a at 13.6v = 1088w, but spec says input of 1300w. 1088/1300 = 83.6%

A PD9260 ---60 x 13.6 = 816 but needs 1000 by spec. 816/1000 = 81.6%

But we all know it takes more VA from the gen to run a 9260 with its 0.7 PF. 100/70 x 1000 = 1429.

A Vector 35amper spec says needs 600w. max when running is when at 35a and 13.9v just before amps taper. = 486. 486/600 = 81%
Don't know what the PF is on those but seems to be more like 0.8 from some results I got.

---I am wondering about that 100amps in the OP now. That is based on the "divide by 10 rule", not a measurement. that rule is 10 vs 12 to take into account inverter losses. So we don't want to count that twice somehow here.

I will get some more numbers and check this some more. eg, I want to run the kettle at 90 amps on the Trimetric and see what the volts and watts on the inverter are, and get the amps from that.

That should help with interpreting what we see when the inverter is running the converter.

12thgenusa

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Posted: 02/11/13 11:25am Link  |  Print  |  Notify Moderator

Not an EE but running the input for the converter through a kill-a-watt gives a PF of .7 at full load. It also takes over 1200 VA in to get about 800 watts out. It may be coincidental that this ratio is about .7 also and the losses may be due to inefficiency rather than PF. We all know that high current output from the converter generates a lot of heat because the fan comes on trying to get rid of it.

Whether due to PF or efficiency, there can be a 30% or greater loss through the converter.

HiTech

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Posted: 02/11/13 11:29am Link  |  Print  |  Notify Moderator

It's not a loss though. It's extra capacity you need to have to run the converter, but you are only actually consuming 800w worth of energy not 1200w. Basically because the load is out of phase with the power source, you need a bigger power source to drive it. 60 times a second it DOES take extra power, and 60 times a second a fraction of a wavelength later, it gives that extra power back. So yes you have to size bigger for a PF <1, but you are not actually using more watts than the rating on the device you are powering.

BFL in your example:

A PD9260 ---60 x 13.6 = 816 but needs 1000 by spec. 816/1000 = 81.6%

***That's the efficiency - 816w out but needs 1000w in***

But we all know it takes more VA from the gen to run a 9260 with its 0.7 PF. 100/70 x 1000 = 1429.

***Yes the genset needs to be sized to compensate for the power factor, but the watts used are 1000w, not 1429w***

Think of it like driving a heavy load up a hill. If you can drive right at the peak torque in the powerband of the engine (power factor = 1) you will pull the hill better. If you have to drive at a speed in that gear that is way out of the powerband (power factor .7) - you'll need an engine with more peak torque, just to pull the same load at the same speed.

But at a given speed in the same gear with a given load, up a given hill, the torque at the wheels is identical in both situations. One just needs a bigger engine because the power the engine makes is not at the right RPM for the speed you want to go.

Jim

* This post was edited 02/11/13 11:36am by HiTech *

BFL13

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Posted: 02/11/13 11:49am Link  |  Print  |  Notify Moderator

More numbers!

"Calibrating" the 1000w PSW inverter- running a kettle that says it needs 1000w input.

Trimetric says 87.5amps. Inverter says 11.6v and 920w. (so 79a)

So the inverter draws 8.5 amps more. Not sure where to take the percentage from. ratio is 87.5/79 doing it that way around.

On 7355 converter again to confirm-

Trimetric says -43.7a Inverter says 12.0v and 1040w (so 86.7a)

86.7 x ratio 87.5/79 = 96 amps So converter was at 96-43.7 = 52.3a ?

Now the plot sickens-- I tried the 400w inverter and the converter load alarmed it off immediately. So no worries about running the converter off the inverter if it won't run it!

Next tried the 2000w inverter. On kettle, Trimetric again showed 87 amps but inverter watts readout (each bar is 200w)-three bars or 600w. 11.9v So 50 amps which is silly. The watt bar meter is no good I think.

So tried the converter on the 2000w inverter (which is MSW---the 1000w one is PSW) and the Tri showed -30 amps. The watts bar was at three bars for 600w and battv was 12.5v (= 48a)

So we have the converter not being like the kettle?. Inductive vs resistive ? MSW/PSW. anyway I don't know from all that what amps the converter was doing when the Trimetric showed -30a on that inverter instead of -44amps on the other one.

But it seems if I were doing whole house on the 2000w instead of the 1000w and forgot the converter was on, it wouldn't hurt as bad. [emoticon]

I know the microwave draws fewer amps on the MSW than on PSW so it seems the converter is like that where the kettle acts the same on either.

* This post was edited 02/11/13 11:55am by BFL13 *

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