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Author Topic: Blowing Inverters!  (Read 3571 times)
pogster
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« on: December 17, 2008, 09:37:29 PM »

I've just set up a supply from my water wheel battery bank to power the lights in the house.

This is how it works - The inverter connects to a 35m cable (1 twisted pair + 2 core 3mm2) the mains output from the inverter via the cable then powers a 16A mains rated relay (with a mains rated coil) this then disconnects the live feed from the mains and connects the feed from the inverter to the lighting circuit.

The idea is that when mains is present from the inverter the relay automatically switches over and provides power to the lighting circuit in the house. When or if the batteries fail the relay automatically reverts to standard utility mains.

All the house lighting is CFL with total power consumption if all on at the same time of 280W

The system works well but I have now blown two inverters, a modified sine 600W and a 300W pure sine - both worked for a day or so. I think the problem may be related to the fact I only switch the live and leave the neutral from the inverter connected straight through. i.e. I'm just connecting the neutral from the inverter to the neutral block in the electricity distribution box, thus the lighting circuit is not truly isolated from the utility mains.

I'm going to isolate the Neutral line using the same method and hope this solves the problem.

Has anyone got any other ideas / comments?

Pete.
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tony.
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« Reply #1 on: December 17, 2008, 10:10:45 PM »

the realy should have another pole that you can use for your neutral.

As you suggest, its probably the lost neutral that causing you to blow your inverters, loosing a neutral can cause a voltage imbalance, and can fry electonic equipment.

If this happens to your electricity supply, they will generally give you compensation, all srts of problems can arise.

tncs systems, loosing neutral can mean that exposed or extraneous metalwork will rise to 230 volt or higher potential!!!

for further advise looks at kirchoffs law!!

in the meantime use a extra pole on your relay to switch the neutral.

regards

tony
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guydewdney
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« Reply #2 on: December 17, 2008, 10:12:09 PM »

are you sure its not make before break relay?

Would the capacitors in the cfl's and other associated trigger circuitry cause some huge back emfs? Maybe a mid point for 1 second?

Neutral = earth in 99% of houses these days... which I only found out t'other day
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billi
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« Reply #3 on: December 17, 2008, 10:25:56 PM »

perhaps this helps  Roll Eyes just a thought

http://www.victronenergy.com/upload/documents/DC%20solar%20system%20with%20a%20MultiPlus%20inverter-charger.pdf


billi
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Adam
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« Reply #4 on: December 17, 2008, 10:26:33 PM »

The idea is that when mains is present from the inverter the relay automatically switches over and provides power to the lighting circuit in the house. When or if the batteries fail the relay automatically reverts to standard utility mains.

I have had this system for a couple of years but i have used 2 pole relays, first one was a 5 amp for the lighting circuit with a navitron 1000w 48v MSW inverter, now i have a Victron 3000va PSW inverter so i had to upgrade to a 2 pole 20amp 240v relay, it works so well that you can't tell when the changeover happens Grin

You need a 2 pole relay, If you are using cheap chinese inverters don't exceed 50% of their rated output especially when changing over Wink
« Last Edit: December 17, 2008, 10:40:49 PM by Adam » Logged
Justme
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« Reply #5 on: December 17, 2008, 11:03:26 PM »

I understand how your system switches back to the mains when the battery fails but how do you you set it so that the switch goes back to the inverter supply when the bank is ready for it but only when no loads (lights) are on? Switching on an inverter under load is not good. Ones like victron cope because they control the switch over them selves after running the inverter up to speed first. Just like they will let the genny be connected & switched to on to the inverter whilst starting as they dont accept the feed till its stable so keeping the load from the genny till its ready.



Justme
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pogster
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« Reply #6 on: December 18, 2008, 07:17:41 AM »

I understand how your system switches back to the mains when the battery fails but how do you you set it so that the switch goes back to the inverter supply when the bank is ready for it but only when no loads (lights) are on? Switching on an inverter under load is not good. Ones like victron cope because they control the switch over them selves after running the inverter up to speed first. Just like they will let the genny be connected & switched to on to the inverter whilst starting as they dont accept the feed till its stable so keeping the load from the genny till its ready.



Justme

I have to manually turn it on from the house (via a high current relay and the twistd pair in with the main cable). Bit of a pain in the backside but I only need to switch it twice a day.

are you sure its not make before break relay?

Would the capacitors in the cfl's and other associated trigger circuitry cause some huge back emfs? Maybe a mid point for 1 second?

Neutral = earth in 99% of houses these days... which I only found out t'other day

The centre pole on the relay is the lighting circuit with the two it poles it switches between being Inverter supply and Mains supply so in theory it has to break before it makes. It does not fail during switchover just some time during use which leads me to think it could be a Neutral problem.


Thanks for all the feedback, I'm going to install a second relay (to essentially make a two pole switchover to include the Neutral also) today and I have purchased a cheap - 20 MSW inverter off e-bay to test it with! looked into repairing the 600W inverter but the transistor cost makes it unfeasable. The 300W PSW inverter is going back to the seller on e-bay if he will refund, if not then a paypal claim...........
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w0067814
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« Reply #7 on: January 28, 2009, 02:56:51 PM »

One other thing to note is that CFL lights have a very poor power factor of around 0.5 (measured myself on two of my desk lamps). This means that a load of 280w which would normally require a current supply of a little over 1 amp (280w/230v = 1.22A) actually requires double that. (280w / 230v / 0.5 = 2.44A).

This means that the transistors on the output of the inverter must be rated to handle that current. My inverter is rated for 750VA or 500 watts. You must ensure that you do not exceed either rating.

Power factor is the ratio between real power (ie watts, or power that actually does work) and the apparent power which is the sum of volts * amps, or VA. Resistive loads have power factor of 1, and inductive / capacitive loads (ie motors, computers, CFLs etc) have power factors of less than 1, and thus the line current is higher than if it was resistive, which increases the cable ohmic losses due to P = IR.

Another things with CFLs is that they do not like running on modified sine wave inverters due to the high harmonic frequencies present. This shortens the life of the bulbs.

For running CFLs, it is best to over rate your inverter. Don't worry about the drain on your battery bank. This is still determined by the real power (280w) not the apparent power (560va), as power factor in DC circuits is always 1.
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Justme
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« Reply #8 on: January 28, 2009, 03:24:46 PM »

1, One other thing to note is that CFL lights have a very poor power factor of around 0.5 (measured myself on two of my desk lamps). This means that a load of 280w which would normally require a current supply of a little over 1 amp (280w/230v = 1.22A) actually requires double that. (280w / 230v / 0.5 = 2.44A).


2,For running CFLs, it is best to over rate your inverter. Don't worry about the drain on your battery bank. This is still determined by the real power (280w) not the apparent power (560va), as power factor in DC circuits is always 1.

!, I dint know that cheers

2, Are you sure? If the cfl is drawing 100w say @ 0.5 PF so consuming 200w via the inverter the inverter must be making that 200w so drawing the 200w rate from the battery?

Justme
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« Reply #9 on: January 28, 2009, 06:22:07 PM »

DC is the future solution..... LEDs or 12v/24v CFLs....... add some 12V sockets and you can also run a laptop.

You will save the 10% inverter loss, and when super-capacitors hit the market most of the 20% battery loss (super-caps are about 97% efficient)

-Paul
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RichardKB
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« Reply #10 on: January 28, 2009, 11:28:21 PM »

1, One other thing to note is that CFL lights have a very poor power factor of around 0.5 (measured myself on two of my desk lamps). This means that a load of 280w which would normally require a current supply of a little over 1 amp (280w/230v = 1.22A) actually requires double that. (280w / 230v / 0.5 = 2.44A).


2,For running CFLs, it is best to over rate your inverter. Don't worry about the drain on your battery bank. This is still determined by the real power (280w) not the apparent power (560va), as power factor in DC circuits is always 1.

!, I dint know that cheers

2, Are you sure? If the cfl is drawing 100w say @ 0.5 PF so consuming 200w via the inverter the inverter must be making that 200w so drawing the 200w rate from the battery?

Justme

No the power is the same its just that the current is taken in a large gulp for a short period.

Rich
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w0067814
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« Reply #11 on: January 29, 2009, 05:42:45 PM »

1, One other thing to note is that CFL lights have a very poor power factor of around 0.5 (measured myself on two of my desk lamps). This means that a load of 280w which would normally require a current supply of a little over 1 amp (280w/230v = 1.22A) actually requires double that. (280w / 230v / 0.5 = 2.44A).


2,For running CFLs, it is best to over rate your inverter. Don't worry about the drain on your battery bank. This is still determined by the real power (280w) not the apparent power (560va), as power factor in DC circuits is always 1.

!, I dint know that cheers

2, Are you sure? If the cfl is drawing 100w say @ 0.5 PF so consuming 200w via the inverter the inverter must be making that 200w so drawing the 200w rate from the battery?

Justme

No the power is the same its just that the current is taken in a large gulp for a short period.

Rich

... now we are really delving into the realm of useless information...  Huh [well nearly]

Not quite although I beleive that is a contributing factor. The issue is that the draw of the current is not in phase with the voltage.



Power is the instantaneous product of volts * current. (Volts are Red, Current is Green) In an AC circuit you can not take the average values as you can in DC. In the above picture the voltage and current are "in phase". That is; at the instant where there is no voltage, there is also no current. The instantaneous power is shown in the dark blue line. Note how it never drops below the "zero" line across the middle of the picture. This is because the power is being transfered from the supply to the load, and never in reverse. Remember -1 * -1 = +1, thus when the voltage and current are both negative the power is still positive.



In the above picture the current and voltage as 45 out of phase. Note how when there is zero voltage there is still current flowing. At this point in time there is no power transfered as the product of anything multiplied by zero will be zero! Also note that the power curve (blue) is now going negative. This is because power from the "load" is being transfered back to the supply from the energy storing devices in the "load" such as capacitors and inductors. +1 * -1 = -1... Over the period of a cycle there is now both positive and negative power. This means that the curves partially cancel out and the average power drops (light blue line). The power available for "real work" is the average power. Any current flowing which is not being used for real work (called wattless or reactive power) is only contributing to IR losses in the cabling & transfromers etc. If cables were super conductors the losses would be low and we wouldn't care about reactive power, but alas, they are not. :-(

Taking it to the extreme, when the voltage and current are 90 out of phase the positive and negative powers are equal and completly cancel each other out over the cycle, and this the power avaliable for real work is zero.



The power taken from the batteries by the inverter is proportional to the average power (light blue line), dispite the current being higher. Thus your 280w load will draw 280 watt hours from the battery bank over the period of 1 hour. (plus losses of course)

Tim
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Justme
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« Reply #12 on: January 29, 2009, 07:01:04 PM »

Not pointless at all.

So if I have this right

1, for mains ac a bad pf results in a higher bill due to being billed for the pf losses

but
 
2, for mains from a dc inverter a bad pf does not draw more power than a good pf

So if I have some bad pf items I am better of running them via an inverter than direct of the mains?

How much extra does a low pf cost to run on mains?

For example a 1000w item with a pf of 1 will cost x per hour
how much would 1000w cost with a pf of 0.5?

x? 1.1x?1.5x? 2x?


Justme
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30 x 58mm panel 259L TS
1200watts solar 120vdc
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« Reply #13 on: January 29, 2009, 07:57:08 PM »

Not pointless at all.

So if I have this right

1, for mains ac a bad pf results in a higher bill due to being billed for the pf losses

but
 
2, for mains from a dc inverter a bad pf does not draw more power than a good pf

So if I have some bad pf items I am better of running them via an inverter than direct of the mains?

How much extra does a low pf cost to run on mains?

For example a 1000w item with a pf of 1 will cost x per hour
how much would 1000w cost with a pf of 0.5?

I don't think domestic meters account for power factor, industrial ones certainly do as a bad pf (<0.95) can upset the balance of 3 phase distribution creating weird neutral currents as well as increasing losses.


Please don't ask for an explanation of unbalanced 3 phase power with different power factors - you will be in electrical engineeing territory and believe me the maths gets scary.   Grin
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w0067814
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« Reply #14 on: January 29, 2009, 08:43:06 PM »

Domestic bills are based purely on kilowatt-hours, thus the power factor of you house is not of concern from a billing point of view. IEE regulations have cables sufficiently over sized that the increased currents through a poor power factor are also not a problem.

Large companies that use a lot of power are concerned with power factor and they are charged on the kilowatt-hours* they draw and the power factor. Thus it is in their interests to ensure that the power factor is as close to 1 as it can be. Companies oftem have large capacitor banks to provide a phase lead in current equal to the phase lag in current from their inductive loads such as motors. As the two cancel each other out the extra line current flows between the capacitor bank and the motors totally on-site, and not from/to the grid.

Tim

* Actually I learned the other day that many companies are charged on the average kilowatts (not hours) drawn from three 30 minute periods over the year. These periods are chosen by the national grid and typically coincide with the highest national demand. Thus if companies have on site generators, running them at times of expected high load can reduce their usage from the grid in these 90 minutes (total) and thus save them a lot of money.

See http://en.wikipedia.org/wiki/National_Grid_UK#Triad_demand and http://www.nationmaster.com/encyclopedia/Control-of-the-National-Grid-(UK)#Triads
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