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How do you wire up a big inverter?


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Hi, I had to start this thread again with a new subject.

I now have an inverter and the suppliers tell me that I wire the incoming mains DIRECTLY into it, and then to the distrubution board. So as the solar and battery drop away the mains takes over again.

I will get an electrician to do this part, but I just want to make sure that this is correct?
ie Mains into inverter to distruibution board?

I have found this diagram and this is the inverter:

https://www.photonicuniverse.com/en/catalog/full/449-Iconica-5000W-48V-hybrid-pure-sine-wave-inverter-with-4000W-solar-input-80A-500V-MPPT-solar-controller-60A-mains-battery-charger-and-inbuilt-Bluetooth-No-battery-required.html

How-to-Wire-UPS-Inverter-to-the-Home-Supply.png

1544014514_60457300.png

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5 minutes ago, NotJustin said:

The manual only refers to the wires in and out. Not how it fits into the house wiring.


Ok so this is top of page 7 - and I’ve only skim read the few pages

 

CAUTION!! Before connecting to AC input power source, please install a separate AC breaker between inverter and AC input power source. This will ensure the inverter can be securely disconnected during maintenance and fully protected from over current of AC input. The recommended spec of AC breaker is 16A for 1.5KW and 32A for 3KW and 50A for 5KW.


CAUTION!! There are two terminal blocks with “IN” and “OUT” markings. Please do NOT mis-connect input and output connectors.


WARNING! All wiring must be performed by a qualified personnel.


So you need to install a breaker between the inverter and the supply. Not rocket science. 
 

But the last point is key - if you’re asking those sorts of questions, you’re not qualified to install this.

 

Electricity doesn’t have a smell, or make your socks wet when you get it wrong, it just kills - instantly.
 

And you’re talking about wiring where then only thing between you and the grid is a 100A supply fuse that will blow long after you’ve become a crisp. 

 


 

 

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Sorry but get the electrician to read the manual - if they screw up then it’s on them and if they don’t know then they ask the supplier. If you advise wrongly and they blow up your £800 inverter then you have no-one to come back on but yourself. 

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I was qualified to fit inverters, really quite simple for a qualified electrician, most will have covered in when they are updating their Part P.

If the electrician does not under stand it, find another one.

 

28 minutes ago, PeterW said:

only thing between you and the grid is a 100A supply fuse that will blow long after you’ve become a crisp. 

And it can draw a lot more current than that for a second or two before it blows.

So let us say it draws 300 A at 230V for 2 seconds, that is the same as a 14 tonne weight resting on your body.

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18 minutes ago, SuperJohnG said:

I'm genuinely interested in how you did this calculation/ know this.

Amps [A] times Volts [V] equals watts [W]

 

A watt is a joule [J] per second.  A joule is the unit of energy.

 

So 300 [A] x 230 [V] = 69,000 [W]

 

I said that it would run for 2 seconds

 

69,000 [W] x 2 = 138,000 [Ws or J]

 

As a joule is the the force [mass x acceleration] to move 1 kg 1 metre, I used the force of gravity, 9.81 m.s2

 

so (and I think I did this step right, but starting to doubt myself now)

 

138,000 [J] / 9.81 [m.s2] = 14,067 kg.

Edited by SteamyTea
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2 hours ago, SteamyTea said:

Amps [A] times Volts [V] equals watts [W]

 

A watt is a joule [J] per second.  A joule is the unit of energy.

 

So 300 [A] x 230 [V] = 69,000 [W]

 

I said that it would run for 2 seconds

 

69,000 [W] x 2 = 138,000 [Ws or J]

 

As a joule is the the force [mass x acceleration] to move 1 kg 1 metre, I used the force of gravity, 9.81 m.s2

 

so (and I think I did this step right, but starting to doubt myself now)

 

138,000 [J] / 9.81 [m.s2] = 14,067 kg.

Looks not bad to me...if I jave a spare 5 mins tomorrow I may just run it through mathcad. Its definitely a good analogy ? 

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