question

The 1600W figure will be peak for 120 seconds.

Case use is for max 800W continuous, if you want it to last I wouldn't go over 0.75*800.

Some quick calcs for you.

1600W peak *1/0.92(worst efficiency)*1/11V(Vbatt under load)= 158A

So the internal 150A fuse is fine.

Now for your answer regarding external fuse and cable.

800W continuous which is all that invertor can supply for continuous load.

800W*1/0.92*1/11V=79A

79A*1.25 derating = 98A (most fuses are not designed to run constantly above 75-80% rated value)

25mm cable is rated @100A for continuous load.

To answer your next question why is this?

The Adiabatic equation as follows.

s = √(i²t)/k

Potential current could be 158A under peak load, lets say the invertor can supply this for 120 seconds at a push.

Your internal fuse is 150A fast blow and say a curve of i*1.6 to 2, so 120 seconds before it blows at 300A continuous load. So this fuse will never blow under normal operation maximum potential draw conditions, as it will supply the 1600W for 120 seconds without blowing.

So that's a current of 158A with a disconnection time of 120 seconds (until the invertor clips) and we will use a K factor of 115 for that cable.

answer=16mm square cable.

So your 25mm cable has been chosen to satisfy both the requirements.

Now lets get anal and look at potential short circuit current from your battery.

Lets say your battery can discharge 4.5kA (12V 200Ah) for 0.2 seconds before your fuse blows.

Now that's 4500A heading towards your invertor at full speed.

Lets do the Adiabatic equation for that scenario to see if your cable is going to be ok.

Fault current of 4500A disconnection time 0.2 seconds with a K factor of 115 for that copper cable.

answer=17.5m squared cable, next size up 25mm, so your still ok.

Regarding the external fuse you would need a 100A to fuse to protect the cable.

25mm cable = 100A

100A fuse is fine as most midi or mega fuses are 2-3*i so they won't blow during 158A surges for 120 seconds anyway.

Hope this clears up how Victron have come to this conclusion, although they may have used different calculations, but it all equates to the same result. The after 5m thing is more to do with voltage drop than anything.

Thanks you very much for your extremely comprehensive answer. Just what I was looking for.