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From: drake.chamber...@redwoodalliance.org
To: RE-wrenches re-wrenches@lists.re-wrenches.org,
Date: 01/02/2015 05:18 PM
Subject:Re: [RE-wrenches] wire sizing
From:
drake.chamber...@redwoodalliance.org
To:
RE-wrenches re-wrenches@lists.re-wrenches.org,
Date:
01/02/2015 05:18 PM
Subject:
Re: [RE-wrenches] wire sizing question
Sent by:
RE-wrenches re-wrenches-boun...@lists.re-wrenches.org
4/0 copper is good for a 24V, 4000 W inverter. What is being
From: drake.chamber...@redwoodalliance.org
To: RE-wrenches re-wrenches@lists.re-wrenches.org,
Date: 01/02/2015 05:18 PM
Subject:Re: [RE-wrenches] wire sizing question
Sent by:RE-wrenches re-wrenches-boun...@lists.re-wrenches.org
Mark,
Dude, it's in Romania! The NEC doesn't apply. Many European
countries have different codes and standards, and in many cases
they are both simpler and equally safe.
Allan
4/0 copper is good for a 24V, 4000 W inverter. What is being used for
multiple inverters?
I've paralleled 4/0 into outback boxes using 2, 2 PVC conduits. It gets
pretty messy trying to put 4 sets of 4/0s into enclosures and battery
boxes (although I have worked on such systems). Does anyone have
Some other things: If you keep the conduit less than 24 long, which is
recommended anyway for main battery connections, the conduit fill
requirements don't apply. (see exception: 310.15(B)3a2)
Using table 310.15(B)16 yeilds 230 amps for 4/0 at 75 C. Then 240.6
lists 250 amp as the next
We just always use 4/0 X flex cable with a 250 amp breaker, and call it
good. (2/0 with a 175 A breaker)
If you want to make life really hard, you can look at the trip curve of
the breaker (Heinnemans have 3 different trip curves depending on the
model I believe) and then do all the normal NEC
Happy new year all.
I have a question on wire sizing for the following.
250 amp breaker, in a UL enclosure ( say midnite or outback etc)
conduit from battery box to enclosure
I'm wondering how folks size this for ampacity not voltage drop.
thanks
jay
peltz power
Jay,
Divide inverter capacity by LVD then divide by full load efficiency.
Example: 4000 Watt inverter shuts down at 42 volts, efficiency 85% =
4000/42 = 95A/.85 = 111 Amps. Then it depends on the insulation and
conductor type and the corrections for ambient temp.
How can you not consider
Here here always include surge and voltage drop.
On Jan 2, 2015 12:10 PM, Larry la...@starlightsolar.com wrote:
Jay,
Divide inverter capacity by LVD then divide by full load efficiency.
Example: 4000 Watt inverter shuts down at 42 volts, efficiency 85% =
4000/42 = 95A/.85 = 111 Amps. Then it
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