Do I have this right?

Re: Do I have this right?

goingtoghana wrote: »

Questions:

1) Series versus parallel on the panels. If I go series, I need smaller guage (cheaper) wire. If I go parallel, I have to buy larger guage (more expensive)wire. Either way, I have to buy a charge controller that can handle either maximum (voltage or amperage). So, does it really matter as long as I size the controller for the current and potential exapnsion of the system?

First, normally, most folks would earth ground the negative battery bank terminal. That means the Breaker in the solar wiring should be in the positive lead.

And, if you have a breaker, you really do not need the disconnect switch too.

If you have Vmp~96 volts (a ~140-150 Volt max input controller typically) and a 12 volt battery bank, you have to use an MPPT charge controller for efficient energy transfer from the solar array to the battery bank.

A PWM controller would be only 20% or so efficient (vs the ~95% efficient MPPT controller.

If you can only get/justify a PWM controller, then all solar panels would have to be in parallel (Vmp-array ~ 17.5 volts to recharge a 12 volt battery bank).

And note, placing a lot of batteries in parallel is not the ideal method of configuring a battery bank--But if you are limited in your sources of batteries (cannot find large capacity / high AH 12 volt battery instead), then you should wire them per this website. Otherwise, the batteries near the +/- bus connection will get charged/discharged more than the batteries at the other end of the string (current flow will favor the batteries with the lowest wiring resistance.

Choosing a charge controller... A ~60-90 Amp output MPPT charge controller is around $500 to $600 (in US). A PWM controller will be 1/3rd the cost.

So, you have a trade off of a lot of copper to parallel connected array with a "cheap" PWM controller, or using much less copper and an "expensive" MPPT charge controller.

In general, for long wiring runs (like your 15 meter run), you probably would be better off with a MPPT charge controller and your Vmp~96 volt array and smaller diameter copper wire needed.

You need to figure out the voltage drop of 15 meter (remember, round trip distance) wire at the Imp-array current--typically, we recommend around 1% to 3% maximum voltage drop for the array wiring.

I can do the voltage drop calculations for you in US Wire Gauge--Don't have a metric calculator handy (and I don't really know standard metric wire sizes--although I am sure I can find a list somewhere).

2) Load Dumps - It seems some controllers have dumps and some don't. I am not sure why. Are load dumps simply the power generated by the panels that is not being used by the batteries? And, if so, can I put an inverter off this dump to power, say a fan for the room?

There are two reasons for Dump Loads.

First, for wind turbines/water turbines, they always need to be attached the battery bank/load to keep them from overspeeding (unloaded turbines can quickly exceed safe RPM unless a brake or other means is used to prevent run-away). So, the "dump controller" + "dump load" is someplace to send excess charging current once the battery bank is full (i.e., turn on dump load at 14.8 volts, turn off dump load at 13.6 volts--repeat).

Dump loads are not needed for solar panels. The charge controller simply "turns off" the current from the solar array when the batteries are full.

The other use for a "dump load/controller" is if you have "excess charging current". You can, for example, once the battery bank is "full" from your solar panels, turn on a relay and start a well pump for irrigation and/or turn on an electric heater for hot water.

The issue with solar electric to heat hot water is that, for many systems, there is simply not enough excess electricity to heat enough hot water to be of much use.

And if you have enough excess power--Then you could over heat the water and have other issues.

3) Anything else I am missing in my diagram?

For every + wire/cable that leaves the + battery bus, there should be a fuse or circuit breaker to protect the wire leaving the battery bus against short circuits.

For example, in the US, typically a 14 AWG wire would have a ~15 amp fuse to protect the wire from overheating if shorted.

If you have a 20 amp load, you should have a x1.25 safety factor for wiring and breaker/fuses:

20 amp * 1.25 NEC safety factor = 25 amp minimum rated wiring + breaker/fuse

If you try to run 20 amps through a 20 amp fuse/breaker, they typically will "open" in minutes or hour (or so) of loading.... That is why a 20 amp circuit should have a 25+ amp fuse/breaker.

You do not show your loads at all--Each load should have its own fuse/breaker. And for some loads (like an AC inverter), you may need a DC disconnect switch too so you can turn off the DC power if the inverter is not needed.

And--Ideally, you should have a fuse/breaker on each parallel battery connection--But not many people do that. With lots of parallel batteries (where you really need the fuse per string), it is expensive and a lot of extra wiring work.

Blue see makes a really nice fuse assembly that can bolt to a battery terminal--But not sure you can easily/cost effectively get them:

http://bluesea.com/products/5191
Attachment not found.
-Bill

Re: Do I have this right?

goingtoghana wrote: »

Thanks so much. With the exception of the fuses on each battery, I have updated my diagram based upon your input.

In reality, for safety you only need a On/Off switch from the solar array to the input of the charge controller. A breaker/fuse is redundant and does not provide any additional protection.

Note: even an On/Off switch from the array to the controller is of questionable use... The handy-ness of being able to turn on/off the solar array (if ever needed) vs the possibility of somebody turning off the array and you not knowing that (and killing the battery bank if nobody is paying attention).

A couple of questions remain:

1) How do I know the amperage from the battery bank to the inverter? I know if is 12 Volts, but is the amperage the accumulation of the AmpHours of the batteries? I'm trying to understand what size fuse I need between batteries and inverter.

First, the "common" point for your battery connection -- The +/- connection points have very high current potential (1,000 amps or more into a dead short from the battery bank). So any connection from the + terminal shall have a fuse or breaker sized to the wire rating (i.e., a 12 AWG wire should be 20 amps fusing or less, if using the very conservative NEC requirements).

Wire Current Ampacities NEC Table 310-16

Other organizations have "their standards" which tend to allow higher current limits:

West Marine

For "low voltage" solar, we generally have very heavy wire vs current flow--This is to keep voltage drops "low" (i.e., a 12 volt system, about 0.5 volts drop max I recommend for loads, 1.0 volts for 24 volt, and 2.0 volts max for 48 volt system).

For battery charging, voltage drop can even be more important... Every 0.1 volt wire drop reduce battery charging voltage/current... For a 12 volt system, I would recommend a charge controller have no more than ~0.05 to 0.1 volt drop (maximum current flow). If you have 0.5 volt or more drop, it will take much longer to recharge your battery bank and you will waste many hours of "full sun" because of this.

Wiring and fusing batteries... This can be done several ways. One is to look at the maximum current from the battery bank that they can supply. A Flooded Cell Lead Acid battery can supply ~C/8 current continuously, and C/2.5 for short surges... You would want wire heavy enough not to overheat and/or too much voltage drop for this current range.

When you have 5 batteries in parallel--The will not evenly share current (not each carrying 1/5th of the current). In general, I would assume that when you have parallel batteries--2 batteries, design such that each battery will carry 100% of the load.

Three or more batteries, assume each battery will carry a maximum of 50% of the load... This will help reduce problems with wiring/overheated connections/etc. down the road.

So, say you have a 1,200 watt 12 volt AC Inverter (1,200 watts is the maximum I would recommend for a 12 volt system--~100 amp nominal current flow).

The branch circuit for such an inverter would need to be rated to carry:

• 1,200 Watt * 1/0.85 inverter eff * 1/10.5 battery cutoff * 1.25 NEC derating fuse/breaker/wiring = 168 amp minimum branch circuit.

So, if you have:

• 1 battery : 175 Amp or larger fuse/wiring rating
• 2 batteries : 175 Amp per parallel battery string
• 3 or more batteries: 168/2=84 amp ~100 amp breaker/fuse/wiring for each parallel battery

Yes, my examples above are conservative--And fusing/breakering each parallel battery string for 3 or more batteries is a common engineering requirement for commerial equipment safety--But not many people do this on home power systems as it is expensive to do.

And this is why I like to recommend getting larger AH batteries and getting back to 1 or 2 strings (maximum) instead.

But, not everyone can get large AH batteries for such an installation (minimizing parallel battery connections).

2) Can the breaker not act as a disconnect between the batteries and inverter just like the breaker between the panels and the charge controller?

Yep--Just make sure you get DC rated breakers... Many/most home AC main breakers are not rated to disconnect DC current (DC power is very good at sustaining Arcs and need heavy contacts/special designs to prevent the breaker/fuse from acting like a DC arc welder).

-Bill

Re: Do I have this right?

ggunn wrote: »

Right. Doubling the cross sectional area halves the resistance and the current, which lowers the power loss in the cable by a factor of eight. Is that right? Something smells funny...

OK ggunn... My snark detector is not always accurate...

But Power=V*I=I²R=V²/R

So--1/2 the resistance only reduces the power loss by 1/2

However, if you go from 12 volts to 48 volts, this reduces current by a factor of 4.

And the I²R loss (with same cable) is reduce by a factor of 16x (1/16th the loss).

Or, you can use a much smaller diameter cable to carry the power with "normal" losses (of ~3% maximum power loss, and ~2 volt (to 4 volt total) maximum wire drop at 48 volts.

-Bill :D