Help ensure my inverter install looks good

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solarman said:
...................FYI.  you shouldn't discharge a flooded battery at more than C/4 rate and an AGM at more than C/2

so for T105 gives a max discharge of 56 Amps and 112 Amps for AGM,  that's  672 and 1344 Watts respectivley....................
Click to expand...
Solarman tossed in a term not everyone may be familiar with "C".  "C" stands for the number of amps that are charged or discharged from a battery based on the size of the battery. 

If a battery capacity is rated at 100AH (amp hours) then 1C would be charge/discharge rate of 100amps. 

As solarman stated, for a standard flooded cell lead acid battery it should not be charged/discharged, for a long duration, for more than 1/4C.
 
AStravelers said:
Drawing didn't get attached.  I see the pictures of the shunt. 

Looking at the shunt, the cable from the battery side of the shunt is going to be replaced with the new 2/0 or 4/0 cable, and the new 2/0 or 4/0 cable from the inverter will attach to the other side of the shunt, correct.

As long as there is only one cable going from the battery side of the shunt to the battery and that is the only cable that attaches to the negative battery post, the Trimetric should read all the current going to the battery. 

Now if the Trimetric plus and minus leads don't attach close to the battery the voltage reading on the Trimetric will be lower by whatever voltage loss is caused by the length of the cables going to battery.  Mainly that will be significant when under a heavy current draw. 

Personally I would relocate the shunt to within a foot or two of the battery.  All you need to do is run your new 2/0 or 4/0 black cable from the junction box to the battery and relocate your Trimetric sense cables to the shunt.
Click to expand...

Yea, looks likes easiest option is to move the shunt and install a new cable from the junction box to the battery.  I have attached the drawing again anyway.
 

Attachments

  • Trimetric 2030 Wiring.pdf
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Dooger54 said:
Yea, looks likes easiest option is to move the shunt and install a new cable from the junction box to the battery.  I have attached the drawing again anyway.
Click to expand...
You are going to run a new cable to the battery anyways, so it shouldn't be so bad. 

I downloaded the drawing and will take a look.
 
Adding two more batteries to the trailer would be best to run the hair dryer and microwave. 

However can your rig take another 120-160 pounds on the tongue and the hitch of your tow vehicle? 

On my former RV, a Class C MH, I only had room for 2 batteries so only installed a 1000 watt inverter and didn't try to run the microwave.
 
AStravelers said:
Adding two more batteries to the trailer would be best to run the hair dryer and microwave. 

However can your rig take another 120-160 pounds on the tongue and the hitch of your tow vehicle? 

On my former RV, a Class C MH, I only had room for 2 batteries so only installed a 1000 watt inverter and didn't try to run the microwave.
Click to expand...

No, my rig cannot take the additional weight in front.  I would have to move my spare tire to the back and haven't quite figured out where I would put it!
 
Your diagram makes sense, nice job of drawing it up.

Back to the location of the shunt.  One big reason I would move the shunt to near the battery is to always be able know exactly what the battery voltage is.  While that is not critical, if there is any question in your mind about the operation, i.e. is the battery not charging as quickly as you think it should, charge voltage is not as high as you think it should be, having the Trimetric ALWAYS showing the exact status of the battery in nice. 

You will also be able to verify the validity of any information you have gleamed from your research and the info provided by various folks on the forum.  Especially early in your use of the batteries you will know for sure how your system is preforming. 

Even today after having used a Trimetric for about 11 years I still like to glance at the readings to be sure the system is preforming as I think it should.
 
AStravelers said:
Your diagram makes sense, nice job of drawing it up.

Back to the location of the shunt.  One big reason I would move the shunt to near the battery is to always be able know exactly what the battery voltage is.  While that is not critical, if there is any question in your mind about the operation, i.e. is the battery not charging as quickly as you think it should, charge voltage is not as high as you think it should be, having the Trimetric ALWAYS showing the exact status of the battery in nice. 

You will also be able to verify the validity of any information you have gleamed from your research and the info provided by various folks on the forum.  Especially early in your use of the batteries you will know for sure how your system is preforming. 

Even today after having used a Trimetric for about 11 years I still like to glance at the readings to be sure the system is preforming as I think it should.
Click to expand...

Funny you should mention the battery voltage and how it?s measured. That?s exactly the problem I was having and what prompted me to draw up the diagram.  So I will move the shunt closer. I just measured and the current negative cable to the shunt is just short of 4?.

Back to the inverter issue.  Is there any reason to go with the 1000w inverter vs. the 2000w if the price is only about $100 less?  I?d prefer to buy the 2000 and see how it works.  If I have to I just won?t use the higher wattage appliances.  I realize I would need to be careful about how far I draw down the batteries.
 
AStravelers said:
I am unclear about your internal cell loss coming up with 2.2V loss. 

It seems like if I have 4 six volt batteries (12 cells) or 8 batteries (24 cells) I would have even more voltage

loss than 2.2 volts.

Using your formula:  ( 0.002 Ohms * 24 Cells ) *  ( 2200 / 12 ) = 8.8 Volts drop and my 12.55V battery pack is now

under 4 volts.  Yet with over 800AH of battery, a 2200 watt (180amp or 1/4C) load is not that significant.

Can you explain what I am missing? 

Also in your example of the end voltage being 10.55V would is that measured at the battery posts?

Is this formula a way to calculate the battery voltage sag under a load?  That is when you have high current draw

from a battery the voltage drops some amount.
Click to expand...


yes, the missing part is parallel resistances..



It's Ohms law at work, in this case Voltage = Current in Amps x resistance in Ohms

if we consider the internal resistance of the cell to be a perfect resistor then the voltage change across the cell
under load is directly proportional to the current.

In the case of the T105, a single cell is approx 0.002 Ohms. if we draw say 100 Amps then the voltage across the

cell will change by 0.002 Ohms * 100 Amps = 0.2 Volts this is subtracted from the cell voltage under no load.

a T105 has 3 cells in series for 6V and so for 12 V we have six cells. each voltage drop from each cell is

additive, so in this example we would have 0.2 V per cell * 6 cells = 1.2 V and our battery voltage
would drop from 12.75 V to 11.55 V


In your example of four batteries, I will assume they are connected as a 2S2P string. so you have 12 Volts and

twice the capacity. the missing piece here is the fact that the two batteries in parallel now have their internal resistance connected in parallel too. ( i'm going to ignore cable resistance for this example )


so...

Parallel resistors: this is where the fun starts..

in plain language: the value of parallel resistors is the reciprocal of the sum of the reciprocals !

whats a reciprocal ? well it's simply 1/R so for a 10 Ohm resistor we get 1/10 = 0.1
get it ?.. good, so now lets have two of these in parallel 1/10 + 1/10 = 0.2
and the reciprocal of that is 1/0.2 = 5 Ohms. !!

we could write it as R = 1 / ( 1/R1 + 1/R2 )

for three 10 Ohms in parallel we get R = 1 / ( 1/10 + 1/10 + 1/10 ) which equals 3.33333333 Ohms


so back to your battery.. we have 2 in parallel and 2 in series so we have 6 cells in series that
gives us 6 * 0.002 = 0.012 Ohms and we have two of these "resistors" in parallel..

so our apparent total internal resistance is now 1 / ( 1/0.012 + 1/0.012 ) = 0.006 Ohms

if we draw 100 Amps from this pack we have 0.006 Ohms * 100 Amps = 0.6 Volts

so 12.75 V - 0.6 V = 12.15 V  this of course is measured at the terminals...

if we had this bank connected to the OP's inverter at 2000W load we would see a voltage drop of
0.006 Ohms * 183 Amps = 1.098 Volts  and at the terminals: 12.75 - 1.098 = 11.652 V


to answer your question, yes, we can calculate the voltage drop based on amperage drawn using the internal resistance if we know it.. we can also calculate the internal resistance if we measure the voltage drop given a known amperage load.. ( approx, it's not exact.. )


has this explanation answered your question ?

 
At somewhere over 200 amps it shows estimated time at about 50-60 minutes.
Click to expand...
 

I'm not sure what graph you are referring to, but no way will a pair of T105's deliver 200A for 60 minutes. They spec a single one for 115 minutes @ 75 amps, but even that has to be a new battery in the lab.  I've owned my share of them and never got that good a result.
http://www.trojanbattery.com/pdf/datasheets/T105_Trojan_Data_Sheets.pdf

A few other observations:
Reducing the size of the inverter isn't really any help, except for inherently limiting the max load.  Keeping the wattage down is what is needed (or more batteries).

I believe the voltage loss formula that solarman used is for cells in series, i.e. a pair of series 6v's has 6 cells in series to produce 12v. Cells in parallel do  not contribute additional resistance.
 
Since I don?t have the weight capacity in front of my trailer to add another set of batteries (120lbs) I guess I will just go with the 1000w inverter and limit the use to lower wattage appliances.  I also won?t add in the microwave circuit.  I can scale back some on the size of the battery cables too.

Thanks all for your help!!
 
AStravelers said:
I don't see a benefit to going to a 1000W inverter.  It won't power the microwave.  Even the 1500W is marginal for the microwave.  It is kind of like "I can always drive my car/truck at max power and max braking".  Yes you can, but it may not last as long.

I remember you have other topics on this forum, but I don't remember the details so I may be asking questions you have already covered.

To charge with your generator you MUST have a 3 stage charger and it must be mounted close to the batteries.  Voltage drop on long wires is significat.  Trojan wants about 14.7 volts for absorb charging, any thing else increases your generator run time.  You also want to be sure your charger has an equalize option.  You may want to go through an equalize cycle every month or two or every year, depending on how much you use your batteries and get them to 100% full.

If you discharge your batteries 40% (60% full) and run your generator about 2 hours with a 3 stage charger, you will get to around 90% full(maybe 85%).  The last 10%-15% may take 4-8 hours (or more) to get to 100% full. 

If you don't bring your batteries to 100% every few days to a week, you shorten the life of the batteries.  You may not be boondocking or dry camping that long so you could be OK.
Click to expand...

Al-

I discussed the option of buying an Iota charger to run off my generator in this post-

http://www.rvforum.net/SMF_forum/index.php/topic,114429.new.html#new

Gary didn?t seem to feel there would be much benefit using the Iota vs. my current Boondocker charger. 
 
solarman said:
yes, the missing part is parallel resistances..



It's Ohms law at work, in this case Voltage = Current in Amps x resistance in Ohms

if we consider the internal resistance of the cell to be a perfect resistor then the voltage change across the cell
under load is directly proportional to the current.

In the case of the T105, a single cell is approx 0.002 Ohms. if we draw say 100 Amps then the voltage across the

cell will change by 0.002 Ohms * 100 Amps = 0.2 Volts this is subtracted from the cell voltage under no load.

a T105 has 3 cells in series for 6V and so for 12 V we have six cells. each voltage drop from each cell is

additive, so in this example we would have 0.2 V per cell * 6 cells = 1.2 V and our battery voltage
would drop from 12.75 V to 11.55 V


In your example of four batteries, I will assume they are connected as a 2S2P string. so you have 12 Volts and

twice the capacity. the missing piece here is the fact that the two batteries in parallel now have their internal resistance connected in parallel too. ( i'm going to ignore cable resistance for this example )


so...

Parallel resistors: this is where the fun starts..

in plain language: the value of parallel resistors is the reciprocal of the sum of the reciprocals !

whats a reciprocal ? well it's simply 1/R so for a 10 Ohm resistor we get 1/10 = 0.1
get it ?.. good, so now lets have two of these in parallel 1/10 + 1/10 = 0.2
and the reciprocal of that is 1/0.2 = 5 Ohms. !!

we could write it as R = 1 / ( 1/R1 + 1/R2 )

for three 10 Ohms in parallel we get R = 1 / ( 1/10 + 1/10 + 1/10 ) which equals 3.33333333 Ohms


so back to your battery.. we have 2 in parallel and 2 in series so we have 6 cells in series that
gives us 6 * 0.002 = 0.012 Ohms and we have two of these "resistors" in parallel..

so our apparent total internal resistance is now 1 / ( 1/0.012 + 1/0.012 ) = 0.006 Ohms

if we draw 100 Amps from this pack we have 0.006 Ohms * 100 Amps = 0.6 Volts

so 12.75 V - 0.6 V = 12.15 V  this of course is measured at the terminals...

if we had this bank connected to the OP's inverter at 2000W load we would see a voltage drop of
0.006 Ohms * 183 Amps = 1.098 Volts  and at the terminals: 12.75 - 1.098 = 11.652 V


to answer your question, yes, we can calculate the voltage drop based on amperage drawn using the internal resistance if we know it.. we can also calculate the internal resistance if we measure the voltage drop given a known amperage load.. ( approx, it's not exact.. )


has this explanation answered your question ?
Click to expand...

Got it. Wasn't thinking about the parallel -- series relationship when dealing with the internal battery cell resistance.

Bottom line, to net it all out and put it simply:  When you put Lead Acid batteries under a heavy load, such as 180 amps for a 225AH battery you can expect the voltage under load to drop to 11V or less.    Or to use the term "C" brought up earlier.  Putting a lead acid battery under a load of 0.8C that will drop the voltage, under load, to 11V or less.

This is one of the advantages of using Lithium batteries.  You don't have this kind of voltage sag, under load, and you can charge at 1C or better all the way to 90-100%. Other advantages as well.  They are expensive.  Unless you are planning on doing a lot of dry camping or boondocking they are not worth the cost.  We have used our lithium batteries for 2 years now and love them. 

 
Dooger54 said:
Al-

I discussed the option of buying an Iota charger to run off my generator in this post-

http://www.rvforum.net/SMF_forum/index.php/topic,114429.new.html#new

Gary didn?t seem to feel there would be much benefit using the Iota vs. my current Boondocker charger.
Click to expand...
Your current Boondocker converter/charger is rated at 55amps I believe.  As long as it is a 4 stage charger, the charger will work fine.  HOWEVER, if I remember correctly, the Boondocker is located 25 or more feet from the batteries and attached with #4 wire and maybe using the frame for ground return.  It this is accurate you will not get a quick charge to your batteries because of the voltage loss in that long relatively light wire. The wire size calculator http://nooutage.com/vdrop.htm shows a 6% voltage drop with that long wire. Probably more voltage loss if the return path is the frame ground.  If you could relocate the Boondocker to near the battery and use copper on both + & - wires, it would work much better. 

If it is easier, buy the Iota and mount it near the batteries. 

One additional word of caution.  When you wire the inverter in, make sure neither the Boondocker nor the Iota sees the 120V from the inverter.  If you do this, the inverter will power the charger(s) which charges the batteries, which are feeding the inverter, which is feeding the charger.  This loop will kill your batteries pretty quickly. 

I did this on an earlier RV.  I quickly noticed the current draw on my Trimetric and figured out I was in the charger, inverter loop.  My inverter was a Magnum 1000W inverter with a charger, so it was simple to just disconnect the unneeded converter.
 
I think you are over-estimating the voltage drop, but obviously it depends on what parameters you put in the calculator.  The chargers are pushing at most 45-50 amps through 6 gauge wire at about 14v, so I'm confident that voltage drop is under 3%.  In any case, it's easy enough to check with a voltmeter at the batteries.  Hopefully it will stay above 14v during the bulk charge phase.
 
Gary RV_Wizard said:
I think you are over-estimating the voltage drop, but obviously it depends on what parameters you put in the calculator.  The chargers are pushing at most 45-50 amps through 6 gauge wire at about 14v, so I'm confident that voltage drop is under 3%.  In any case, it's easy enough to check with a voltmeter at the batteries.  Hopefully it will stay above 14v during the bulk charge phase.
Click to expand...

Highest I have seen at the batteries using a multimeter was 14.1. Trimetric read 14.15 and 6.8 amps.  Amps started dropping almost immediately and where down to 4.9 in about two minutes. I ran this test after I had "drained" the batteries down about 20-30%.  The Boondocker is rated for something like 14.6-14.7 in bulk mode.
 
Gary RV_Wizard said:
I think you are over-estimating the voltage drop, but obviously it depends on what parameters you put in the calculator.  The chargers are pushing at most 45-50 amps through 6 gauge wire at about 14v, so I'm confident that voltage drop is under 3%.  In any case, it's easy enough to check with a voltmeter at the batteries.  Hopefully it will stay above 14v during the bulk charge phase.
Click to expand...
I gave a link to the Wire size calculator, or use a calculator of your choice and verify.  Let us know what you find. Note, that some calculators don't automatically calculate for the round trip distance.  If so the wire length you enter becomes 50', not the 25'. 
 
AStravelers said:
Got it. Wasn't thinking about the parallel -- series relationship when dealing with the internal battery cell resistance.

Bottom line, to net it all out and put it simply:  When you put Lead Acid batteries under a heavy load, such as 180 amps for a 225AH battery you can expect the voltage under load to drop to 11V or less.    Or to use the term "C" brought up earlier.  Putting a lead acid battery under a load of 0.8C that will drop the voltage, under load, to 11V or less.

This is one of the advantages of using Lithium batteries.  You don't have this kind of voltage sag, under load, and you can charge at 1C or better all the way to 90-100%. Other advantages as well.  They are expensive.  Unless you are planning on doing a lot of dry camping or boondocking they are not worth the cost.  We have used our lithium batteries for 2 years now and love them.
Click to expand...


for those that stay at 12 V then AGM is a good for C/2 and a lot less expensive than LFP

the main issue with this topic is that high currents and 12 Volts don't really go well together.. that's why
in off grid we always go for high voltage and low amps.. it's easier all round on cabling and converters.

volts thrill, current kills.. LOL

I would only consider LFP for weight or space constraints.
in my case.. both !



 
Hi Please forgive me - NEWBIE ALERT - joined the forum in 2012 but don?t use them (reasons too boring).
Just wanted to mention I have recently installed a 500AH Elite lithium system in our 5th wheel that has 6 solar panels so if anyone was considering doing the same and wants to read the story, here?s a link if I?m allowed to put it here?!  Sorry if I?m breaking any rules, or off-topic, I just hope it might be of help to people!  http://www.kimartanddesign.com/Daves/JourneyToLithium.htm
(And I?m certainly no expert and can?t claim to know all about it!)
 
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