Wire size?

[PaulRainbow]

Except that as the battery charges the current will reduce. You may have 3% volts drop at the max charger output current but as this reduces so will the volts drop . By the time you have reached say 5 amps it will only be 1/5th of what it was at 25 amps. Approximately 0.086 volts. At lower currents it will be even less.
This is why the Ctek chargers will work even though the battery connecting cables are relatively small compared with what you calculate to be necessary if you base your wire size calculation on the max charger output.

The Ctek chargers are basically compact portable chargers for which having several metres of AWG 4 or AWG 6 as battery connections would be a complete nonsense. The cables would be bigger than the charger.

I thought I had already explained this.

Just because you have offered an explanation that does not mean that it's correct, it certainly doesn't mean that it's correct for all chargers in all installations. To even type that smacks of arrogance.

The Ctek charger is closer to 6% loss at max output.

Anyway, what has the Ctek charger got to do with anything ? It's just a red herring thrown in to justify suggestions of light weight cabling. We don't even know what the OP has.

I have thicker cables going to my Sterling charger than the Ctek and my cables are less than 1m long.

 

[PaulRainbow]

Excuse my complete ignorance - but given that there is a recommended size, and I understand why one should not use smaller cables, but is there anything wrong with using cables that are too big, other than its nonsense, they cost more, they might not fit in tight spaces and they are heavier? I have not looked at cable prices and maybe bigger ones are factorially higher, especially ones the terminal is crimped on - but anyway??

Jonathan

I can't think of a real downside to having cables that may be a little over spec Jonathan. Other than those that you suggest, none of which will actually harm nothing other than your wallet. Why spec a charging system that's borderline ?

 

[Neeves]

Paul,

That was, in total ignorance, my gut feel - but gut feel is hardly supportable (though often used as justification).

Many thanks

 

[lw395]

Except that as the battery charges the current will reduce. You may have 3% volts drop at the max charger output current but as this reduces so will the volts drop . By the time you have reached say 5 amps it will only be 1/5th of what it was at 25 amps. Approximately 0.086 volts. At lower currents it will be even less.
This is why the Ctek chargers will work even though the battery connecting cables are relatively small compared with what you calculate to be necessary if you base your wire size calculation on the max charger output.

The Ctek chargers are basically compact portable chargers for which having several metres of AWG 4 or AWG 6 as battery connections would be a complete nonsense. The cables would be bigger than the charger.

I thought I had already explained this.

All you've done is trot out some muddle-headed arm waving.
I there is significant voltage drop, the charger will go into voltage limiting much earlier.
It will take longer to charge the battery because the current is less.
Yes it will still get there given enough time.

If you don't want to put 25A into the battery, why have a 25A charger in the first place? (there may be valid answers to this, such as I want to run say a fridge off the charger at the same time of course)

Fundamentally long leads between charger and battery are a very bad idea.
Long thin leads an even worse idea.

You can get over this with remote voltage sensing, or some wizzy sensing of the cable resistance. I don't believe the chargers in question do that, but I'm happy to be corrected on this point.

 

[PaulRainbow]

Except that as the battery charges the current will reduce. You may have 3% volts drop at the max charger output current but as this reduces so will the volts drop . By the time you have reached say 5 amps it will only be 1/5th of what it was at 25 amps. Approximately 0.086 volts. At lower currents it will be even less.
This is why the Ctek chargers will work even though the battery connecting cables are relatively small compared with what you calculate to be necessary if you base your wire size calculation on the max charger output.

The Ctek chargers are basically compact portable chargers for which having several metres of AWG 4 or AWG 6 as battery connections would be a complete nonsense. The cables would be bigger than the charger.

I thought I had already explained this.

You offered your theory, which i don't agree with.

Whilst you're explaining things, can you explain why you think 5mm cable is OK for a 5m run of cable from a 25a charger, when you recently insisted that a 15a solar array had to have 20mm cable from the controller to the battery ?

http://www.ybw.com/forums/showthrea...sizing-for-solar-panels&p=6259858#post6259858

 

[GeorgeLlewellin]

I have now had time to read the many replies, thank you all. I answer here the questions raised by forumites about this particular installation.

The charger is for my friend’s boat which is a Southerly 110, 10.8m LOA and has a Yanmar 3YM30 diesel engine. The battery charger purchased is a “Dometic” Perfect charge, model MCA1225, 12v 25amp for 2 battery banks.
The battery banks are as follows.
Domestic bank: 2 * 2013 Elecsol Yellow Service EL115. Sealed wet lead acid & AGM Leisure. 115Ah
Engine battery: 1 * Varta 830. Professional Deep Cycle AGM 830 115 060. 115Ah

The primary reason for the question was to make sure the wires do not get overloaded.
The secondary reason is to make sure the charger senses the battery level without being unduly distorted by wire resistance. My friends want to use it continually sometimes, while living onboard in port for maybe a week or so.

I appreciate that the distance run is twice that of the length of cable, that is why I used the distance between charger and batteries, most of you know what you are commenting about and noticed that.

I am grateful for all responses and it just confirms in my own mind that I did the right thing to ask.

Many thanks George

 

[pvb]

The battery charger purchased is a “Dometic” Perfect charge, model MCA1225, 12v 25amp for 2 battery banks.
The battery banks are as follows.
Domestic bank: 2 * 2013 Elecsol Yellow Service EL115. Sealed wet lead acid & AGM Leisure. 115Ah
Engine battery: 1 * Varta 830. Professional Deep Cycle AGM 830 115 060. 115Ah

The domestic batteries are either wet lead acid or AGM, they can't be both. He needs to find out which and set the charger appropriately.

Using a good quality deep cycle AGM battery for engine starting is perhaps an odd choice.

 

[GeorgeLlewellin]

The domestic batteries are either wet lead acid or AGM, they can't be both. He needs to find out which and set the charger appropriately.

Using a good quality deep cycle AGM battery for engine starting is perhaps an odd choice.

I have quoted what was written on the side of the batteries, the choice also was not mine but that of the Northshore Yachts or the previous owners.

 

[PaulRainbow]

At 25a voltage drop will be :

6mm cable 5.92%
10mm cable 3.5%
16mm cable 2.25%

Example prices :

6mm £30
10mm £45
16mm £80

 

[GeorgeLlewellin]

At 25a voltage drop will be :

6mm cable 5.92%
10mm cable 3.5%
16mm cable 2.25%

Example prices :

6mm £30
10mm £45
16mm £80

Thanks Rainbow Marine, that really is a concise and definitive answer that I can show my friends. regards, George

 

[Centesimus]

The calculation if you’re still interested is:

R=V/I

V is the volt drop in volts.
I is the maximum current in Amps.
R is the resistance on ohms.
R for copper = (0.017 * Length in meters * 2)/Cross Sectional Area in mm2

Using your figures:
25A
5m
3% volt drop = 0.36V

R=0.36/25 = 0.0144 ohms

CSA = (0.017*5*2)/0.0144 = 11.8 mm2

 

[GeorgeLlewellin]

The calculation if you’re still interested is:
R=V/I
V is the volt drop in volts.
I is the maximum current in Amps.
R is the resistance on ohms.
R for copper = (0.017 * Length in meters * 2)/Cross Sectional Area in mm2

Using your figures:
25A
5m
3% volt drop = 0.36V

R=0.36/25 = 0.0144 ohms
CSA = (0.017*5*2)/0.0144 = 11.8 mm2

Thank you very much Centesimus, that is really what I was looking for in the first place. I must remember to be more specific with my questions in future. I can now create my spreadsheet, but I have learned a lot from this discussion.
Best regards, George

 

[lw395]

Bear in mind that 3% drop being acceptable is a commonly used rule of thumb for loads like lights and motors.
When you are charging a battery, dropping the voltage by e.g. 3% can drop the charging current by e.g. 30%, looking at some arbitrary points on the graph.

This kind of thing can add up to the battery never getting fully charged, if charging time is limited.

 

[pvb]

Bear in mind that 3% drop being acceptable is a commonly used rule of thumb for loads like lights and motors.
When you are charging a battery, dropping the voltage by e.g. 3% can drop the charging current by e.g. 30%, looking at some arbitrary points on the graph.

This kind of thing can add up to the battery never getting fully charged, if charging time is limited.

As the battery approaches full charge, the current will be much lower and so the voltage drop will be much lower too. However, I can't see any point in skimping on the cable size, which is maybe Sterling's view as well.

 

[lw395]

As the battery approaches full charge, the current will be much lower and so the voltage drop will be much lower too. However, I can't see any point in skimping on the cable size, which is maybe Sterling's view as well.

That's just more arm waving!
Take as an example, the op's batteries at 90% charge. 25A going in through zero resistance cables, that close to a C/10 charge current. My graph suggests this is the point where voltage limiting would start.
At the same state of charge, with 0.36V drop, the charger is voltage limiting at 14.4V, the battery is seeing 14.04V and taking closer to C/20 current, I interpolate this as about 30% less current going in.

Depending on exactly what algorithm the charger uses to detect 'end of charge', it may also go into float mode before the battery is fully charged.

 

[pvb]

Take as an example, the op's batteries at 90% charge. 25A going in through zero resistance cables,

You reckon 230Ah of batteries at 90% charge would accept a 25A charge current?

 

[lw395]

You reckon 230Ah of batteries at 90% charge would accept a 25A charge current?

It is a point on the standard C/10 V/Q curve.

But the point is, you need 14.4V right at the battery terminals, not inside the charger at the other end of 5m of wire. Actually I suspect the standard graph is multiplied up from the volts right on the plates of a cell.
In reality many chargers are not capable of giving their rated output into a real battery. But adding resistance is only going to make that worse.