Preventing current in specific cable

[Bob666]

Hi!

I've asked fuzzy questions about the electrical system before, so this is a similar question, but having read up a bit, I will try to be exact and precise and cut out the irrelevant.

Device A consumes 90 A.
Battery B provides electricity. 105 Ah.
Cables connects A to B. The plus cable via fuse C.
Cables connects DC Source D to battery B.

Cannot simplify any further

A visual drawing:
(See attachment)

Questions:
1. Must there be a fuse between Battery B and DC source D? Source D is a 50 A alternator.
2. When device A consumes 90 A. Will it take all current from battery B or, if DC Source D is providing 50 A, also from DC source D?
3. How can I extend this diagram to make sure Device A takes all or nearly all current from battery B? Does it exist a device to put on the plus cable between DC Source D and Battery B?

(That should suffice, but I will add some info that might be relevant:
The cable run from DC Source D and Battery B is 10 m one-way. If I buy 10 mm2 cable, I need 20 m, plus and minus. The max fuse on that cable is 60 A, which is 10 A lower than the max of the DC Source D.

However, I do not want to run into a situation where Device A decides to get all 90 A over the 10 mm2 cable, and get a blown fuse. But why would it ever, when the voltage drop is much lower from Battery B, and Battery B is providing the power it needs?

The cable between Battery B and Device A is 2 m one-way and 35 mm2).

Cheers

[DeepFrz]

1. Yes
2. 90 amps for how long? Yes, you figure it out.
3. Not that I know of.

Relationship between Voltage Current and Resistance

[jeepbluetj]

Originally Posted by Bob666

Questions:
1. Must there be a fuse between Battery B and DC source D? Source D is a 50 A alternator.
2. When device A consumes 90 A. Will it take all current from battery B or, if DC Source D is providing 50 A, also from DC source D?
3. How can I extend this diagram to make sure Device A takes all or nearly all current from battery B? Does it exist a device to put on the plus cable between DC Source D and Battery B?

1. Yes. You actually need a fuse on both ends, as you have two sources of current. (Typically this can be handled by feeding through a big fuse on the battery itself)

2. It's a split, depending on the resistance of the wires and voltage of each source. The alt won't produce more than 50A because it can't.

3. You can't, nor do I know why you would wish to.

[smac999]

A yes,. you should have a 75a fuse for a 50a alt. close to the battery. you do not need one on the alt side the above post is wrong. Fuse C should be over 100a.

2 it will take 50a from the alt and 40a from the battery. if all cable is good. if cable is long / bad / small it will vary. and may take more from battery and less from alt. it will never take more then 50a from alt if it's a 50a alt. however it sounds like you are sizing cable for max load and not voltage drop which you need to do. if your 50a alternator is 10m away it should be sized for 3% drop which is 50mm2 cable.

3 you can't. it'll take up to 50a from alt and that's the way you want it.


also keep in mind that 105ah battery will power a 90a load for about 20mins before dying. and about 40mins while running the engine. so hopefully you are not running it long... (windlass?)

[ramblinrod]

Originally Posted by Bob666

Hi!

I've asked fuzzy questions about the electrical system before, so this is a similar question, but having read up a bit, I will try to be exact and precise and cut out the irrelevant.

Device A consumes 90 A.
Battery B provides electricity. 105 Ah.
Cables connects A to B. The plus cable via fuse C.
Cables connects DC Source D to battery B.

Cannot simplify any further

A visual drawing:
(See attachment)

Questions:
1. Must there be a fuse between Battery B and DC source D? Source D is a 50 A alternator.
2. When device A consumes 90 A. Will it take all current from battery B or, if DC Source D is providing 50 A, also from DC source D?
3. How can I extend this diagram to make sure Device A takes all or nearly all current from battery B? Does it exist a device to put on the plus cable between DC Source D and Battery B?

(That should suffice, but I will add some info that might be relevant:
The cable run from DC Source D and Battery B is 10 m one-way. If I buy 10 mm2 cable, I need 20 m, plus and minus. The max fuse on that cable is 60 A, which is 10 A lower than the max of the DC Source D.

However, I do not want to run into a situation where Device A decides to get all 90 A over the 10 mm2 cable, and get a blown fuse. But why would it ever, when the voltage drop is much lower from Battery B, and Battery B is providing the power it needs?

The cable between Battery B and Device A is 2 m one-way and 35 mm2).

Cheers

These questions indicate a very limited understanding of marine electrical systems, basic electrical theory, and good electrical installation practice.

Before you burn your boat to the waterline, familiarize yourself with ABYC E11 or seek a professional, to avoid responsibility for setting all the boats in the marina on fire.

Mucking with electrical systems you don't understand to save some time or a couple hundred bucks, is false savings should anything go wrong (and it most likely will, based on your questions, and at least half the responses I expect you'll receive).

[Bob666]

Originally Posted by ramblinrod

These questions indicate a very limited understanding of marine electrical systems, basic electrical theory, and good electrical installation practice.

Before you burn your boat to the waterline, familiarize yourself with ABYC E11 or seek a professional, to avoid responsibility for setting all the boats in the marina on fire.

Mucking with electrical systems you don't understand to save some time or a couple hundred bucks, is false savings should anything go wrong (and it most likely will, based on your questions, and at least half the responses I expect you'll receive).

Hi!

Please skip ahead to BOLD face. Or read my reply to this as well.

A bit boring to get that kind of answer. I don't work with it no, it's my hobby. But it's quite logical and I understand P=U*I etc. I scored pretty good in physics in School. But, that is different from pragmatic hands-on experience, I know. Still, many knowledgeable people state that electricity cannot be comprehended, unless you're Einstein, or maybe Maxwell.

I try to think of it like water, with pressure in hoses etc. So you're right about following standards, it's a good way to not start a fire, if you (like most people do not fully do) do not understand electricity.

I ask simple questions to get the simple building blocks in place. I'm +40 now and often I used to know, but forgot as years have passed since I worked with similar stuff. Different for somebody like you working with it all the time.

So, asking a simple question revives the memory a bit and I understand again.

I have fused everything, it's a promise I tried to simplify my question and left out things on purpose. There are main switches on the battery banks etc.

We don't have ABYC in Europe. In Sweden the so called 12 V gurus are against the U.K / U.S. way of having two banks = Start + Leisure banks and the 1-2-ALL switch. They think there should be only one bank and a battery brain that disconnects if voltage too low.

I try to be open-minded. What you use over there seems more clever perhaps.

Another possibly stupid question, perhaps, but I need to know:

If I measure the voltage at the Device A when DC source D is off.
With the fat cable between Battery B and Device A, there will be hardly any voltage drop and everything will be PERFECT.

By adding a DC Source D and a cable between DC Source D and Battery B things get pretty messed up (read: not the way I want them to be). Unfortunately. But it's nature's law.

Please explain this: will the voltage measured at the windlass be less than what is measured when DC Source D is connected? If so, I appreciate that it will run worse when DC Source D provides 50 A along with 40 A from the battery. If not, I don't understand, sorry.

Explain, with theoretical physics terms please, what happens when I use a 10 mm2, or perhaps 16 mm2 cable between DC Source D and Battery B. As I cannot afford 50 mm2 so I will probably not buy that kind of cable.

And yes, Device D is a windlass.

DC Source D is behind a Sterling alternator to battery charger 80 A. The START output of the Sterling goes to Battery B. The Sterling will first charge Battery B on its START output fully, before giving priority to the house bank.

The START output of the Sterling will be shut off completely when engine is not running. It's like the alternator is not connected.

I'd really appreciate if somebody who understand's Ohms law etc could explain this to me (rather than telling me that I should sell the boat and give up sailing, btw I am not rich and cannot use the ultra-expensive yards to help out + 2/3 yards in the vicinity don't understand Ohm's law so I stand better chances of sorting the electrical system myself + previous owner was an electrician and messed things up pretty badly. ).

As a side note, by taking the engineering approach to e.g. changing a window in the house. I did a better job than the local carpenter, because it is not HIS HOUSE. (Whereas I put my soul and heart into the job). And good luck trying to find a painterman who can paint your house using a brush, by hand, thin strokes, and with linseed oil!.

2. Battery B is at the Bow. If I'd like to have it as an emergency starting battery. Then I should go up in mm2. With the Sterling and all. I'd like to, in the most smart way, accomplish this, I'd like a switch, something like 1-2-ALL. I'll think this through properly when I get time, perhaps somebody has some tip in the meantime

Cheers

[Bob666]

Sorry for getting into a pissing contest.

Previous post a bit hard to read.

I am looking for an answer to:

Scenario 1:
Battery B connects to Device A with negligible voltage drop.
Device A started, 90 A consumed continously (theoretically, simplifies the example and perfectly in order here).
At time 1 s after start, 30 s after start and 60 s after start. What voltage can be measured at Device A side?

Scenario 2:
Battery B connects to Device A.
Battery B connected to DC Charging Source D with large voltage drop, but large enough cable to avoid burning the fuse (or the cable).
At time 1 s after start, 30 s after start and 60 s after start. What voltage can be measured at Device A side?

Will the voltage measured differ at Device A, at 1 s, 30 s and 60 s, in Scenario 1 vs. 2, please?

If I want to start my engine with Battery B, what is the acceptable voltage drop max for the starting motor? It's a Volvo Penta 2003B from 1994. 3-cyl, 29 hp.

I am currently researching a solution with minimum fancy electronics (already got enough: Sterling etc) involved.

Moreover, the Sterling START will never charge the windlass battery fully as it is not voltage compensated and adjustable to battery chemistry etc (its output to house bank is fully regulated in all the ways). I am accepting this, as I am dedicating a 50 W solar panel via its own high-end solar regulator to that battery only, will put everything close to the battery with adequate mm2 wiring. So the battery will eventually get topped-up.

Have a nice day!

[fallingeggs]

On the alternator fuse question, let me quote Maine Sail...

Originally Posted by Maine Sail

The alternator, solar is considered a current limited source and should not ever be able to ignite the wire, if properly sized. There is no failure mode of an alternator than can cause it to deliver more than its output. The wire should be comfortably sized to handle its full output for hours on end. Even if the alternator developed a dead short internally the battery bank is driving the dangerous current into that short, not the alternator, so the fuse goes as close to the battery as possible..

The battery bank, on the other hand, can provide in excess of 20,000A into a dead short. A single Odyssey Group 31 battery can deliver 5000A of short circuit current... This is why OCP is installed at the battery end or whenever a drop in wire gauge occurs. The battery bank is your dangerous current source that needs the OCP not the alternator, charger or solar system....

http://www.cruisersforum.com/forums/...ml#post1747312

If the battery wire and the alternator wire are sized correctly for the current they will carry and both are the same size, one should be able to fuse once at the battery.

On that note, that Fuse C should cover both the alternator and the windlass. If the windlass drops a wire size, you'll need another fuse behind Fuse C specifically for that windlass wire.

[Cap Erict3]

These links might be of help..


http://dl.kashti.ir/ENBOOKS/The%20Ma...ble%201993.pdf

https://www.uscgboating.org/regulati...ELECTRICAL.pdf

[smac999]

Originally Posted by Bob666

Please explain this: will the voltage measured at the windlass be less than what is measured when DC Source D is connected? s

or course not. the voltage will be higher with the engine running. but with 10mm cable not by much. with bigger cable it'll be higher.

to answer your other question. in a totally thery way as it totally denpands on bank size and condition.

1s, 30s 60s meaused at A

b only. 12.6v, 12.4v, 12,2v : total battery drain from B ~2% : amps from D 0a

D on, 10mm cable. 12.8v, 12.6v, 12.4v : total battery drain from B~1.5% :amps from D 25a

D on, 50mm cable. 13v,12.9v, 12.7v. : total battery drain from B ~1% amps from D 50a

your engine would have battery B back fully charged in a few mins. (you took 90a out for 1 min and can put 50a back into it for 2mins once windlass stopped)


120a source D with 50mm 14v, 14v, 14v. no dischange of battery B, all power from alternator : amps from D 90a


to start your engine from 10m away you'd probablay want bigger cable then 50mm. though 50 might due

so you don't have a start battery? you have a house bank and a windlass bank?

[Bob666]

MIGHTY interesting!

Originally Posted by smac999

or course not. the voltage will be higher with the engine running.

Yes of course, what was I thinking.

Originally Posted by smac999

but with 10mm cable not by much. with bigger cable it'll be higher.

to answer your other question. in a totally thery way as it totally denpands on bank size and condition.

1s, 30s 60s meaused at A

b only. 12.6v, 12.4v, 12,2v : total battery drain from B ~2% : amps from D 0a

D on, 10mm cable. 12.8v, 12.6v, 12.4v : total battery drain from B~1.5% :amps from D 25a

D on, 50mm cable. 13v,12.9v, 12.7v. : total battery drain from B ~1% amps from D 50a

I don't mind more battery drain, it's a 105 Ah so no problem.

Let me see if I get the above right. I realise it's theoretical figures but they hint at something, and the difference shows what's going on.

The 10 mm2 cable seems to be ok here? It limits the charging, but the windlass will be getting its 90 A and with OK voltage. Am I correct, please?

If 25 A from 10 mm2 cable, there will be going out 65 A from the battery, right? I'd like the battery to take the major part in running the windlass after all. Otherwise it doesn't make much sense to put the windlass battery in the bow, near the windlass.

The good thing about putting it near, if it takes major part, is that the cables can be kept short. I got some tinned copper 35 mm2 lying around, 6.6 m to be precise, that I can use for this, I expect max 3 m from windlass battery to windlass, one-way.

Originally Posted by smac999

your engine would have battery B back fully charged in a few mins. (you took 90a out for 1 min and can put 50a back into it for 2mins once windlass stopped)

It won't be fully recharged as it's a 10 mm2 cable, long-run, and the Sterling doesn't do its magic over the START output. But the solar panel will take care of that (see below).

Originally Posted by smac999

120a source D with 50mm 14v, 14v, 14v. no dischange of battery B, all power from alternator : amps from D 90a


to start your engine from 10m away you'd probablay want bigger cable then 50mm. though 50 might due

so you don't have a start battery? you have a house bank and a windlass bank?

I have a house bank of 300 Ah = 4 x 75 Ah GEL batteries, from 2013. I want them to last beyond 2019. They are well maintained behind the Sterling House Bank output. They are used also to start the engine.

Second bank would be the Windlass battery, behind the Sterling START output.

The House bank will be charged by min 450 W (it is what I am putting horisontally on the arch that is being welded as we speak). I might put more solar panels for the house bank. I will get a Morningstar MPPT 45 A or a Outback Power FlexMax FM60 Charge Controller. I don't expect them to break. If they do I will survive somehow while I get them replaced under warranty, Global service offered from factory (checked).

I have a pretty good Naps Maxpower MPPT regulator that I will use for the Windlass battery. A 50 W solar panel in the vicinity to keep cable runs short. All in bow section of boat. Have a spare PWM regulator, can't remember the brand right now, that we might bring along as spare for the windlass battery.

If I go for the 10 mm2 cable and the worst happens, then I could move the Windlass battery near the engine, connect it and start.

The Windlass is a Lewmar V2.

The alternator is original 50 A.

[boatbod]

Way too much theory being quoted here.

Put simply, fuse the leads that connect directly to a battery per ABYC E12. Fuse size based on ampacity of conductor.

Alternator will provide current and attempt to keep battery voltage constant until it can output no additional amps.

Load drawn from battery will depress battery voltage when there is a net current outflow.

In your scenario the alternator will provide most of the power (50A) unless voltage drop in undersized conductors prevents it doing so.

[Bob666]

Originally Posted by boatbod

Way too much theory being quoted here.

Put simply, fuse the leads that connect directly to a battery per ABYC E12. Fuse size based on ampacity of conductor.

Alternator will provide current and attempt to keep battery voltage constant until it can output no additional amps.

Load drawn from battery will depress battery voltage when there is a net current outflow.

In your scenario the alternator will provide most of the power (50A) unless voltage drop in undersized conductors prevents it doing so.

Ok. I will buy 16 mm2 from DC Source D to Battery B. That's the long run cable.