Automotive dual battery question

[Tparker]

A little more information. According to the Service manual max. load of the starter is 2400 amps. I would assume that this is the starting surge amperage to get the starter spinning. One thing that I need to add is the 700 amps DOES NOT include the glow plug current draw which is over 100 amps or under certain cold conditions, the intake air heating element in the intake manifold. I am not sure what it draws current wise but it has a 6 gauge wire driving it and controlled by a starter solonoid type relay. One reason for the high current draw is that it has a 17-1 compression ratio and also a piston/swashplate type hydralic pump that must generate 3000 psi before the fuel injectors will operate.
Its hard for me to understand why this battery system was designed the way it was with that 4 gauge interconnection.

[Will]

Wow! I'm never wrong - I said that my post might show how little I know and that's exactly what I did. I screwed up badly - how come someone didn't do the realtively simple math and jump on me ? One of the problems with doing things off the top of your head instad of consulting the books, is that it's easy to screw up - which I did. To calculate the voltage drop I was using I^2.R (I squared R) which is totally wrong because that is the power dissipation in Watts. the drop is, of course I.R Accepting for the moment that, as a contributor said. the internal battery resistances are about equal to that of the 4 ft cable then, if they are all 0.001 ohm and both batteries have the same internal resistance i.e. 0.001 ohm then, with a steady state starter load of 99 amps (100 nominal) then the 4 ft cable, hence the non-starter side battery, will carry 33 amps and the starter aide battery load will be 66 amps. Similarly when the initial cranking amps are 700 the starter side battery would carry 2/3 rds i.e. 466.6 amps and the non starter side battery will carry 233.3 amps. This again off the top of my head - now I will have to go check the math and see if I screwed up again.
It would be interesting if a battery man or anyone who knows, joined this thread and gave some actual examples of battery internal resistance. I suspect it may be more than 0.001 ohm but have no real way of knowing - Sorry about the screw-up guys.

[Edd]

Tparker…paging Tparker…….or has he left the building…<p><<My question is, does the pass. side battery take more of a load than the drivers side battery>>
The two like batteries are going to want to try to balance the load imposed upon them by the starter. However,the interconnect of that five feet of #4 wire (~1/4 in dia) imposes a .434875 VDC drop within the line at the prescribed worst case value of the starter requiring 700 amps at initial start of cranking. (considering the wire to be a common 49 of #21ga wire build).With that loss, the end result is the drivers side battery just providing 96.3% of the halved currentage value that the pax side was providing, with the 3.7% balance having to be imparted upon the passengers side battery. Actually, not to much of a current imbalance at all, considering the heavy potential initial current involved.
After the motor settles into its crank speed and goes down to ~ 125A….[if I can plug in the same figures that my old ’94 Mercedes diesel pulled]….those figures drop on the inequity of load sharing and there is only a 2% difference then.
Guessing that you also might be curious on a #00 wire(~1/2in dia) interconnect in place of the #4 …that would result in only a 2% imposition on the pax side battery at initial crank and only1/2% at crank running speed.
For grins….Sir Josephs….jestingly suggested #0000 (~3/4 in dia) use would only leave an .09% differentiality……but possibly impose the use of a conduit bender for routing/shaping and having one he@# of a connectivity/interfacing problem.
BTW whats the typical crank time on that unit in 70 deg F weather….…wondering what it is now on current day models ?<p>73's de Edd
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<p>[ February 25, 2004: Message edited by: Edd Whatley ]</p>

[Chris Smith]

The Idea behind two wire sizes is distance. The further away from the starter the larger the cable, so that by the time resistance is accounted for, the apx same current is drawn from both batteries, including loss of voltage. The Idea is not the current draw, but rather keeping both batteries at the same voltage while discharging. If the voltages vary too much, then like water pressure, voltage will flow in either direction, charging one battery, instead of drawing current to starter from it.[in theory it will slow down] As long as the voltages match, the current what ever it may be, will flow in one direction, only. When I say the “same” I mean each battery is approximately delivering the same voltage and the current will flow acording to that, also approximately the same. <p>Current "follows" voltage.

[haklesup]

Do they still put big diodes between the main battery and the auxillary battery. A properly connected diode will prevent one battery from becoming a load if the other one should become too weak to fully charge or if a cell shorts out and the voltage drops.<p>In such a configuration, the Aux battery will supply little or no current unless the voltage in the system drops low enough (because of current load) to forward bias the diode.<p>Or have I misunderstood the purpose of a battery isolator. Does your truck have one of these?<p>Also, the resistivity of actual copper may be much higher if you have cheap wire. This is because in some parts of the world, Copper is routinely recycled to make new wire or motor windings. This copper is less pure and has higher resistivity. I once saw an estimate of how much power is wasted in china due to high resistivity conductors and motor windings and it was a very big number.<p>You can measure the actual current in each leg by measuring the differential voltage at both ends of the wire. The measured voltage (although small) will represent the IR drop across the cable. You can use this like an ammeter if you measure the actual resistance of each wire first. (the battery cable takes the place of the ammeter shunt). You will need to make the measurement while someone cranks the engine for you.<p>This technique is similar to the Kelvin resistance measurement method (R=(V2-V1)/I) except you are solving for I instead of R

[Chris Smith]

No, they NEVER put diodes on diesels. Camper batteries with two batteries yes, they draw a small charge back and forth, but with diesels, they draw enough to power a 20 ton crane on cold crank, in cold weather, and then some.

[Joseph]

Sir Ed, I guess I could not help the humor.

[Bernius1]

Chris, I think the 'relatively equal voltage drop despite uneven loading and unequal distance' concept is very good, and probably right. I wouldn't go math crazy on this one, because the cables,connections,terminals, and internal battery resistance all can add that 1/10th ohm that costs you 100amps. i.e., this is one time that 10 minutes outside with a voltmeter will yield more than 2 hrs of math.
That said, I'll stir the pot:
Mack trucks 20+ yrs ago had 4-batt's, & a series-parallel solenoid (mucho $ )to allow 24V starting. Some diesel mechanics still use these in TParker's case, saying that the starter can take 24V in short bursts. But it's HEARSAY on my part.
NOW, here's a goodie: A truck with 4 batteries and no wiring errors keeps blowing alternators. Why? (I'll answer tomorrow if y'all don't)
Have a great day; Do your very best!!!

[Tparker]

Thanks guys for all the input. All of your responses make sense. On the F-rd Diesel truck forum that I belong to, an "Auto engineer" feels the pass. side cable is 2/0 becasue it carrying the full current load and the drivers side battery is only carrying approx. 1/2 the current.

[Will]

This has developed into an interesting thread in spite of addressing what seems to be a relatively simple subject - Can someone post a reasonable battery internal resistance for a 700 cranking amps battery ? I would go out and measure it myself except that I have no one to watch the meter.

[josmith]

I have heard that batteries are current tested by loading them down to 9 volts. Three volts drop at 700amps is around .004 ohms.

[Crowbar]

I may be off base by not noticing, however I didn't notice anyone make mention that the current rating of a conductor is based on the highest temperature its insulation can withstand. The starting period is short enough that the cable's insulation would be able to withstand a substantial overload. Thereby the starter would receive ample current, not caring in the least which battery supplied it for under 60 seconds.

[Bernius1]

All very good perspectives. Now to answer my posed problem. 1) loose belt. An alternator's output curve is hyperbolic with RPM, so it rises quickly as Alt RPM breaks 2000, & levels off asymptotically as RPM climbs. So, at idle, the instantaneous RPM of the Alt varies as the belt slips, and the regulator can't give a solid output. A rough idle, or too large a pulley can have the same effect, because they keep the Alt operating on the 'steep' part of the curve. 2) Mismatched batteries. If you have 3 new batteries, & 1 old, the older battery will keep 'pulling down' the newer ones, & the Alt will respond by trying to charge all. So the Alt never gets a rest, & the newer batt's cook.
In either case, the symptom turns out to be a burnt diode bridge, because the regulator is trying to turn the fields on/off too rapidly, and at times when conduction angle is too short for a very large transient current to pass. Moral; load-matching re-charged batt's prior to installation, and a tight belt will prevent an 'instantaneous-rate-of-change' problem from costing you $175.00.

[Will]

Jo Smith - Good - now we have it - Thank you. Crowbar - the question advanced to the point where what was in question was the relative drain on each battery - not whether or now the starter motor would be starved of current. In fact at 700 A starter current the cable would consume about 116 Watts which, according to my (Sometimes wrong !) math would take (I don't know the specific heat of copper but assuming it to be 0.5 and no heat loss through the insulation during the period) about two minutes for the cable to herat up by 25 deg Celsius./
Now ! with 0.0043 ohms internal battery resistance the drivers side current can be calculated from Id = It*Ri/(2Ri + Rc) where Id = driver side current, Ri = battery internal resistance and Rc = cable rssistance. For 700 A starter current this produces Id = 305.7 A and Ip = 394.7 A i.e. Id = 43.7% and Ip = 56.3 % the proportion is exactly the same for any other starter current so that, at 125 starter current the respective currenrs are 54.6 and 70.4 amps

[Will]

BTW Thank you for the IN34 info Edd I will get one, play with it and see what I get.