Engine run time to keep battery charged

If you turn over an engine periodically to keep it charged, how long do you run it to make up for the charge lost in starting?

In this case it's my neighbor's 87 Buick Regal while he's in the hospital.

Reply to
Tom Del Rosso
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72 seconds

Having said that, here's how I arrived at 72 seconds, bearing in mind there's a complexity to your question which, outside of the engineering specs of both the battery & engine (and parasitics), we can only help you guess at it mathematically, where empirical results would seem to be more accurate than our guestimates.

Starting with the basics, a quick search for a Buick Regal Alternator nets

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which says the alternator outputs 100 amps at idle (if needed) and 150 amps output at max rpm (again, if needed as alternators adjust output based on "B" sensing).

Running a direct search for the power needed to start an 87 Buick Regal, it's easy to find the vehicle, but hard to find the power needed to start the engine:

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We're kind of stuck with the "generic" stuff, such as this: o How Many Amps Does It Take to Start a Car?

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Which says an average car needs 400 to 500 amps but doesn't say how long.

Let's assume it takes five to ten seconds to start it, at 500 amps, where the maximum power would be 10 seconds times 500 amps, which means you sucked out 5,000 Coulombs (i.e., 5000 amp seconds) if the math is right.

If I did the math right, that's less than 1.5 amp hours, and since we guessed high, I'd say the amount used is roughly about 1 amp hour to 1.5 amp hours, but since we want to "be safe" and have "easy math", I'd use 2 amp hours as the amount to add back.

If you put back two amp hours (to cover for inherent losses, mostly in heat), you're back to where you started, where we have to "assume" that the battery sense circuit allows the alternator to output enough current to charge the battery after just one start.

At idle, if we assume the battery sense allows you to get those 100 amps we saw in the spec, to generate 2 amp hours would take only about 0.02 hours, or about 72 seconds (if I did the quick math right) - which - coincidentally - is about how long it took to run the quick math. :)

If that 72 second answer is wrong, I welcome someone who can tell us how to arrive at the better answer.

Reply to
arlen holder

Ooops... Coulombs ... not power... (power would be via P=IV or I^2R but not amp seconds)...

(I hacked that out in a minute on the run, so, please correct where I err.)

Reply to
arlen holder

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It's behind a pay-wall but I can probably get my hands on a copy

Reply to
bitrex

It's just an equation. (I suppose a 12V car battery).

3s of starter, means 3*(900/12) C = 225C = 62.5mAh=0.0625Ah (900W is the starter power). Just add this to the loss of the battery.

Knowing that the alternator charges at 13.8V about 500W (500/(13.8-12)  i.e. 200As=0.55Ah , it's easy.

Tom Del Rosso a écrit le 04/02/2019 à 22:45 :

Reply to
Look165

Having authored peer-reviewed papers myself (in a different field), I took a quick peek at the abstract of that paper, titled: o Computer Simulation of an I.C. Engine During Cranking by a Starter Motor

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"A mathematical model is developed to study the transient behavior of a two stroke or four stroke, single cylinder I.C. engine during cranking and starting by a starter motor. The engine model includes forces due to inertia of reciprocating and rotating parts of engine, gas pressure, frictional loss while starter motor dynamics is determined by the motor's torque versus speed behavior. The numerical results of the analysis when compared with the experimental results showed close correlation. Engine starting by three models of starter motor is presented for a given battery. Effect of different parameters like engine inertia and reduction ratio between engine and a starter motor is described. It is shown here how this analysis can be effectively used as a first step by an engine designer for determining a suitable starter motor characteristic and its related transmission parameters."

Hmmm... they _might_ cover the charge payback component, but I suspect likely it will only be an ancillary input to the mathematical model, and certainly it won't apply _directly_ to an 87 Buick Regal.

We should note that the given "battery" is seemingly incidental in this paper, which seems to be aimed more toward designing starter motors, and, specifically between choosing among three different types of fundamental starter motor designs.

Still, it may be an interesting read, where, I'd be curious how the three types of starter motors affected the model - but - I hazard a guess that we won't find a direct answer for our charge component in that paper.

Reply to
arlen holder

Here's a modern review of a 1989 Buick Century:

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"Boomers were in their early 40s and at the height of their power - they understood the world, and the world turned according to their whims. They understood computers because floppy disks were goddamn floppy, and they knew the HIV virus was out there doing the good Lord's work."

Reply to
bitrex

I have a '96 Chevy Astro van. It's about a 1/4 mile from the house to shop. If all I do is go back and forth, by the end of the week, I have to put the battery on a charger to "refill it" properly. If I drive to Walmart, 10 miles each way, its more than happy. I also have a '02 Dodge Dakota. Which ever vehicle I'm not driving has a Harbor Freight "top off" charger on it to maintain things while they are parked. It's that simple.

Reply to
Fox's Mercantile

It's juuuuunk throw it in the garbage

Reply to
bitrex

I bought it used in 2000. It's been a daily driver for past 19 years. Still going strong.

Reply to
Fox's Mercantile

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