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Electronics Repair (sci.electronics.repair) Discussion of repairing electronic equipment. Topics include requests for assistance, where to obtain servicing information and parts, techniques for diagnosis and repair, and annecdotes about success, failures and problems. |
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Posted to rec.autos.tech,sci.electronics.repair
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On 02/04/2019 08:30 PM, arlen holder wrote:
On Mon, 4 Feb 2019 16:45:34 -0500, Tom Del Rosso wrote: 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. 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 https://www.partsgeek.com/catalog/1987/buick/regal/engine_electrical/alternator.html 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: https://en.wikipedia.org/wiki/Buick_Regal#Grand_National,_Turbo-T,_T-Type,_and_GNX We're kind of stuck with the "generic" stuff, such as this: o How Many Amps Does It Take to Start a Car? https://www.reference.com/vehicles/many-amps-start-car-e35b6f3d4d8bf426 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. https://www.jstor.org/stable/44611429?seq=1#page_scan_tab_contents It's behind a pay-wall but I can probably get my hands on a copy |
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