Home |
Search |
Today's Posts |
![]() |
|
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. |
Reply |
|
LinkBack | Thread Tools | Display Modes |
#11
![]() |
|||
|
|||
![]()
On Tuesday, 5 February 2019 01:30:57 UTC, 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. Best answer so far. 100A would only be delivered to a flat battery, it'll charge much slower than that. There's also the parasitic loads to make up for, the electronics that eats power when the vehicle is off and on. You're better off testing battery voltage and not doing anything until it drops enough to warrant charging. Leads acids don't like sitting even half discharged, keep it near full. NT |
#12
![]() |
|||
|
|||
![]()
"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. -- A real test at idle, turn on lights. If revving the engine makes them brighter, you have little reserve power to charge battery. For a fast charge you need 14 or more volts. Greg |
#14
![]() |
|||
|
|||
![]()
On Monday, February 4, 2019 at 4:45:38 PM UTC-5, 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. -- God Help Us! This was given in Drivers'Ed. I must be VERY old. The rules of thumb are as follows. All times at ~1,000 rpm. a) For a pre-catalyst car, a minimum of 15 minutes. This will bring every part of the system above the temperatures necessary to boil water out of the oil and the exhaust. This will also re-coat the cylinders with oil - which tends to be rinsed off by the very rich mixture on starting, especially when the outside temperatures are below freezing. b) Post-Catalyst, carburetor: About the same, maybe only 10 - 12 minutes, as the catalyst will do a fine job of heating the exhaust. A V8 or other large-displacement engine will take longer. c) Fuel-injected, 7 - 12 minutes. This directly related to engine displacement. Big = more time. The physics of removing moisture from the oil becomes the driver (pun intended). The system must reach full operating temperature and stay there for a couple of minutes. Pretty much when moisture (steam) stops coming out of the tailpipe - and then a few minutes. If you open the oil-fill cap and find a milky foam, you haven't been doing it long enough. |
#15
![]() |
|||
|
|||
![]()
On Monday, February 4, 2019 at 8:43:17 PM UTC-5, arlen holder wrote:
a bit steaming pile of bull**** You don't have a clue how to answer that question. You've left out the variables that *must* be included to calculate the recharge time. Car batteries have an internal resistance that changes from the minute it's installed until it's finally dead. As it ages, the car battery's internal resistance rises and it won't draw the same current as it did when it was new. Batteries slowly sulfate over time and the more sulfated the car battery is, the larger the battery's capacitor effect is and the longer it takes to charge. Car alternators often don't provide their rated current, particularly when they are older. You left out the temperature of the components and the quality of wiring involved. Is that old Buick even idling at it's programmed rpm? You also left out the self discharging of the battery and the standby parasitic drag from the car's computer and accessories, so that must be added to the charge time. OP asked about replenishment charge time but he also said he wants to start the car to keep the battery from self discharging. His question was two-fold, and in the real world, the best advice is keep a float charger on it or, better yet, run the car on the road to charge the battery, boil off condensation and contaminants in the fluids, and keep moving parts moving. If OP starts that car, runs it 72 seconds and shuts it off, that battery won't last. |
#16
![]() |
|||
|
|||
![]()
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)Â* i.e. 200As=0.55Ah , it's easy. Tom Del Rosso a écrit le 04/02/2019 Ã* 22:45Â*: 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. |
#17
![]() |
|||
|
|||
![]()
Arlen Holder and its various socks and clones is a genuinely dangerous individual and should be muzzled, blind-folded, have its fingers in mittens, encased in epoxy and super-glued to the opposite shoulders, feet tied, crossed and bent back at the knees, tied to its legs.
Only then will it be rendered very nearly harmless. Peter Wieck Melrose Park, PA |
#18
![]() |
|||
|
|||
![]() |
#19
![]() |
|||
|
|||
![]()
On Tue, 5 Feb 2019 00:03:14 -0500, bitrex wrote:
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 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 https://www.jstor.org/stable/44611429?seq=1#page_scan_tab_contents "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. |
#20
![]() |
|||
|
|||
![]()
On 02/04/2019 04:45 PM, 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. Here's a modern review of a 1989 Buick Century: https://www.youtube.com/watch?v=VKYMgfjCd7E "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 |
Thread Tools | Search this Thread |
Display Modes | |
|
|
![]() |
||||
Thread | Forum | |||
How can it keep charged in use, but be unable to recharge? | Electronics Repair | |||
Lib Obama voter charged with battery over $1 worth of gas | Metalworking | |||
Freezers: keep warm or keep cold? | UK diy | |||
Furnace low heat run time = A/C low cool run time | Home Repair | |||
Can you fix a lead-acid battery that's charged backwards? | Electronics Repair |