Engine run time to keep battery charged

T

Tom Del Rosso

Guest
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.


--
 
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.

5 min once every two weeks should be sufficient assuming the
battery and charging system is good. Turn everything off except
engine. It will need to be run at least 1,000 rpm for that time
period.

OTH, Harbor Freight has cheap little float chargers but 120v power
will be needed. A friend uses one of these when out of the country:
https://www.harborfreight.com/automatic-battery-float-charger-42292.html
 
On Monday, February 4, 2019 at 5:57:03 PM UTC-5, Paul in Houston TX wrote:
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.

5 min once every two weeks should be sufficient assuming the
battery and charging system is good. Turn everything off except
engine. It will need to be run at least 1,000 rpm for that time
period.

I've always been told that short run times creates condensation and acid in the motor oil from incomplete warmup - get her good and hot to drive off moisture. I realize the OP was asking about charging times, but he'd be better served by letting that old Buick idle for a good half an hour every few weeks, or better yet, have OP take the old girl for a blast.


OTH, Harbor Freight has cheap little float chargers but 120v power
will be needed. A friend uses one of these when out of the country:
https://www.harborfreight.com/automatic-battery-float-charger-42292.html

NOOOOOOOOOOO!!!!!! Don't suggest a solution (even a cheap one) that comes from a credit card lest Arlen Holder or one of his socks pitches a fit!!!
 
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.
 
On Tue, 5 Feb 2019 01:30:54 -0000 (UTC), arlen holder wrote:

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.

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.)
 
On Mon, 4 Feb 2019 17:05:13 -0800 (PST), John-Del wrote:

OTH, Harbor Freight has cheap little float chargers but 120v power
will be needed. A friend uses one of these when out of the country:
https://www.harborfreight.com/automatic-battery-float-charger-42292.html

NOOOOOOOOOOO!!!!!! Don't suggest a solution (even a cheap one)
that comes from a credit card lest Arlen Holder or one of his socks
pitches a fit!!!

First off, neither of you appeared to have _comprehended_ the question,
which was how much time does it take to recharge the battery to compensate
for the charge lost in starting, which I roughly calculated at about 72
seconds, meaning, a couple of minutes "should" recharge the battery if the
assumptions I made were reasonable.

Even if you did _comprehend_ the question, you made no attempt to _answer_
the OP's question, which, is par for the course since all you _can_ do is
off-topic chit-chat drivel.

NOTE: I don't need to prove that statement since you prove it for me.

Moving on though, assuming the OP is satisfied with the 72 second
assumption, there _is_ a question of how often he _needs_ to charge the
battery.

I wonder if you, John-Del, have the brains to answer _that_ question?
(HINT: I don't think you do ... but maybe you'll prove me wrong.)

HINT: I already calculated it in the same minute (or so) that I calculated
the answer I provided - but I didn't post it because, unlike you, I
actually _comprehended_ the question that the OP had asked.

HINT: If you can't take a hint - you always prove to be _stupid_, John-Doe.
DOUBLEHINT: The proof will be in exactly what you write in response.
 
On 2/4/19 7:43 PM, arlen holder wrote:
Nothing of interest as usual.

Quack quack quack, ding, reverse direction.

--
"I am a river to my people."
Jeff-1.0
WA6FWi
http:foxsmercantile.com
 
On Mon, 4 Feb 2019 20:59:38 -0600, Fox's Mercantile wrote:

> Quack quack quack, ding, reverse direction.

Two points which _adults_ will comprehend.

1. Snit here acts like a child _all_ the time, and,
2. Snit here didn't even _attempt_ to answer the OP's question.

We really shouldn't fault him as his brain _is_ that of a child.
o He proves that fact in _every_ post - as he did here.

Meanwhile, I at least attempted to faithfully answer the guy's question.
o And, yet, Snit (aka Fox's Mercantile), calls everyone but himself, a troll.

I don't even have to prove these two statements since he proves it himself.
o What Snit (aka Fox's Mercantile) wries, proves these two facts.

1. Snit (aka Fox's Mercantile) is _never_ purposefully helpful, and,
2. Snit (aka Fox's Mercantile) _always_ proves to own the brain of a child.

The funny thing is that it's not even an ad hominem attack!
o It's simply pointing out what Fox's Mercantile proves himself to be a fact.
 
In article <072fc2e7-dc4d-48ac-b069-4fe7224adbac@googlegroups.com>,
ohger1s@gmail.com says...
I've always been told that short run times creates condensation and acid in the motor oil from incomplete warmup - get her good and hot to drive off moisture. I realize the OP was asking about charging times, but he'd be better served by letting that old Buick idle for a good half an hour every few weeks, or better yet, have OP
take the old girl for a blast.


OTH, Harbor Freight has cheap little float chargers but 120v power
will be needed. A friend uses one of these when out of the country:
https://www.harborfreight.com/automatic-battery-float-charger-42292.html

I am using one of those HF float chargers to hold up a lawn mower
battery over the winter. I really use that battery to power a sprayer
instead of the lawn mower.

My mower is in a shed with out power. I have a small solar cell about
the size of a mouse pad to keep it charged over the winter months.

If starting the mower, I would run it long enough to get the motor and
exhaust system hot. Maybe a 5 ot 10 mile trip to the store and back
once every 2 weeks or so. It will probably do the other parts of the
car some good also.
 
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
 
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
 
"Tom Del Rosso" <fizzbintuesday@that-google-mail-domain.com> 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
 
Ralph Mowery <rmowery28146@earthlink.net> wrote:
In article <072fc2e7-dc4d-48ac-b069-4fe7224adbac@googlegroups.com>,
ohger1s@gmail.com says...

I've always been told that short run times creates condensation and acid
in the motor oil from incomplete warmup - get her good and hot to drive
off moisture. I realize the OP was asking about charging times, but
he'd be better served by letting that old Buick idle for a good half an
hour every few weeks, or better yet, have OP
take the old girl for a blast.


OTH, Harbor Freight has cheap little float chargers but 120v power
will be needed. A friend uses one of these when out of the country:
https://www.harborfreight.com/automatic-battery-float-charger-42292.html




I am using one of those HF float chargers to hold up a lawn mower
battery over the winter. I really use that battery to power a sprayer
instead of the lawn mower.

My mower is in a shed with out power. I have a small solar cell about
the size of a mouse pad to keep it charged over the winter months.

If starting the mower, I would run it long enough to get the motor and
exhaust system hot. Maybe a 5 ot 10 mile trip to the store and back
once every 2 weeks or so. It will probably do the other parts of the
car some good also.

My HF floaters make too much bubbling on a charged battery.

Greg
 
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.
 
On Monday, February 4, 2019 at 8:43:17 PM UTC-5, arlen holder wrote:
a bit steaming pile of bullshit


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.
 
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 Êcrit le 04/02/2019 à 22:45 :
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.
 
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
 
pfjw@aol.com <peterwieck33@gmail.com> wrote:
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.

Those times are probably not long enough. Even in my small car
(4-cyl 1.6L engine) I find that after a lot of 15-minute runs the oil
temperature caps at 100C when driving a bit longer, and it requires a
45 minute drive or so before it rises to 110-120C.
 
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.
 
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."
 

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