Formula for charging time of 12v battery

  • Thread starter Northern Night Sky
  • Start date
John G wrote:
on 10/02/2011, George Herold supposed :

On Feb 9, 4:57 pm, Northern Night Sky <headbangerstu...@gmail.com
wrote:

On Feb 9, 5:37 pm, "Phil Allison" <phi...@tpg.com.au> wrote:





"Northern Night Sky"


I have a Powerpack 300, which is a 12v battery at 12Ah.
Using the AC adapter with the house outlet (20amp at 120v) it takes 17
hours to fully charge the battery. How would I calculate the
(approximate) time it would take to charge the battery, using a solar
panel that draws 1amp at 15v?


** Don't solar panels only work when the sun shineth brightly ??
Any such predictions would have to involve detailed knowledge of the
local
weather, cloud cover and solar illumination levels.
Wouldn't it ?


.... Phil


You still can get an approximate time of how long it would take to
charge the battery, giving the variables I've provided...Duh! If I
get 5 hours of sunshine/day, then I'll do the math in consequence.
I'm still interested on getting the FORMULA, as per my original
question.- Hide quoted text -

- Show quoted text -


The 12Ah is the capacity of the battery. 12 amp x hours. So if your
solar charger provides 1 amp, then it will take about 12 hours.
(Assuming the battery is fully discharged at the start.) What was
the current rating on your AC adapter that took 17 hours?

george h.


The charger according to the sales blurb supplies 500ma.
The 20 amp 120 volt house outlet is useless and probably wrong
information. Is not the normal US outlet limited to 15 amps?
Click to expand...
Regarding your question, both 15 and 20 amp branch circuits are
permissible in the US, and some of those branch circuits _must_
be 20 amps. (That doesn't change your correct observation that
the 20 amp outlet information is useless with respect to the
op's post.) Those branch circuits provide power to "normal"
receptacles (often referred to as "outlets".)

Ed
 
On Feb 10, 8:25 am, John Fields <jfie...@austininstruments.com> wrote:
On Thu, 10 Feb 2011 07:23:37 -0800 (PST), George Herold



gher...@teachspin.com> wrote:
On Feb 10, 7:33 am, John Fields <jfie...@austininstruments.com> wrote:
On Wed, 9 Feb 2011 14:15:42 -0800 (PST), George Herold

gher...@teachspin.com> wrote:
On Feb 9, 4:57 pm, Northern Night Sky <headbangerstu...@gmail.com
wrote:
On Feb 9, 5:37 pm, "Phil Allison" <phi...@tpg.com.au> wrote:

"Northern Night Sky"

I have a Powerpack 300, which is a 12v battery at 12Ah.

Using the AC adapter with the house outlet (20amp at 120v) it takes 17
hours to fully charge the battery. How would I calculate the
(approximate) time it would take to charge the battery, using a solar
panel that draws 1amp at 15v?

** Don't solar panels only work when the sun shineth brightly ??

Any such predictions would have to involve detailed knowledge of the local
weather, cloud cover and solar illumination levels.

Wouldn't it ?

.... Phil

You still can get an approximate time of how long it would take to
charge the battery, giving the variables I've provided...Duh! If I
get 5 hours of sunshine/day, then I'll do the math in consequence.
I'm still interested on getting the FORMULA, as per my original
question.- Hide quoted text -

- Show quoted text -

The 12Ah is the capacity of the battery. 12 amp x hours. So if your
solar charger provides 1 amp, then it will take about 12 hours.

---
Sorry, but since there's no free lunch, that's just not true.

Rule of thumb, as I recall, is that the charging efficiency of a
lead-acid cell/battery is about 60%, so to get 12Ah into the battery
using a 1A constant-current source which will take it to its terminal
voltage will take about 20 hours.

As bad as 60%?  I didn't know that.  The only time I've used/measured
battery capacity was in rapid charging of NiCad's.  And there the
efficiency was pretty good.

George H.

---http://photovoltaics.sandia.gov/docs/PDF/batpapsteve.pdf

---
JF
Click to expand...
I don't get it. The article suggests battery efficiency drops off near
full charge, however figure 1 shows about 68 AH input for about the
same out, while an input of 116 yields only about 97 out. That would
indicate less efficiency at deeper discharges, and higher efficiency
near full charge.

What did I miss?

-Bill
 
On Fri, 11 Feb 2011 19:49:07 -0800 (PST), Bill Bowden
<bperryb@bowdenshobbycircuits.info> wrote:

On Feb 10, 8:25 am, John Fields <jfie...@austininstruments.com> wrote:
On Thu, 10 Feb 2011 07:23:37 -0800 (PST), George Herold



gher...@teachspin.com> wrote:
On Feb 10, 7:33 am, John Fields <jfie...@austininstruments.com> wrote:
On Wed, 9 Feb 2011 14:15:42 -0800 (PST), George Herold

gher...@teachspin.com> wrote:
On Feb 9, 4:57 pm, Northern Night Sky <headbangerstu...@gmail.com
wrote:
On Feb 9, 5:37 pm, "Phil Allison" <phi...@tpg.com.au> wrote:

"Northern Night Sky"

I have a Powerpack 300, which is a 12v battery at 12Ah.

Using the AC adapter with the house outlet (20amp at 120v) it takes 17
hours to fully charge the battery. How would I calculate the
(approximate) time it would take to charge the battery, using a solar
panel that draws 1amp at 15v?

** Don't solar panels only work when the sun shineth brightly ??

Any such predictions would have to involve detailed knowledge of the local
weather, cloud cover and solar illumination levels.

Wouldn't it ?

.... Phil

You still can get an approximate time of how long it would take to
charge the battery, giving the variables I've provided...Duh! If I
get 5 hours of sunshine/day, then I'll do the math in consequence.
I'm still interested on getting the FORMULA, as per my original
question.- Hide quoted text -

- Show quoted text -

The 12Ah is the capacity of the battery. 12 amp x hours. So if your
solar charger provides 1 amp, then it will take about 12 hours.

---
Sorry, but since there's no free lunch, that's just not true.

Rule of thumb, as I recall, is that the charging efficiency of a
lead-acid cell/battery is about 60%, so to get 12Ah into the battery
using a 1A constant-current source which will take it to its terminal
voltage will take about 20 hours.

As bad as 60%?  I didn't know that.  The only time I've used/measured
battery capacity was in rapid charging of NiCad's.  And there the
efficiency was pretty good.

George H.

---http://photovoltaics.sandia.gov/docs/PDF/batpapsteve.pdf

---
JF

I don't get it. The article suggests battery efficiency drops off near
full charge, however figure 1 shows about 68 AH input for about the
same out, while an input of 116 yields only about 97 out. That would
indicate less efficiency at deeper discharges, and higher efficiency
near full charge.

What did I miss?

-Bill
Click to expand...
---
Let's say we have a 12V lead-acid battery which has been discharged to
the point where its terminal voltage is 10.5V

Then let's say the battery has a capacity of 97Ah, but to get that out
of it we have to put in 116.

Since efficiency is equal to output divided by input, that gives us an
efficiency of about 0.84, or 84%.

Now, with the battery discharged to 10.5V, (since we took 97Ah out of
it) let's pump 68Ah back into it.

We find that in this case we can get, say, 65Ah out of it until it
gets discharged to 10.5V, so thats's an efficiency of 65Ah/68Ah, or
about 96%, so efficiency decreases as the battery approaches full
charge.

---
JF
 
On Feb 12, 4:45 am, John Fields <jfie...@austininstruments.com> wrote:
On Fri, 11 Feb 2011 19:49:07 -0800 (PST), Bill Bowden



bper...@bowdenshobbycircuits.info> wrote:
On Feb 10, 8:25 am, John Fields <jfie...@austininstruments.com> wrote:
On Thu, 10 Feb 2011 07:23:37 -0800 (PST), George Herold

gher...@teachspin.com> wrote:
On Feb 10, 7:33 am, John Fields <jfie...@austininstruments.com> wrote:
On Wed, 9 Feb 2011 14:15:42 -0800 (PST), George Herold

gher...@teachspin.com> wrote:
On Feb 9, 4:57 pm, Northern Night Sky <headbangerstu...@gmail.com
wrote:
On Feb 9, 5:37 pm, "Phil Allison" <phi...@tpg.com.au> wrote:

"Northern Night Sky"

I have a Powerpack 300, which is a 12v battery at 12Ah.

Using the AC adapter with the house outlet (20amp at 120v) it takes 17
hours to fully charge the battery. How would I calculate the
(approximate) time it would take to charge the battery, using a solar
panel that draws 1amp at 15v?

** Don't solar panels only work when the sun shineth brightly ??

Any such predictions would have to involve detailed knowledge of the local
weather, cloud cover and solar illumination levels.

Wouldn't it ?

.... Phil

You still can get an approximate time of how long it would take to
charge the battery, giving the variables I've provided...Duh! If I
get 5 hours of sunshine/day, then I'll do the math in consequence.
I'm still interested on getting the FORMULA, as per my original
question.- Hide quoted text -

- Show quoted text -

The 12Ah is the capacity of the battery. 12 amp x hours. So if your
solar charger provides 1 amp, then it will take about 12 hours.

---
Sorry, but since there's no free lunch, that's just not true.

Rule of thumb, as I recall, is that the charging efficiency of a
lead-acid cell/battery is about 60%, so to get 12Ah into the battery
using a 1A constant-current source which will take it to its terminal
voltage will take about 20 hours.

As bad as 60%?  I didn't know that.  The only time I've used/measured
battery capacity was in rapid charging of NiCad's.  And there the
efficiency was pretty good.

George H.

---http://photovoltaics.sandia.gov/docs/PDF/batpapsteve.pdf

---
JF

I don't get it. The article suggests battery efficiency drops off near
full charge, however figure 1 shows about 68 AH input for about the
same out, while an input of 116 yields only about 97 out. That would
indicate less efficiency at deeper discharges, and higher efficiency
near full charge.

What did I miss?

-Bill

---
Let's say we have a 12V lead-acid battery which has been discharged to
the point where its terminal voltage is 10.5V

Then let's say the battery has a capacity of 97Ah, but to get that out
of it we have to put in 116.

Since efficiency is equal to output divided by input, that gives us an
efficiency of about 0.84, or 84%.

Now, with the battery discharged to 10.5V, (since we took 97Ah out of
it) let's pump 68Ah back into it.

We find that in this case we can get, say, 65Ah out of it until it
gets discharged to 10.5V, so thats's an efficiency of 65Ah/68Ah, or
about 96%, so efficiency decreases as the battery approaches full
charge.

---
JF
Click to expand...
Yes, I was looking at it from the top down rather than the bottom up.
So, it appears efficiency is better when the state of charge (SOC) is
less than maximum. But that brings up another issue of battery life
operating at lower SOCs. Maybe it's incorrect, but I thought lead acid
battery life is extended by keeping the battery fully charged. So, the
trade-offs might be better efficiency with reduced life, or visa
versa?

-Bill
 
On Sun, 13 Feb 2011 18:37:49 -0800 (PST), Bill Bowden <bperryb@bowdenshobbycircuits.info> wrote:

On Feb 12, 4:45 am, John Fields <jfie...@austininstruments.com> wrote:
On Fri, 11 Feb 2011 19:49:07 -0800 (PST), Bill Bowden



bper...@bowdenshobbycircuits.info> wrote:
On Feb 10, 8:25 am, John Fields <jfie...@austininstruments.com> wrote:
On Thu, 10 Feb 2011 07:23:37 -0800 (PST), George Herold

gher...@teachspin.com> wrote:
On Feb 10, 7:33 am, John Fields <jfie...@austininstruments.com> wrote:
On Wed, 9 Feb 2011 14:15:42 -0800 (PST), George Herold

gher...@teachspin.com> wrote:
On Feb 9, 4:57 pm, Northern Night Sky <headbangerstu...@gmail.com
wrote:
On Feb 9, 5:37 pm, "Phil Allison" <phi...@tpg.com.au> wrote:

"Northern Night Sky"

I have a Powerpack 300, which is a 12v battery at 12Ah.

Using the AC adapter with the house outlet (20amp at 120v) it takes 17
hours to fully charge the battery. How would I calculate the
(approximate) time it would take to charge the battery, using a solar
panel that draws 1amp at 15v?

** Don't solar panels only work when the sun shineth brightly ??

Any such predictions would have to involve detailed knowledge of the local
weather, cloud cover and solar illumination levels.

Wouldn't it ?

.... Phil

You still can get an approximate time of how long it would take to
charge the battery, giving the variables I've provided...Duh! If I
get 5 hours of sunshine/day, then I'll do the math in consequence.
I'm still interested on getting the FORMULA, as per my original
question.- Hide quoted text -

- Show quoted text -

The 12Ah is the capacity of the battery. 12 amp x hours. So if your
solar charger provides 1 amp, then it will take about 12 hours.

---
Sorry, but since there's no free lunch, that's just not true.

Rule of thumb, as I recall, is that the charging efficiency of a
lead-acid cell/battery is about 60%, so to get 12Ah into the battery
using a 1A constant-current source which will take it to its terminal
voltage will take about 20 hours.

As bad as 60%?  I didn't know that.  The only time I've used/measured
battery capacity was in rapid charging of NiCad's.  And there the
efficiency was pretty good.

George H.

---http://photovoltaics.sandia.gov/docs/PDF/batpapsteve.pdf

---
JF

I don't get it. The article suggests battery efficiency drops off near
full charge, however figure 1 shows about 68 AH input for about the
same out, while an input of 116 yields only about 97 out. That would
indicate less efficiency at deeper discharges, and higher efficiency
near full charge.

What did I miss?

-Bill

---
Let's say we have a 12V lead-acid battery which has been discharged to
the point where its terminal voltage is 10.5V

Then let's say the battery has a capacity of 97Ah, but to get that out
of it we have to put in 116.

Since efficiency is equal to output divided by input, that gives us an
efficiency of about 0.84, or 84%.

Now, with the battery discharged to 10.5V, (since we took 97Ah out of
it) let's pump 68Ah back into it.

We find that in this case we can get, say, 65Ah out of it until it
gets discharged to 10.5V, so thats's an efficiency of 65Ah/68Ah, or
about 96%, so efficiency decreases as the battery approaches full
charge.

---
JF

Yes, I was looking at it from the top down rather than the bottom up.
So, it appears efficiency is better when the state of charge (SOC) is
less than maximum. But that brings up another issue of battery life
operating at lower SOCs. Maybe it's incorrect, but I thought lead acid
battery life is extended by keeping the battery fully charged. So, the
trade-offs might be better efficiency with reduced life, or visa
versa?
Click to expand...
Yes, like that. It's very hard to push the top 20% of capacity back into
the battery, that's why car batteries quickly fail when the car is used for
mostly short trips, they're never fully charged.

If you look at voltage (or energy) while charging the battery, you need to
push 30 to 50% more into the thing than what you got on discharge. The
industry talks in AH and expect about 1:1 AH charge/discharge for a good
battery, the failure modes for LA batteries is a race between corrosion and
sulphation, difficult to get right at times.

Other day I noticed a 12V 5AH SLA had split it's case. They do that sometimes,
all by themselves, just crap out -- that battery was on standby float charge
for several years, had no useful capacity, output voltage dropped to about 8V
on load, which is 2 of the 6 cells shorted? I floated them at the low end of
the range too (13.5V).

Grant.
-Bill
Click to expand...
 
On Sun, 13 Feb 2011 18:37:49 -0800 (PST), Bill Bowden
<bperryb@bowdenshobbycircuits.info> wrote:

On Feb 12, 4:45 am, John Fields <jfie...@austininstruments.com> wrote:
On Fri, 11 Feb 2011 19:49:07 -0800 (PST), Bill Bowden



bper...@bowdenshobbycircuits.info> wrote:
On Feb 10, 8:25 am, John Fields <jfie...@austininstruments.com> wrote:
On Thu, 10 Feb 2011 07:23:37 -0800 (PST), George Herold

gher...@teachspin.com> wrote:
On Feb 10, 7:33 am, John Fields <jfie...@austininstruments.com> wrote:
On Wed, 9 Feb 2011 14:15:42 -0800 (PST), George Herold

gher...@teachspin.com> wrote:
On Feb 9, 4:57 pm, Northern Night Sky <headbangerstu...@gmail.com
wrote:
On Feb 9, 5:37 pm, "Phil Allison" <phi...@tpg.com.au> wrote:

"Northern Night Sky"

I have a Powerpack 300, which is a 12v battery at 12Ah.

Using the AC adapter with the house outlet (20amp at 120v) it takes 17
hours to fully charge the battery. How would I calculate the
(approximate) time it would take to charge the battery, using a solar
panel that draws 1amp at 15v?

** Don't solar panels only work when the sun shineth brightly ??

Any such predictions would have to involve detailed knowledge of the local
weather, cloud cover and solar illumination levels.

Wouldn't it ?

.... Phil

You still can get an approximate time of how long it would take to
charge the battery, giving the variables I've provided...Duh! If I
get 5 hours of sunshine/day, then I'll do the math in consequence.
I'm still interested on getting the FORMULA, as per my original
question.- Hide quoted text -

- Show quoted text -

The 12Ah is the capacity of the battery. 12 amp x hours. So if your
solar charger provides 1 amp, then it will take about 12 hours.

---
Sorry, but since there's no free lunch, that's just not true.

Rule of thumb, as I recall, is that the charging efficiency of a
lead-acid cell/battery is about 60%, so to get 12Ah into the battery
using a 1A constant-current source which will take it to its terminal
voltage will take about 20 hours.

As bad as 60%?  I didn't know that.  The only time I've used/measured
battery capacity was in rapid charging of NiCad's.  And there the
efficiency was pretty good.

George H.

---http://photovoltaics.sandia.gov/docs/PDF/batpapsteve.pdf

---
JF

I don't get it. The article suggests battery efficiency drops off near
full charge, however figure 1 shows about 68 AH input for about the
same out, while an input of 116 yields only about 97 out. That would
indicate less efficiency at deeper discharges, and higher efficiency
near full charge.

What did I miss?

-Bill

---
Let's say we have a 12V lead-acid battery which has been discharged to
the point where its terminal voltage is 10.5V

Then let's say the battery has a capacity of 97Ah, but to get that out
of it we have to put in 116.

Since efficiency is equal to output divided by input, that gives us an
efficiency of about 0.84, or 84%.

Now, with the battery discharged to 10.5V, (since we took 97Ah out of
it) let's pump 68Ah back into it.

We find that in this case we can get, say, 65Ah out of it until it
gets discharged to 10.5V, so thats's an efficiency of 65Ah/68Ah, or
about 96%, so efficiency decreases as the battery approaches full
charge.

---
JF

Yes, I was looking at it from the top down rather than the bottom up.
So, it appears efficiency is better when the state of charge (SOC) is
less than maximum. But that brings up another issue of battery life
operating at lower SOCs. Maybe it's incorrect, but I thought lead acid
battery life is extended by keeping the battery fully charged. So, the
trade-offs might be better efficiency with reduced life, or visa
versa?

-Bill
Click to expand...
---
It appears from:

http://en.wikipedia.org/wiki/Lead%E2%80%93acid_battery#Sulfation

that battery life will be decreased by keeping it less than fully
charged due to the action of sulfation.

That being the case, it'll cost more, in electricity, to keep a
battery fully charged than it would to keep it partially charged, but
it'll cost more, in terms of $ spent on batteries, if less electricity
is used than is required to keep the battery fully charged.

I don't know where the crossover point is, but I'd guess that the
electricity used to keep a battery fully charged costs less than
replacing the battery would.

---
JF
 
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