Buying a Multimeter

[DHutch]

The table of battery charge vs voltage is as follows:

12.7v = 100% charge

12.6v = 90% charge

12.5v = 80% charge

12.4v = 70% charge

12.3v = 60% charge

12.2v = 50% charge

Yeah, thats bascialy what we do too.

- However the table we use is very slightly diffrent to that one (which we also have on the boat) and was the table given with the prevous set of batterys.

- However, the diffrences are only really to the second decimal place and get smaller towards the '50%' readings.

 

As ive mentioned before, basicaly all we have is an large analogy ammeter mesuring the current in/out of the domestic batterys, and a voltmeter messuring the voltage accross the domestic batterys. The latter being read in the morning before using any electrical appliences, and used with the above table to give an indication the charge left. The former (ammeter) being used soley to give an indication on instantainous drain or charge.

 

We then also have a crude ammeter the mesure the output of the alternator, an ammeter to messure the current going into (but not out of) the bow batterys (liek starter batterys, but for the bowthruster).

- And plans to fit a voltmeter to the bow batterys also, one day.... (for now we jsut wave the fluke at it)

 

 

Daniel

[Keeping Up]

Also, the above readings MUST be carried out with NO LOAD on the battery or the result is meaningless. Do not take a reading immediately after charging because it will be artificially high - switch on your tunnel headlight for 10-15 minutes first, switch the light off again and then take the reading.

Chris,

 

Is there a similar recovery period after switching the load off? I'm planning to check the capacity of my batteries by applying a steady load to my previously-charged batteries (say 15 or 20 Amps for my 330Ah bank) and taking a voltage reading every few hours (with that load switched off) until they have discharge to 50%. But would the readings be valid immediately after switching the load off, or do I need to wait 10-15 minutes each time then too?

 

Allan

[chris w]

Chris,

 

Is there a similar recovery period after switching the load off? I'm planning to check the capacity of my batteries by applying a steady load to my previously-charged batteries (say 15 or 20 Amps for my 330Ah bank) and taking a voltage reading every few hours (with that load switched off) until they have discharge to 50%. But would the readings be valid immediately after switching the load off, or do I need to wait 10-15 minutes each time then too?

 

Allan

 

Allan

 

Yes, there will be a similar "equalisation" period. In both cases, it's caused by water and acid not being mixed properly. On charge there is an abundance of unmixed acid around the terminals immediately after charging and, on discharge, an abundance of unmixed water around the terminals. If you leave the batteries for a while after discharge, their terminal voltage will increase to a truer reading. This is entirely due to the proper remixing of the water and acid.

 

If you take two lots of discharge readings, say one at 20 amps and another at 5 or 10 amps, you will be able to deduce, by calculation, the Peukert Factor for your own set of batteries and therefore know a lot more about their performance under load.

 

The equation you need is:

 

n = [Log R₂ /R₁] / [Log (C₁/R₁)- Log (C₂/R₂)]

 

 

where:

n = the Peukert Factor for your batteries

R1 = the first discharge rate (say 20 amps)

R2 = the second discharge rate (say 5 amps)

C1 = Ampere-Hour figure from your experiment @ the R1 rate

C2= Ampere-Hour figure from your experiment @ the R2 rate

 

If you discharge to 50%, don't forget to double the AH figures derived, before entering them into the formula above.

 

If it's of any interest I can send you the derivation of the formula as I worked it out recently from first principles for inclusion on Gibbo's "Smartgauge" webpage on Peukert maths.

 

The answer for "n" will be somewhere around 1.2 - 1.3 probably. If you can derive this figure, then you will be able to calculate the true Peukert capacity of your batteries and therefore the real time to discharge to a certain point at various currents.

 

Chris

Edited February 15, 2007 by chris w

[chris w]

Allan

 

Doing a quick calculation and assuming your batteries are quoted as 330AH at the 20 hour rate, then the true capacity is around 765AH. The discharge times to 50% charge versus discharge current will be approx as follows (by calculation):

 

5 amps: 47 hours

10 amps: 19 hours

15 amps: 11 hours

20 amps: 8 hours

30 amps: 4.5 hours

50 amps: 2.5 hours

 

Chris

[Gibbo]

If it's of any interest I can send you the derivation of the formula as I worked it out recently from first principles for inclusion on Gibbo's "Smartgauge" webpage on Peukert maths.

 

And a damned fine job you did of it too. I had a mathematical question about it recently. I gave him your email address

 

Now here's one for you.

 

If you charge a battery, or discharge a battery, when you "let go" of it, the voltage starts to return to the natural "resting voltage" for its current charge status. It's exponential, so technically it never *actually* gets there. The longer you wait, the better the accuracy

 

Measuring the voltage after a "quick discharge with a headlamp to get rid of the surface charge" doesn't actually work. You might not have *quite* reached it (so the voltage reading is too high), or you might have gone past it (so the voltage reading is too low). And you have *no* way of knowing. If you get what I mean. 24 hours rest seems to work within about 5%. Obviously this isn't practical if you actually want to *use* the batteries.

 

But...... As it's exponential, we can work it out from the slope of dv/dt during say the first couple of minutes after removing the charge or discharge.

 

This is one of the tricks my battery monitor uses, but that's a computer.

 

Is there a simple "maffs" way of doing it with a pen and paper?

 

Gibbo

[chris w]

I do indeed "get what you mean" and agree with you. As the voltage change is a decaying exponential, it will have the form:

 

V = A.e^-(sT)

 

where V = battery voltage, T = time, e = 2.718 etc and A and s are constants to be derived by experiment. By measuring dv/dt at various points, we could solve for A and s. However, and this is where bloody Mr Peukert sticks his oar in again, I suspect that A and s will be different for different batteries and different for different banks of batteries.

 

It would be a neat experiment to set up though. It never fails to amaze me how little info is actually available generally on the true "mechanics" of such a supposedly simple device as a lead-acid battery.

 

Possibly a more convenient way to solve your enigma would be to charge the batteries and wait 24 hours before measuring the terminal voltage, then comparing this to the same scenario but switching on the headlight for 10mins, 20 mins, 30mins etc etc until the terminal voltage was seen to be stabilising at or near to the 24 hour figure. This way at least one could gauge (or even smartgauge!) a more correct figure for the headlight ON time.

 

Chris

PS: just had a thought. I wonder if there might be a correlation between "sT" and the Peukert factor. Intuitively I bet there is.

 

PPS: another thought. What if "sT" turned out to be the Peukert factor! That would be interesting and then we would only need to solve for A.

Edited February 16, 2007 by chris w

[scribe]

You guys are legends.

I think I have some cheap stuff on board and in time I will upgrade for reasons you all make clear.

A long time ago I did my 2 year "nashos" in signals, and battery management was serious stuff then.

I should explain all this (believe me-there is heaps of it!!!) is being copied to my lap top so that on board I can refer to it when I meet a prob.( Here i have broadband-but once underway I am on my Todd)

An adventure on UK waterways has me amazed I can still make use of maths lectures some 50+ years ago.

Thanks.

Doug

 

PS The battery condition meter shows 11 volt, and its "semi" charged!!!!!!!

Edited February 16, 2007 by scribe

[Gibbo]

I do indeed "get what you mean" and agree with you. As the voltage change is exponential, it will have the form:

 

V = A.e^(sT)

 

where V = battery voltage, T = time, e = 2.718 etc and A and s are constants to be derived by experiment. By measuring dv/dt at various points, we could solve for A and s. However, and this is where bloody Mr Peukert sticks his oar in again, I suspect that A and s will be different for different batteries and different for different banks of batteries.

 

The way I do it in the monitor is to take 3 voltage readings at short intervals, check to see if there's enough difference between them to get meaningful results (if there isn't, do it again at longer intervals) then "curve fit" to the three points and predict the voltage after 12 hours. This is merely one of the many strategies used. This task is trivial for a computer but not so for a brain.

 

It keeps a log of the minimum time required between readings and uses that as the starting point next time. This remains roughly constant as the battery ages. Having also "sucked" the info out of units in installations it's pretty constant for all batteries of any one type. You need to wait about 30 mins after removal of the load or charger (not saying how I detect that) then about 15 mins between each reading to get meaningful results. Obviously the longer you wait between readings, the more accurate the final result will be. But you never know when something might get switched on so it keeps going and keeps a record of the previous readings, finally settling on using the longest time interval it got before something loused it up.

 

It *does* follow a perfectly exponential curve, there's no deviation at all that I could detect. I can take the three readings at any time, and the same algorithm gives the same answer every time.

 

So given three voltage readings, what's the simple maths to predict the voltage in say 12 hours time?

 

Gibbo

 

PS The battery condition meter shows 11 volt, and its "semi" charged!!!!!!!

 

?

 

Gibbo

[Keeping Up]

PS The battery condition meter shows 11 volt, and its "semi" charged!!!!!!!

?

Some of these meters are pretty expensive. Maybe Scribe has only half-finished paying for his?

[chris w]

So given three voltage readings, what's the simple maths to predict the voltage in say 12 hours time?

 

Gibbo

 

Right, the maths:

 

since for a decaying exponential curve v = Ae^-sT , where A and s are unknown constants, then at time T = 0 (ie: your initial reading) v(0) = A since e^-s0 = e^0 = 1 so we have one unknown, viz: A.

 

Take a second voltage reading v(1) at a time T(1). Since we know A, we may write:

 

v(1) = Ae^-sT(1), therefore e^-sT(1) = v(1)/A

 

therefore sT(1) = -Ln[v(1)/A] (where Ln means the natural logarithm to the base e)

 

therefore s = -1/T(1) x Ln[v(1)/A] so we can now solve for s

 

Therefore to predict the voltage v(n) at time T(n) in the future:

 

v(n) = Ae^-sT(n) and since A, s and T(n) are now known, we may solve for v(n)

 

 

 

An example may help:

 

Suppose we measure the battery voltage initially at time T = 0 and it's 12.6v.

 

2 hours later we measure it and it's reading 12.4v. (ie: @ T = 2)

 

We want to know the voltage at time T = 12 hours.

 

Assuming exponential decay, ie: v = Ae^-sT then at time T = 0, v = A = 12.6 since e^0 = 1

 

therefore rearranging the equation s = -1/T x Ln(v/A)

 

therefore at time T = 2, s = -1/2 x Ln(12.4/12.6) = 0.008

 

Therefore at time T = 12, v = 12.6 x e^(-0.008 x 12) = 11.44v

 

Chris

Edited February 16, 2007 by chris w

[chris w]

this post appended to bottom of post above

 

Chris

Edited February 16, 2007 by chris w

[Gibbo]

Right, the maths:

 

since for a decaying exponential curve v = Ae^-sT , where A and s are unknown constants, then at time T = 0 (ie: your initial reading) v(0) = A since e^-s0 = e^0 = 1 so we have one unknown, viz: A.

 

Take a second voltage reading v(1) at a time T(1). Since we know A, we may write:

 

v(1) = Ae^-sT(1), therefore e^-sT(1) = v(1)/A

 

therefore sT(1) = -Ln[v(1)/A] (where Ln means the natural logarithm to the base e)

 

therefore s = -1/T(1) x Ln[v(1)/A] so we can now solve for s

 

Therefore to predict the voltage v(n) at time T(n) in the future:

 

v(n) = Ae^-sT(n) and since A, s and T(n) are now known, we may solve for v(n)

An example may help:

 

Suppose we measure the battery voltage initially at time T = 0 and it's 12.6v.

 

2 hours later we measure it and it's reading 12.4v. (ie: @ T = 2)

 

We want to know the voltage at time T = 12 hours.

 

Assuming exponential decay, ie: v = Ae^-sT then at time T = 0, v = A = 12.6 since e^0 = 1

 

therefore rearranging the equation s = -1/T x Ln(v/A)

 

therefore at time T = 2, s = -1/2 x Ln(12.4/12.6) = 0.008

 

Therefore at time T = 12, v = 12.6 x e^(-0.008 x 12) = 11.44v

 

Chris

 

Nicely done..... smartar*e

 

I smell a webpage coming on......

 

Gibbo

[Keeping Up]

Allan

 

Doing a quick calculation and assuming your batteries are quoted as 330AH at the 20 hour rate, then the true capacity is around 765AH. The discharge times to 50% charge versus discharge current will be approx as follows (by calculation):

 

5 amps: 47 hours

10 amps: 19 hours

15 amps: 11 hours

20 amps: 8 hours

30 amps: 4.5 hours

50 amps: 2.5 hours

 

Chris

Thanks.

 

Unfortunately on an 11 Amp load, starting from full charge the results were:

 

After initial rest: 12.50 volts

After 1 hour: 12.25

After 2 hours: 12.13

After 3 hours: 12.00

After 4 hours: 11.78

 

I know they'd gone below 50% by then - but they're heading for the skip next so it doesn't really matter!

 

Allan

[chris w]

Thanks.

 

Unfortunately on an 11 Amp load, starting from full charge the results were:

 

After initial rest: 12.50 volts

After 1 hour: 12.25

After 2 hours: 12.13

After 3 hours: 12.00

After 4 hours: 11.78

 

I know they'd gone below 50% by then - but they're heading for the skip next so it doesn't really matter!

 

Allan

 

 

Yep, they seem totaly kn*ckered. Interestingly, if one plots your results, the resultant discharge curve is approximately linear rather than bearing any resemblance to an exponential decay. This would seem to imply that the decay time constant is being swamped by a high internal resistance.

 

Chris

[Gibbo]

Yep, they seem totaly kn*ckered. Interestingly, if one plots your results, the resultant discharge curve is approximately linear rather than bearing any resemblance to an exponential decay. This would seem to imply that the decay time constant is being swamped by a high internal resistance.

 

Chris

 

The exponential curve is the recovery curve back to the "at rest" voltage for a particular charge state after charge or discharge. It's "perfectly" exponential.

 

The discharge curve is *roughly* a straight line between 90% and 40%. 100% to 90% is steeper and slightly exponential. 40% down to zero percent is steeper and "inverse" exponential. Usually, under basic tests, the graph comes out almost completely straight because the load is a resistance hence it draws less current as the charge status (and thus voltage) becomes lower. If the load is a constant current the graph is an s-bend. Can you tell I have gigabytes of test results on this ?

 

Gibbo

[chris w]

The exponential curve is the recovery curve back to the "at rest" voltage for a particular charge state after charge or discharge. It's "perfectly" exponential.

 

Usually, under basic tests, the graph comes out almost completely straight because the load is a resistance hence it draws less current as the charge status (and thus voltage) becomes lower.

Gibbo

 

Because the Peukert effect is caused by the inability of the battery to instantly mix the additional water, produced by discharge, with the remaining acid in the battery, the voltage under load drops exponentially not linearly. Therefore the current through a fixed load resistor will also drop exponentially. If this were not so, the Peukert effect would be the Peukert non-effect. Similarly on recharge the inabilty to instantly mix the fresh acid produced with the remaining water means that the terminal voltage increases exponentially.

 

Discuss

 

Chris

Edited February 19, 2007 by chris w

[Gibbo]

Because the Peukert effect is caused by the inability of the battery to instantly mix the additional water, produced by discharge, with the remaining acid in the battery, the voltage under load drops exponentially not linearly. Therefore the current through a fixed load resistor will also drop exponentially. If this were not so, the Peukert effect would be the Peukert non-effect. Similarly on recharge the inabilty to instantly mix the fresh acid produced with the remaining water means that the terminal voltage increases exponentially.

 

Discuss

 

Chris

 

If I must......

 

That would be a good argument (and correct), if that was what caused Peukert's effect. But it isn't

 

It's a *tiny* part of it. The inability to mix is a very high part of surface charge and surface discharge (battery recovery). The biggest part of Peukert's effect is that the resistance of the elecrolyte isn't linear. It is proportional to the square of the current passing through it. Ditto the resistance of the interface between electrolyte and plates. Both these are also proportional to the square of the *depth* if discharge. This is what causes the largest proportion of Peukert's effect.

 

Then you have to consider that drawing more current, warms up the battery (due to internal resistance), which speeds up the reactions and allows it to produce more current. It also produces bubbles which mixes up the electrolyte. Both these effects have a time delay on them.

 

All the effects combine in a *very* complicated way.

 

And the discharge graph I described is indeed correct for *all* types of lead acid batteries. The start and end points vary slightly, the extent of the variations vary slightly, but the overall graph is indeed s-shaped. And it is indeed a straight line between *roughly* 90% and 40%

 

During charge, at a constant current, the voltage graph is almost *perfectly* straight. Once acceptance is reached the falling current is almost *perfectly* exponential.

 

Gibbo

[chris w]

Now you've confused me. So why did I derive the v= Ae^-sT equation for you the other day? That's definitely exponential and you wanted to be able to predict the terminal voltage at a particular time, given the start point and one other point. If it's linear (as you seem to be suggesting) the calculation is trivial... if it's exponential you need the equation above.

 

Chris

[Gibbo]

Now you've confused me. So why did I derive the v= Ae^-sT equation for you the other day? That's definitely exponential and you wanted to be able to predict the terminal voltage at a particular time, given the start point and one other point. If it's linear (as you seem to be suggesting) the calculation is trivial... if it's exponential you need the equation above.

 

Chris

 

After *removal* of a charge source or load, the terminal voltage will be higher or lower than the "at rest" voltage for the particular charge state. After *removal* of the charge source or load the terminal voltage returns to the "at rest" voltage exponentially. The equation you derived was to "get at" the "at rest" voltage given two voltage readings taken a short time after *removal* of the load or charge source.

 

Gibbo

[chris w]

As an analogy, I have always presumed (maybe incorrectly) that a lock's emptying through a paddle is exponential - ie: the "current" through the paddle decays exponentially. Maybe it's linear - any thoughts?

 

Chris

[scribe]

As an analogy, I have always presumed (maybe incorrectly) that a lock's emptying through a paddle is exponential - ie: the "current" through the paddle decays exponentially. Maybe it's linear - any thoughts?

 

Chris

 

 

Chris,

Its exponential-you are safe on that one. I would draw you a diagram recalled from physics lectures yonks ago. There is this empty baked bean tin with a row of holes down the side and you fill it with water and watch the pattern of the water flows.

Oh, that everything else on a narrowboat could be that easy.

Doug

[chris w]

Thanks. Yes, I thought as much because the water pressure (the depth of the water in the lock) in forcing water out through the paddle actually reduces its own pressure through the water loss - hence exponential decay.

 

I would have thought the same analogous situation woud occur inside a lead-acid battery in that in driving current through a load, the voltage decays and thereby reduces the current which in turn reduces the rate of voltage decay which means the rate of decay is therefore not linear. I was genuinely surprised by Gibbo's data.

 

Chris

[Gibbo]

Thanks. Yes, I thought as much because the water pressure (the depth of the water in the lock) in forcing water out through the paddle actually reduces its own pressure through the water loss - hence exponential decay.

 

I would have thought the same analogous situation woud occur inside a lead-acid battery in that in driving current through a load, the voltage decays and thereby reduces the current which in turn reduces the rate of voltage decay which means the rate of decay is therefore not linear. I was genuinely surprised by Gibbo's data.

 

Chris

 

Right I see your point now.

 

That would be the case with a capacitor (which merely stores charge in an electric field) running into a load. And it does indeed decay exponentially into a fixed resistance load. However a battery doesn't "store charge", it "makes" it as a result of chemical reactions. Hence the difference. Hence the voltage does not fall linearly with a constant current load (as it would with a capacitor).

 

Gibbo

[chris w]

Right I see your point now.

 

That would be the case with a capacitor (which merely stores charge in an electric field) running into a load. And it does indeed decay exponentially into a fixed resistance load. However a battery doesn't "store charge", it "makes" it as a result of chemical reactions. Hence the difference. Hence the voltage does not fall linearly with a constant current load (as it would with a capacitor).

 

Gibbo

 

 

But the battery does have a finite amount of stored charge - its capacity (Peukert or otherwise). It doesn't make it on the fly - it starts off with whatever capacity it was charged to and that capacity gets used up on discharge.

 

Chris

[Gibbo]

But the battery does have a finite amount of stored charge - its capacity (Peukert or otherwise). It doesn't make it on the fly - it starts off with whatever capacity it was charged to and that capacity gets used up on discharge.

 

Chris

 

But it isn't "stored charge". It is "converted charge" that has been converted into chemical potential energy. That behaves nothing like "stored charge".

 

Gibbo