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2.4 Battery upgrade

S

StuH

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Location
Lancashire
Car
Superb Estate 3.6 V6
The short and infrequent journeys I'm currently doing are taking their toll on the battery. I had a new one fitted a few months back at the local car electrical garage and he was laughing at how small the battery was. The mechanic next door couldn't believe the battery size when he popped his head under the bonnet yesterday. I've been looking at larger batteries but having real difficulty finding something to fit. Has anyone ever managed to fit a bigger capacity battery to the 2.4?
 
not me, but if you find something I'd like to know, I agree that the battery is a bit small
 
It got me thinking ....

There are two factors for a car battery: the Ampere-hours (Ah) ; and the Cold Cranking Amps (CCA).

For a car battery, Ah gives you the amount of time that a small to moderate current can be maintained at 12V. So a 30 Ah battery would be able to maintain 30A for 1 hour or 15A for 2 hours.

CCA is the amount of current a battery can provide at 0 °F for 30 seconds while maintaining at least 1.2 volts per cell (7.2 volts for a car battery).

However, there is no formula that relates Ah to CCA. So if you know the Ah of a car battery, you cannot calculate the CCA of that battery. And vice versa.

But, CCA is related mainly to plate surface area in the battery, while Ah is related mainly to the amount of material in the plates. This means that two batteries with the same amount of plate material, one with a lot of thin plates and the other with a few thick ones, could easily have the same Ah rating but very different CCA.

As a guess looking at the shape of the battery in the 2.4 Accord, it has a high CCA but not a very high Ah.

The battery is Panasonic 46B24L(S)-MF so I've googled that and found that its capacity is 45 Ah with a CCA of 330 Amps.

45Ah is not a lot for an Accord with all that on-board electronics, although you are unlikely to leave the ignition on and everything else on without the engine running. But if you went on holiday for 4 weeks over xmas into January, and left the car in airport parking with the alarm on, then I would think that there is a chance that the car won't start when you come back (4 weeks = 672 hours, so a current of 67mA or more will drain enough from the battery to seriously affect the CCA Volts, and I bet the alarm draws more than 67mA).

But otherwise, the battery should be OK for normal use in UK climate.
 
TBO i would stick with a genuine Honda battery give HH a call as there were some good deals on these a wile back.
 
Panasonic 46B24L(S)-MF is the battery that Honda supply for the 2.4 btw (45 Ah, CCA 330 Amps)

The original, and a replacement from HH, both that one (my car 2.4)

EDIT
Anyone know the battery in the diesel, and googled it to get the Ah and CCA ?
 
I'd go along with Brian's comments. Unless the journeys you're doing are stupidly short the battery should cope. Before changing the battery I'd be checking that the alternator's charging OK and that you do not have a problem with something not shutting down properly when the engine's switched off.

Higher capacity batteries (60 Ah) of the same physical size are avaiable, at a price. However they won't really solve the problem. The battery will still go flat but it'll just take longer to do so.
 
I've done quite a bit of asking around this morning with no joy. Whilst I do agree the battery should be okay, I just think that a 45aH battery is too small but typical of Jap cars.

One thing I have noticed is that I can move the positive clamp on the battery terminal, it wont come off but I can spin it even though the clamp is as tight as possible. I'm sure that can't be helping matters.
 
No that certainly won't help, it introduces a voltage drop when cranking, and also possibly when charging.

V = I x R

A lose connection has some "R". So if "I" is big (cranking) then the "V" across the "R" is big. If "I" is not so big (charging) then the "V" across the "R" is not so big, but it still exists.

If the battery is not Panasonic 46B24L(S)-MF then that might be why the clamp is lose, because the pillar diameter might be a couple of mil too small. Alternatively, and anyway, maybe the clamp was not tight previously, and someone has been spinning it about and worn some of the lead from the pillar.

The reason why this used to happen with lead clamps, was that the lead clamp would stretch. I don't think that these clamps will stretch, but I could be wrong.

EDIT: I think someone told me a looooong time ago that the resistance across a slightly lose clamp is inversely proportional to the current being drawn. So the higher the current, the smaller the resistance across a lose clamp, due to heat being created. The smaller the current, the larger the resistance.
If this is true, then take a hypothetical example: Resistance across the loose clamp is 0.1 Ohm, when charging at 10 Amps, 1 Volt is dropped. If resistance drops to 0.01 Ohm when 300 Amps is drawn when cranking, then dop is 3 Volts, and 9V 300A into starter motor is still enough to crank.
No idea how true this is, but if there is any truth to it, then one would never quite reach full charge across a loose clamp, yet still be able to crank ...most times that is.
 
Well the battery wouldn't come off last night but after brief drive this morning it came clean off with little pressure. No wonder I've been having problems! A quick call to a mechanic I know asking if he know where I can get shims resulting in a quite bizarre solution; he said forget shims just loosen the clamp, slip a small screw between the clamp and battery post and tighten. Not only does it give a tight connection the thread beds into the lead terminal and clamp and gives more resistance to slipping,or something. At first I wasn't sure he was serious but decided to do it and got to say it seems to have worked. Starting is easier, the engines idles lower much quicker and it seems to be driving much better - not sure if that's a placebo...

Never heard of the latter Brian but sounds feasible.
 
/\ Sounds good now at least.
I can't remember when/where I got that point from, it might be an old wife's tale passed around and just based on experience of many people. Maybe something for Mythbusters, or Bang Goes The Theory :lol:
 
StuH said:
Well the battery wouldn't come off last night but after brief drive this morning it came clean off with little pressure. No wonder I've been having problems! A quick call to a mechanic I know asking if he know where I can get shims resulting in a quite bizarre solution; he said forget shims just loosen the clamp, slip a small screw between the clamp and battery post and tighten. Not only does it give a tight connection the thread beds into the lead terminal and clamp and gives more resistance to slipping,or something. At first I wasn't sure he was serious but decided to do it and got to say it seems to have worked. Starting is easier, the engines idles lower much quicker and it seems to be driving much better - not sure if that's a placebo...

Never heard of the latter Brian but sounds feasible.
Click to expand...

A nice simple fix. Alright, it's a bit of a bodge but so what. it works. Resistance between the clamp and terminal post can have a big effect.
 
The difference is beyond belief. It feels lively again, the climate is cycling in and out again and the air is freezeing (thought I was due a air con recharge). It now takes one crank to starthe and the sat navime starts up first time - it still doesn't like an interrupted power supply mind. Two lessons learnt; One, don't trust the work garages do and two, regularly check the terminals!
 
/\ absolutely, and good news all round :D

I just worked out that, ***uming the CCA cranking current is initially 300 amps on the Accord and that there is initially no voltage drop to the starter motor, then this means that the starter motor resistance is 12 ÷ 300 = 0.04 Ohms :eek:
 
freddofrog said:
/\ absolutely, and good news all round :D

I just worked out that, ***uming the CCA cranking current is initially 300 amps on the Accord and that there is initially no voltage drop to the starter motor, then this means that the starter motor resistance is 12 ÷ 300 = 0.04 Ohms :eek:
Click to expand...

Yup, the combination of high current and low voltage demands that the whole of the starter circuit is in very good electrical condition. It's actually even worse than that, partly because of the drop in battery voltage with cranking but also because the leads and the windings of the starter motor heat up with high currents running through them and this increases their resistance. Think about it too long and you'll wonder how the whole thing manages to work reliably - well most of the time.
 
freddofrog said:
/\ absolutely, and good news all round :D

I just worked out that, ***uming the CCA cranking current is initially 300 amps on the Accord and that there is initially no voltage drop to the starter motor, then this means that the starter motor resistance is 12 ÷ 300 = 0.04 Ohms :eek:
Click to expand...
The starter motor is not a purely resistive load, there is a significant inductive element to it. It therefore has impedance along with resistance, therefore Ohm's law (for a resistive circuit, as per your calculations) doesn't properly apply. Once the starter motor windings generate a magnetic field flux (and therefore start spinning), the impedance rises. There is also another effect called back EMF whereby the motor generates its own electricity... http://en.wikipedia.org/wiki/Back_emf ...which partially offsets the demand on the battery (so appears as if the starter resistance is increasing).

As mentioned by Alan, the starter motor leads have their own resistance, but also a small inductance. The starter solenoid will also have a resistance and - most importantly - the battery has its own internal resistance. For info, the maximum current that can be drawn from any battery is when the total (series) load resistance equals the battery internal resistance.
 
Jon_G said:
The starter motor is not a purely resistive load, there is a significant inductive element to it. It therefore has impedance along with resistance, therefore Ohm's law (for a resistive circuit, as per your calculations) doesn't properly apply. Once the starter motor windings generate a magnetic field flux (and therefore start spinning), the impedance rises. There is also another effect called back EMF whereby the motor generates its own electricity... http://en.wikipedia.org/wiki/Back_emf ...which partially offsets the demand on the battery (so appears as if the starter resistance is increasing).

As mentioned by Alan, the starter motor leads have their own resistance, but also a small inductance. The starter solenoid will also have a resistance and - most importantly - the battery has its own internal resistance. For info, the maximum current that can be drawn from any battery is when the total (series) load resistance equals the battery internal resistance.
Click to expand...

I was trying to keep it simple!
 
I'm not sure that inductance will have an effect on DC current in the DC part of the circuit driven by a lead acid battery i.e the battery and cable act purely resistively, so the current down them cannot be affected by the inuctance in the motor
 
freddofrog said:
I'm not sure that inductance will have an effect on DC current in the DC part of the circuit driven by a lead acid battery i.e the battery and cable act purely resistively, so the current down them cannot be affected by the inuctance in the motor
Click to expand...
You are quite right about not having to consider impedance in a DC circuit. But the way in which the starter motor operates (brushes acting upon a rotating commutator) results in pulses of energy flowing, so no longer a DC condition. I should have mentioned this!

And the back EMF from the motor armature rotating in a magnetic field will also further complicate ****ysis by Ohm's law.
 
This could get interesting !!

If I took 60 car batteries to my local substation, opened the circuit breakers on the outputs, disconnected the feeds to the houses, connected 20 car batteries in series 3 times, and wired one bank of batteries to each phase of the circuit breakers, then closed the circuit breakers, well my money would be on the substation losing the ensuing battle, because DC will go through the substation windings, but AC will not go through the car batteries. Or maybe the whole lot would blow up. Nice job for Mythbusters to try :lol:

What I'm saying is that there can be no reactive current circulating round the loop with the car battery in that loop, so the car battery, connecting cables, and motor can only circulate active DC current in that loop. That means that, as a "black box", the starter motor is resistive. I have never taken a starter motor apart, but I presume that there are capacitors inside to circulate the reactive current internally.

The only counter argument I can think of, is that the back-emf would result in the resistance of the "black-box" appearing to increase, so less DC current would be drawn from the battery. If so, then this means that the intitial 300 Amp draw, would reduce almost immediately. Even if this is so, I doubt if it would reduce by very much at all. :(
 
freddofrog said:
If I took 60 car batteries to my local substation, opened the circuit breakers on the outputs, disconnected the feeds to the houses, connected 20 car batteries in series 3 times, and wired one bank of batteries to each phase of the circuit breakers, then closed the circuit breakers, well my money would be on the substation losing the ensuing battle, because DC will go through the substation windings, but AC will not go through the car batteries. Or maybe the whole lot would blow up. Nice job for Mythbusters to try
Click to expand...
No. The substation AC current would flow through (and thus destroy) the chain of batteries... the batteries have very low internal resistance and - importantly for your example - very low inductance. Therefore very little to prevent a huge AC current from flowing.

freddofrog said:
That means that, as a "black box", the starter motor is resistive. I have never taken a starter motor apart, but I presume that there are capacitors inside to circulate the reactive current internally.
Click to expand...
The starter motor, as a collection of electrical coils, is very highly inductive (the heavy duty internal wiring has very low resistance). There are no capacitors present in a starter motor... perhaps you are thinking of the 'power factor correction' capacitors fitted to inductive loads (such as fluorescent lamp fittings) on AC power distribution circuits (mostly in industrial premises) for phase correction purposes, thereby ensuring that energy consumption is properly measured.
 
Well all I know is that my car is miles better with a screw between the clamp and the battery post!
 
Jon_G said:
No. The substation AC current would flow through (and thus destroy) the chain of batteries... the batteries have very low internal resistance and - importantly for your example - very low inductance. Therefore very little to prevent a huge AC current from flowing.
Click to expand...
What you're saying is that the substation has far lower internal resistance than the car batteries have, so that the substation's equivalent AC source is supeior to the car batteries' equivalent DC source. Obviously I didn't intend it to be literal, so if you want to get literal, let's just say you connected a 12 V rms, 300 VA, transformer across a single car battery, now the example is real. No chance for the transformer.

Jon_G said:
The starter motor, as a collection of electrical coils, is very highly inductive (the heavy duty internal wiring has very low resistance). There are no capacitors present in a starter motor... perhaps you are thinking of the 'power factor correction' capacitors fitted to inductive loads (such as fluorescent lamp fittings) on AC power distribution circuits (mostly in industrial premises) for phase correction purposes, thereby ensuring that energy consumption is properly measured.
Click to expand...
Well if they don't have capacitors inside, then where does the reactive current flow (it can't go through the car battery) ?
If it doesn't go through the car battery, then as I stated, you're implying that the resistance (as seen by the car battery) appears to vary, though at what "frequency" it would vary, I'd like to know.

The point is : if you're saying that rective current flows through the battery, then say so, and provide a URL that says so. If you're not saying that, then the circuit is seen as resistive.
 
Rather than use EDIT:
Transformer in above should not be 300 VA, but 3600 VA, 12 V rms output, 240 V rms input, with a big 30 Amp fuse on the input .... to be sure that the transformer catches fire before the fuse blows :lol:
 
freddofrog said:
Well if they don't have capacitors inside, then where does the reactive current flow (it can't go through the car battery) ?
If it doesn't go through the car battery, then as I stated, you're implying that the resistance (as seen by the car battery) appears to vary, though at what "frequency" it would vary, I'd like to know.

The point is : if you're saying that rective current flows through the battery, then say so, and provide a URL that says so. If you're not saying that, then the circuit is seen as resistive.
Click to expand...
I'm not sure, but have I said something to annoy you? I really didn't mean to and I'm sorry if I have... I'll try to answer your latest point, but then I'm going to slowly back away from this thread!

Both the reactive and real power will flow through the battery, I haven't suggested otherwise. The 'effective resistance' - as seen by the battery - will appear to vary, mostly due to back EMF (link already provided that explains this). The "frequency" issues are hard to resolve in a pulse power circuit (as opposed to a proper AC circuit), as we are not dealing with nice sine waves, but instead have a chain of discreet pulses with a certain rise time, a peak DC level for a certain duration, then the pulse falling edge (and then the next pulse and so on). The rising and falling edges (i.e. high frequency component) of these pulses will cause reactive currents in ***ociation with the inductive load present in the starter windings (but the flat peak will be a DC condition, so will not). To properly ****yse current flow through the starter motor circuit would require time and phase domain consideration of both current and voltage, rather than simple Ohm's Law.

Enjoy the rest of your evening.
 
I did enjoy the rest of my evening thanks :D

And thanks for finally devoting a bit more time to the explanation rather than saying originally "The starter motor is not a purely resistive load, there is a significant inductive element to it. It therefore has impedance along with resistance, therefore Ohm's law (for a resistive circuit, as per your calculations) doesn't properly apply. "

This is the first time that you've said that reactive current flows through the battery, which it must do if your original statement is true.

As I've been saying, if it doesn't then the circuit is purely resisistive, whatever goes on in the coils in the starter motor.

If it does flow through the battery, then the circuit is no longer purely resistive, but it would have to be significant in comparison to the main active DC current for your statement "Ohm's law (for a resistive circuit, as per your calculations) doesn't properly apply", to be true.

Sorry, but I don't believe that reactive current from a car's starter motor can flow through the car's battery, unless you can find a paper, or a URL on the internet that says that it does (in a car's starter circuit).

Alternatively, if you have access to the equipment, do this:
On a Spark Ignitrion engine (such as the Accord 2.4), disable the ignition circuit, then crank the engine and take some screenshots that show waveforms that a massive DC current of 300 amps circulating, also has sufficient reactive currents to be able to say that "there is a significant inductive element to it. It therefore has impedance along with resistance, therefore Ohm's law (for a resistive circuit, as per your calculations) doesn't properly apply." If you don't fancy using a modern car (I wouldn't really) then use an older 2.0-ish SI car without ECU etc.

I genuinely would like to see the waveforms (otherwise you're speculating) :D
 
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