[Healeys] Electrical problem NON HEALEY

[john harper]

Micheal

No real answer but just a thought

I assume that the batteries are nominally 12V.

How do you know that all the batteries are fully charged?

AGM batteries have different changing characteristics that are not the same
as lead acid. I admit that I have not studied this in detail but there
appear to be different chargers for AGM and Lead Acid

There also might be some confusion in the current flow when charging two
banks in parallel. One bank might show that it is fully charged and turn
off the charging whereas the other bank may still need more input. If
possible try charging each bank separately and before connecting them back
together measure the voltage difference. If significant one bank would
maybe try to discharge the other.

If one of the 12V batteries is starting to have problems then check after
charging each bank and measure the nominal 12V of each battery. These
voltages should be very similar. If one is different it could cause an
overall problem.

As I say just some thoughts

All the best

On Sat, 30 Sept 2023 at 17:15, Michael Salter <michaelsalter at gmail.com>
wrote:

> This group has been a huge help in the past and I am wondering if
> there are any electrical guys out there who can help me with a DC
> electrical problem.
>
> I'm trying to help my BIL with the propulsion system for a 30 ton
> wooden boat on our lake.
> This 100 year old boat was originally steam powered but in keeping
> with modern times it has been converted to electric propulsion.
> It is fitted with a 106 volt DC motor
> At normal cruise the motor draws 45 amps.
> It has eighteen 220 amp hour AGM batteries.
> The batteries are arranged in 2 banks of 9 in parallel so each bank
> produces 108 VDC
> By my calculations ...
> For a total of 45 amps each bank would deliver 22.5 amps.
> Consisting of 2.5 amps from each of the 9 batteries.
> Each battery is 220 amp hours.
> Presuming that you want to avoid discharging below 50% the usable
> capacity of each is 110 amp hours.
> So my calculation for the running time is 110/2.5 = 44 hours !!!!
> It only runs for about 20 minutes before the amperage starts dropping
> off rapidly.
>
> Are my calculations correct?
> If not, where are they wrong?
>
> Thanks
>
> M
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