O.K. hot news, what do you think of this ? 24v bank of lead batteries = 12 cells, fully charged = 2.12v/cell = 25.4 volts.
LEAD ACID BATTERY CHARGING INFORMATION, BATTERY TYPES AND OPERATION - SOLAR NAVIGATOR WORLD ELECTRIC NAVIGATION CHALLENGE, NELSON KRUSCHANDL, BLUEBIRD ELECTRIC LAND SPEED RECORD CARS
At 70% depth of discharge, cell volts = 2.05 = 24.6v this is where I need the supplemental power supply needs to kick in. Going below this voltage level shortens battery life.
So the regulated dc voltage supplies, rated 20amps each @ 24v, are adjusted up to 25v. Now, taking the battery bank as a zero-ohm impedence, 24.6vdc constant voltage source.. at this discharge time.. the current flow from the power supply with a .05 ohm series resistor would be (delta V)/R = (25-24.6)/.05 = 8 amps.
The power supply(s) can handle this, but the total draw down by the motors which is much 3x more than the electronic power supplies can provide, will eventually bring the battery bank down to 50% depth of discharge = 2.01 v/cell = 24.1v for the bank.
In this reduced battery condition, the regulated 24vdc power supplies, with .05ohm resistors in series will see (delta V)/R = (25-24.1)/.05 = 21 amps
So, it looks like with a .05ohm resistor in series with the 25v set-point power supplies, they will be working withing their standard ratings.
Feed back, please !
The max resistive power loss would be IxIxR = 400x.05 = 20 watts ??? I can live with that !
That resistor could be #14 coiled copper wire ( @ 2.58 ohms/1000ft) = 20ft wire = .0515 ohms
With extended runs that draw batteries below this voltage, the electronic power supplies would have to be disconnected. Thats no biggie.
So, I guess Im answering my own posting. If you have observations on this , they are greatly appreciated. My goals for using these switch-mode power supplies is to save cost and weight on the boat.