Voltage drop question

Futbalfantic

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Aug 5, 2006
Location
Charlotte
So I'm wiring up 2 pumps and a bilge pump on my boat. I want all 3 to be able to run at the same time with <5%* voltage drop. The run for the wire is 15-20 feet. The pumps draw 20amps and the bilge draws 15 amps. Based on all of the charts that I have found the two 20amp pumps call for 10-8ga wire and the 15 amp pump calls for 10ga wire. What I want to do it combine the grounds for all of these pumps into one 4-2ga wire. How can I calculate the voltage drop for this? How do you calculate voltage drop for multiple wire sizes in a circuit?
 
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Why do you want greater than 5% voltage drop?

Just kidding. I'll write out an answer soon after I eat, I'm all over this topic.
 
Voltage drop is really just a function of wire resistance for a certain wire size with a certain current through it. Ohm's Law says that V=I x R, so voltage = current x resistance.

So if you have a fairly generic copper wire of a certain gauge, that wire gauge is a fairly standard cross sectional area of copper, and that area has a certain amount of resistance per length (there's charts for resistance of standard gauge at a standard temperature, etc). Larger wire has larger cross sectional area, and less resistance per foot. Anyway, the resistance part of Ohm's law will be the resistance per foot of your wire gauge, times the number of feet.
There are some other things that add resistance, like terminals, etc., but you can fairly well neglect those for a back-of-the napkin calculation like you're doing.

So you know the wire length, and you know the current draw per device, and you know the max voltage drop that you want.

For the ground, you don't need to calculate voltage drop for multiple wire sizes, because you're just dealing with individual devices that happen to all be connected to the same voltage source and the same ground. All the devices are common to that ground wire, so you have a total current (I), a wire length and a resistance per length of wire of that gauge (that gives you the R). If you want less than 5% voltage drop, (V) then becomes 5% of 14.4vdc (or the operating condition you're looking for; engine running, not running, whatever).
Do that first, because then you can include that ground voltage drop in your other calcs (your max acceptable voltage drop is now 5% minus the ground voltage drop).

If you want to calc the voltage drop for each device and each wire size, that's easy too. You have a pump with a certain current on a length of wire of a certain size and length. You have a bilge pump with a different current on another piece of wire of a certain size and length. They're different devices on different wires.


DC voltage drop is much simpler than AC voltage drop, especially when you're going on the conservative side and can ignore the small bits that don't really matter.

You can do the rough calcs, pick the closest wire size and call it good, or go one wire size up for extra margin, or just accept that (for example) 4.1% drop really isn't any better or worse than 5.1% voltage drop because it's a just a boat. Basically, calculate the big important stuff, ignore the rest.

You're not trying to run the smallest wire possible for temperature rise to save weight (racing), or trying to save money by using the least copper possible, etc.
 
This is actually very simple to do, you just need to realize that it can be broken apart into chunks that don't really affect each other. Then the math becomes very easy, unless you're allergic to basic algebra. :p It's really just one formula, applied to each part by itself.
 
Guessing since you said "combine all grounds" that it's either 12 or 24vdc. So assuming you're using 12vdc, assuming all pumps are actually pulling Max amperage and assuming you're using 4ga copper at .308ohm/Kft, the last pump will see a 5.65% drop.

: Edit: 2 late
 
Guessing since you said "combine all grounds" that it's either 12 or 24vdc. So assuming you're using 12vdc, assuming all pumps are actually pulling Max amperage and assuming you're using 4ga copper at .308ohm/Kft, the last pump will see a 5.65% drop.

: Edit: 2 late

If they're wired in parallel (all parallel between power and ground), how is there a last pump? Don't understand....
 
So how do you add those together? Calculate the voltage drop for the 55 amps on the ground wires then calculate the 2-20 amp and 1-15 amp voltage drop then.. what average them together, add them together?

Also the motors themselves have a 2 foot leader on them. Calculate the voltage drop on those and same as above?
 
So how do you add those together? Calculate the voltage drop for the 55 amps on the ground wires then calculate the 2-20 amp and 1-15 amp voltage drop then.. what average them together, add them together?

Yep, you've got it. They add together for each device. So each device has the voltage drop from its individual wire between the battery and device, plus the voltage drop from the ground wire.

For rough calc, just assume the 2 foot leader is the same gauge as the smallest wire feeding the device. The difference in such a short length will be small enough to not worry about.
So if you've got 8 feet of 12 gauge, plus a 2 foot leader, just say it's 10 feet of 12 gauge.
 
so from the internets I found a calculator. Figures I got.

For a 2 foot leader (total of 4') a voltage drop @ 20 amps of 0.4031 V or 2.7609%
10' of 10ga @ 20 amps --------------------------------------- 0.3986 V or 2.73%
10' of 2ga @ 55 amps (ground combined) ------------------- 0.1715 V Or 1.1745%
Totaling---------------------------------------------------------- 0.9732 V or 6.6654%

Is that right? Seems high?
 
Do the pumps actually draw 20a each, or is that just the recommended fuse size?
 
They draw upto 20 amps (recommended is actually 25 amps.) but it's based on head. It's an open system with max 3' of head
 
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