DC voltage drop over long wires to bilge pump

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G'day guys,

I have just set up a bilge pump (1150GPH) with about 8 Metres (fused) of cable between the pump and the Anderson plug that I can connect to my 100 Ampere hour deep cycle battery. I wanted long lead so I can connect to my battery in my 4WD and not have to be right on top of the water supply.

When I connect it all ...it runs OK but I reckon the bilge pump doesn't pump as much volume as when it just had the 1 metre wire coming out of the bilge pump.(before I added the 8 metres and the fuse)

I am wondering If I am suffering from voltage drop because of the 8 Metres wire.

1150 GPH bilge seemed to be OK to run into my mini McKirk sluice.

Is voltage drop my problem...if so..how do I work out what's going on. ( I am not a real techo but I can dig sh*t loads of dirt = digger/grunt)
thanks in advance
Mike
 
Could well be..

And the losses will become bigger with any load. V= I x R

So with no load you may measure the correct voltage at the output of the run. But under load is very different as a vew variables come into play.
 
While the pump is running measure the voltage over the battery and the voltage over the pump. The difference is the voltage loss over the wire. Note that this loss will be over both wires as the positive and negative wires extend an equal distance from your vehicle, the loss over a single wire is half what you measure.
Jon
 
1482058012_dc_wire_selection_chartlg.jpg


This might help
 
Hi mate sounds like you have voltage drop 8metres is a decent run for a 12v system. Voltage drop doesn't change when connected to a load u can measure the voltage at the battery then go to the end of the cable and measure the differenc there without it connected to a loadd. What you ll find is when you are running a load ur actual battery voltage will decrease slightly this is perhaps where the confusion lies. What I would suggest is either increase cable size or change to a 24volt system where u can have longer runs (most trucks are 24v being far longer then a car which is 12v) .
 
Meroo the battery will decrease slightly due to internal resistance under load. V=IR if there is 0 current (I) then there is 0 voltage (V) drop on the wire.
Jon
 
Your supply voltage at the end of the 8m is what he's worried about not how much voltage he ll get on the other side of his load which is the voltage drop your referring to. If you measure your battery voltage and get say 13v at the terminals then connect say a 5 m cable and measure the voltage at the end of the cable you will get less voltage depending on the size of the cablee irrespective if it's connected to a load or not. this is due to the resistance of the cable. bigger cable less resistance equals less voltage drop and can draw a higher current. Cable selection is what he needs to know there are heaps of online calculators that will automatically calculate the correct cable size based on voltage ,distance,load etc
 
To answer the question you need to know firstly how much current your pump draws. Should say it on it if it's a watt value simply divide this by 12 to get your current example 100w light divided by 12v equals 8.3 something amps off the top of my head . If the pump draws more then what the cable is rated for then there's your problem. Most cable will have a rating on it when you buy it if you got the cable from the back of the shed then good luck unless u know what to look for lol. So if the cable is rated to the current then your easiest bet is to use an online cable selector to give you an idea of what to use. If you want to get adventurous and have a multimeter you could join the 8 m cable at one end then go the the other end and put ur multimeter on ohms. This reading will give you the resistance of your cable. Now you can use ohms law to work out how much voltage you are losing across your cable say your pump draws 10amps. Say for argument sake the resistance of the cable is 0.1 ohms. 10ampsx0.1ohms equals 1 volt. So if your voltage is 12v from your battery then at the end of your 8m run you ll have 12-1 equals 11volts. Your pump may need a minimum 11.5 volts to work efficiently
 
If you get an 5m length of wire and measure the voltage over it you'll get 0 volts because there is no current running through it.
Jon
 
I have the same issue with my 12v fridge in the car, it is a companion. What happens is the high in rush current on fridge startup causes an increased voltage drop over the length of the lead and the fridge will not start because it thinks the battery voltage is too low.

My solution was to buy from china a buck boost regulator. this regulator is a little different in that as the input voltage reduces it will always maintain a constant output to your device (settable via a small pot on the cct board) , in other words it draws a little more current from the battery to maintain that constant output voltage to your device. it is 98% efficient according to my tests. It is only good up to 10A or so. It can be used down to 5v on the input and up to 24 volts on the output and the thing to remember is that instead of your device showing degrading performance as the battery voltage decreases it will go fine up to the point that the regulator input voltage is too low and will turn itself off and that is between 5 and 8V.

https://www.aliexpress.com/snapshot/6348951071.html?orderId=64654330751044&productId=1690655672

Cheers

Barry
 
Meroo said:
Thats not right mate if your not convinced get yourself a battery andd some cable and see for yourself.

You first, you're saying there's a measurable voltage drop when there is no current running through the wire. That defies the laws of physics.

You have now described a load is required to cause a voltage drop over the wire in your own post #13 with a 10amp load:

"how much voltage you are losing across your cable say your pump draws 10amps."

Let's not confuse the OP, he's not technical. Happy to take it offline through PM.

Jon
 
Lol I have no need to I've done it a million times. No pms you don't need a load connected to a circuit to measure voltage drop plain and simple. get a battery and it's 13v. Get 8m of cable hook up the positive and negative measure at the other end of the cable with a voltmeter you may have say 12.6v. the loss is the resistance of the cable only very miniscule .Basically what your saying is you can't measure voltage off a battery terminal cas it's not connected to a load.. absolute bollocks. So when you have cables connected and measure the voltage at the other end of the cables it's exactly the same thing only you will register a slightly lower reading as you are getting a resistance loss over the cables and that is your voltage drop. :D
 

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