Use two cables in parallell to double area?

[jonasaberg]

Is this the right sub forum for this discussion?
I was thinking of using two 35 mm^2 (AWG 2) in parallell to get a total of 70 mm^2 to use in a windlass installation since I have the cable from another project already.

A guy I talked to adviced against this but did not explain exactly why. He talked about different load in the two cables, risk of one cable having a bad connection, inductive coupling between cables...
I can understand the risk of loosing one cable through a bad connection, but not any inductive coupling in a DC system.

So, I don't see this as a problem as long as each cable is large enough not to start a fire if the other is cut.

Can someone elaborate on this?

Some info:
The windlass is 1000 W mounted about 10 m from the battery, so at 12 V = 83 A in load.
In domestic installations in scandinavia we use 16 A / 2.5 mm^2 cable, so 6.4 A/mm^2.
This means, 83 A/6.4=12 mm^2 should be enough to avoid a cable fire. However, the voltage drop needs a larger diameter, and using a voltage drop calculator I found that 10 m between battery and load, 70 mm^2 of cable and 83 A gave a 4.5% = 0.5 V drop in voltage.

I checked in Nigel Calders book "Boat owner's mechanical and electric manual" and this fits the tables in the book for an expected voltage drop of 10% using 35 mm^2 (AWG 2) or 3% using 70 mm^2 (AWG 000).

Comments?

[Wotname]

Here is a recent similar discussion
http://www.cruisersforum.com/forums/...re-179794.html

[jonasaberg]

Originally Posted by Wotname

Here is a recent similar discussion
http://www.cruisersforum.com/forums/...re-179794.html

Thanks, good thread!

Summary:
Doubling is recommended practise at large required cable dimensions by Victron and commonly used in electronic circuitry design and in industrial installations.

However, several posters warn about the case if one cable breaks and all current flows through the remaining cable possibly causing a fire. On the other hand, it is no risk of fire as long as the one remaining cable is large enough, and the doubling is just there to reduce the voltage drop.

Conclusion:
As long as each individual cable can accept the current of the fuse then it is acceptable and can be recommended if that simplifies the installation.

[Wotname]

^^ yes, pretty good summary

[john61ct]

Yes no problem, as long as

Originally Posted by jonasaberg

several posters warn about the case if one cable breaks and all current flows through the remaining cable possibly causing a fire.

proper fusing

[jonasaberg]

Originally Posted by john61ct

Yes no problem, as long as
proper fusing

Yes, but proper fusing is obviously a general requirement. Nothing special to double-cabling.

[john61ct]

Just saying the problem cited is not a problem, in other words there is no problem, as long as. . .

But increasing # of connections is increasing odds of failure, perhaps only slightly more risk, just makes proper crimping tools/supplies/technique more critical.

Which of course just like CP should be done that way anyway. . .

[sailon46]

Set up the double cable run the windlass and check if heat is generated if it is bring up the anchor with 30 sec breaks. That should last you a long time without too much trouble.
Ernie on the Mary Jane

[KP44]

Originally Posted by jonasaberg

Is this the right sub forum for this discussion?
I was thinking of using two 35 mm^2 (AWG 2) in parallell to get a total of 70 mm^2 to use in a windlass installation since I have the cable from another project already.

A guy I talked to adviced against this but did not explain exactly why. He talked about different load in the two cables, risk of one cable having a bad connection, inductive coupling between cables...
I can understand the risk of loosing one cable through a bad connection, but not any inductive coupling in a DC system.

So, I don't see this as a problem as long as each cable is large enough not to start a fire if the other is cut.

Can someone elaborate on this?

Some info:
The windlass is 1000 W mounted about 10 m from the battery, so at 12 V = 83 A in load.
In domestic installations in scandinavia we use 16 A / 2.5 mm^2 cable, so 6.4 A/mm^2.
This means, 83 A/6.4=12 mm^2 should be enough to avoid a cable fire. However, the voltage drop needs a larger diameter, and using a voltage drop calculator I found that 10 m between battery and load, 70 mm^2 of cable and 83 A gave a 4.5% = 0.5 V drop in voltage.

I checked in Nigel Calders book "Boat owner's mechanical and electric manual" and this fits the tables in the book for an expected voltage drop of 10% using 35 mm^2 (AWG 2) or 3% using 70 mm^2 (AWG 000).

Comments?

Your math is off because squares (of numbers) are not straight-line functions.

35 squared is 1,225

You have two cables of 1,225 square millimeters each and added together you have 2,450 square mm of cable. That does not come close to 70 squared which is 4,900 square mm (and approximates a 3/0 cable.)

Your doubled #2 wire will handle twice the current of a #2 wire.

[KP44]

I think my windlass needed #2 wire and I installed 1/0 just to insure no voltage drop and no problems. These installations are a one-time expense. To me, it does not make sense to install an expensive windlass and then save $50 by using small wire.

[Frankly]

If the main purpose of paralleling cables is to reduce voltage drop (often the case in 12 Volt DC circuits), then there is no harm assuming that the overcurrent protection is sized for a single cable. Try to make the length/ connections identical to get the current to divide equally.

[Cadence]

Two may serve your purpose. However adding diameters does not equal the circular mil. area as if it were one cable thus the current capacity.

[Cadence]

Originally Posted by Cadence

Two may serve your purpose. However adding diameters does not equal the circular mil. area as if it were one cable thus the current capacity.

I think it is a factor of four?

[noelex 77]

Originally Posted by KP44

Your math is off because squares (of numbers) are not straight-line functions.

35 squared is 1,225

You have two cables of 1,225 square millimeters each and added together you have 2,450 square mm of cable. That does not come close to 70 squared which is 4,900 square mm (and approximates a 3/0 cable.)

35 mm2 is an area. So the cross sectional area of the conductor in two 35 mm2 cables is the same as one 70 mm2.

It is not 35 squared. It is 35 square mm.

Think of a plot of land 17.5m x 2m. The area is 35m2. Two identical plots of land would have a total area of 70m2.

[sailormed]

1000W at 10 m. For calculations you need to include the return line as well. So 20 meter. my calculator for DC wire sizing gives me for 1008W, 84A at 12V DC and 3% voltage loss a minimum wire size of 95 mm2. Hope this helps.