24 guage wire

if you tie "four" 24 gauge conductors together, what is the process for determining equivalent wire gauge?

Reply to
rflowers
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24 AWG = 404 circular mills

404 X 4 = 1,616 circular mills. Almost 18 AWG (1620 circular mills).

Reply to
Lewis Gardner

The usual goal of doubling the conductors is to increase ampacity, which has a more linear relationship to diameter than to cmil area. IOW, two 24-gauge wires bonded together give you approximately the same ampacity as one 21-gauge wire. Four 24-gauge wires will carry approximately the same current as one 18-gauge wire.

Reply to
Robert L Bass

Robert is correct- approximately18 guage.

If you are just carrying line voltage, tying them together isn't a problem. If you are carrying a signal, line attenuation could be a problem.

Reply to
Stanley Barthfarkle

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Reply to
Dave Houston

Increasing ampacity is a poor reason to use multiple conductors of this size. I assume the OP was referring to something like cat5 cable. I am hesitant to allow more that 500 mA on communications cable like cat5.

If you need additional current carrying capacity then use the proper wire for that purpose.

In any case the overall ampacity of more than 3 insulated conductors in a cable or raceway must be derated.

A good reason to "tie four 24 gauge conductors together" is to reduce voltage drop.

Here are some nice tables:

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xtronics.com/reference/wire_gauge-ampacity.htm

Reply to
Lewis Gardner

"Stanley Barthfarkle" wrote ...

Robert's method (adding diameters) may be close this time but it is wrong.

2+2 = 2x2 but that doesn't mean the 10+10 = 10x10.

There are a lot of things involved in a equivalent wire gauge. The resistance is a significant factor. It is related to the area of the wire, not the diameter. Lewis Gardner had the correct method for this. }} 24 AWG = 404 circular mills }} 404 X 4 = 1,616 circular mills. Almost 18 AWG (1620 circular mills).

This works for all combinations of wires. Not just fortunate numbers. You don't need to use circular mils, but charts for them are easy to find and save calculating areas from diameters.

Are you primarily interested in equivalent current carrying capacity. If not, all this is for nothing. You need to decide what specs you are interested in and compare them.

For max current carrying capacity, you are looking at temperatures in the wires. Small wires with extra layers of insulation in a bundle could build higher temperatures than a single wire. If you were stringing the wires with flow between them, they could be an advantage.

Without knowing details, I would recommend under rating any combo of wires.

Reply to
B Fuhrmann

I don't know where the 10+10=10*10 logic came from but the simple fact is that doubling the number of conductors does double the ampacity. We can consider whether this is a *good* practice or not but it is

*common* practice.

Actually, AWG is a measure of the cross-sectional area of the wire, nothing more. Two identical conductors can carry twice as much current as one -- plain and simple.

Area of a wire is a function (pi R^^2) of diameter. However, the ampacity is more closely related to diameter than area. On long runs wire resistance will be an issue. In most home automation apps, it will not matter.

Reply to
Robert L Bass

Sorry Bob, but I have to disagree with you here. With all other factors being constant, the current rating of a wire is directly related to its cross-sectional area. As you point out above, that increases with the square of diameter.

How well the wire dissipates heat can be as important as cross-sectional area. While a factor for multi-conductor cables, this becomes a major consideration for transformer and inductor design.

Jeff

Reply to
Jeff Volp

Reply to
rflowers

Oh, yes. I was thinking of the difference between diameter and area. The following was borrowed without permission from a handy website. I had the AWG issue right but I erred about cmil area vs diameter. Thanks.

For the benefit of anyone I may have confused (as well as anyone who was already confused:)). Size Diameter AWG inches cmils Amps18 -------- 0.04030 ------- 1,624

------ 2.300

20 -------- 0.03196 ------- 1,022 ------ 1.500 22 -------- 0.02535 ------- 642.5 ------ 0.920 24 -------- 0.02010 ------- 404.0 ------ 0.577If we use four 24-gauge wires we can carry 4 x 0.577 amps. That is just a tad over 2.3 Amps total, the same as dropping the gauge by 6. Many installers use a 2:3 rule of thumb when doubling up on conductors -- 2 wires:3 gauge decrease.
Reply to
Robert L Bass

Not safely.

This sort of brute force math can get you into trouble. There are so many variables involved in this sort of calculation that the only real answer is that the ampacity of "four 24-gauge wires" is LESS than 2.3 amps.

As I stated in another fork of this thread my guess is that the OP was referring to something like a cat5 cable since it is the most common 24 AWG cable used in HA applications. The twisted pairs in a cat5 cable and the insulation used are not designed for current carrying applications. The issue is that closely twisted pairs in a PVC jacket are not able to shed heat easily. Cat5 is communications cable not a power cable. The construction of power cables is quite different.

To give you an idea of some derating factors refer to:

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The closest correction factor to this case is at the bottom of the page titled "Group Correction Factors", "NUMBER OF CABLES IN AIR". The table gives a derating factor of .84 for 4 conductors horizontally. So the 2.3 amps turns into 1.9 amps. I think this is optimistic since the conductors are not in air but instead are bundled in a jacket and additional derating would likely apply

The only good reason I can see joining conductors in a cat5 cable is to reduce voltage drop.

A good example of this (using the dandy calculator in Dave Houston's post) is a 12 VDC 250 mA cordless phone located 100 feet from the wiring closet. If you had a spare cat5 and used two sets of 4 conductors to power the phone remotely the voltage at the phone would be 11.7 volts (2.7% drop). Compare that to 10.7 volts (10.9% drop) with two single conductors.

On 18 AWG wire, 12 VDC at 2.3 Amps the longest distance you can go with a 5% voltage drop is 20 feet. At low voltages ampacity is NOT the issue, voltage drop is.

If you want to run several amps from one place to another use a properly sized power cable. In HA work (low voltage) the sizing calculations will tend be voltage drop not ampacity.

Reply to
Lewis Gardner

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