Current capacity of multi-strand versus single-strand

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Would a multi-strand flexible wire have the same current carrying capacity as a solid single stranded cable of the same cross section.

For example ... a multi-strand with a cross section for its wire portion of 0.75 mm squared and a single strand cable also of 0.75 mm squared.

• posted

Any particular frequency in mind?

Skin effect is the obvious thing that will make a stranded wire better than a solid of the same csa. But this kicks in at higher frequencies.

Here is a brief simplified guide:

However, if the actual csa of the solid wire is compared with actual csa of the stranded wire (rather than the sum of the csa of individual strands), then a solid wire will have more copper and hence lower resistance. This is frequency independent.

So, at low frequencies or DC, a solid wire of the same core diameter will be better than stranded. As frequency rises, a point will come where stranded wire will be better; it will be able to take a higher current than a solid core of the same diameter.

There are obviously exceptions.

Hope that is what you were after...

• posted

Ya, until you get into higher frequencies where the slightly greater surface area of the stranded wire represents more cross section. Probably above

10MHz...

Tim

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Skin effect is easily measurable at a few tens of kHz. It comes into effect at regular mains frequences for very large conductors. An alternative to making them stranded is to pick a different shape which avoids significant depth of metal, such as a flat strip (thin rectangular profile), or tubular.

• posted

In most cases, they are exactly the same. When the frequency gets high enough that skin effect starts to crowd the current toward the surface, stranded wire has a slight advantage, but this is greatly enhanced if all the strands are insulated from each other (look up litz wire). This effect can also be used ot advantage when winding magnetic devices with enameled wire, and two or more parallel strands can be used in place of an equivalent cross section single wire. But for ordinary hook up wire, they are usually assumed to have an ampacity based on their cross sectional area, not the number of strands.

• posted

A No, 12 AWG wire has a diameter of 0.78 mm.

For wire sizes 2.47 mm and under (No. 2 AWG and under) the DC resistance nearly equals AC resistance. Therefore the ampacity for stranded conductors is nearly equal to that for solid conductors for these small conductors (also demonstrated using the N-M equations.) For building wire types, conductors size No. 8 AWG and larger that are installed in raceways are required to be stranded (NEC Section 310.3.) Therefore, we seldom see conductors larger than No. 8 that are solid, except for the No. 4 bare grounding conductors carried on line trucks and sold for services by supply houses that are used as grounding electrode conductors so that no additional protection is required. The standard ampacity table in the NEC used for building wires, 310.16, does not distinguish between solid and stranded conductors. This is probably because almost all building wire No. 8 and larger is stranded. The N-M equations also do not distinguish between solid and stranded, but uses DC resistance multiplied by (1+YC) where YC becomes measurable for wire sizes above No. 2 because of skin effect.

• posted

For No. 12 AWG Table 8 of the NEC does list two different DC resistances for stranded and solid copper.

DC resistance per 1000 feet for solid is 1.93 ohms

For stranded the DC resistance is given as 1.98 ohms per 1000 feet.

So solid should have a slightly higher ampacity.

If we use Table 310.16 of the NEC to determine RCA and substitute into the Ampere calculation we can find the approximate differences in ampacity.

From Table 310.16 using 75 degrees C as the ambient.

I = 25 amperes, TC = 75 degrees C, and TA = 30 degrees C and RDC = 1.98 ohms per 1000 feet or 0.00198 ohms per foot.

This converts to 1980 microhms.

From I (in kiloamperes) = SQRT(( TC-TA)/RDC*RCA))

Or

RCA=(TC-TA)/RDC*I*I

Or RCA = (75-30)/1980*0.025*0.025

RCA = 36 thermal ohm feet

For stranded, I = 0.025 kiloamperes from the table

For solid No. 12 copper

I (in kiloamperes) = SQRT ((75-30)/1930*36)

or I = 0.0254 kiloamperes

Then the solid No. 12 copper would have a 0 .4 ampere increase in ampacity.

This is a 0.4/25 *100 or only a 1.6 per cent increase.

Considering that ampacity tables are approximations, this increase in ampacity does not exceed the error of approximation.

• posted

If indeed the conducting areas are the same the capacity will be the same for all practical purposes. The AWG (American Wire Gauge) numbers take into account the actual total cross-sectional area. Some charts will compare stranded vs. solid for actual wire outer diameter though. If you start talking about impedance in some applications the story will change. Regards, Tom

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As far as I am aware, the amount of copper being the same should mean the same current flow - but the more strands should mean it is easier to work with, albeit more expensive?

H.

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I just reviewed the Samuel Rosch Paper from 1938, "The Current Carrying Capacity of Rubber-Insulated Conductors." He used a slightly different formula where N represented the number of current carrying conductors. Since Table 310.16 is for three current carrying conductors in a raceway to find RCA the following should be used:

I (in kiloamperes) = SQRT(( TC-TA)/N*RDC*RCA)) RCA=(TC-TA)/N*RDC*I*I RCA = (75-30)/3*1980*0.025*0.025 RCA= 12 Thermal Ohm Feet

Then for solid No. 12 AWG copper: I (in kiloamperes) = SQRT ((75-30)/3*1930*12) I = 0.0254 kiloamperes

And there is no difference in the answer.

• posted

I should also note that Samuel Rosch did his calculation for an ambient of

30 degrees C and TC = 50 degrees. He came up with the following ampacity table for copper (only sizes up to No. 2 are shown):

Wire Size - Amperes

14 - 15 12 - 19 10 - 26 8 - 35 6 - 47 4 - 61 2 - 78 To convert these values to ampaeres for a TC (insulation temperature) equal to 75 degrees C. We need to derive an equation: Let I1 equal ampacity for 50 degree insulation and I2 equal ampacity for 75 degree insulation, Then: I1 (in kiloamperes) = SQRT(( TC1-TA)/N*RDC1*RCA)) I2 (in kiloamperes) = SQRT(( TC2-TA)/N*RDC2*RCA)) Where TA = 30 degrees C. and RCA is equal in both equations. RDC1 is DC resistance in microhms at 50 degrees C RDC2 is DC resistance in microhms at 75 degrees C. TC1 = 50 degrees C. TC2 = 75 degrees C. Using the proportionality rule where two ratios that are equal can be cross multiplied: I2*(SQRT(( TC1-TA)/N*RDC1*RCA))) = I1*(SQRT(( TC2-TA)/N*RDC2*RCA))) and squaring both sides: I2*I2*(( TC1-TA)/N*RDC1*RCA)) = I1*I1*(( TC2-TA)/N*RDC2*RCA)) I2*I2 = (I1*I1*(( TC2-TA)/N*RDC2*RCA))) / (( TC1-TA)/N*RDC1*RCA)) I2*I2 = I1*I1*( TC2-TA) * RDC1/ ( TC1-TA) * RDC2 I2 = I1* SQRT (( TC2-TA) * RDC1 / ( TC1-TA) * RDC2)

Next we use the N-M equation for finding DC resistance:

rdc = ohms pc = circular mil ohms per foot of conductor at 20 degrees C. (10.371 ohms for 100% IACS copper, 17.002 ohms for 61% IACS aluminum) tah = absolute value of inferred temperature of zero resistance. (234.5 degrees C. for copper and 228.1 degrees C. for aluminum) cma = circular mil area of conductor from Chapter 9 Table 8 of NEC tc = conductor temperature in degrees C.

At 50 degrees C. RDC1 = (1.02 * 10.371 / CMA) * (234.5 + 50) / (234.5 + 20) At 75 degrees C. RDC2=(1.02 * 10.371 / CMA) * (234.5 + 75) / (234.5 + 20)

If we feed this into a spreadsheet we come up with the following ampacity table for TC = 75 degrees C. We get the following table: Wire size - amperes

14 - 21 12 - 27 10 - 37 8 - 50 6 - 67 4 - 88 2 - 112 If we round these to the nearest 5 amperes: 14 - 20 12 - 25 10 - 35 8 - 50 6 - 65 4 - 90 2 - 110 How do these compare to table 310.16? wire size - calculated amperes - Table 310.16

14 - 20 - 20

12 - 25 - 25 10 - 35 - 35 8 - 50 - 50 6 - 65 - 65 4 - 90 - 85 2 - 110 - 115

This shows that Table 310.16 in the NEC has a general round off error that exceeds the difference in ampacities between solid and stranded conductors (at 60 Hz) for small wire sizes.

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• posted

In very high voltages, 400 kV and over, it's not the same due to corona effect (ionization of the air around the conductor).So, the utility puts two conductors in parallel.That's the only example I can think of;in every day applications it has no meaning.

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• posted

What I mentioned before, is actually the strength of the electric field around the conductor, that is reduced by the use of two (conductors) and ionizes the air, making a noise like humming bees.

-- Tzortzakakis Dimitriïs major in electrical engineering, freelance electrician FH von Iraklion-Kreta, freiberuflicher Elektriker dimtzort AT otenet DOT gr Ï "Tim Williams" Ýãñáøå óôï ìÞíõìá news:QI4sd.429\$ snipped-for-privacy@fe05.lga...

• posted

Here, we use the following gauges (in residence and small industry):1.5,

2.5, 4, 6, 10 mm^2.The former 3 are stranded or solid, the latter 2 always stranded.It's very difficult to bend the thick wires when they are solid, so they stopped producing solid 10 mm^2 wires, used to wire old-fashioned fuses and circuit breakers.The ampacity of these conductors is 10,16,20,25,35 amperes respectively and should be properly fused, according to local rules, with circuit breakers.Actually, the max.current of a copper wire depends on the max.allowed temperature, hence whether the insulation is PVC or something else (usually PVC here).All formulas have the gauge in mm^2 here, my book says nothing about the mentioned issue, maybe it's different in USA. the voltage drop in a (loaded) cable is: ÄU/U=2 I Ø' I cos ö /U (the greek letters are delta, psi and phi respectively)

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• posted

"Dimitrios Tzortzakakis" wrote in message news:covp5o\$4pe\$ snipped-for-privacy@usenet.otenet.gr...

Are your areas the total area of stranded or solid or are they the combined area of the copper only. We use Circular Mil Area (CMA) to define the copper area.

CMA is the diameter of a conductor in thousandths of an inch squared and represents the total copper area. For instance, a No. 12 AWG has a CMA equal to 6530 for both stranded and solid No. 12 conductors, but the cross sectional areas are different. The cross sectional area of a solid No. 12 is 3.31 MM^2 and 4.25 mm^2 for stranded. Cross sectional areas do little for calculating ampacity. We use CMA because resistance is inversely proportional to CMA because CMA represents actual copper. Also you give your ampacities, but what maximum operating temperature are these for and what is the ambient temperature? Also, are your ampacities for in cable, 3 conductors in raceway, etc.

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------------------------ Utilities use bundled conductors at lower that 400KV. The main reasons for this is to a)make construction easier and cheaper b)reduce the series inductance of the line c) reduce conductor surface fields and corona. So you have hit one reason out of three.

• posted

That means I am not passing the exam?

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• posted

The ampacities are for normal conditions, thus residence.It's a thumb rule which cable to use for what load.A cooking range needs 6 mm^2, a water heater 4 mm^2, a washing machine and dishwasher and most space heaters 2.5 mm^2 and incadescent lighting 1.5 mm^2.I suppose these calculations are for the climate of Greece, and for three wires (live, neutral, earth) in a conduit.

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• posted

Area of the copper only, in Europe.

Assumes ambient of 30ºC and maximum operating temperature of 70ºC. For ambient temperatures above 30ºC, max current is reduced by tables, and reaches zero at 70ºC. Circuit protection for overload and fault current conditions is designed to prevent cable exceeding

160ºC during a fault or overload (i.e. a maximum 90ºC temperature rise from operating temperature), or it would be permanantly damaged and need replacing. Higher temperature cable is available for situations which require it.

We have different maximum ampacities for a number of different installation methods. The highest is for a cable in a cable tray, going through to the lowest which is for a cable embedded in a thermally insulating wall. We also have 'grouping factor' which reduces the ratings of multiple cables when closely spaced.

• posted

"Dimitrios Tzortzakakis" wrote in message news:cp1kid\$o2r\$ snipped-for-privacy@usenet.otenet.gr...

"Normal conditions" is a broad term. In the USA the NEC (National Electrical Code) has many specifics for ampacity tables. For instance, the main Table 310.16 that we use for building wire ampacity defines the ambient as 30 degrees C. and has derating factors where the temperature exceeds this for long runs. If the temperature is excessive for no more than 10 per cent of the circuit length to a maximum of ten feet whichever is less, then no excessive temperature derating is required. The ampacities are also defined for no more than three current carrying conductors in a raceway or cable. If conductors are bundled or cables with more than three current carrying conductors are bundled for longer than 24 inches or if there are more than three current carrying conductors in a raceway or cable there are additional derating factors. In some cases where there is excessive temperatures and more than three current carrying conductors double derating is required. We also have a rule for continuous loads requiring a 80 per cent derating. There are also rules defining a current carrying conductor. If a neutral only carries unbalanced current it is not counted as a current carrying conductor. But where the majority of current in a neutral is from discharge lighting, third harmonics, or computer loads it is counted. Also, the grounding conductor is not counted as a current carrying conductor. Also, ampacities are listed for three temperatures for copper and for aluminum. One column is for 60 degrees insulation, one for 75 degrees C. and one column for 90 degree C. The maximum operating temperature of a conductor cannot exceed the maximum allowed rated operating temperature for the terminals, equipment or insulation whichever is the lesser. We can use the

90 degree column for derating purposes. For example if there are 9 No. 12 AWG (American Wire Gauge) 90 degree C. rated current carrying conductors in a raceway we can multiply the derating factor of .7 times the 90 degree ampacity of 30 amperes to get 21 amperes and use this as the ampacity but other rules still apply. Our rules are so complicated that it took me about 5 weeks to write an Excel program to determine ampacity. This program is accessible at