C factor for 4C 185 sq mm cu/xlpe/swa/pvc & 4C 240 sq mm cu/xlpe/swa/pvc
C factor for 4C 185 sq mm cu/xlpe/swa/pvc & 4C 240 sq mm cu/xlpe/swa/pvc
(OP)
C factor for 4C 185 sq mm cu/xlpe/swa/pvc & 4C 240 sq mm cu/xlpe/swa/pvc-
Anyone can share the conductor factor of the above cable to be layed underground.
I need to compute the fault current unforetunately I dont have the book/table.
Thanks for your help
Anyone can share the conductor factor of the above cable to be layed underground.
I need to compute the fault current unforetunately I dont have the book/table.
Thanks for your help






RE: C factor for 4C 185 sq mm cu/xlpe/swa/pvc & 4C 240 sq mm cu/xlpe/swa/pvc
http://www.legrand.com/files/fck/File/pdf/Power_gu...
page 30.
This factor may be used in order to check the cable short-circuit withstand. But –I guess-you need the ampacity [maximum continuous load current permissible]. The ampacity depends on a few factors as depth of the earth above the cable run, earth thermal resistance and temperature. Following IEC 60364-5-52 recommendation you could verify the [approximate] ampacity.
For instance: from page 11 table “Protection Against Overloads” let’s chose (D) for: “Direct in the ground with added mechanical protection.”
From page no.20 following row D up to PR3[for xlpe insulation 3 active conductors] we get for 185 sqr.mm 304 A [per one conductor] and for 240 sqr.mm 351 A. This in conditions of 2.5 k.m/w earth thermal resistance and 20 dgr.C earth temperature. On page no.18 there are factors for other thermal resistance and temperature. The cable bury depth is 0.8 m.
RE: C factor for 4C 185 sq mm cu/xlpe/swa/pvc & 4C 240 sq mm cu/xlpe/swa/pvc
Sorry I did not explain well all the detail. I use the below:
f = 1.73 x L x Isc/C x N x VLL
M= 1/(1+F)
Isc = m x Isc
Given:
Main Distribution board
Ik = 50kA -upstream SC
Vll = 415
A. SC # 1 ???
S1 = 4c 240 sq mm cu/xlpe/swapvc, Length is 100m
B. SC # 2 ????
S2 = 4C 185 sq mm cu/xlpe/swapvc, Length is 40m
where:
: Cable is to be installed direct underground
L =100m = Length of conductor relative to the immediate upstream fault point
C =???? Conductor factor in short in circuit calculation
f =??? f - factor
n =1 number of parallel run conductor
m = Multiplier to be used against the value of fault current
Isc sym = I short circuit Symmetrical value
I use this formula, since everything is known and the origin of fault is known, I need only is the conductor factor.
I saw one table(BUSSMANN) the but installation is Busway in-case in non metallic conduit, I guess I get wrong C factor since my installation is direct buried undergroud.
RE: C factor for 4C 185 sq mm cu/xlpe/swa/pvc & 4C 240 sq mm cu/xlpe/swa/pvc
http://www.cooperindustries.com/content/dam/public...
First of all you have to transform square mm in awg [or kcm-mcm].Since 1mcm=0.5067 sqr.mm then 350 mcm it is close to 185 sqr.mm and 500 mcm it is close to 240 sqr.mm
From Table 4 for 3 conductors cable in nonmagnetic conduit column you'll get:
for 350 mcm c=22525 and for 500 mcm c=30096.
Since this is a way to calculate short-circuit current then factor “f” is proportional with cable impedance[reactance] so what is important it is the conductor diameter and the distance between conductors center-lines. It is not important how the cable will run: underground or in air or the ambient details-thermal resistance or temperature.
RE: C factor for 4C 185 sq mm cu/xlpe/swa/pvc & 4C 240 sq mm cu/xlpe/swa/pvc
So the installation is not a matter? Wither its underground or air?
If that the case bussmann method is easy.
Thanks
RE: C factor for 4C 185 sq mm cu/xlpe/swa/pvc & 4C 240 sq mm cu/xlpe/swa/pvc
It is, indeed, a very simple [approximate] method, but for very simple connection diagram.
RE: C factor for 4C 185 sq mm cu/xlpe/swa/pvc & 4C 240 sq mm cu/xlpe/swa/pvc
Big thanks
RE: C factor for 4C 185 sq mm cu/xlpe/swa/pvc & 4C 240 sq mm cu/xlpe/swa/pvc
BASED ON BUSSMANN CALCULLATION THE Isc = SYMMETRICAL RMS,
WHAT DOES IT MEAN SYMMETRICAL RMS?
IS IT INITIAL SYMMETRICAL SHORT CIRCUIT CURRENT???
THANKS
RE: C factor for 4C 185 sq mm cu/xlpe/swa/pvc & 4C 240 sq mm cu/xlpe/swa/pvc
RE: C factor for 4C 185 sq mm cu/xlpe/swa/pvc & 4C 240 sq mm cu/xlpe/swa/pvc
Hi 7anoter4,
How to culcukate the resistance @ 90degrees c??
RE: C factor for 4C 185 sq mm cu/xlpe/swa/pvc & 4C 240 sq mm cu/xlpe/swa/pvc
What i mean is how to calculate c?
C = 1/Zo
Zo= sq rt(ro^2 + xo^)
at 90dgr for cu/swa/xlpe/pvc
Thanks
RE: C factor for 4C 185 sq mm cu/xlpe/swa/pvc & 4C 240 sq mm cu/xlpe/swa/pvc
http://www.nexans.co.uk/eservice/UK-en_GB/fileLibr...
For 3*185 sqr.mm ro=0.128 ohm/km xo=0.091 ohm/km [you have to multiply by:0.3048/10^3 in order to obtain the result in ohm/ft]
RE: C factor for 4C 185 sq mm cu/xlpe/swa/pvc & 4C 240 sq mm cu/xlpe/swa/pvc
RE: C factor for 4C 185 sq mm cu/xlpe/swa/pvc & 4C 240 sq mm cu/xlpe/swa/pvc
Below is my sample computation, however the correction is not yrt consider:
I cant understand the formula above.
Consider Upstream = I"k =19 kA
Point A
f = 1.73 x L x Isc/C x N x VLL
f = 1.73 x 100 x 50/30028 x 1 x 415
f = 0.432
m= 1/(1+f)
m = 1/1.432
m = 0.70
Isc sym = m x Isc
0.70 X 19000
13.27 kA
Consider Point B
f = 1.73 x L x Isc/C x N x VLL
f = 1.73 x 40 x 29541/26915 x 1 x 415
f = 0.135
m= 1/(1+m)
m = 1/1.135
m = 0.88
Isc sym = m x Isc
11694.81706
11.69 kA
Consider Point C 1 Ph
f = L x Isc/C x N x VLL
f = 30 x 24951/2466 x 1 x 230
f = 1.014
m= 1/(1+m)
m = 1/1.014
m = 0.496
Isc sym = m x Isc
5805.417061
5.81 kA
RE: C factor for 4C 185 sq mm cu/xlpe/swa/pvc & 4C 240 sq mm cu/xlpe/swa/pvc
http://www.omancables.com/brochure/Low_Voltage_Pow...
page 59 Table 73
for 4*185 sqr.mm ro=0.128 xo=0.071 zo= 0.14686 ohm/km = 4.48E-05 ohm/ft C= 22339.8
for 4*240 sqr.mm ro=0.0989 xo=0.071 zo= 0.1217465 ohm/km= 3.711E-05 ohm/ft C= 26948.1
No data for the third cable [from point C]. Could be 2*16 sqr.mm?
How the short-circuit current drops does from 50 kA to 19 kA? Is a transformer in point A?
A one line diagram would help here.
RE: C factor for 4C 185 sq mm cu/xlpe/swa/pvc & 4C 240 sq mm cu/xlpe/swa/pvc
How did you get the below? What is the formula you used?
= 4.48E-05 ohm/ft C= 22339.8
= 3.711E-05 ohm/ft C= 26948.1
I cant upload the one line diagram.
Thanks
Don
RE: C factor for 4C 185 sq mm cu/xlpe/swa/pvc & 4C 240 sq mm cu/xlpe/swa/pvc
RE: C factor for 4C 185 sq mm cu/xlpe/swa/pvc & 4C 240 sq mm cu/xlpe/swa/pvc
Thanks
RE: C factor for 4C 185 sq mm cu/xlpe/swa/pvc & 4C 240 sq mm cu/xlpe/swa/pvc
Do you have any idea where did the bussman take the furmula? What standard/methology?
I saw similar appoachin in youtube and states on the lecture heading is from NEC 2011?
Thank yuo very much.
f= 1.73 x L x I/C x N x Vll
m=1/(1+f)
Isc sym rms= Isc x m
BR,
Don
RE: C factor for 4C 185 sq mm cu/xlpe/swa/pvc & 4C 240 sq mm cu/xlpe/swa/pvc
In a simple circuit –as attached-if we take the circuit formed by “source” and transformer we get:
Is.c.=VL_L/sqrt(3)/Ztrf [symbols used in the article].
From here Ztrf=VL_L/sqrt(3)/Is.c.
The total circuit contains 2 impedances: Ztrf+Zcable.
Zcable=zcable*L/n zcable=cable impedance per unit length[m,km,ft or 1000 ft] n=number of parallel cables.
For 3 phases Z3ph=Zcable
IS.C. sym. RMS=VL_L/sqrt(3)/(Ztrf+Zcable) or if we divide the parentheses (Ztrf+Zcable) by Ztrf:
IS.C. sym. RMS=VL_L/sqrt(3)/Ztrf/(1+Zcable/Ztrf)
M=1/(1+Zcable/Ztrf)=1/(1+f)
Zcable/Ztrf=f and Ztrf=VL_L/sqrt(3)/Is.c. then :
Zcable/Ztrf=L/C/(VL_L/sqrt(3)/Is.c.)=sqrt(3)*Is.c.*L/(C*n*VL_L)=f
For 2 phases [live-live or live-neutral] Z2ph=2*Zcable and the supply voltage is VL-L or VL-N:
f=2*Is.c.*L/(C*n*VL_L) for live-to-live short-circuit or
f=2*Is.c.*L/(C*n*VL_N) for live-to-neutral short-circuit.