Whether multi core cable can be used as a single core??

Status
Not open for further replies.
In a panel board there are three multi core incoming cables . Is it possible to use each multi core cable as single core csble meaning terminating all three cable terminated at each phases to avoid haphazard connection. Is there any code violations.
 

Besoeker

Senior Member
Location
UK
In a panel board there are three multi core incoming cables . Is it possible to use each multi core cable as single core csble meaning terminating all three cable terminated at each phases to avoid haphazard connection. Is there any code violations.
I don't know about code but generally you should keep all three phases in close proximity. Your suggested arrangement could make that difficult.
 

augie47

Moderator
Staff member
Location
Tennessee
Occupation
State Electrical Inspector (Retired)
Generally it's not done especially if there are any ferrous raceways or enclosures involved. As with many Code items, there are exceptions. Be sure to look at 300.3(B) and 300.20(B)
 

don_resqcapt19

Moderator
Staff member
Location
Illinois
Occupation
retired electrician
While it is my opinion that 300.3(B)(3) would permit that installation, Code Making Panel 3 insists that the only isolated phase installation permitted by the NEC is the one in the Exception to 300.3(B)(1).
 

Julius Right

Senior Member
Occupation
Electrical Engineer Power Station Physical Design Retired
If I well understand your problem you intend to use 3 three phase cables and at each one all 3 conductors short-circuited at both ends and use as one phase: three cables ,three separate phases.
I don't see any advantage of this: the ampacity will be the same but the voltage drop will be approximate double[if the cable are in touch] and at a feet distance will be triple.
 
If I well understand your problem you intend to use 3 three phase cables and at each one all 3 conductors short-circuited at both ends and use as one phase: three cables ,three separate phases.
I don't see any advantage of this: the ampacity will be the same but the voltage drop will be approximate double[if the cable are in touch] and at a feet distance will be triple.
I would appreciate if you kindly elaborate on the voltage drop issue?? As far as I understand voltage drop is function of current following through the conductor and not spacing between the conductor.
 

Julius Right

Senior Member
Occupation
Electrical Engineer Power Station Physical Design Retired
According to IEEE 141/1993 cap. 3.11 Calculation of voltage drops
V = IRcosf + IX sinf where:
V is the voltage drop in circuit, line to neutral
I is the current flowing in conductor
R is the line resistance for one conductor, in ohms
X is the line reactance for one conductor, in ohms
f is the angle whose cosine is the load power factor
cosf is the load power factor, in decimals
sinf is the load reactive factor, in decimals
For three-phase you have to multiply by 1.73
Let's take an example.
From NEC Table 9 500 kcmil copper R=0.027 ohm/kft
dcond=0.789
dins.core=1”
overall dia=2.13”
However, the reactance is lower for three core cable than three single core cables in cradle configuration as per NEC. Then we have to calculate it.
X=2*pi*f*(0.1404*log10(DMD/DMR)/10^3 ohm/kft
For 3 core 3 phase cable the reactance is
X=2*pi*f*(0.1404*log10(2*S/dia)+0.0153)/10^3 =0.0271 ohm/kft
where S=ins.core dia and dia=dcond.
For 3 core cable used as single phase
the distance=overall dia [when touching] but usually the distance is 1 ft=12”.
DMD=(2^(1/3)*distance[geometric mean of distances in flat configuration]=2.68 or 15.11 at 12”.
DMR=(ins.core dia^2*0.7788*cond.dia/2)^(1/3)= (1^2*0.7788*0.789/2)^(1/3)= 0.675
X=0.0317 when touching and X=0.07147 at 12” apart.
If there are three three phase cables parallel then the resistance and reactance will be 1/3
If there are three one phase three core cables the resistance will be 1/3 but the reactance does not.
So eventually in first case R=0.027/3 and X=0.0271/3
and in the second case
R=0.027/3 and X=0.0317 [or 0.07147]
If the total current will be 1000 A cosf=0.85 and sinf=0.5267 then in first case
V=1.73*1000*(0.85*0.009+0.5267*0.00903)=21.46 V
In second case
V=1.73*1000*(0.85*0.009+0.5267*0.0317)=42.12 V[or 78.4 V]
 
Status
Not open for further replies.
Top