- Occupation
- Licensed Electrician
That's the way I see it.
Why De-Rated
- Thread starter Toros
- Start date
- Status
ggunn
PE (Electrical), NABCEP certified
- Location
- Austin, TX, USA
- Occupation
- Consulting Electrical Engineer - Photovoltaic Systems
I don't believe the function of heat containment in a conduit is a stairstep one but a continuous curve of some sort. The breakpoints in CCC count are a stairstep function designed to stay beneath the curve. The code could direct that every added CCC bump up the derate, but I am glad it doesn't.
ggunn
PE (Electrical), NABCEP certified
- Location
- Austin, TX, USA
- Occupation
- Consulting Electrical Engineer - Photovoltaic Systems
Rethinking...
Of course the function of heat containment is a stairstep function, not a smooth curve - you cannot have a fraction of a CCC in a conduit - but the breakpoints in the code are coarser and designed to stay below the more granular function.
Badams
Member
- Location
- St. Paul, MN
Heat containment in a wireway is a fairly dramatic stairstep. According to 376.22(B), 30 current carrying conductors in a wireway require no adjustment to ampacity. As soon as you add the 31st wire, an adjustment factor of 40% applies.
ggunn
PE (Electrical), NABCEP certified
- Location
- Austin, TX, USA
- Occupation
- Consulting Electrical Engineer - Photovoltaic Systems
But I don't think that it was the 31st wire that made all the difference. The 31st wire just raised the heat containment to the point that some threshold was passed and the derate kicked in. The 30th wire raised it incrementally just as much, or close to it.
Julius Right
Senior Member
- Occupation
- Electrical Engineer Power Station Physical Design Retired
On NEC 310.15 Ampacities for Conductors Rated 0–2000 Volts.
(C)-Engineering Supervision. is stated:
Under engineering supervision, conductor ampacities shall be permitted to be calculated by means
of the following general equation:
I=SQRT((Tc-Ta)/Rdc/(1+Yc)/Rca)
where:
Tc = conductor temperature in degrees Celsius (°C)
Ta = ambient temperature in degrees Celsius (°C)
Rdc = dc resistance of conductor at temperature Tc
Yc = component ac resistance resulting from skin effect and proximity effect
Rca = effective thermal resistance between conductor and surrounding ambient
No sign of number of conductors.
Actually the analogy of thermal circuit with electric circuit [See IEEE 835] will take us to IEC 60287-1-1
where is stated:
The permissible current rating of an a.c. cable can be derived from the expression for the
temperature rise above ambient temperature:
Δθ = (I^2R + ½ Wd) T1 + [I^2R (1 + λ1) + Wd] n T2 + [I^2R (1 + λ1 + λ2) + Wd] n (T3 + T4)
where:
I is the current flowing in one conductor (A);
Δθ is the conductor temperature rise above the ambient temperature (K);
R is the alternating current resistance per unit length of the conductor at maximum operating temperature (Ω/m);
Wd is the dielectric loss per unit length for the insulation surrounding the conductor (W/m);
T1 is the thermal resistance per unit length between one conductor and the sheath (K.m/W);
T2 is the thermal resistance per unit length of the bedding between sheath and armour (K.m/W);
T3 is the thermal resistance per unit length of the external serving of the cable (K.m/W);
T4 is the thermal resistance per unit length between the cable surface and the surrounding medium, 2 (K.m/W);
as derived from 2.2 of part 2 (K.m/W);
n is the number of load-carrying conductors in the cable (conductors of equal size and carrying the same load);
λ1 is the ratio of losses in the metal sheath to total losses in all conductors in that cable;
λ2 is the ratio of losses in the armouring to total losses in all conductors in that cable.
From here the actual formula of I it is as follows:
I=SQRT((Δθ - Wd*(0.5*T1 + n(T2 +T3+T4))/(R*T1+n*R*(1+λ1)+n*R*(1+λ1+λ2)*(T3+T4))
That means-exception what happened in a single conductor insulation- the temperature drop in sheath and the
surrounding medium is direct proportional with number n of conductors as the copper losses of each conductor
participate at total losses.
(C)-Engineering Supervision. is stated:
Under engineering supervision, conductor ampacities shall be permitted to be calculated by means
of the following general equation:
I=SQRT((Tc-Ta)/Rdc/(1+Yc)/Rca)
where:
Tc = conductor temperature in degrees Celsius (°C)
Ta = ambient temperature in degrees Celsius (°C)
Rdc = dc resistance of conductor at temperature Tc
Yc = component ac resistance resulting from skin effect and proximity effect
Rca = effective thermal resistance between conductor and surrounding ambient
No sign of number of conductors.
Actually the analogy of thermal circuit with electric circuit [See IEEE 835] will take us to IEC 60287-1-1
where is stated:
The permissible current rating of an a.c. cable can be derived from the expression for the
temperature rise above ambient temperature:
Δθ = (I^2R + ½ Wd) T1 + [I^2R (1 + λ1) + Wd] n T2 + [I^2R (1 + λ1 + λ2) + Wd] n (T3 + T4)
where:
I is the current flowing in one conductor (A);
Δθ is the conductor temperature rise above the ambient temperature (K);
R is the alternating current resistance per unit length of the conductor at maximum operating temperature (Ω/m);
Wd is the dielectric loss per unit length for the insulation surrounding the conductor (W/m);
T1 is the thermal resistance per unit length between one conductor and the sheath (K.m/W);
T2 is the thermal resistance per unit length of the bedding between sheath and armour (K.m/W);
T3 is the thermal resistance per unit length of the external serving of the cable (K.m/W);
T4 is the thermal resistance per unit length between the cable surface and the surrounding medium, 2 (K.m/W);
as derived from 2.2 of part 2 (K.m/W);
n is the number of load-carrying conductors in the cable (conductors of equal size and carrying the same load);
λ1 is the ratio of losses in the metal sheath to total losses in all conductors in that cable;
λ2 is the ratio of losses in the armouring to total losses in all conductors in that cable.
From here the actual formula of I it is as follows:
I=SQRT((Δθ - Wd*(0.5*T1 + n(T2 +T3+T4))/(R*T1+n*R*(1+λ1)+n*R*(1+λ1+λ2)*(T3+T4))
That means-exception what happened in a single conductor insulation- the temperature drop in sheath and the
surrounding medium is direct proportional with number n of conductors as the copper losses of each conductor
participate at total losses.
Carultch
Senior Member
- Location
- Massachusetts
Good question. I do not know the answer, and I want to know too.
Similarly, if you put a physical barrier in a wireway, does each side of the partition count separately for the number of conductors? I know you don't count derate factors in wireways up to 30 wires, but when the 31st wire is added, you are already down to the derate factor for 31 wires.
I'm wondering if I were to split those wires 16 to 15 across the partition, if I could take advantage of that.
ggunn
PE (Electrical), NABCEP certified
- Location
- Austin, TX, USA
- Occupation
- Consulting Electrical Engineer - Photovoltaic Systems
I would think not, since the wireway is vented, but that's just a guess on my part.
- Location
- Illinois
- Occupation
- retired electrician
Julius Right
Senior Member
- Occupation
- Electrical Engineer Power Station Physical Design Retired
If the wiring duct is located in an enclosure, in my opinion, for cables carrying continuous load NFPA 79
Electrical Standard for Industrial Machinery ch. 12.5 Conductor Ampacity. has to be followed.
So one has to derate the conductor ampacity for more than 3 loaded conductor.
Electrical Standard for Industrial Machinery ch. 12.5 Conductor Ampacity. has to be followed.
So one has to derate the conductor ampacity for more than 3 loaded conductor.
- Status