Eddy Currents Generated in a Ferrous Conduit 2

[hildeew2]

Hey all,

I'm looking for the equation/s to determine the heat energy created in a galvanized rigid steel conduit (10 inches long 3.5in diameter)

The cables running through the conduit are 3 500MCM THHN Copper (all the same phase) at 1500A.

Ultimately I am trying to take that heat energy and transfer it back to the conductor core to determine the time it would take to melt the cable.

 

[waross]

The insulation will fail and the cable will short to the conduit long before the cable melts.

Bill
--------------------
"Why not the best?"
Jimmy Carter

 

[rcw retired EE]

Too many variables to get an answer accurate within 50%:
-Heat conduction from conduit to material around outside of conduit (air, ground, concrete or other). Depends on thermal resitivity of materials and their interfaces.
-Starting temperature of conduit, surrounding material and wire.
-Heat loss to cooling air flow through conduit.
-Radiation heat loss from exposed conduit.
-Heat transfer along wires to outside the conduit.
-Resitivity of the conduit.
-Method of attachment at each end. If the conduit terminates on a metal panel, the panel will also have significant heating, raising the ambient temperature.

500 amps per conductor exceeds the 430 Amp 90C rating of the THHN per NEC, so there may already be overheating problems. Also, my experience is that the metal can get hot enough to blister paint and smoke insulation within 10 minutes. We had to repaint switchgear steel after high current testing a large molded case breaker with improper routing of our test cables and bus bars.

 

[waross]

I have seen serious heat damage from 200 Amps umbalanced. More than once.

Bill
--------------------
"Why not the best?"
Jimmy Carter

 

[jghrist]

1500 A single phase through rigid conduit? Not much time before the insulation melts. To get any kind of accuracy for a short section of conduit, you would need some kind of finite element analysis. Is it worth the effort to determine exactly how much time? It's against code anyway.

 

[hildeew2]

Thanks for the response.

Is it worth the time? Absolutely not. Unfortunately I have been tasked by my VP with trying to back up a field error that occurred and is being disputed.

I am aware of the results as we validated them via experimentation.(And your right, it took about 20-30 minutes to start smoking.) I am trying to determine the calculations involved.

Thank you for the posts.

 

[waross]

The code states words to the effect that there must be an equal number of conductors of each phase in each conduit. Look it up and quote the code. Then add a picture of the damage with the caption;
"THIS IS WHY!!"

Bill
--------------------
"Why not the best?"
Jimmy Carter

 

[7anoter4]

I agree with rcwilson, of course, there are too much parameters to know.
I'll try to overcome this by supposing that it is an adiabatic process and no energy will be wasted outside .
I'll neglect the hysteresis losses also and I'll follow the Steinmetz formula for eddy current.
The steel pipe of 3.5" has inner diameter of 3.548 and the outside dia 4 inches. So the thickness will be 0.226 inch= 5.74 mm
The average dia =95.86 mm then the length of circumference will be pi()*9.586=30.12 cm
The field intensity will be H=1500/30.12=49.8 A/cm. From B=F(H) curve B=1.65 Wb/m^2[=16500 Gs]
Peddy=(pi()*B*t*f)^2/(6*10^7*d*ro) w/kg
where flux density (B) is in Gauss, frequency (f) in Hz, laminate thickness (t) in cm, density (d) in g/cm3, and electrical
resistivity (ρ) in micro-ohm-cm.
Peddy=(pi()*16500*0.574*60)^2/(6*10^7*7.8*10) =681 w/kg
conduit weight[10 inches long]=3.4 kg
Pfe=Peddy*weight=3.4*681=2315 w
Let's say all the conduit iron losses will penetrate inside and will heat the cables.
dQinput=dQoutput
{I^2*Ro*[1+alpha*(T-To)]+Pfe}*dt= TCAP*volume*dTemp [ if we shall neglect the outside heat dissipation.]
Integrate we'll get:
t*Kb/TCAP/volume=1/Kb*LN((Kb*Tf+Ka)/(Kb*Ti+Ka))
where Ka=I^2*Ro*(1-alpha*To) +Pfe and Kb=I^2*Ro .
Ro=1/58*long[m]/sqmm 3*500 MCM=910 SQR.MM
For copper Ro=1/58*0.254/910=4.81E-06
TCAP=3.42[ J/cm^3/dgr.C]
Extracting Tf from the above equation:
Tf=(exp(t*Kb/TCAP/volume)*(Kb*Ti+Ka)-Ka)/Kb
volume =439 cm^3
For t=232 sec Tf=1089 dgr.C