Hi,
The two responses above surprise me. I think you would not need (or indeed want) to crossbond cable sheaths if the phase conductors are fully transposed.
My reasoning is described below without a great deal of rigor. If people disagree, perhaps I'll try harder!
Suppose we look at the cable end-on:
[tt]
(A) x (B) (C)
[/tt]
Where (A), (B), (C) represent the phase conductors, and x is a point between two phases, in the same horizontal plane. Assume that all current flows into the page, which means that each conductor produces circular lines of flux which flow in a clockwise direction. At point x, the lines of flux from all phase conductors are vertical and can be summed algebraically. The magnetic field at x due to the current in A-phase is k1*Ia, the magnetic field due to B-phase is k2*Ib, and the magnetic field due to C-phase is k3*Ic where k1, k2, k3 are constants.
The total magnetic field at x is:
Bx = k1*Ia + k2*Ib + k3*Ic
For a balanced three phase system Ia + Ib + Ic = 0, however, k1, k2 and k3 are different, because x is not the same distance from all conductors. An non-zero magnetic field
therefore exists at x.
The cable sheaths form loops which are shown below:
[tt]
+-----------A phase sheath-------------+
| A-C loop x |
+-----------B phase sheath-------------+
| B-C loop |
+-----------C phase sheath-------------+
^^ ^^
Local end Remote end
[/tt]
CASE 1 - conductors not transposed:
The field at point x will be the same when x is shifted along the cable lengthwise. A total flux cuts the A-C loop which induces a voltage and causes current to circulate in the sheaths.
CASE 2 - conductors fully transposed:
We now need to pick three x points, since three cable arrangements exist.
For the first third of the cable run:
[tt]
(A) x (B) (C)
[/tt]
Bx_first_third = k1*Ia + k2*Ib + k3*Ic
For the second third of the cable run:
[tt]
(B) x (C) (A)
[/tt]
Bx_second_third = k1*Ib + k2*Ic + k3*Ia
For the final third of the cable run:
[tt]
(C) x (A) (B)
[/tt]
Bx_final_third = k1*Ic + k2*Ia + k3*Ib
If we sum the magnetic field at the three x points:
Bx_first_third + Bx_second_third + Bx_final_third =
k1*Ia + k2*Ib + k3*Ic +
k1*Ib + k2*Ic + k3*Ia +
k1*Ic + k2*Ia + k3*Ib +
which is equal to zero, because Ia + Ib + Ic = 0.
We can see that for every x point along the cable run, there will be two other x points which provide a cancelling magnetic field. The total flux in the sheath loop is zero, and no circulating current flows.
Am I correct?
Thanks,
submonkey