Answer / 7anoter4

You intend to run 3 cables of 3*50 mm^2 and 2 cables of
3*35 mm^2 type AYFY, I guess.
It depends on type and width of the cable trays. Let’s say
you want to use an open top and ventilated tray [ladder or
well perforated plate] and you want to carry the maximum
possible currents. Then, according to IEC 60364-5-52 Table
A.52-17 Note 2-you have to keep two cable diameters
clearance between cables and at least 0.3 time cable
overall diameter from the wall-both ends.
A 3*50+25 0.6/1 kV [according to IEC 60502-1] aluminum
conductor [compact], pvc insulated, steel tape armored and
pvc serving overall diameter 34 mm and for 3*35+16 overall
dia= 30 mm.
Here is the width calculation:
|0.3*34|O|2*34|O|2*34|O|2*34|o|2*30|o|0.3*30| where O =one
3*50+ cable diameter and o=one 3*35+ cable dia.
That means:
0.3*34+34+2*34+34+2*34+34+2*34+30+2*30+30+0.3*30=445.2
Then the tray width will be more than: 445.2 mm.
In 30 dgr.C air permissible current-carrying capacity will
be:
[See Table A.52-11 – Current-carrying capacity in amperes
system E aluminum]
[Or Tableau B.52-1 for aluminum col.7]
3*50 117 A
3*35 96 A
You may reduce the cable tray width but you have to reduce
the current in order to limit the conductor temperature to
70 dgr.C.

Answer / 7anoter4

For open top solid bottom cable tray and without any
clearance between cables the cable carrying capacity
has to be reduced to 73%. However the tray width will be
only:
0.3*34+34+34+34+30+30+0.3*30=181.2 mm
3*50+ 117*.73= 85.4 A ; 3*35+ 96*.73=70 A

Answer / 7anoter4

The cable tray is an assemble of tens or hundreds parts.
There are a big number of types and materials involved. You
may get a look in IEC 61537 or NEMA VE-1 or VE-2.

The simple way is to take the construction drawings and to
make a sketch of the cable tray run including the all
derivation on this drawing and then to send this to some
tray manufacturer as B-Line [for instance, you can find
many of them in web or Google].

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