in this Q and A video we'll be considering the
question can i use steel conduit as the CPC or the earth in a circuit
the first thing to suggest in this video is that we know the difference between an
earth and a CPC or circuit protective conductor however if we didn't include the word earth in
the title of the video no one will watch it now this content relates to the recent review that
Gary and Gordon made on the Metpro Conlok system so if you're interested in a quicker and easier
way to install steel conduit then please do go and watch that video after this one of course
everything that we're about to discuss applies equally to threaded conduit connections and the
Conlok grub screw system because that system has been tested to meet the relevant requirements for
earth continuity so can we use metal conduit as the CPC in a circuit well the simple answer is yes
providing you meet certain requirements make sure you stay on to the end of the video for the
last piece of the puzzle in this process so let's look at the conditions that allow us to use steel
conduit as a CPC we find the first part of the answer in Regulation 543.2.1 in BS7671 where we
find a list of items that can be acceptably used as a protective conductor and six on the list we
find that we can use a metal conduit metallic cable management system or other enclosure or
electrically continuous support system for conductors so if we stopped reading there
we'd have our answer so why don't we stop there well that list also says that we can use a single
core cable as the CPC but we know that installing a one millimeter squared conductor as the CPC for
a 150 mil three-phase sub-main probably won't cut it so let's keep digging there's more detail
further down the page in Regulation 543.2 which reads the metal covering including the sheath bare
or insulated of a cable in particular the sheath of a mineral insulated cable trunking and ducting
for electrical purposes and metal conduit may be used as a protective conductor for the associated
circuit if it satisfies both requirements of items 1 and 2 of regulation 543.2.2 so
there's some requirements that we need to meet in regulation 543.2.2 in order to be able to use the
steel conduit for our CPC however it's only the first two items in that Reg let's read those now
where a metal enclosure or frame of a low voltage switch gear or control gear assembly or bus bar
trunking system is used as a protective conductor it shall satisfy the following three requirements
its electrical continuity shall be assured either by construction or by suitable connection in such
a way as to be protected against mechanical chemical or electrochemical deterioration
so this part is saying that you really need to make sure that all the connections between your
pieces of conduit and accessories like boxes and socket outlets are properly made as the
system will rely on those for both a mechanical and an electrical connection the second part
of that Reg is really interesting and is what we're going to focus on for a few moments now
two its cross-sectional area shall be at least equal to that resulting from the application of
Regulation 543.1 or verified by test in accordance with the appropriate part of BSEN61439 series
so if the conduit system is put together properly and tightly with the right accessories
we should be able to meet the first item easily enough but what about that second item
have you ever thought about the cross-sectional area of steel conduit let's do it now regulation
543.1 which is just one page back in the regs discusses the cross-sectional areas of protective
conductors and starts in 543.1.1 but by giving us two options for making sure the CSA of the steel
conduit is the right size first we can calculate using Regulation 543.1.3 which if we glance down
the page involves using what's generally referred to as the adiabatic equation which is not as scary
as it looks it's just a way of figuring out what size your protective conductor should be
using the actual conditions it will experience when a fault occurs but it can be a bit more
long-winded so we'll cover that in a future video the second option is to use regulation 543.1.4
which is what we're going to do just a quick note before we do that though just after outlining
these two options the following paragraph points out that if we've selected the CSA of our line
conductors to make sure they can cope with the expected short circuit current and we expect the
earth fault current to be lower then we do need to use the adiabatic method however generally
speaking we size our line conductors to make sure they can cope with either the design or nominal
current in the circumstances they're installed in and sometimes we may beef them up a bit to reduce
vault drop but only occasionally do we size them to make sure the cable can cope with short circuit
current because if all the other conditions for sizing a cable are met then it should be okay
one other thing to think about is if you're running a couple of circuits through the conduit
say one for lighting and one for power then Regulation 543.1.2 comes into effect this states
where a protective conductor is common to two or more circuits its cross-sectional area shall be
1 calculated in accordance with Regulation 543.1.3 for the most onerous of the values
of fault current and operating time encountered in each of the various circuits or 2 selected
in accordance with Regulation 543.1.4 so as to correspond to the cross-sectional area of the
largest line conductor of the circuits so because we're using the second method of those two values
we need to use the largest line conductor in the conduit to check our conduit is okay
now let's move on to Regulation 543.1.4 to see how we're going to do this check
it states where it is desired not to calculate the minimum cross-sectional area of a protective
conductor in accordance with regulation 543.1.3 the cross-sectional area may be
determined in accordance with table 54.7 this is the polite British Standard way of saying
if you can't be bothered with the adiabatic equation you can use this easier way so why
are there two different methods well it's because while the adiabatic method is a bit more involved
it's also more specific meaning that you could get away with using a smaller cross-sectional area
for your CPC it won't apply in this situation but it could make a big difference if you're
installing very large singles the regulation concludes where the application of table 54.7
produces a non-standard size a conductor having a larger standard cross-sectional area shall be
used this is just saying that if after the process we're about to do you end up with a weird number
for your CPC like 3.1 mil or something then you can go to the next size up so that would be
four mil in that case okay so let's look at table 54.7 as directed and see what it's telling us
in the first column we've got the cross-sectional area of the line conductor in square millimeters
this value is given the identifying letter s so if s the size of your line conductor is less than or
equal to 16 mil you'll use the top row if it's bigger than 16 mil up to and including 35 mil
you'll use the second row and if it's bigger than 35 mil you'll use the bottom row the second and
third columns are used to find what the size of our CPC must be the second column doesn't apply
to the situations it's referring to a situation where your CPC is the same material as the line
conductor and in this case of course we're using steel conduit not copper conduit the third column
is for when the protective conductor is of a different material to the line conductor which is
applicable in this situation so just to make this a little more relevant let's say we're feeding a
socket outlet in 2.5 mil singles with 70 Degrees C thermoplastic insulation but we want to use the
conduit as the CPC is that going to be acceptable well our line conductor is smaller than 16 mil so
we'll be using the top row here we're instructed to use the formula k1 over k2 multiplied by s
and performing this calculation will tell us what the CSA of our conduit needs to be
so we know s is the CSA of our line conductor so that'll be 2.5 in this case but what about these k
values well in the Regs a k value is a factor that can be applied to calculations regarding fault
current and the effects of the heat it generates in conductors it's based on the resistivity heat
coefficient and heat capacity of the conductor material in question beneath the table we find out
specifically what they are for in this calculation k1 is the value of k for the line conductor
selected from table 43.1 in chapter 43 according to the materials of both conductor and insulation
k2 is the value of k for the protective conductor selected from tables 54.2 to 54.6 as applicable
so let's find our value for k1 in table 43.1 you can see here we'll find our value in this
column because we've got a thermoplastic insulated conductor rated at 70 Degrees C
smaller than 300 square millimeters because it's a copper conductor we get a k value of 115
so that's our k1 value that we can put into our formula but what about k2 well for this one
we've got a number of tables to consult namely 54.2 to 54.6 but the one we're interested in
is 54.5 which is values of k for steel conduit ducting and trunking as the protective conductor
now we'll use this column because the insulation is 70 Degrees C thermoplastic so our k value is 47
and putting that into the k2 slot of the formula we now have the following calculation
115 divided by 47 and then multiply the whole thing by 2.5 and this gives us an answer of 6.12
now remember that value of 6.12 is the minimum cross-sectional area that our CPC should have
in square millimeters so the next logical question is what's the cross-sectional area of our conduit
well we can figure it out fairly easily no doubt you remember how to find the area of a circle
which is simply pi times the radius squared if we're using 20 mil conduit it will have a radius
of 10 mil so that's pi times 10 squared which gives us 314.16 square millimeters now that would
be great if the conduit was a solid cylinder but it would make it harder to get the cables in so
we need to subtract the area of the hollow section now what's the radius of the hollow section well
we could measure it but it's better to consult the manufacturer's data in this case Metpro state that
for 20 and 25 mil steel conduit the wall thickness is at least 1.37 millimeters so subtracting that
value from the 10 mil outer radius of the tube gives us 8.63 for the radius of the hollow section
so pi times 8.63 squared gives us 244.95 square millimeters subtracting the second value from
the first will give us the area of the solid section which is 69.21 square millimeters
so can we use the steel conduit as the CPC well we needed a minimum of 6.12 square millimeters
from our calculation from table 54.7 and we got 69.21 so yeah should be okay one final point to
emphasize is found in 543.2.7 and it states that where the protective conductor is formed by metal
conduit trunking or ducting or the metal sheath and or armor of a cable the earthing terminal of
each accessory shall be connected by a separate protective conductor to an earthing terminal
incorporated in the associated box or other enclosure so if you feed a socket outlet or a
switch with conduit you need to connect a fly lead from the earthing terminal in the socket or the
switch to the earthing terminal in the back box to ensure that it's properly earthed so there we go
you can indeed use steel conduit as the CPC for a circuit as long as you've met the requirements
discussed in this video but as always we want to hear from you do you use steel conduit as a CPC
or do you always run a separate one in for good measure do you ever use the calculation from table
54.7 to check that your conduit CSA is acceptable whatever your thoughts whatever your questions
please leave them in the comments section below and as always thank you very much for watching