How were the cable sizes calculated for a 618 KW load in a construction project?

[KhalidAbu]

Hi guys
I received electrical drawings for a construction project as a part of invitation for tendering.
The project includes a number of SMDBs connected to two MDBs which are connected to two transformers. Looking at the schematic design, I found that a SMDB is connected to one of the two MDBs using 4 cables of 4 cores of 240 and for each cable there is 120mm2 ECC.

4x (4CX240mm2 XLPE/SWA/PVC+1CX120mm2 ECC)

The total connected load of the SMDB is 618 KW and it is 3 phase load, 400 V and Power Factor =0.8 . The distance between the SMDB and the MDB is 110 meters.

My question is how on Earth did he find this result? How did he make the cable size calculation? Be aware that the regulation used here is BS 7671.
Many thanks in advance

 

[DEvens]

Probably he didn't calculate it. Probably he looked it up in a handbook of some kind that has a table of cable-vs-load-vs-distance for satisfactory operation.

Probably the calculation would go something like so. The resistance per meter of cable is this function of area. The cable is so long. You need the loss in the cable to be no more than so much. You could express that as either a voltage drop or a power dissipated in the cable. And you have an estimate of the power to be used in the building. You plug in the values and get out the area.

But probably this is not the whole story. Probably after that there are things like the physical requirements on the cable, such as being able to withstand being strung between supports. And to be able to withstand temperature or weather. And to provide appropriate levels of insulation. And to meet the local building codes, fire codes, and so on. The result is probably that you need a cable with somewhat more area than the minimal area required to carry the expected load.

And that is probably in a handbook rather than being calculated every time. Possibly the cable manufacturer puts out such tables to help their clients decide what cable to buy.

 

[William White]

Hi guys
I received electrical drawings for a construction project as a part of invitation for tendering.
The project includes a number of SMDBs connected to two MDBs which are connected to two transformers. Looking at the schematic design, I found that a SMDB is connected to one of the two MDBs using 4 cables of 4 cores of 240 and for each cable there is 120mm2 ECC.

4x (4CX240mm2 XLPE/SWA/PVC+1CX120mm2 ECC)

The total connected load of the SMDB is 618 KW and it is 3 phase load, 400 V and Power Factor =0.8 . The distance between the SMDB and the MDB is 110 meters.

My question is how on Earth did he find this result? How did he make the cable size calculation? Be aware that the regulation used here is BS 7671.
Many thanks in advance

copper?
aluminium?
in insulated materials (beneath lagging, soil etc?)
conduit?
tray
free air?

618KW is a hefty load and 4x 240mm four core seems very reasonable.The calculation starts with the maximum possible current carrying capacity of the cable (in) then you apply correction factors which derates the cable to get the actual capacity , itit = in / cc x ci x cg x ca x cr

cc - derating for buried cable
ci - derating for thermal insulation
cg - derating for cable grouping
ca - derating for ambient temperature
cr - derating when wired fuses are usedthen you have to consider volt-drop which is

cable length x design current x (mV/A/m) / 1000

so if you have a cable of 100 m and a current of 350A and the volt drop per amp per metre of of 0.5mV then

100 x 350 x 0.5 / 1000 = 17.5 V

which is unacceptable, so a bigger cable is needed so as to give a lower volt drop
you can pick up guides to BS7671 which have the relevant tables giving cable sizes.

The best (I think) it Unite Union's 'The Electrician's Guid to Good Electrical Practice' which is somewhat cheaper than buying BS7671

 

[KhalidAbu]

Probably he didn't calculate it. Probably he looked it up in a handbook of some kind that has a table of cable-vs-load-vs-distance for satisfactory operation.

Probably the calculation would go something like so. The resistance per meter of cable is this function of area. The cable is so long. You need the loss in the cable to be no more than so much. You could express that as either a voltage drop or a power dissipated in the cable. And you have an estimate of the power to be used in the building. You plug in the values and get out the area.

But probably this is not the whole story. Probably after that there are things like the physical requirements on the cable, such as being able to withstand being strung between supports. And to be able to withstand temperature or weather. And to provide appropriate levels of insulation. And to meet the local building codes, fire codes, and so on. The result is probably that you need a cable with somewhat more area than the minimal area required to carry the expected load.

And that is probably in a handbook rather than being calculated every time. Possibly the cable manufacturer puts out such tables to help their clients decide what cable to buy.

Probably he didn't calculate it. Probably he looked it up in a handbook of some kind that has a table of cable-vs-load-vs-distance for satisfactory operation.

Probably the calculation would go something like so. The resistance per meter of cable is this function of area. The cable is so long. You need the loss in the cable to be no more than so much. You could express that as either a voltage drop or a power dissipated in the cable. And you have an estimate of the power to be used in the building. You plug in the values and get out the area.

But probably this is not the whole story. Probably after that there are things like the physical requirements on the cable, such as being able to withstand being strung between supports. And to be able to withstand temperature or weather. And to provide appropriate levels of insulation. And to meet the local building codes, fire codes, and so on. The result is probably that you need a cable with somewhat more area than the minimal area required to carry the expected load.

And that is probably in a handbook rather than being calculated every time. Possibly the cable manufacturer puts out such tables to help their clients decide what cable to buy.

Thanks for your reply. I have been told that there was a calculation using tables from the BS 7671

 

[KhalidAbu]

copper?
aluminium?
in insulated materials (beneath lagging, soil etc?)
conduit?
tray
free air?

618KW is a hefty load and 4x 240mm four core seems very reasonable.The calculation starts with the maximum possible current carrying capacity of the cable (in) then you apply correction factors which derates the cable to get the actual capacity , itit = in / cc x ci x cg x ca x cr

cc - derating for buried cable
ci - derating for thermal insulation
cg - derating for cable grouping
ca - derating for ambient temperature
cr - derating when wired fuses are usedthen you have to consider volt-drop which is

cable length x design current x (mV/A/m) / 1000

so if you have a cable of 100 m and a current of 350A and the volt drop per amp per metre of of 0.5mV then

100 x 350 x 0.5 / 1000 = 17.5 V

which is unacceptable, so a bigger cable is needed so as to give a lower volt drop
you can pick up guides to BS7671 which have the relevant tables giving cable sizes.

The best (I think) it Unite Union's 'The Electrician's Guid to Good Electrical Practice' which is somewhat cheaper than buying BS7671

Thanks you William for your reply .
The cables are copper and they will run on a perforated cable tray on air (they are inside the building) and they will run together with other 5 cables of the same size (in total there are 9 cables on the cable tray). Could you please make the calculation more specific ? consider Ca=55 C . What ACB do you select for that? and Why ACB and not MCCB? please bear in mind that he choose and ACB 1600 TP.
Thank again

 

[William White]

Well, I've shown you HOW to do the calc, and where to get the information to do the calc. The choice of acbs and mccbs is to do with discrimination.
Not being funny, but this question is about a commercial tender, not an academic exercise (are we getting near the rules for this forum?)

I am loathe to be doing a job for you that an electrician should be getting paid for. Nor willing to give advice that could be used in an argument with your electrician (and me possibly ending up getting the blame for advice that you misinterpreted).

If you think what the electrician has specified is unreasonable you need to take it up with them, or compare it to another tender. It is your job as a customer to make that choice! I'm not going to do commerical evaluations for free.

OR

Buy BS7671 and do the calcs yourself, they are not difficult.

 

[Babadag]

I agree with William White: it is a job for an electrical engineer. However, in my opinion ,the voltage drop is not more then 1.6% then less than maximum permissible .So, I think the cable is suitable for the job –some one has to check the temperature drop of entire circuit and the actual length of the cables.

 

[William White]

I was not prepared to calculate the volt drop - I just showed how to do it

 

[KhalidAbu]

Well William, I'll try to figure out by myself and I will come back to you if I fail. Anyway thank you for your clew
Regards

 

[anorlunda]

The limiting thing for a cable is likely short-circuit current, not load-carrying current.

If the cable is destroyed every time there is a short circuit, it is not of much use.

 

[William White]

Thats why there is protection!

You choose your cable for load; your choose your protection for faults.
Just think about the simpe case of house wiring.
A 2.5mm^2 ring-main is rated at 32A continuous, whilst the short circuit protection is in the order of 6000A.

 

[Babadag]

I don't think it's a problem with short-circuit current: 240 mm^2 copper withstands 60 kA short-circuit current in 0.3 second- from 90oC steady state load.
Four cables withstand 60 kA [ up to 250oC] in more than 5 sec.