Large cable termination

[saucyjack]

Hi chaps. I have a potential install which is a 56kW micro array. The supply is three phase, but the array is going to be 380M from the house (yes that is the correct number!). Working to a 1% volt drop over that distance, we'd be looking at 240mm2 4 core SWA (£42,000 worth), and I'm told this isn't an issue for the customer! My problem is that I don't normally deal with this kind of physical scale etc., so can anyone offer some advice on what I should do to actually terminate this size cable? I'm assuming the cores themselves would have to be lugged, but I'm not sure what ancillary gear (MCCB's etc) that I will need to physically come out from? The intention at the array end would be to feed a large-ish 3P DB, with the 12 or 13 strings radiating out from there, but it's the house end that I'm unsure of. Any and all advice would be much appreciated

 

[snowhead]

Have you also considered handling the weight of the cable and drum and also pulling / laying the cable in the ground?
It's going to be around 5 tonne.

If you've not handled this size / amount of cable before it may not be the best one to start learning with.

I beleive it's usual to use 2 smaller parallel cables in situations like yours.

 

[davesparks]

First thing I'd do is look at getting that cable size down by running multiple cables in parallel.
Off the top of my head I'd guess that maybe 3 x 70mm in parallel may work.

Either way you'll want to look at reducing the cable size at each end to something that is suitable for the equipment being installed. You can achieve this in multiple ways such as using a busbar chamber, enclosure and din rail connectors (yes they go that big) or taking it straight in to a switch/isolator (sized for the cable, not the current)

How is the cable being installed? Is it in a trench or something else?

Working with/terminating big cables is one part of the job I really enjoy but don't get to do as often as I'd like.

You'll probably want to get the cable in place in its final location, strip it, fit the glands and then slide whatever enclosure it's going into onto the cable. Trying to wrangle the cable into a pre-fixed enclosure will be somewhat tedious.

Whatever you do don't cut the cores short until you have set them in to their final shape/position as far as possible, the longer the cores are the easier they are to work with.

Spreader boxes are essential, either one designed to work with your switchgear or make your own from 12x12 trunking.

 

[saucyjack]

Thanks for the advice guys; I do appreciate that the cable weight would be in tonnes rather than kilos, which was one of my logistical concerns. The customer is apparently trenching the run himself, but I'll certainly work out if the parallel cabling is a better (i.e. easier!) alternative. Once again, thanks for the help!

 

[saucyjack]

Having worked the volt drop out on 70mm2 over the immense distance the customer has in mind, (with half of the total array being supplied by this, and a 63A demand/phase), the Vd is something like 13 volts. This is unfortunately way off the 2.3V max I need to allow for; back to 5 tonnes of 240mm2!

 

[Rockingit]

Having worked the volt drop out on 70mm2 over the immense distance the customer has in mind, (with half of the total array being supplied by this, and a 63A demand/phase), the Vd is something like 13 volts. This is unfortunately way off the 2.3V max I need to allow for; back to 5 tonnes of 240mm2!

Er.... three runs of parallel 70 gets you 210mm2. You'll need to work out the maths manually by calculating the overall resistance of that as it won't be tabulated.

20 secs online just gave me a figure of 0.000860098 V /Km @ 1A

 

[Rockingit]

Having worked the volt drop out on 70mm2 over the immense distance the customer has in mind, (with half of the total array being supplied by this, and a 63A demand/phase), the Vd is something like 13 volts. This is unfortunately way off the 2.3V max I need to allow for; back to 5 tonnes of 240mm2!

Walk us through your maths on this, please.

 

[saucyjack]

Re the calcs:

135 x 350W inverters, so 47,25kW.

47.25kW = approx. 190A (ish), over three phases, so around 63A/phase (max). I know there's a slightly more accurate formula to get the current flow on three phase, but this is close enough.

To achieve no more than a 1% volt drop (4V) with a 380M cable run at 63A = 240mm2 (both with cable calc software, then doing it via the table(s) in the big brown book to double check).

(Having done a similar exercise with 70mm2 and a correspondingly lower current demand, the result is still something like 150mm2)

If I've messed up, I'm more than happy to take the egg on face if it means an easier way of doing this!.
Thanks guys.

 

[James]

When calculating parallel cables, devide the current by the amount of cables.
i.e. 60A total would be 30A per cable if you run 2 or 20A if you ran 3 cables

 

[saucyjack]

When calculating parallel cables, devide the current by the amount of cables.
i.e. 60A total would be 30A per cable if you run 2 or 20A if you ran 3 cables

100% understand my friend, so my other calculation was on the basis of 32A/phase over the same distance (380M) though still requiring a 1% limit. Unfortunately, 70mm would still (individually) have a drop of something like 7V+. Further digging has led me to the possibility of two 95mm2 cables though, which is obviously much better!
Thank you for all you help though

 

[davesparks]

Having worked the volt drop out on 70mm2 over the immense distance the customer has in mind, (with half of the total array being supplied by this, and a 63A demand/phase), the Vd is something like 13 volts. This is unfortunately way off the 2.3V max I need to allow for; back to 5 tonnes of 240mm2!

I think you may have made an error in your calculations.

If you are planning 3 cables in parallel then you need to divide the load current by 3 to calculate the drop in the individual cables.

I make the volt drop in a parallel set of 3x70mm to be 4.4V at a conductor operating temperature of 70C.

 

[davesparks]

Re the calcs:

135 x 350W inverters, so 47,25kW.

47.25kW = approx. 190A (ish), over three phases, so around 63A/phase (max). I know there's a slightly more accurate formula to get the current flow on three phase, but this is close enough.

To achieve no more than a 1% volt drop (4V) with a 380M cable run at 63A = 240mm2 (both with cable calc software, then doing it via the table(s) in the big brown book to double check).

You're mixing up voltages here, you've calculated the current at 240V and your volt drop limit for 400V and then earlier in the thread you mentioned a 2.3V volt drop limit which implies a 230V nominal voltage.

What voltage does this actually need to be designed for?

 

[saucyjack]

You're mixing up voltages here, you've calculated the current at 240V and your volt drop limit for 400V and then earlier in the thread you mentioned a 2.3V volt drop limit which implies a 230V nominal voltage.

What voltage does this actually need to be designed for?

Ooops! 400V 3 phase supply, 47.25kW max load (roughly 63A/phase), 380M run, 1% volt drop (as you correctly point out, 4V!). I originally based the cable size on one cable to achieve this (240mm2), but I appreciate that splitting this into 2 or three smaller sizes may be far more practical. Sorry for the earlier confusion!