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my surge protection 'cpd', such as it is...

psychicwarrior
Good morning all


I have obtained the following [I feel as the] 'best we can do for now' information from the DNO. They were helpful in my test case request for info.


"...address 1:

LV underground - 362m

HV underground to primary-  2000m

No Surge protection devices


address 2:

LV underground to substation - 110 m

HV UG from substation to HV pole is 823m then 301m 11kV overhead then 190m HV UG to the primary substation.

Distance to cable termination with surge protection device from secondary substation is 1124m (823m 11kV UG + 301m 11kV Overhead).  The nearest cable

termination does not have surge protection installed. ..."


In relation to the risk assessment equation variables dealing with distribution cable lengths  ie. the Lpal Lpcl Lpah Lpch, how would one fit the given values into the variables ?


For addr1, ignore the HV (e.g. use zero for Lpah Lpch),  then take Lpcl as 362 and Lpal as 638 ?


For addr2, i'm not sure on this one ?


Thanks for your input.

Habs


Parents
  • 0
    You only take into account the LAST kilometer of distribution cabling (nearest to the premises), or the distribution cabling distance to the first upstream overvoltage protective device in the distribution network (if any), whichever is closer.


    In this case, there is no overvoltage protection in the network (if you can't get information, assume there are none).


    Therefore (hopefully my maths head is working today and I've not made any schoolboy errors - but I'm sure others will be quick to chip in if I have), working back from the premises a cable distance 1 km:

    Address 1
    LV underground, LPCL = 0.362 km

    HV underground, LPCH = 0.638 km (because you're working back from the installation a maximum distance of 1 km)

    Which also means:


    HV overhead, LPAH = 0 km

    LV overhead, LPAL = 0 km

     


    So, LP = 2 × 0 + 0.638 + 0.4 × 0 + 0.2 × 0.362 = 0.7104


    Address 2

    LV underground, LPCL = 0.110 km

    HV underground, LPCH = 0.823 km

    HV overhead, LPAH = 0.067 km (because you're working back from the installation a maximum cable distance of 1 km)

    Which also means:


    LV overhead, LPAL = 0 km

     


    So, LP = 2 × 0 + 0.823 + 0.4 × 0.067 + 0.2 × 0.110 = 0.8718




    Just for the sake of examples:

    High Ng - If we use the worst-case Ng in Figure 44.2, Ng = 1.4


    If Address 1 is rural/suburban, CRL = 85/(0.7104 × 1.4) = 85.5, and as this is less than 1000, protection against transient overvoltages of atmospheric origin is required

    If Address 1 is urban CRL = 850/(0.7104 × 1.4) = 855, and again protection is required


    Since for Address 2, LP is greater, CRL will be smaller (as it's a divisor), and hence protection will be required there in either case also.


    Low Ng- If we use a lower value of Ng = 0.8 (which if you look at the map of the country, probably much of the landmass of the UK has Ng ≤ 0.8)

    If Address 2 is rural/suburban, CRL = 85/(0.8718 × 0.8) = 121.9, and as this is less than 1000, protection against transient overvoltages of atmospheric origin is required

    If Address 2 is urban CRL = 850/(0.8718 × 1.4) = 1219, and this time, protection is NOT required


    Similarly, Address 1 would require protection in rural/suburban (CRL = 149.6) but not urban (CRL = 1496)

Reply
  • 0
    You only take into account the LAST kilometer of distribution cabling (nearest to the premises), or the distribution cabling distance to the first upstream overvoltage protective device in the distribution network (if any), whichever is closer.


    In this case, there is no overvoltage protection in the network (if you can't get information, assume there are none).


    Therefore (hopefully my maths head is working today and I've not made any schoolboy errors - but I'm sure others will be quick to chip in if I have), working back from the premises a cable distance 1 km:

    Address 1
    LV underground, LPCL = 0.362 km

    HV underground, LPCH = 0.638 km (because you're working back from the installation a maximum distance of 1 km)

    Which also means:


    HV overhead, LPAH = 0 km

    LV overhead, LPAL = 0 km

     


    So, LP = 2 × 0 + 0.638 + 0.4 × 0 + 0.2 × 0.362 = 0.7104


    Address 2

    LV underground, LPCL = 0.110 km

    HV underground, LPCH = 0.823 km

    HV overhead, LPAH = 0.067 km (because you're working back from the installation a maximum cable distance of 1 km)

    Which also means:


    LV overhead, LPAL = 0 km

     


    So, LP = 2 × 0 + 0.823 + 0.4 × 0.067 + 0.2 × 0.110 = 0.8718




    Just for the sake of examples:

    High Ng - If we use the worst-case Ng in Figure 44.2, Ng = 1.4


    If Address 1 is rural/suburban, CRL = 85/(0.7104 × 1.4) = 85.5, and as this is less than 1000, protection against transient overvoltages of atmospheric origin is required

    If Address 1 is urban CRL = 850/(0.7104 × 1.4) = 855, and again protection is required


    Since for Address 2, LP is greater, CRL will be smaller (as it's a divisor), and hence protection will be required there in either case also.


    Low Ng- If we use a lower value of Ng = 0.8 (which if you look at the map of the country, probably much of the landmass of the UK has Ng ≤ 0.8)

    If Address 2 is rural/suburban, CRL = 85/(0.8718 × 0.8) = 121.9, and as this is less than 1000, protection against transient overvoltages of atmospheric origin is required

    If Address 2 is urban CRL = 850/(0.8718 × 1.4) = 1219, and this time, protection is NOT required


    Similarly, Address 1 would require protection in rural/suburban (CRL = 149.6) but not urban (CRL = 1496)

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