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Posted By Topic: Volt drop

D.ele
Mar 27 2013 06:24

Can someone explain the difference between vc & vd values and when you would use each equation in a practical situation,
I am told vc is actually an impedance value (found in tables which I have) and not a voltage value.
I understand the working and maths but needing help on which equation is used where or if both are used?
   

AlecK
Mar 27 2013 09:28

As clause 4.2 of AS/NZS 3008.1.2 defines it;
V(c) is a voltage, not an impedance.
It's the voltage, in mV, dropped for every Amp for every metre of cable route length.
Though the cause of it is impedance of the cable.

Simply mltiply the mV value for the cale, x current, x metres to get total volt drop V(d)

Clause 4.3 arrives at V(d) by a different method, using the impedance of the cable, calculated from reactance and resistance.

If you look at Tables 30 to 39 you'll see that the reactance varies depending on layout of the cables, so this method suits situations where single-core cables are being laid.

The mV method is simpler, and is all you'll ever need for multicore cables.

really it's a case of "same dog, different bit of string": what matters is the total amount of volt drop in the circuit; or more commonly the total volt drop over several sections of cable from point of supply to end of any circuit.

Which is where simplified systems like Gencalc fall over, because they only look at one cable length at a time. Very useful, but because the working is all hidden you just get an answer without seeing how you got there.

The old way was to allow 2.5% of nominal voltage from main switch, the current rule is 5% from POS.
Rule-of-thumb splits that 5% to allocate 2.5% to mins, and 2.5% for submains & subcircuits (similar to old rule).

But when things get tight, it's often worth increasing mains size so that more of the "available" VD can be allocated to final subcircuits.

Or using a bigger cable for feeds to light switches, or the first few sockets on a circuit.

For each cable section. you need to do a max-demand calc / assessment to determine the current. The first section cable in a subcircuit should normally be taken at protection rating, but as you get further away you can use a lower figure (eg 10 A for last length to furthest away socket).

And doing the calcs by hand will give you a better feel for the effect of the change.
It's not all all hard, and you only have to calculate for longest circuit of each type .

I generally reverse it, and start with 11,500 mV, then subtract from that for mains, then submains, then each type of subcircuit.
When it gets down towards zero, time to start increasing cable sizes.
Most I've ever spent was a day, but that was a big house with 300m of mains, including 230m from meterbox, and at the end I could not only prove I was going to comply with volt drop under worst-case conditions, but could also be sure that EFLI would be fine.

Also that those long mains were cheaper than putting in poles and ruining the view.

Without those calcs, either my price wouldn't have been right or the finished installation wouldn't have complied.





   

D.ele
Mar 27 2013 14:52

Thanks aleck, awesome explanation.
   

D.ele
Apr 01 2013 21:19

One more thing,
I've been looking through as/nz 3008.12:2010 and need help.
Looking at table 10 and in my case dealing mostly with
Either partially or completely surrounded by thermal insulation unenclosed (copper tps).
Let's take process for 1mm of partially and completely covered scenario.
Partially = 13 amps & completely = 8 amps.
So this represents the current carrying capacity (also referred to in regs as 'continuous current carrying capacity' which I assume is the same thing?)
If it is continuous then I assume its referring to maximum amperage in calculated conditions without cable failure or loss of insulation integrity.
Secondly that would mean circuit breaker would have to rated below this value but what figure would be used when volt drop becomes a factor?
I am guessing you would use maximum demand and most often with final sub circuits you're trying to calculate whether 4 mm or 6 mm is required for range circuit.
I see lb < ln < lz which I take to mean calculated current must be equal or less to protective device which must be equal or less than current carrying capacity.
Assuming rule of thumb is breaker chosen to ensure circuit operates within designed circuit current (max demand) but is not subjected to cables amperage capacity for unsafe periods.
Sorry for the massive question aleck and others but this is bread and butter and I want to get my head around it.



   

Rob
Apr 01 2013 22:00

You have to do both calcs, volt drop and current carrying capacity. Whichever is most restrictive (biggest csa) determines which cable you use.
   

Wheelspanner
Apr 02 2013 10:39

Q: but what figure would be used when volt drop becomes a factor?

Ans:
Refer to AS/NZS 3000 3.6.2
The value of current used for the calculation of VD on a circuit need not exceed the
a) Total of the connected load supplied through the circuit
OR
b) Maximum demand of the circuit
OR
c) Current rating of the circuit protective device.

   

AlecK
Apr 04 2013 11:09

"So this represents the current carrying capacity (also referred to in regs as 'continuous current carrying capacity' which I assume is the same thing?)
If it is continuous then I assume its referring to maximum amperage in calculated conditions without cable failure or loss of insulation integrity."

Yes CCC is continuous current, ignoring short-term events such as fault currents.
Reason being that what we're trying to avoid is heating the conductor beyond the temperature limits of the insulation.
Which is why we have different tables for PVC (70 deg C)than for XLPE (90 deg C), etc.


"Secondly that would mean circuit breaker would have to rated below this value "

YES

"but what figure would be used when volt drop becomes a factor?
I am guessing you would use maximum demand and most often with final sub circuits you're trying to calculate whether 4 mm or 6 mm is required for range circuit.
I see lb < ln < lz which I take to mean calculated current must be equal or less to protective device which must be equal or less than current carrying capacity.
Assuming rule of thumb is breaker chosen to ensure circuit operates within designed circuit current (max demand) but is not subjected to cables amperage capacity for unsafe periods."

Wheelspanner's got the official answer; but to keep things simpler for most cases the process would bealong these lines:

1 establish the likely load on the circuit
2 select a cable size to carry that current (in the worst-case installation conditions of the cable route)
3 select circuit protection at setting not higher than cable's CCC
4 guestimate (or measure) the route length, including both horizontal and vertical components (eg lighting, 1 m up from swbd, 5 m to first drop 1.5 m down, 1.5m up, 6m to next drop... etc).
Usually just do this for longest circuit of each type.
5 do a VD calc using mcb rating and length to end of circuit (last socket / light / whatever.
If req'd, incrase cable size to cope.

That is conservative, but will keep you on the right side.

If you are pushing the edge, use 10 A at last socket and connected load for last room's furthest lightlight.

On a bigger instalation the VD component may result in using say 1.5 mm for ltg feeds and 1.0 for the load run from switch to light; or 4 mm for the first part of a socket circuit and 2.5 for the further away sockets.

Remember to count in the length for 2-way straps




   

D.ele
Apr 04 2013 23:39

Awesome, it all makes sense now.
Thanks aleck.