do you use the multipliers from tab 13 in OSG when calulating volt drop?

[juice]

I've recently done a C&G 2396 Electrical Design Course and unfortunately the tutor who taught the course did so for the first time and in my opinion didn't do a good job of teaching the material. I'm now doing my revision for an exam this Thursday and trying to make sure all the formula I use are correct, so I'm cross referencing across my IET design book, Regs, OSG and the Unite book (if you guys know it) they sold us, in order to clear up any confusion from the tutor. I daren't call him as what he says will probably confuse me further

My query is in calculating the voltage drop on a circuit. The formula I was given in class was;

volt drop = length (l) x design current (Ib) x mV/A/m x 1.2 (multiplier from table 13 OSG)
1000

After looking in the books and on electricians forum it seems that the formula is;

volt drop = length (l) x deign current (Ib) x mV/A/m
1000

Do I need to apply the multiplier from table 13 OSG to this equation? As I see it the multiplier is applied due to operating temperature of the conductor. Does this have an effect on volt drop?


Thanks in advance

Juice

 

[Darkwood]

Welcome to the forum

 

[Richard Burns]

Voltage drop is calculated as part of the design process using the volt drop values in the tables in BS7671.
These values are given for the normal operating temperature as stated at the top of each table.
Therefore there is no need to correct the values to that temperature.

It would only be in the case where you were performing local measurements that you need to correct values to the normal operating temperature.
This is not normally undertaken at any time.

Table I3 is for when you are measuring resistances for R1 and R2 values or such like.

 

[Leesparkykent]

You would apply the multiplier if using the measured resistance of line and neutral. VD= RLNx1.2xIb

 

[juice]

Cheers gents I thought what the tutor had taught was wrong. Can't believe it! Feel like asking for my money back. He'd even put the formula on a handout he gave us.

Been looking at the books trying to wrap my head round it. I see it now in Appendix 4 table 4D1A (for example) conductor operating temperature 70 C at ambient temp 30 C so the mV/A/m has allowed for the max operating temp of conductor.

In reality what temperature do cables reach? Do they get close to 70 C? As the higher the conductor temp the greater the volt drop right? Then if the volt drop value exceeds the max value permitted (eg 3% or 5% of 230V) we increase the size of the conductor to allow for the increase in temp due to increase in amps to prevent deterioration of conductors and possible risk of fire. So volt drop tells us (measures) how hot the cable/conductor is getting and if it presents a risk. Hey boys I think I've got it! Unless you shoot me down in flames that is.

I'd still like to ask more but that'll do for now, still a little unsure how the ambient temp gets factored into the conductor temp if it ever does. I know we do with the protective device which I guess governs the amps in the cable by governing Zs with max values. So much to learn so little time till payday..

 

[Risteard]

and the Unite book (if you guys know it)

When I did the 2391-20 the Amicus Guide was probably the most useful book for it. (It is also permitted material for the exam, or at least was for many years. I presume this hasn't changed since the transition to 2396 for the course.)

 

[Richard Burns]

Cables can reach any temperature, this is dependent on the current flowing through the conductor.
The intent of designing a circuit such that Iz > In > Ib is that the conductors can never carry enough current on a long term basis to allow the conductors to exceed (for a PVC insulated conductor) 70°C.
I.e the design current is such that this is well within the current carrying capacity of the cable. Therefore in normal use the conductor is never at or close to the maximum temperature permitted.
If there is a fault such that the current increases beyond the design current then the intent is that the circuit protective device will operate before the conductors overheat. This is why the In is less than the Iz.

The calculation for voltage drop is there to ensure that devices that are dependent for their operation on receiving a 230V supply do not fail to operate because the voltage has reduced beyond the equipment's limits.
If you start with 230V and 12V is lsot through the resistance of the supply cable then the equipment will only see 218V and this may be too low to allow the device to operate.
Therefore in the design stage the expected voltage drop is calculated to ensure that the level is not too great (i.e 3% of the supply voltage for lighting and 5% of the supply voltage for other uses).
If the voltage drop is too high then this means the resistance of the conductors is too high.
In order to reduce the resistance the size of the conductor can be increased (obviously if the circuit can be made shorter this will also work but the distances are normally set by the supply location and the equipment installation location) is the cross sectional area of a conductor is increased keeping the length the same the overall resistance will be lower.


The ambient temperature around a conductor affects how fast the cable loses heat and therefore, over a long time period, how hot the conductor gets.
If a conductor is carrying enough current in an ambient temperature of 30°C to raise the conductor to (for PVC) 70°C then this is the limit specified in the the current carrying capacity tables in BS7671.
However if the cable is located in a cold area at say 10°C then the rate of heat loss from the cable will be higher as the temperature differences are greater, therefore the cable could carry slightly more current before it reached 70°C.
In order to take account of these variations from the tabulated values in BS7671 there is a table of correction factors for ambient temperature, this is table I2 in BS7671.

The other correction factor (in table I3) is applicable to resistance values measured on a dead circuit (that will not be running hot as it is off) to account for the fact that when the conductor is in use it will be hotter because of the current flowing through it and so the resistance will increase.

 

[DPG]

Another cracking reply from Richard. Daz

 

[happysteve]

As well as correcting for temperature for measured values of R1+RN (as per Leesparkykent's post above), you could also use the 1.2 multiplier from I3 of the OSG if, instead of the volt drop factors (in mV/A/m) from the tables in Appendix 4 of BS7671, you used table I1 of the OSG instead, which lists resistance of copper cable, in mΩ/m.

If you think about it (Ohm's law), R = V/I; if you express this in units, it's Ω = V/A, and also mΩ = mV/A. So a value in mΩ/m is the same as a value in mV/A/m.

For example, line 2 of table I1 lists a situation with a line and cpc of 1mm, and lists "R1+R2" as 36.20mΩ. This could equally be R1+RN of 1mm. As stated at the top of the table, it's at 20C. If you apply the temperature correction factor of 1.2, you would get 43.44mΩ/m. If you look at the volt drop of 1mm copper cable in appendix 4 (any sort should do) if memory serves me right it will say 44mV/A/m. There's a bit of rounding up or down somewhere, but you get the gist!

You can also use this relationship if for some reason you find yourself with BS7671 but without the OSG to work out the expected resistances of cables. Use the volt drop values from Appendix 4 in BS7671, apply the inverse correction factor (divide by 1.2) and that gives you the reistance per metre - in total, "there and back" so to speak - for the same size cable.

 

[Apeglar]

When I did the 2391-20 the Amicus Guide was probably the most useful book for it. (It is also permitted material for the exam, or at least was for many years. I presume this hasn't changed since the transition to 2396 for the course.)

Hi
Is this the book you are referring to:
Electrician's Guide to Good Electrical Practice - Based on BS7671 : 2008 (17th Edition)
Cheers