2391 exam question. pls give an answer

[IQ Electrical]

Are you using 1.2 as a temperature correction factor when ambient temperature is not given?

 

[morph]

1.2 because its the multiplier

no 1.20 because the resistance value is at 20 degree C , and when vd occurs its under load so at a given temp of 70 degree C

 

[ringer]

Why do you measure resistance of both L and N when calculating volt drop? I thought the purpose of establishing if volt drop is in acceptable limits is to ensure that equipment will receive adequate voltage/current to function correctly. So we ought to then only be concerned with volt drop between CU and the load, not from CU then through the load and over the complete return path.

 

[morph]

Why do you measure resistance of both L and N when calculating volt drop? I thought the purpose of establishing if volt drop is in acceptable limits is to ensure that equipment will receive adequate voltage/current to function correctly. So we ought to then only be concerned with volt drop between CU and the load, not from CU then through the load and over the complete return path.

because if at DB1 you have 230V and at DB2 you have a 4.5 VD and then another 7.6 V at the end of a circuit fed from DB2 then the total VD is now 12.1 v

 

[IQ Electrical]

no 1.20 because the resistance value is at 20 degree C , and when vd occurs its under load so at a given temp of 70 degree C

I've never done a voltage drop calculation using the 1.2 factor, this assumes a fully loaded circuit at maximum operating temperature, a scenario so rare that even the IET's own Design Guide book does not consider it necessary.

 

[morph]

I've never done a voltage drop calculation using the 1.2 factor, this assumes a fully loaded circuit at maximum operating temperature, a scenario so rare that even the IET's own Design Guide book does not consider it necessary.

1.2 is correct when using resistance at 20 degrees though , are you using the other method where you times the length and IB and mv/m ?
which is different to the figures in table 9a

 

[ringer]

because if at DB1 you have 230V and at DB2 you have a 4.5 VD and then another 7.6 V at the end of a circuit fed from DB2 then the total VD is now 12.1 v

Sorry, I still don't get it. I see no reference to a sub-board in the original question, and even if there was, surely the VD will only be along the L that supplies the sub-board. I still don't see how VD along the neutral comes in to play.

 

[morph]

N eutral comes into play because vd is under load, the question say 4.5 v at DB

 

[IQ Electrical]

1.2 is correct when using resistance at 20 degrees though , are you using the other method where you times the length and IB and mv/m ?
which is different to the figures in table 9a

No, the tabulated values of volt drop already include a factor for increased conductor operating temperature, I'm talking about the figures in table 9A and the rare scenario whereby a circuit will be fully loaded and operate at it's maximum temperature.

For that reason, on my designs, the 1.2 factor is not used.

 

[Chr!s]

A single phase power circuit supplying an item of equipment,having a full load current of 25 A,is to be checked to confirm voltage drop compliance.The circuit is installed using 6.0 mm2 cooper conductors,with a circuit length of 35m.
a) Calculate the maximum permissible voltage drop for this circuit.

230 x 5% = 11.5v

b) The line and neutral combined resistance for the circuit is 0.26 Ohms . Determine whether the circuit meets the voltage drop requirements if the voltage drop to the local distribution board 4.5 V.

25 x .26 x 1.2 = 7.8 v + 4.5 v distribution = 12.3 v, i would assume this was a measured value at 20 C so temp factor needs applying.


c)State one other method which could be used to confirm voltage drop.

It may be evaluated using site design information for example

 

[Chr!s]

I've never done a voltage drop calculation using the 1.2 factor, this assumes a fully loaded circuit at maximum operating temperature, a scenario so rare that even the IET's own Design Guide book does not consider it necessary.

And that caught out many taking the June 2010 2391-10 exam, the conductor resistances were given, not the tabulated values from the reg's.

IQ if you remember the exam, the resistance were given at 20 C so correction was required

 

[IQ Electrical]

Yes, for exam purposes, you need to use the 1.2 factor but I'm talking about my design calculations,-If I have a cable with a CCP of say 87 Amps, my Ib will be well below this level and as a result, my conductor operating temperature will be nowhere near maximum.

It's the difference between a rigid factor and some real world commercial acumen/experience to supplement the regulations.

"Where it is known that the actual operating temperature under normal load is less than the maximum permissible value for the type of cable insulation concerned (as given in the tables of current carrying capacity) the multipliers given in Table 9C may be reduced accordingly"

OSG Page 167

 

[Chr!s]

Yes, for exam purposes, you need to use the 1.2 factor but I'm talking about my design calculations,.

Its required for testing purpose where the conductor temp is not at 70 C (or its operational temp).

-If I have a cable with a CCP of say 87 Amps, my Ib will be well below this level and as a result, my conductor operating temperature will be nowhere near maximum

If you were designing, you may calculate the operating temp at a given load, then you may apply a ct to the mV/A/m values.This will give you a lowered volt drop, and also decrease your i2r losses.

It's the difference between a rigid factor and some real world commercial acumen/experience to supplement the regulations

Not really, if you know the measured temp is lower than the operational temp then a factor need applying, I agree it may not be 1.2 but a factor needs applying .

"Where it is known that the actual operating temperature under normal load is less than the maximum permissible value for the type of cable insulation concerned (as given in the tables of current carrying capacity) the multipliers given in Table 9C may be reduced accordingly"

OSG Page 167

Not disagreeing, it may not be 1.2, though a measured value will in most instances need correcting.

 

[Chr!s]

Sorry, I still don't get it. I see no reference to a sub-board in the original question, and even if there was, surely the VD will only be along the L that supplies the sub-board. I still don't see how VD along the neutral comes in to play.

Does current flow in the neutral in a single phase circuit?

 

[telectrix]

Does current flow in the neutral in a single phase circuit?

if it didn't then we could do away with it and just use single core cable.