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VD Quiz Question 1

VD Quiz Question 1

by Deleted user -
Number of replies: 8

Determine the minimum conductor size for a final sub-circuit supplying a single-phase 230V power circuit protected by a 20A Circuit breaker. The circuit consists of TPS cable with copper conductors in a roof space. The route length of the circuit is 38m, and the voltage drop in the existing consumer's mains is 3V. 
Voltage drop permissible in the final subcircuit: 
11.5 volts- 3 volts=
8.5 Volts

Find the maximum Vc Value for the cable that you can look up in the table.
1 Phase Vc= Vdx1000/(LxI)
                  = 8.5 x1000/(38x20)
                  = 11.18m.v/a.m
3 Phase Vc= 1 phase Vc/1.155
                 =11.18/1.155
                 = 9,68m.v/a/m

TPS cable with copper conductors should use Table 42; Column 6 (75 Degrees-max)
Answer: 4mm2 with a 6.49m.v/am rating.

Correct answer on the system is: 2.5mm2.

Is there and error?




Minimum size cable as per table


In reply to Deleted user

Re: VD Quiz Question 1

by Deleted user -
Disregard. Clause; 3.6.2. (1) has been brought to my attention.

1 Phase Vc= Vdx1000/(LxI)
= 8.5 x1000/(38x10)
= 22.36m.v/a.m
3 Phase Vc= 1 phase Vc/1.155
=22.36/1.155
= 19.36m.v/a/m

Giving 2.5mm2 perfectly fine to use.
In reply to Deleted user

Re: VD Quiz Question 1

by God. The creator of gunsparky -
That's what I like to see.
That's a great moment right there. smile
In reply to God. The creator of gunsparky

Re: VD Quiz Question 1

by Deleted user -
Hey God or someone,

Can you please explain step by step how to workout this question thankyou
In reply to Deleted user

Re: VD Quiz Question 1

by Deleted user -

Hi Brad,

The Total permissible Voltage Drop is 5%. Therefore 5% of 230V is 11.5V.

We've been given that the voltage drop in the existing consumer mains is 3V. Therefore the permissible Vd in the final subcircuit is:

Vd = 11.5V - 3V = 8.5V

Vc = 1000xVd/(LxI)

Normally for max demand questions, you would use the max demand for the final subcircuit (if provided). As only the circuit protection value is provided and this is a power final subcircuit (we're going to assume that these are socket outlets), we can use half the value of the circuit protection value. I think it's clause 3.6.2 Exceptions.

Therefore we get:

Vc = 1000x8.5/(38x(20x0.5))

= 22.4mV/A.m

This is the single phase Vc so we need to multiply this by 0.866 to get the 3 phase Vc

Vc = 22.4x0.866

= 19.4mV/A.m.

Assuming this is V90 TPS, we have a normal operating conductor temp of 75 degrees.

Looking up Table 42, column 6, our Vc value sits between 1.5mm2 and 2.5mm2. Always use the larger cable size so this gives us 2.5mm2

In reply to Deleted user

Re: VD Quiz Question 1

by Deleted user -
Hi Ashley, why does the VC formular involve the CB rating x 0.5 ?
My AS/NZS:3008 does not apply the 0.5 in the formular.
In reply to Deleted user

Re: VD Quiz Question 1

by Deleted user -

Hi Jai,

It really depends on the Voltage Drop question you’ve got. 

For questions where the max demand is provided, you would use the max demand value in your calculations. These would normally be for dedicated circuits where max demand can be determined. For final sub- circuits consisting of lighting or GPOs, as max demand is virtually impossible to determine, you can use half the value of the circuit protection device. Thus you get the CB rating x 0.5. 

Look up Clause 3.6.2 Exceptions 1 in AS3000 and that will give you the details