"Voltage Drop" Please help!

[johnohagan]

Hi folks Iknow this topic has been beaten to death on this forum but after reading some of the posts and speaking to some of my co workers I am still a littleconfused, I am aware there is several ways of calculating voltage drop but amstill unsure of when to use the correction factor of 1.2 and when to leave itout I have some examples below could you folks look at them and tell me if theyare the correct methods and when to use the 1.2 and when to leave it out. Thankyou

1. Calculate the voltagedrop for a 2.5 cable supplying a radial socket 22 meters long operating at anambient temperature of 30°C and having a operating temperature of 70°C and adesign current of 16 Amps the voltage drop is 18 mV/A/m, calculate the overallvoltage drop
   Voltage Drop= (mV/A/m) x Ib x L
   1000
   Where mV/A/m= milivolts per amp per meter.
   IB- is the design current.
   L -is the length

Voltage Drop​

18x 16 x22 = 6336 mV ​

6336/ 1000 = v ​

6.33V​

1. Ifa 10mm 3 core cable has a length of 80meters and is supplying a 9KW shower at avoltage of 230volts, operating at 20°c. Conductor resistance for 10mm copperconductor is 1.83mΩ/mtr
   
   Voltage Drop = (r1 +rn)x L x Ib
   Voltage Drop = voltagedrop
   (r1 + rn) = Conductorresistance
   L = Correctiontemperature (30°) (1)
   L= Correction temperature(20°) (1.2)
   Ib = Circuit RatedCurrent

9000/230 = 39.13A so theBreaker size 40A​

1.83mΩ/ mtr x 80 = 146.4m Ω​

146.4/1000 = 0.146Ω and 0.146 x 2​

(r1 + rn) = 0.2928 Ω​

Vd = (r1 +rn) x L x Ib​

(0.2928) x 1.2 x 40A​

= 14.05 V​

1. A radial circuit is fed from a 16A mcb the value of the combined conductors are 0.29ohms,calculate the volts drop for the circuit, and determine if the value isacceptable?

Ohms lawI X R
16 X .29 = 4.64 v

 

[Pete999]

Homework???????

 

[Richard Burns]

Generally when calculating volt drop you are using design values of volt drop that are given in BS7671 as at the maximum operating temperature (70°C or 90°C as appropriate) therefore there is not normally any reason to make adjustments for the temperature.

However if you were using measured values of the resistance of a conductor (that would likely be around 20°C) then in order to get this value to be equivalent to the tabulated values you would then multiply the measured resistance by 1.2 to get the resistance at 70°C and use that in your calculation.

In your calculations above it is not defined which values are being used.
The first question is using the 70°C values for 2.5mm² conductors so no correction should be applied.
(I would round 6336/1000 up to 6.34 V)

The second question is using the 20°C values for a single 10mm² conductor.
So 1.83mΩ * 1.2 = 2.2mΩ @70°C

r1 + rn = 2.2 x 2 =4.4mΩ

Vd = 4.4 * 80 *39.13 /1000 = 13.77 V

Using the design current not the breaker size.

The third question does not specify and since there is no length then the reference temperature cannot be determined by back calculation.
There is no design current specified so taking the worst case scenario of 16A as the design current performing your calculation may well be correct.
If it were specified that the 0.29 Ω was measured at 20°C then multiplying by 1.2 to get the 70°C value would adjust the resistance to 0.348 Ω.

 

[johnohagan]

I appreciate your feedback thank you that has clarified that for me.