Maximum Cable Run Question.

[staedtler]

Hello all,
Im not an electrician or really have much background in electrical theory but I was Just wondering if someone could show me how I would go about answering the following question.

A power ring is to be installed using a 2.5mm twin and earth PVC cable. The circuit will be protected by a 32A type B circuit breaker. The cable will be directly clipped to the walls and joists. Calculate the maximum cable run for the ring if the external earth loop impredance is 0.8Ω?

Take into account electrical resistances for 2.5mm and 1.5mm copper cables are 7.41 mΩ/m and 12.1 mΩ/m. The maximum earth loop impedaqnce for a type B circuit breaker is 1.44Ω and the correction factor for PVC cable is 1.2.

Thank you for your help,

Luke

 

[mickys86]

How come you wanna answer the question?

 

[Jimmyb]

Sorry joking thought it was a windup

 

[staedtler]

I'll be honest im a mathematics student doing a module on basic electrical theory and I dont understand how to complete this problem.

 

[staedtler]

Jimmyb is that a joke yes

 

[mickys86]

I'm sure it is and my calc is in the van

 

[Richard Burns]

Hello all,

A power ring is to be installed using a 2.5mm twin and earth PVC cable. The circuit will be protected by a 32A type B circuit breaker. The cable will be directly clipped to the walls and joists.

Calculate the maximum cable run for the ring if the external earth loop impredance is 0.8Ω?

Take into account electrical resistances for 2.5mm 7.41 mΩ/m
and 1.5mm copper cables 12.1 mΩ/m.
The maximum earth loop impedaqnce for a type B circuit breaker is 1.44Ω and the correction factor for PVC cable is 1.2.

Total resistance of the loop (from line to earth) of twin and earth cable plus the external loop impedance must be less than or equal to the earth loop impedance for the breaker (as corrected).

You have the resistance of the cables and with a ring final circuit there will be two cables of equal length both sharing the current.
resistance of resistors in parallel is 1/R = 1/R1+1/R2+1/R3+1/R4....

Take external loop impedance off the corrected max ELI and the remaining resistance is the maximum allowed for the cables.
Back calculate to obtain the maximum length of the cable permitted.

 

[staedtler]

I'm sure it is and my calc is in the van

electricians are suprisingly funny

 

[Silly Sausage]

Luke, post your answer & workings.
I'm sure someone will check it out for you.

 

[telectrix]

OK, start with the formula Zs(max.) = Ze + (R1+R2), where R1 is the resistance of the L. and R2 that of the E.

 

[staedtler]

Thank you.

Maybe an even easier one first;

The turbine for the system will provide a maximum voltage of 200V DC to the inverter at a current of 5A. The specification of the inverter shows the input voltage must not fall below 170V else the inverter will shut down. Calculate the maximum run that can be used if 2.5mm[SUP]2[/SUP] cable with a loss of 19mV / A / m is installed to connect the turbine to the inverter.

Would this simply be;

0.019 x 30 = 0.57
0.57 x 5 = 2.85
30/2.85 = 10.53m​

 

[telectrix]

be interesting to see what answers come up. i know what i get.

 

[Silly Sausage]

Thank you.

Maybe an even easier one first;

The turbine for the system will provide a maximum voltage of 200V DC to the inverter at a current of 5A. The specification of the inverter shows the input voltage must not fall below 170V else the inverter will shut down. Calculate the maximum run that can be used if 2.5mm[SUP]2[/SUP] cable with a loss of 19mV / A / m is installed to connect the turbine to the inverter.

Would this simply be;

0.019 x 30 = 0.57
0.57 x 5 = 2.85
30/2.85 = 10.53m
​

No.
volt drop = 19mv/A/m
you're allowed 30 volts @ 5A
so......

 

[staedtler]

volts drop per m run with a current of 1A

 

[Richard Burns]

You have a current running of 5A so if you calculate the value of volts dropped for a current of 5A over one meter.
Then you are allowed a maximum of 30V drop do the division and you will get the number of metres permitted.