Bit confused

[Rich306]

I'm currently doing the 2357 City and Guilds coursework at this precise moment, but i cannot seem to get my head round what should be a simple process in the design area.
Hopefully you guys will be willing to help me get round this first question so i can understand and complete the remaining questions.
Its regarding the minimum size CPC for 'each circuits' (one only provided).
I need to give consideration to disconnection times and thermal constraints.
Here is the details i have so far.

Prospective fault current - 50 kA -(doesn't seem right)?
400v Supply three phase
Ze of 0.008
i need to work out the CPC size for a 5kW 400v 3 phase machine.

I've looked in the Wiring regs and even looked at the adiabatic equation but still cannot get my head around it.
I know the following:

S is the nominal CSA of the conductor in mm[SUP]2[/SUP]
I is the value in amperes of fault current
t is the operating time of the disconnecting time in seconds. (But would that be type C circuit breaker? and what sort of time would i need to give for the above circuit?? 0.1 or 5?)
k being the factor of resistivity and temperature of the material.


A walk through talk through would be appreciated?

Thanks in advance.

 

[Sparky Ninja]

What type of protective device had been given?

 

[Rich306]

What type of protective device had been given?

I had to choose the protective device.

These are the calculations i had made a while back relating to this machine
i). 5kW, 400V three-phase machine
*Design current = 12.5A
*Overcurrent device = 16A
*Wiring enclosed in Metallic trunking
*Ambient temperature of 25degrees C
*Tabulated current = 23.98A
*Size Cable required = 4mm[SUP]2[/SUP]
*Length of cable required –19.4m
*Voltage drop = 2.3V

 

[spark 68]

I had to choose the protective device.

These are the calculations i had made a while back relating to this machine
i). 5kW, 400V three-phase machine
*Design current = 12.5A
*Overcurrent device = 16A
*Wiring enclosed in Metallic trunking
*Ambient temperature of 25degrees C
*Tabulated current = 23.98A
*Size Cable required = 4mm[SUP]2[/SUP]
*Length of cable required –19.4m
*Voltage drop = 2.3V

You might want to re-check your previous calculations, the bits highlighted in red first.

 

[Rich306]

You might want to re-check your previous calculations, the bits highlighted in red first.

Where am I going wrong here? Design current I gained from doing I = P/V so 5000/400

 

[spark 68]

Where am I going wrong here? Design current I gained from doing I = P/V so 5000/400

I know you did, but you have missed something out that is required for 3-phase calcs.

 

[Rich306]

I know you did, but you have missed something out that is required for 3-phase calcs.

I think it may have just clicked.... Square root of 3 x answer?

- - - Updated - - -

So 6.12amps

 

[spark 68]

Depends on how you calculate it,

kW / (Line Voltage X root 3)

or (kW / line voltage) / root 3

When you are given the kW value initially, it simplifies the calculation somewhat,as the PF and EF is already taken care of in the final kW value.

It is not as simple working the calculation forward. for where a motor/machine is involved then: Power (Watts) = 1.73 × Line Volts × Line Amps × PF × Efficiency.


Calculate it again, you are nearer.

 

[Engineer54]

I'm waiting to see what you're going to do about the protective device you choose!! lol!!

 

[telectrix]

I'm waiting to see what you're going to do about the protective device you choose!! lol!!

nails are rated at 1/2" per KVA, so a 3" should suffice.