2396 Design exam help on processes

[chrisjw3330]

Hi
I am resitting my 2396 tomorrow and going through my past papers I have a question that I stumbled on last time out. The training provider I failed with are no longer in business, they weren't very good, so I am on my own.

The questions are ;

A 230v single phase circuit is wired using a multicore 70c thermoplastic cable having 1.5mm2 conductors and 49m in length.
It is protected by a 16a BS88c,PScc reading at the origin of the circuit is 110a.

Q. Determine if this circuit has adequate short circuit protection?

And -

Q. How do you determine if a circuit breaker will protect the live conductors in line with 434.5.2?

Not necessarily looking for answers but a breakdown of the process would be greatly appreciated.

Thanks

Chris

 

[Zdb]

Think the first one is calculating Zs and making sure that it meets disconnection time and the second one you need to use the adiabatic equation.

 

[chrisjw3330]

Think the first one is calculating Zs and making sure that it meets disconnection time and the second one you need to use the adiabatic equation.

Thanks for the response, didn't think I would get a reply as I have left it late. For the first question do you know how I might calculate the ZS without the Ze or even a clue of earthing arrangement?

 

[SparkyChick]

Read the first question again carefully... short circuit protection... you're not interested in Zs... you're interested in the line/neutral loop impedance at the origin (which you have information from which it can be derived) and the R1+Rn of the circuit.

 

[chrisjw3330]

Read the first question again carefully... short circuit protection... you're not interested in Zs... you're interested in the line/neutral loop impedance at the origin (which you have information from which it can be derived) and the R1+Rn of the circuit.

Thanks also. I did query this with the original course provider and he said just use the adiabatic equation, it came up and was worth 20 marks so knew he told me incorrectly.

So I am looking for the milliohms per metre for 2x 1.5mm conductors x LOR (49) x Cr (1.2)?

And then the adiabatic equation from there? Which is (?)

sqrt 110x110x0.2\115 ?

 

[SparkyChick]

For the first question about whether the circuit will be adequately protected from short circuit, I'd do this...

Calculate R1+Rn, calculate the LN loop impedance (R=V/I) and sum them, then compare that to the maximum loop impedance tables for the OCPD for the circuit. If the value you got is less than the table (adjusted as required) and the OCPD should meet the required disconnection times for the circuit.

 

[chrisjw3330]

For the first question about whether the circuit will be adequately protected from short circuit, I'd do this...

Calculate R1+Rn, calculate the LN loop impedance (R=V/I) and sum them, then compare that to the maximum loop impedance tables for the OCPD for the circuit. If the value you got is less than the table (adjusted as required) and the OCPD should meet the required disconnection times for the circuit.

I'll give that a go, thanks again.

I wish I hadn't let the gaffer talk me into this.

 

[Ian1981]

Adiabatic equation is I2t square root top answer divide by K
K in most cases is 115 70 degree thermoplastic conductors if conductor is not bunched or incorporated in a cable
143 for conductor incorporated into a cable or bunched.
For say steel like swa it’s 51 (using memory here) and so on.
So the fault current needs to be known either by direct measurement or by calculation.

Worth noting that for disconnection times quicker than 0.1 seconds for protective devices given in appendix 3 then the manufacturers let through energy is needed and can be used in the equation so that I2t is equal to or less than K2S2

 

[chrisjw3330]

Adiabatic equation is I2t square root top answer divide by K
K in most cases is 115 70 degree thermoplastic conductors if conductor is not bunched or incorporated in a cable
143 for conductor incorporated into a cable or bunched.
For say steel like swa it’s 51 (using memory here) and so on.
So the fault current needs to be known either by direct measurement or by calculation.

Hi Ian

Cheers for that. Does the fault current of 110a at the circuit origin need to be in the adiabatic or does the fault current at the end of the circuit need working out?

 

[Ian1981]

Hi Ian

Cheers for that. Does the fault current of 110a at the circuit origin need to be in the adiabatic or does the fault current at the end of the circuit need working out?

Surely the 110amps is at the end of the circuit?
And is your fault current for the circuit.
Sorry didn’t read earlier post about pssc.

Edit the fault current will be for the circuit at the most furthest point ie end of the circuit for a radial for example

 

[Richard Burns]

The questions appear to be very similar.
To determine short circuit protection for a 16A BS88-3c the modified adiabatic equation t= (k²S²)/I² should be used to determine the maximum disconnection time permitted and this should be compared to the actual disconnection time of the protective device at that fault current.
With 110A PSCC at origin then the resistance would be 2.09Ω.
Add the 1.42Ω of the circuit to give 3.51Ω, so a short circuit current at the extremity of the circuit of 65.5A.
t= (115²*1.5²)/65.5² =6.94s.
From graph 3A1 at 65.5A a 16A BS88-3c will disconnect in 2s which is less than the 6.94s above so the circuit has adequate short circuit protection.

The second question asks about circuit breakers and 434.5.2 which is the modified adiabatic equation t= (k²S²)/I². Circuit breakers are current limiting so the I²t for the circuit breaker needs to be obtained (35000 type B, 42000 type C) from the manufacturer or product standard and this value must be less than the k²S² (29756) for the circuit.

 

[chrisjw3330]

Surely the 110amps is at the end of the circuit?
And is your fault current for the circuit.

Edit the fault current will be for the circuit at the most furthest point ie end of the circuit for a radial for example

The question states at the origin of the circuit in question.

The questions appear to be very similar.
To determine short circuit protection for a 16A BS88-3c the modified adiabatic equation t= (k²S²)/I² should be used to determine the maximum disconnection time permitted and this should be compared to the actual disconnection time of the protective device at that fault current.
With 110A PSCC at origin then the resistance would be 2.09Ω.
Add the 1.42Ω of the circuit to give 3.51Ω, so a short circuit current at the extremity of the circuit of 65.5A.
t= (115²*1.5²)/65.5² =6.94s.
From graph 3A1 at 65.5A a 16A BS88-3c will disconnect in 2s which is less than the 6.94s above so the circuit has adequate short circuit protection.

The second question asks about circuit breakers and 434.5.2 which is the modified adiabatic equation t= (k²S²)/I². Circuit breakers are current limiting so the I²t for the circuit breaker needs to be obtained (35000 type B, 42000 type C) from the manufacturer or product standard and this value must be less than the k²S² (29756) for the circuit.

These were the methods I originally started working both questions out with but then started to doubt myself.
Thanks for the input, a lesson learned for me here is don't do home study and choose providers more carefully. Thanks.

 

[Richard Burns]

Good luck with the exam.

 

[Spottydogs]

Best of luck to all doing it!

I'm also doing the exam tonight, keeping my fingers crossed it isn't as bad as I'm expecting it to be! Long time since I've sat and done a 3 hour exam and my 2.5 year old kindly passed on his cold and cough to me!!

 

[Spottydogs]

How did you find it? I seemed to start off ok enough but time began to beat me and I'm not sure I answered some of the later questions as well as I might have. Not really sure if I did enough, blooming hope so though, I'd rather not have to do it all again!!