Old Exam Question

[Blueday]

Hi,

I was hoping someone could help me with following question from an old exam paper I found in the house:

Aring final circuit is fed by pvc cablesin steel conduit from a 240 V a.c. supply. The phase conductor develops a fault of negligible impedance to earth at a socket-outlet. The previously tested value of Zs at the socket-outlet was 0.9 ohms. Assume that a conduit joint develops a resistance of 5 ohms due to local corrosion. State the
(i) probable earth fault current
(ii) probable power developed acrross a joint in the conduit
(iii) likely effects of the fault being sustained.

Many thanks in advance.

 

[telectrix]

ohm's law might work here.

 

[trev]

My thoughts exactly Tel

 

[Silly Sausage]

and I[SUP]2[/SUP]R

 

[daiplayer]

first thought : 0.9+5 =5.9 ohm. so 230/5.9 - 38A

HOWEVER :the Zs is no longer 0.9 as the joint is changing it. eg, if you assume a tncs to be 0.35 , the r1+r2 remaining will be 0.55,.

how do we find the R1 alone to find old R2 reading ? ( which needs to be deducted )
we cant use the old 1.67 x r1 as ( i can assume ) the cpc is the metal conduit (not t+e) .

example : old Zs = Ze +(R1+R2) = 0.35 +(0.2+0.35) = 0.9 Zs
new Zs ( with altered r2) 0.35+(0.2 +5) = 5.2 ohm Zs
or 0.35 +(0.2 +5+0.35) ..........old calc plus new high 5 ohm joint = 5.9 ohm Zs

Or am i overthinking it, lol

 

[buttonmoon]

I think you may have over thunk this one.

I would go with:

(i) Probable Earth Fault Current = U/Z = 240/(0.9+5) = 40A ish
(ii) probable power developed across a joint in the conduit. If they're talking about the joint with the 5 ohm resistance then the power dissipated across that joint would be P = U^2/Z = 11.5 kW
(iii) likely effects of the fault being sustained. Meaning no effective protective device = melting, burning, exploding and all that.

I have just noticed the age of this post but I've gone and written all this now and perhaps it will be useful to someone.

 

[buttonmoon]

Oh my mistake its not an old post I was reading the wrong date.

 

[Silly Sausage]

(ii) probable power developed across a joint in the conduit. If they're talking about the joint with the 5 ohm resistance then the power dissipated across that joint would be P = U^2/Z = 11.5 kW

You might want to re-think that one!

 

[telectrix]

You might want to re-think that one!

before the joint melts. (P= Isq x R) = 40 x 40 x 5 = 320watts ( approx)

 

[Chr!s]

You might want to re-think that one!

He's only missed Zs

 

[Chr!s]

before the joint melts. (P= Isq x R) = 40 x 40 x 5 = 320watts ( approx)

40 x 40 x 5 = 320, i don't like your calculator lol

 

[Guest111]

before the joint melts. (P= Isq x R) = 40 x 40 x 5 = 320watts ( approx)

thats odd I get it to 8000, 40 x40 =1600x5 =8000, or am I having a senior moment?

 

[Chr!s]

thats odd I get it to 8000, 40 x40 =1600x5 =8000, or am I having a senior moment?

Yes, that's right

 

[telectrix]

bloody small keys. i pressed divide by 5 .instead of times by 5. DOH. :6:

 

[Chr!s]

The fault is 5.9, so 9762.713

 

[Silly Sausage]

He's only missed Zs

You mean he's included Zs?

 

[telectrix]

something tells me it's beer o'clock.

 

[Chr!s]

You mean he's included Zs?

No, missed it

 

[Chr!s]

5 = 11520

5.9 = 9762.713

 

[Silly Sausage]

something tells me it's beer o'clock.

Already on the case!

I get...230/5.9=40.7
P at the dodgy joint is 40.7[SUP]2[/SUP] * 5 = 8273W


or is the beer acting quick today?