Exported earth

[biker mike]

you really have no concept of earthing have you.?place a short from PHASE to EARTH on any circuit and instantaneously the cpc will rise to 230v, before the protective device kicks into action.

you dont seem to have grasped the theory of what the 50v touch voltage is measured from

i think a real time experiment is required,,,, as it seems itll be the only way to solve the argument ......... ]

 

[Nicky Tesla]

I bet you really are steptoe with your holier than thow replys to everybody's posts.

Its you who doesnt have any concept of ohms law. You cannot grasp the concept that a resistor MUST have a current flowing through it for a voltage to exist across it, and that voltage, can be found by multiplying the current flowing through it by its own resistance.

AND can you also grasp the concept that what flows in the cpc also flows in the live conductor. So the 230v will be shared across the two conductors so this meens if you are still with me steptoe, that the cpc CANNOT REACH 230v

The current in the cpc before the fault occurs is 0A that is why it is at earth potential.

When the fault occurs there will be if you have any vision, be able to realise that the current MUST start from zero and rise at an alarming rate giving a graph of current verses resistance(which also rises during the fault) giving the voltage across the cpc.

This voltage wants to go way past 50v, but the characteristics of the mcb prevent it if the conditions of low cpc values are met.

So Steptoe stop being so arrogant and put some theory and maths down instead of your Im superior to you type answers you give everyone not just me.

 

[steptoe]

i think a real time experiment is required,,,, as it seems itll be the only way to solve the argument ......... ]

well there is someone on this thread that 'KNOWS' it will only be 50v MAX,

if I was that way inclined(read stoopid) then Id do it,

but I think that man there has just volunteered....

step forward young man,

and welcome NICKY

round of applause peeps,,,,

B)

 

[Nicky Tesla]

Go and buy a trailer load of old false teeth or something. ALBERRRRRT

 

[welshwizard]

Welcome Nicky.

Nothing wrong with a "Healthy Argument".

 

[steptoe]

Welcome Nicky.Nothing wrong with a "Healthy Argument".

until you realise your rubbish isnt being taken in by anyone cos most of the peeps here actually understand the principles of wiring, and earthing arrangements.

 

[Nicky Tesla]

Thats right constructive criticism yes bring it on, but sheer arrogance no

 

[steptoe]

Thats right constructive criticism yes bring it on, but sheer arrogance no

so Nicky,

Im sorry if I have missed something here,

but you have had this arrogance about being right with your postings about exporting an earth from a PME system,

yet you have failed to come up with any verifiable explanation as to how it would be safe to do so.

you have tried to state that under a phase earth fault condition that the cpc will not reach 230v potential,

that is just a ridiculous statement.

just how do you come about that conclusion?

that no prospective fault current capable of giving someone a severe electric shock will flow through an earth rod if you also connect to the MET of a PME system, yet your calcs are all wrong,

even though the fault current in PME could be up to 16kA, (DNO figure)

I have no idea how you can possibly even still consider making a dual earth system with TNCS and TT

or perhaps its the old scenario of a little knowledge is more dangerous than none at all.

 

[kme]

Thats right constructive criticism yes bring it on, but sheer arrogance no

Evenin` all.

Sorry I`m late. May I join in?

Nicky bud. Arrogance? I`ve just carefully read all postings since your first, and cannot honestly see ANY "out of order" comments from Steptoe or SL. Further, if my opinion is of any value, you have missed the basic premise of the indirect fault scenario; and I would suggest that if you were to back-track, and re-read the postings, you`ll find that these gents really DO know what they`re talking about. I`m not going to re-hash the argument within this post; as the topic has been comprehensively covered already.

Of course the EC/cpc rises towards Uo. No one suggests otherwise. But, although it won`t ever reach Uo; it could becove very (as in mV) close.

A PSC in excess of 10KA is not unheard of. As steptoe points out, DNO can, if determined by enquiry, quote 16KA as fault current.

You cannot work on the MCB`s rated figure. Otherwise, why would they be provided with a breaking figure (which is invariably in the KA realms)?

If the indirect fault, of negligable impedance, appeared on a 10mm T+E, 2M from the CU, in a domestic scenario (for arguments sake, lets call it TN-S).

2M 6mm cpc - approx 0.03 ohm (sorry, OSG in van)

Ze=0.01 ohm (measured)

IRRESPECTIVE of the overcurrent protective device.

I=V/R

=230/0.04=5700A

stop there a second, and confirm calculation.

Uo= phase/earth voltage=230V

R=0.01+0.03=0.04

ergo I=5750A.....ok so far?

That would be the current flowing (admittedly for a short duration, but we are talking instantaneous here, ok?)

Now lets introduce a parallel TT rod, which, due to the length of cable feeding it, has the following characteristics:

Rod impedance:99.87 ohms (allow for easy calcs. It`s late, and I`m tired.)

Earthing conductor impedance:0.1 ohm.

so we now have our two parallel resistances.

n.b. BOTH have Uo across them.

resistance 1: 0.04 ohms ( as described above)

resistance 2:

=0.03(cpc to CU)+0.1( 2nd E.C.)+99.87(Ra) =100 ohms

Yes........?

2 resistances in parallel. Voltage constant, current divides.

so fault current (5750A)/100.04 ohms=57.47 Amps

multiplied by 100(Ra)=2.29A

and by 0.04(R2+Ze)=5747A

Now, your comment r.e. resistors in series. You are correct. Lets look at that then, on the TT section:

0.03; 0.1; and 99.87 ohms respectively.

Series resistance - voltage divider.

Uo is still 230V

across a total of 100 ohms.

Uo per ohm =230/100=2.3

X 0.03 = 0.069V

X 0.1 = 0.23V

X 99.87 = 229.7V

All these are voltage across resistor.

So.... At the top of the rod, where connection is made, you will have Uo- 0.069 -0.23 =229.7

and at the bottom - You have the "general mass of earth".

If anyone sees a flaw in my description, please feel free to correct me. I would hope I know what I`m on about by now!!!

I really hope this helps; as I can`t think of a way to make it easier m8.

 

[Nicky Tesla]

Evenin` all.Sorry I`m late. May I join in?

Nicky bud. Arrogance? I`ve just carefully read all postings since your first,

HELLO KM THANK YOU FOR YOUR MORE CIVILIZED REPLY

You cannot work on the MCB`s rated figure. Otherwise, why would they be provided with a breaking figure (which is invariably in the KA realms)?

LOOK IN THE REGS BOOKS AND GUIDES AND SEE HOW MANY GRAPHS AND CHARTS CONSIDER THE MCB BREAKING AMPS

If the indirect fault, of negligable impedance, appeared on a 10mm T+E, 2M from the CU, in a domestic scenario (for arguments sake, lets call it TN-S).

2M 6mm cpc - approx 0.03 ohm (sorry, OSG in van)

Ze=0.01 ohm (measured)

IRRESPECTIVE of the overcurrent protective device.

I=V/R

=230/0.04=5700A

stop there a second, and confirm calculation.

Uo= phase/earth voltage=230V

R=0.01+0.03=0.04

ergo I=5750A.....ok so far?

OK SO FAR ....NO

V=230V(at the incoming supply ie at the db)

R=0.01(Ze)+0.03(cpc)+0.03(live conductor if the same csa of the cpc)

ergo I=V/R =230/0.07=3285A

That would be the current flowing (admittedly for a short duration, but we are talking instantaneous here, ok?)

NO THIS WILL BE THE CURRENT THE CONDITIONS WILL BE TRYING TO ACHIEVE...A CAR CANNOT GO FROM OMPH TO 100MPH AS IF IN A TIME WARP OR SOMETHING, IT WILL NEED TO ACCELERATE UP TO THAT SPEED, THE SAME PRINCIPLE HOLDS WITH THE FAULT CURRENT.

Now lets introduce a parallel TT rod, which, due to the length of cable feeding it, has the following characteristics:

Rod impedance:99.87 ohms (allow for easy calcs. It`s late, and I`m tired.)

Earthing conductor impedance:0.1 ohm.

so we now have our two parallel resistances.

WHERE?

n.b. BOTH have Uo across them.

HOW DO THEY?

resistance 1: 0.04 ohms ( as described above)

resistance 2:

=0.03(cpc to CU)+0.1( 2nd E.C.)+99.87(Ra) =100 ohms

Yes........?

NO....YOU DONT WORK IT OUT LIKE THAT

2 resistances in parallel. Voltage constant, current divides.

so fault current (5750A)/100.04 ohms=57.47 Amps

multiplied by 100(Ra)=2.29A

and by 0.04(R2+Ze)=5747A

Now, your comment r.e. resistors in series. You are correct. Lets look at that then, on the TT section:

0.03; 0.1; and 99.87 ohms respectively.

Series resistance - voltage divider.

Uo is still 230V

NO IT IS NOT....THE ONLY WAY 230V WOULD BE AT BOTH POINTS WOULD BE IF THERE WAS NO CURRENT FLOWING ...IE THE MET DISCONNECTED FROM EARTH

across a total of 100 ohms.

Uo per ohm =230/100=2.3

X 0.03 = 0.069V

X 0.1 = 0.23V

X 99.87 = 229.7V

All these are voltage across resistor.

So.... At the top of the rod, where connection is made, you will have Uo- 0.069 -0.23 =229.7

and at the bottom - You have the "general mass of earth".

If anyone sees a flaw in my description, please feel free to correct me. I would hope I know what I`m on about by now!!!

YOU SAY THERE IS A VOLTAGE OF 229.7 VOLTS AT THE TOP OF THE ROD+0.23 VOLTS BACK AT THE MET GIVING 230V BUT FROM THERE TO THE GENERAL MASS OF EARTH IS THE OTHER RESISTOR IN PARALLEL NAMELY Ze

SO WHY NOT RECALCULATE CURRENT AT THAT POINT THEN

I=V/R =230/0.01=23000A

SO USING YOUR METHOD THERE IS MORE CURRENT AT THE MET TO GND THAN AT THE FAULT....SOMETHING WRONG DONT YOU THINK.

I really hope this helps; as I can`t think of a way to make it easier m8.

HOW ABOUT THINKING OF HOW YOU CALCULATE VOLT DROP TO STAY ABOVE 96%

 

[SPECIAL LOCATION]

had writ me post & copied it in

then got this message...

The text that you have entered is too long (16449 characters). Please shorten it to 10000 characters long.

:^O :^O:^O:^O:^O:^O:^O:^O:

I shall split it modiy and re-post in sections...

(a word at a time would catch biker mike an WW lol)B)

 

[SPECIAL LOCATION]

Hi guys... HERE IS PART 1

 

[Mr Sworld]

Uh oh.....

It's getting into a shouting match!

We'll be getting 100pt fonts next! :^O

 

[SPECIAL LOCATION]

Hi guys...

 

[SPECIAL LOCATION]

Uh oh.....It's getting into a shouting match!

We'll be getting 100pt fonts next! :^O

No ones shoutin m8 well I'm not anyway:^O:^O:^O

just need to make sure trainee sparks don't go thinking it is impossible to get hazardous voltages due to inappropriate earthin arrangements.

No i reckon smaller fonts are needed

get more on a page..

LOL :^O

Still watin to see your BUTT you keep promisin!

 

[SPECIAL LOCATION]

I reckon thats about a week and a halfs worth of posts crammed into two posts!

 

[Nicky Tesla]

Hi guys... HERE IS PART 1

 

[kme]

HOW ABOUT THINKING OF HOW YOU CALCULATE VOLT DROP TO STAY ABOVE 96%

LOOK IN THE REGS BOOKS AND GUIDES AND SEE HOW MANY GRAPHS AND CHARTS CONSIDER THE MCB BREAKING AMPS

I`m sorry - relevance to this scenario is?

NO THIS WILL BE THE CURRENT THE CONDITIONS WILL BE TRYING TO ACHIEVE...A CAR CANNOT GO FROM OMPH TO 100MPH AS IF IN A TIME WARP OR SOMETHING, IT WILL NEED TO ACCELERATE UP TO THAT SPEED, THE SAME PRINCIPLE HOLDS WITH THE FAULT CURRENT.

Maybe my explaination was at fault. I`m not talking about the instant that the fault occurs. A few milliseconds after will be fine.

so we now have our two parallel resistances.

WHERE? Cable sheath & TT rod

n.b. BOTH have Uo across them.

HOW DO THEY? Because there is a fault to earth of negligable impedance.

THE ONLY WAY 230V WOULD BE AT BOTH POINTS WOULD BE IF THERE WAS NO CURRENT FLOWING ...IE THE MET DISCONNECTED FROM EARTH

Sorry m8 - daft comment.

I think we are probably going to have to "agree to disagree" on this; unless SL`s in-depth & excellent posts have helped you understand.

SL - full marks on that. have that other scooby snack now m8

 

[Nicky Tesla]

Hi guys...

 

[Nicky Tesla]

so we now have our two parallel resistances.

WHERE? Cable sheath & TT rod

CABLE SHEATH...WOT THE PVC ON THE OUTSIDE OF THE CABLE, THATS WOT THE SHEATH IS

IF YOU MEAN THE CABLE JOINING THE ROD TO THE MET, THEN THAT IS IN SERIES WITH THE ROD

WOT IS IN PARALLEL IS THE ROD AND THE MET (which has a distributors earth in these examples) JOIN THESE TOGETHER TO GET AN EQUIVALENT RES SLIGHTLY LOWER THAN THE ZE VALUE THEN CONNECT THE LOWER PART TO -VE (battery vision again) THE TOP PART TO THE BOTTOM OF THE NEXT RESISTOR BEING THE RESISTANCE FOR THE CPC (from the met to the fault) AND LASTLY THE TOP OF THAT RESISTOR WOULD BE JOINED TO THE +VE

TWO RESISTORS 1 ABOVE THE OTHER IN SERIES THATS SERIES AND ONCE MORE FOR GOOD LOOK IN SERIES, AND NOT SIDE BY SIDE IN PARALLEL

I