Adiabiatic & voltage drop

[aaelectric]

Hi all,
been trying to work this out now and carnt get my head round some of the calculations.
please see where im going wrong.
voltage drop:
so 63amp mcb supplying electric boiler, run is approx 25 meters in 16mm twin & earth.
so Mv/a/m x length x amps divided by 1000= 2.8x25x63 divided by 1000=4.41%
which means its under 5% so is ok?

also trying to work out the adiabiatic equation to see if earth is big enough:
so i worked out zs first: zs=ze+r1+r2 so ohms per meter for 16mmtwin & earth is 0.00423 per meter.
so 0.00423x25meters=0.11.
next is 230v divided by 0.11=2,090
this is where ia am stuck any help would be great

also trying to work out adiabiatic on main cable. Soo.
tncs system with ze of 0.15 at main and 1.5Ka. The flat is being feed from a 5419 60amp isolator in 16mm twin and earth. Zdb at fuseboard is 0.19 and 1.2Ka. Its about 10 meter run from main fuse, how would i go about this one.
cheersguys hopes this clears things up and someone shows an easy way of calculating this.
also when filling out the eicr where is says main earth would this be the earth in the 16mm t&e or the 16mm earth going to MET at service head.

 

[Geoffsd]

so 63amp mcb supplying electric boiler, run is approx 25 meters in 16mm twin & earth.
so Mv/a/m x length x amps divided by 1000= 2.8x25x63 divided by 1000=4.41%

Use the actual amperage of the load and you need to multiply by 2 (for the circuit).

Also the result - actually 8.82 is Volts not percentage. 230 x 5% = 11.5V so it's ok.

also trying to work out the adiabiatic equation to see if earth is big enough:
so i worked out zs first: zs=ze+r1+r2 so ohms per meter for 16mmtwin & earth is 0.00423 per meter.
so 0.00423x25meters=0.11.
next is 230v divided by 0.11=2,090

You wrote Zs = Ze + R1 + R2 but you have not done it.
You need to use Zs so add on Zdb, 0.11 + 0.19 = 0.30
230/0.30 = 767A

also trying to work out adiabiatic on main cable. Soo.
tncs system with ze of 0.15 at main and 1.5Ka. The flat is being feed from a 5419 60amp isolator in 16mm twin and earth. Zdb at fuseboard is 0.19 and 1.2Ka. Its about 10 meter run from main fuse, how would i go about this one.

(sq.rt.(I² t)) / k.

 

[brman]

Use the actual amperage of the load and you need to multiply by 2 (for the circuit).

Also the result - actually 8.82 is Volts not percentage. 230 x 5% = 11.5V so it's ok.


You wrote Zs = Ze + R1 + R2 but you have not done it.
You need to use Zs so add on Zdb, 0.11 + 0.19 = 0.30
230/0.30 = 767A


(sq.rt.(I² t)) / k.

Why are you doubling the voltage drop? the 2.8mVolts/meter/amp is taken from the onsite guide so is for R1+R2 and including the adjustment for 70DegC.
EDIT: actually it is R1+Rn for voltage drop. Must turn my brain on before typing.....

 

[D Skelton]

Hi all,
been trying to work this out now and carnt get my head round some of the calculations.
please see where im going wrong.
voltage drop:
so 63amp mcb supplying electric boiler, run is approx 25 meters in 16mm twin & earth.
so Mv/a/m x length x amps divided by 1000= 2.8x25x63 divided by 1000=4.41%
which means its under 5% so is ok?

The result you have (4.41) is volts, not percentage. 230V x 5% = 11.5V so 4.41V is within 5%. Other than that, contrary to what Geoff said (sorry pal), you are correct.

also trying to work out the adiabiatic equation to see if earth is big enough:
so i worked out zs first: zs=ze+r1+r2 so ohms per meter for 16mmtwin & earth is 0.00423 per meter.
so 0.00423x25meters=0.11.
next is 230v divided by 0.11=2,090
this is where ia am stuck any help would be great

You're working out your equation wrong.

You first need to measure the PFC at the point where the 63A MCB is to be utilised which you've already done and got a value of 1.2kA, this is 'I'.
You then need to use data from the manufacturer of the MCB or the time/current curve graphs in Appendix 3 of the BGB to ascertain the breaking time of that MCB under the load measured (because the breaker is above 32A the time needs to be below 5 seconds). In this case, the PFC is 1.5kA, so the 63A MCB (assuming B type) would trip in about 0.01s, this value is 't'.
Now find out what the 'k' value is by using tables 54.2-6 in Chapter 54 of the BGB. For T&E you would use the value of 115 for 'k' as found in table 54.3.

These are the only figures you need. The adiabatic equation is; The square root of I x I x t, then divide that by 'k'.

In this example that would be (1200 x 1200 x 0.01 square rooted) divided by 115. This gives you a minimum csa of CPC as 1.04mm (or 1.5mm)

also trying to work out adiabiatic on main cable. Soo.
tncs system with ze of 0.15 at main and 1.5Ka. The flat is being feed from a 5419 60amp isolator in 16mm twin and earth. Zdb at fuseboard is 0.19 and 1.2Ka. Its about 10 meter run from main fuse, how would i go about this one.
cheersguys hopes this clears things up and someone shows an easy way of calculating this.
also when filling out the eicr where is says main earth would this be the earth in the 16mm t&e or the 16mm earth going to MET at service head.

Use the same equation as above but your PFC is now 1.5kA instead of 1.2kA and your main fuse I'm assuming is a 60A BS88 as you mentioned a 60A isolator. 1.5kA would blow a 60A BS88 in roughly 0.05 seconds. Also, your 'k' value is still the same, although you will almost certainly have a seperate main earthing comductor from the service head to the MET, your main earthing conductor to the flat is in T&E.

Your equation for the CSA of the main earth is (1500 x 1500 x 0.05 square rooted) divided by 115. This gives you a minimum CSA of main earthing conductor as 2.92mm (4mm for TN-S or 10mm for TN-C-S)

Your main earthing conductor will be the one that goes between your MET and your service head and also the main earth supplying the flat.


After all this however, my main question to you would be how are you achieving discrimination by having a 63A MCB in a DB protected by a 60A fuse?


Edit:

Use the actual amperage of the load and you need to multiply by 2 (for the circuit).

Not quite right, you use the size of the protective device. Also there is no need to double the mV/A/m value as the value given takes account of the fact that there are two conductors.

 

[Geoffsd]

Why are you doubling the voltage drop? the 2.8mVolts/meter/amp is taken from the onsite guide so is for R1+Rn and including the adjustment for 70DegC.

Oh, sorry, my mistake. I was thinking it was just for one 16mm. conductor.

 

[brman]

In this case, the PFC is 1.5kA, so the 63A MCB (assuming B type) would trip in about 0.01s,

I'm in a picky mood today I think. surely 0.1s not 0.01? That being the minimum time on the graphs for 60898 MCBs........

 

[D Skelton]

I'm in a picky mood today I think. surely 0.1s not 0.01? That being the minimum time on the graphs for 60898 MCBs........

It's a rough guess, the graphs given in the BGB are rough guesses also. The most accurate way to find out would be to use manufacturers data, hovever I can assure you that 1.2kA would trip a 63A B type a lot faster than 0.1s.

 

[Deleted member 26818]

To conduct the adiabatic equation, you need to work out the PFC that will flow through the overcurrent protective device.
Using the values quoted in the 'Time/Current characteristics' in Appendix 3, or obtained from the device manufacturer.
Conduct the equation.
If for instance you have a PFC of 2,090A and a Type B 63A MCB, the device will operate (according to Fig 3A4) within 0.1s (in truth it will probably operate within 0.01s).
However using the values in the chart:
315V X 315V = 99225
99225 X 0.1s = 9922.5
√9922.5 = 99.6
99.6 / 115(k) = 0.866mm²
99.6 / 143(k) = 0.7mm².

 

[brman]

It's a rough guess, the graphs given in the BGB are rough guesses also. The most accurate way to find out would be to use manufacturers data, hovever I can assure you that 1.2kA would trip a 63A B type a lot faster than 0.1s.

Fair enough but I was taught not to extrapolate the graphs below those shown. i.e. never use a t<0.1s as there is no data to support it. (unless you can find it in specific manufacturers specs of course)

 

[D Skelton]

Fair enough but I was taught not to extrapolate the graphs below those shown. i.e. never use a t<0.1s as there is no data to support it. (unless you can find it in specific manufacturers specs of course)

You're right, however through experience and also having the manufacturers data of many types and brands of OCPD to hand, I can say quite confidently that 0.01 is much more accurate than 0.1.

 

[brman]

To conduct the adiabatic equation, you need to work out the PFC that will flow through the overcurrent protective device.
Using the values quoted in the 'Time/Current characteristics' in Appendix 3, or obtained from the device manufacturer.
Conduct the equation.
If for instance you have a PFC of 2,090A and a Type B 63A MCB, the device will operate (according to Fig 2A4) within 0.1s (in truth it will probably operate within 0.01s).
However using the values in the chart:
315V X 315V = 99225
99225 X 0.1s = 9922.5
√9922.5 = 99.6
99.6 / 115(k) = 0.866mm²
99.6 / 143(k) = 0.7mm².

You've confused me spin, where did the 315V x 315V come from? you say just above a PFC of 2090 although there you appear to have ignored Ze....

 

[D Skelton]

You've confused me spin, where did the 315V x 315V come from? you say just above a PFC of 2090 although there you appear to have ignored Ze....

He has used the table on the right hand side of the time/current graph, 315A is the lowest amount of fault current needed to trip the MCB within its required 5s.

And yes, well spotted, it should be 315A not 315V

 

[Deleted member 26818]

The problem with using the instantaneous operating time of 0.01s, is that you would also have to use the value of current which would cause instantaneous tripping.
Just using the measured/calculated value of PFC will result in a CSA greater than required.
In practice this is often immaterial, as the calculation will generally produce a non standard size and the nearest available standard size with a greater CSA than that produced by the equation will work for both methods.
However, to my mind whenever the adiabatic equation is conducted, it should be conducted correctly.
Either use the values in Appendix 3, or actual values, measured and determined from manufacturers' data.
Not a combination of both.

 

[Deleted member 26818]

You've confused me spin, where did the 315V x 315V come from? you say just above a PFC of 2090 although there you appear to have ignored Ze....

I'm using the information provided by the OP.
I don't know how the OP has worked out the value of 2090A, but it is the figure they have.
Yes it should be 315A not 315V.

 

[brman]

You're right, however through experience and also having the manufacturers data of many types and brands of OCPD to hand, I can say quite confidently that 0.01 is much more accurate than 0.1.

Agreed, I've got a couple of specs to hand and the match what you say. I think I would still use 0.1 unless I was desperate to get a smaller cable though just to stop someone who uses the tables arguing with me :lol:

He has used the table on the right hand side of the time/current graph, 315A is the lowest amount of fault current needed to trip the MCB within its required 5s.

Ah, but should you do that? I thought I was the Ipf. i.e. what fault current will flow on a low impedence fault. Not the minimum MCB trip current.

 

[Deleted member 26818]

Seems our posts are crossing, so I'll just attempt to reiterate.
Either use the measured values, and those supplied by the manufacturer.
Or use the values in Appendix 3.
Do not use a mishmosh of values, some measured, some extrapolated and some from Appendix 3.

 

[D Skelton]

An easy way of summing it up

 

[brman]

Seems our posts are crossing, so I'll just attempt to reiterate.
Either use the measured values, and those supplied by the manufacturer.
Or use the values in Appendix 3.
Do not use a mishmosh of values, some measured, some extrapolated and some from Appendix 3.

Still not sure I agree with you though

Starting again.....
1st - figures from appendix 3: 320A will trip in 0.1s. Ok, fine that gives 0.87mm2.
but if you put a short on that cable the current will be limited by Zs. that is 2090A (lets assume) not 320A. So ok, lets use Ipf (2090A) and the manufacturers data of 0.01s. Gives 1.83mm2.
Ok. So "real" figures are showing that if you had used the table figures your cable would fry in a fault condition

I use the Ipf plus the Appendix 3 graph times (which stop at 0.1s). So 2090A & 0.1s gives 5.8mm2. So yes, I have got a cable much bigger than necessary (but actually still less than the minimum normally fitted for a main earth.....)

So I can see Mr Skelton's use the real data is better (i.e. more accurate and saves copper) but I can't see how your using the table current figures can work as it gives a cable 1/2 that actually required.
What is the flaw in my logic? I've seen the table current figures used before in the adiabatic but I just don't see it....

 

[D Skelton]

Still not sure I agree with you though

Starting again.....
1st - figures from appendix 3: 320A will trip in 0.1s. Ok, fine that gives 0.87mm2.
but if you put a short on that cable the current will be limited by Zs. that is 2090A (lets assume) not 320A. So ok, lets use Ipf (2090A) and the manufacturers data of 0.01s. Gives 1.83mm2.
Ok. So "real" figures are showing that if you had used the table figures your cable would fry in a fault condition

I use the Ipf plus the Appendix 3 graph times (which stop at 0.1s). So 2090A & 0.1s gives 5.8mm2. So yes, I have got a cable much bigger than necessary (but actually still less than the minimum normally fitted for a main earth.....)

So I can see Mr Skelton's use the real data is better (i.e. more accurate and saves copper) but I can't see how your using the table current figures can work as it gives a cable 1/2 that actually required.
What is the flaw in my logic? I've seen the table current figures used before in the adiabatic but I just don't see it....

If the PFC was 2090A, then you would use the manufacturers data and realise that the tripping time would then be closer to 0.001s.

The higher the PFC, the faster the breaker will trip. That is the slight flaw in your logic

 

[Geoffsd]

(The posts are also confusing the various opds., Ipfs. and Zss. - nevertheless -)

The overcurrent protective device breaking does not limit the Ipf to 315A.

The full IPF will flow before the opd breaks the circuit - however quick.

 

[brman]

If the PFC was 2090A, then you would use the manufacturers data and realise that the tripping time would then be closer to 0.001s.

The higher the PFC, the faster the breaker will trip. That is the slight flaw in your logic

Ok. I am a catching up slowly

The only trouble I have with that is the data I have just looked at (MK sentry) stops at 0.01s. I do see that the time would be less than that but I am not convinced I could reliable guess how much lower as it will be limited mechanically (just thinking that a lot of RCDs I test go at around 0.01s regardless of current). Plus the graph for MK definitely shows the curve tending to horizontal below 0.1s. Doing to sums in reverse and using the cable size from the "appendix 3" results (0.87mm2) then the real trip time at 2090A would have to be 0.0023s. Feasible but I'm still (edit) not convinced.

The thing is (and I know you guys are going to disagree with me) I think this is a misuse of the tables in appendix 3. I believe these tables are giving the minimum current required to achieve a certain trip time. i.e they are to calculate the max Zs. I don't believe they hold up in this case where you are effectively assuming the breaker will limit the current to 320A for 0.1. I know you guys don't believe this is actually what happens but it is what would have to happen for the adiabatic results to make any sort of sense.

 

[brman]

(The posts are also confusing the various opds., Ipfs. and Zss. - nevertheless -)

The overcurrent protective device breaking does not limit the Ipf to 315A.

The full IPF will flow before the opd breaks the circuit - however quick.

that is my point. I get what DS is saying about t being much less that the tables (or even the manufacturers specs) but...........

 

[Guest55]

If the PFC was 2090A, then you would use the manufacturers data and realise that the tripping time would then be closer to 0.001s.

The higher the PFC, the faster the breaker will trip. That is the slight flaw in your logic

you are not obliged to use the manufacturers data beyond that which is given in the regs tables.
i'm with brman on this , for all intent and purposes 0.1 secs or less can be considered "instantaneous" in operating a protective device and i wouldnt really look for any quicker times unless the cable was huge lol.

 

[D Skelton]

you are not obliged to use the manufacturers data beyond that which is given in the regs tables.
i'm with brman on this , for all intent and purposes 0.1 secs or less can be considered "instantaneous" in operating a protective device and i wouldnt really look for any quicker times unless the cable was huge lol.

i agree

 

[brman]

Just one last thing (and then I will shut up for the sake of the OP!). I've just realised I have been ignoring the let through energy (I2t) specs of the breakers. I'm not very clear how they actually spec these (and how that relates to the real world) but if I have read a couple right then the breaker will limit I2t to less than that expected from Ipf and the t-I curves. That could be a flaw in my argument.........

 

[Deleted member 26818]

Instananeous operation is determined by the a.c. frequency (50Hz).
50Hz means that the voltage cycles form 0V up through +230V, back down through 0V to -230V and back up to 0v again.
In effect the voltage reaches + or - 230V 100 times a second. 1s / 100 = 0.01s or 10ms.
There are no figure below 0.01s because it is not known at which part of the cycle the fault will occur.
It could be that the fault may occur at 0v, thereby taking one half cycle before peak voltage is reached, or it could be that the fault occurs half way between 0V and +230V taking only a quarter of a cycle for peak voltage to be reached.
Conversely the fault could occur after peak voltage (+230V) is reached, and it will take 3/4 of a cycle before peak voltage of -230V will be reached.
Of course, depending upon the value of current, the device may operate before peak current is reached.
It's just not possible to determine, so the value of 0.01s is used.

 

[Silly Sausage]

230 x root 2 = 325V peak.

 

[Deleted member 26818]

Here is a little example for you where using values determined from different sources willl cause problems.
A standard 32A Radial uses 4mm² T&E with a 1.5mm² CPC.
With a fault current of just 1.6kA, the adiabatic equation can show that the minimum CSA for the CPC should be greater than 4mm².
In fact the circuit conductors would not even be adequately sized for the fault current.
1600A² = 2560000
2560000 X 0.1s = 256000
√256000 = 505.9644
505.9644 / 115(k) = 4.4mm².
However if we conduct the adiabatic equation correctly using the values from Appendix 3 for a Type B 32A MCB:
160A² = 25600
25600 X 0.1s = 2560
√2560 = 50.5964
50.5964 / 115(k) = 0.44mm².
The CPC in 4mm² T&E will be sufficient and so will the circuit conductors.

You can use whatever values you wish.
However using a mishmosh of values from different sources will often result in an oversized CPC being determined.
In some cases it will even mean that the circuit conductors are not adequately sized.
To my mind it is always better to either use the tabulated values, or actual values, not a mishmosh of both.

 

[brman]

yes, I can see that. I've been doing my own sums and it does appear to work as long as you are happy to extrapolate curves beyond that give. For example I found some curves for hagar MCBs which show clearly that t is never less than 0.01s (the line is horizontal at that point). However if I then look at the I2t curves for those devices it implies that actually the let through energy will indeed be limited to the level that makes the sum work.
I am not sure I understand it but I reckon I might have to accept it :lol: