Need Help Learning to Compute Adiabatic Method

[Cookie]

How would I go about calculating it for any given circuit?

Lets say I had a 60 amp feeder, 13.30mm2 phase and 5.261mm2 earth. How long could I run the circuit before the 90*C insulation on the earth wire overheats for a fault? At what point would I need to up size my earth wire?

 

[Pete999]

How would I go about calculating it for any given circuit?

Lets say I had a 60 amp feeder, 13.30mm2 phase and 5.261mm2 earth. How long could I run the circuit before the 90*C insulation on the earth wire overheats for a fault? At what point would I need to up size my earth wire?

Where did you get those cable sizes from? sorry you are quoting US cables

 

[pc1966]

You need to know the characteristics of the OCPD in some detail, and also to have in mind a starting temperature (is it 30C ambient, or 70C operating, etc).

 

[davesparks]

How would I go about calculating it for any given circuit?

Lets say I had a 60 amp feeder, 13.30mm2 phase and 5.261mm2 earth. How long could I run the circuit before the 90*C insulation on the earth wire overheats for a fault? At what point would I need to up size my earth wire?

We don't normally calculate cable size that way.

We start with the details of the circuit, design current, length etc and then calculate what size cable is required from that.
We don't pick a cable size and then work out how long we can make the circuit.

 

[Vortigern]

As you know adiabatic equations relate to heat loss and in the case of a cable the idea is to work out when it will melt/vapourise or take the fault current envisaged. So the first item you need is the fault current which we work out by U/Zs=I from there we can use the adiabatic equation to assay the required S(ize)

The t/time value and K value are read from tables.

 

[Cookie]

You need to know the characteristics of the OCPD in some detail, and also to have in mind a starting temperature (is it 30C ambient, or 70C operating, etc).

Here is the OCPD:

https://library.industrialsolutions.abb.com/publibrary/checkout/GES-6203B?TNR=Time%20Current%20Curves%7CGES-6203B%7CPDF&filename=GES-6203B.pdf

30*C ambient, 90*C rated wire.

 

[Cookie]

We don't normally calculate cable size that way.

We start with the details of the circuit, design current, length etc and then calculate what size cable is required from that.
We don't pick a cable size and then work out how long we can make the circuit.

Right, but you need to know if any particular run is safe or not.

 

[Cookie]

As you know adiabatic equations relate to heat loss and in the case of a cable the idea is to work out when it will melt/vapourise or take the fault current envisaged. So the first item you need is the fault current which we work out by U/Zs=I from there we can use the adiabatic equation to assay the required S(ize)
View attachment 60498
The t/time value and K value are read from tables.

Two things here- my understanding is that heat loss can be ignored. 1) The wire is treated as a "closed cylinder" in that heat applied via I2R will not radiate into space during the short circuit. 2) Our limitation is the final temperature of the copper, not the at what point the metal will anneal or begin to melt. An EGC should in theory survive multiple short circuits through the in service life of the system.

I'm assuming T is time, but where or how do I obtain "k"?

 

[pc1966]

If you really wanted to to it backwards, this is how you would go about it:

Start with your cable characteristics, and if you look up the 18th wiring regs Table 54.2/3/4/5 you can pick as 'k' value for your start temp and cable types.

Let us assume you are running fully loaded at 70C and have 90C thermopastic insulation so the short term max conductor temperature is 160C before damage. Then for copper you have k = 100

Reversing the usual equation we have:

sqrt(I2t) = S * k = 5.261 * 100

So I2t = 526.1 ** 2 = 27.7k (units of A2s)

Now we need to find what current for our OCPD will result in this let-through energy. For the usual graph of log(time) against log(PFC) there are straight lines of constant I2s we can plot. So we can find two points and then use a ruler to draw it. So

1 second => I = sqrt(I2t / t) = 526.1 / sqrt(t) = 526.1 Amps
10 seconds => I = 526.1 / sqrt(10) = 166.4 A
100 seconds => I = 526.1 / sqrt(100) = 52.6 A

Alternatively

10A => t = 27.7E3 / 10**2 = 277 seconds
100A => t = 27.7E3 / 100**2 = 2.77 seconds
1000A => t = 27.7E3 / 1000**2 = 0.0277 seconds
10kA => t = 27.7E3 / 10000**2 = 0.000277 seconds

So what OCPD do we have? Let us assume a 63A BS88 fuse of the 22x58mm size:

From this graph we can see that the 63A fuse curve intercepts our constant I2t line at around 1kA fault current. So if our PFC is above 1kA then the cable is safe as the fuse disconnects fast enough, if below 1kA then the cable overheats.

If you know your supply voltage range then from the minimum voltage you can compute your maximum Zs at the cable end for 1kA PFC, and then from your supply Ze you know your maximum R1+R2 to have the prvious Zs value, and from R1+R2 and the cable resistance per unit length, you can compute the maximum length.
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If you are using a MCB or MCCB you really need to look up the manufacturers data as they all vary a bit. For example if you had a Hager industrial MCB of the B-curve type they provide this graph of let-through energy:

Going back to the previous numbers, we see that 27.7k A2s actually has two solutions, at around 250A (when it just enters the magnetic trip region) and again at around 4-5kA fault current. So here you need to have a low enough Zs to hit 250A (probably OK) but not so low that you hit 4-5kA or more.

So there might be a short cable range where it would fail!

 

[Cookie]

Here is my OCPD:

https://library.industrialsolutions.abb.com/publibrary/checkout/GES-6203B?TNR=Time%20Current%20Curves%7CGES-6203B%7CPDF&filename=GES-6203B.pdf

What do you make of it?

 

[pc1966]

Looks like a typical simple MCCB or similar. That graph is normalised so you would need to compute points for the constant I2t in units of the 1 = 60A or whatever and see where it intercepts it.

With MCB/MCCB if you get down in to the "instantaneous" region you can't really use the time very well - in fact often it is not really specified as such - so you need to get a plot of let-through energy instead.
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Two things here- my understanding is that heat loss can be ignored. 1) The wire is treated as a "closed cylinder" in that heat applied via I2R will not radiate into space during the short circuit. 2) Our limitation is the final temperature of the copper, not the at what point the metal will anneal or begin to melt. An EGC should in theory survive multiple short circuits through the in service life of the system.

I'm assuming T is time, but where or how do I obtain "k"?

Yes, the adiabatic assumption is no significant heat is lost. So for trip times under 10-ish seconds that is a fair assumption.

Temperature limit is usually that of the insulation as much lower than the conductor limit, but depends on specific class of cable type, etc.

There are various physics models for this, but normally you just look at the tables in various wiring regulations for values of 'k' to use.
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Doh! Made a numerical mistake I2t = 276.8 kA

Good job this is not for a £100M+ hospital.

 

[Cookie]

Why do you use the starting temp and not the final temp for the R1+R2 values? And why 160*C?

 

[Cookie]

Looks like a typical simple MCCB or similar. That graph is normalised so you would need to compute points for the constant I2t in units of the 1 = 60A or whatever and see where it intercepts it.

With MCB/MCCB if you get down in to the "instantaneous" region you can't really use the time very well - in fact often it is not really specified as such - so you need to get a plot of let-through energy instead.
[automerge]1598818008[/automerge]

Yes, the adiabatic assumption is no significant heat is lost. So for trip times under 10-ish seconds that is a fair assumption.

Temperature limit is usually that of the insulation as much lower than the conductor limit, but depends on specific class of cable type, etc.

There are various physics models for this, but normally you just look at the tables in various wiring regulations for values of 'k' to use.

What do I do if my regs do not have a "K" value? This is my big hold back.

 

[Vortigern]

We take the K value from a table (54.2) in BS7671 which is a list of different materials and their K value. I am not sure where else you would find them and how the values are derived.
EDIT: I see @pc1966 has already alluded to that.

 

[pc1966]

Trying again:
10A = 276.8E3 / 10**2 = 2768 seconds
10kA = 0.00278 s
Plot looks like:

So cross over is around 400A for the 63A BS88 fuse.

From Hager plot for B-curve MCB then the solution is only the one place at around 150-200A, above which it is OK until the MCB's break limit of 10kA.
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What do I do if my regs do not have a "K" value? This is my big hold back.

k depends on the material (copper, etc) and the temperature rise that is acceptable (more or less, R varies a bit with t etc).

So you can use the tabulated values from, for example, UK regs provided the type of conductor and insulation are comparable. Most likely they are as Cu & PVC or XLPE is pretty much common use world-wide.
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Why do you use the starting temp and not the final temp for the R1+R2 values? And why 160*C?

Starting temp is what you design for normal operation.

Final temp is what won't damage the insulation short-term, and 160C is the value used in the BS regs for cables up to 300mm section, with 140C above 300mm (not sure why, maybe more pysical forces rising insulation movement, etc, when hot).
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It is 160C for thermoplastic and 250C for thermosetting.