I can’t get my Iz greater than In

[Mike2themoon]

Hi all, this is driving me mental, if I have a Ca of 1.03 (25degrees ambient) & a In of 6A or any number you like then Iz will never come out greater than In unless I add another grouping factor just to make it work, which shouldn’t be necessary.

There has to be an explanation to this but I can’t figure out how to make it work.

Please help

 

[timhoward]

Can you explain your process (in other words show your working!)

To apply a correction factor to a cable's current-carrying-capacity (lz), the starting point is which cable and which installation method.
e.g. 1 sq mm twin and earth PVC, clipped direct is 16 amps from table 4D5

The ambient air rating factor from table 4B1 for the appropriate cable should then be applied to lz

In this example it will definitely be greater than 6 amps.

 

[Mike2themoon]

Can you explain your process (in other words show your working!)

To apply a correction factor to a cable's current-carrying-capacity (lz), the starting point is which cable and which installation method.
e.g. 1 sq mm twin and earth PVC, clipped direct is 16 amps from table 4D5

The ambient air rating factor from table 4B1 for the appropriate cable should then be applied to lz

In this example it will definitely be greater than 6 amps.

Appreciate the reply but if In is 16 & 25degrees Ca is 1.03 then to calculate Iz =In16/Ca1.03 = Iz of 15.53 which does not work, so how does one get past this?

 

[timhoward]

I'm not sure what formula you are trying to use.
I would have written that relationship as ln < lz * Ca

The hotter the cables get via ambient air, the less current they can carry before they reach their design limit (or melt!)

See appendix 4 - 2.1 (page 421) - the values in the tables are for 30 degrees.
You are wanting 25 degrees. As the temperature goes down, the conductors have greater ability to carry current.
So we are expecting the number to get bigger.

In a nutshell, you should be multiplying not dividing. 16 * 1.03 = 16.48 amps
We can carry 0.48 amps extra because it's colder.

Sorry if I'm missing your point, but I'm not understanding how you arrived at that calculation.

 

[Julie.]

Just what result are you expecting?

If you are running a 16A cable in an elevated temperature environment you would expect the rating to drop.

And that's exactly what you are getting.

If you are running it at a lower ambient then normal you would expect the rating to be higher. In which case you have the formula upside down.

 

[FlashBangWallop]

Appreciate the reply but if In is 16 & 25degrees Ca is 1.03 then to calculate Iz =In16/Ca1.03 = Iz of 15.53 which does not work, so how does one get past this?

As the tables are for cables at 30 degrees, the ambient temperature of 25 degrees is giving a value of a lower Iz by being in a cooler environment. I would then work on the basis that Iz is equal to In and calculate the cable to the 16A. Iz has to be equal to or greater than In so it seems the logical next step.

 

[Mike2themoon]

I'm not sure what formula you are trying to use.
I would have written that relationship as ln < lz * Ca

The hotter the cables get via ambient air, the less current they can carry before they reach their design limit (or melt!)

See appendix 4 - 2.1 (page 421) - the values in the tables are for 30 degrees.
You are wanting 25 degrees. As the temperature goes down, the conductors have greater ability to carry current.
So we are expecting the number to get bigger.

In a nutshell, you should be multiplying not dividing. 16 * 1.03 = 16.48 amps
We can carry 0.48 amps extra because it's colder.

Sorry if I'm missing your point, but I'm not understanding how you arrived at that calculation.

Hi thanks for preserving with me, I’m sure it’s my mistake but I have been taught
Ib ≤ In≤ Iz ≤ It this is correct?

 

[timhoward]

Hi thanks for preserving with me, I’m sure it’s my mistake but I have been taught
Ib ≤ In≤ Iz ≤ It this is correct?

I now understand the original question!
Personally I'd focus on:
design current ≤ protective device rating ≤ cable current carrying capacity

The last bit is generally true as a lot of the time the cable's current carrying capacity will go down when rating factors are applied. The example you chose above was one of the exceptions to this. If you are only applying the factor for a lower temperature than 30 degrees it won't be true.

 

[Mike2themoon]

I now understand the original question!
Personally I'd focus on:
design current ≤ protective device rating ≤ cable current carrying capacity

The last bit is generally true as a lot of the time the cable's current carrying capacity will go down when rating factors are applied. The example you chose above was one of the exceptions to this. If you are only applying the factor for a lower temperature than 30 degrees it won't be true.

Thank you I appreciate your time, at least I know I wasn’t wrong now & will investigate it further

 

[timhoward]

Thank you I appreciate your time, at least I know I wasn’t wrong now & will investigate it further

No worries - glad you found this forum - there are much cleverer people that me around and you can learn a lot by lingering here!

 

[davesparks]

In a nutshell, you should be multiplying not dividing. 16 * 1.03 = 16.48 amps
We can carry 0.48 amps extra because it's colder.

Yes if calculating the current carrying capacity of a given cable size.

But the standard cable size calculation in BS7671 is designed for selecting a cable size for a known current.
So when calculating the cable size we modify the nominal current using rating factors and then select a cable size based on that modified value.

So the OP's In/1.03 is correct for carrying out a calculation to select a cable size.

 

[davesparks]

Appreciate the reply but if In is 16 & 25degrees Ca is 1.03 then to calculate Iz =In16/Ca1.03 = Iz of 15.53 which does not work, so how does one get past this?

That does work, if you are doing the standard calculation to select a cable size for circuit design.

This can take a bit of thinking to get your head around as the way this calculation work isn't immediately obvious, and when working with temperatures less than 30C it doesn't give the result you immediately expect.

To select a cable size we take the nominal current of the circuit and artificially modify it using ratings factors before selecting a cable size. This modified value doesn't exist in the real world, it is just a number used to select the cable size.

The majority of ratings factors are less than 1 so when you divide by them it increases the value. With temperatures below 30C it actually increases the value because the rating factor is greater than 1.

So for a situation where the temperature is less than 30 and no other rating factors apply Iz will be less than In.


I assume this has come from a college question? In which case it has probably been deliberately done this way to see whether you actually understand how the calculation works.

 

[Mike2themoon]

That does work, if you are doing the standard calculation to select a cable size for circuit design.

This can take a bit of thinking to get your head around as the way this calculation work isn't immediately obvious, and when working with temperatures less than 30C it doesn't give the result you immediately expect.

To select a cable size we take the nominal current of the circuit and artificially modify it using ratings factors before selecting a cable size. This modified value doesn't exist in the real world, it is just a number used to select the cable size.

The majority of ratings factors are less than 1 so when you divide by them it increases the value. With temperatures below 30C it actually increases the value because the rating factor is greater than 1.

So for a situation where the temperature is less than 30 and no other rating factors apply Iz will be less than In.


I assume this has come from a college question? In which case it has probably been deliberately done this way to see whether you actually understand how the calculation works.

Hi thank you for that in-depth answer, what do I do when Iz comes out less than In please

 

[davesparks]

Hi thank you for that in-depth answer, what do I do when Iz comes out less than In please

Do exactly the same as you would when it's higher than In, select a cable size which is suitable for that current.

 

[timhoward]

Do exactly the same as you would when it's higher than In, select a cable size which is suitable for that current.

My understanding is that his calculation ends up with a number which is the minimum he could use, taking into consideration the protective device and the rating factors. (Appendix 4 section 5.1.1)

His question is whether that gives him license to ignore the general rule: Ib ≤ In ≤ Iz

I've never had this situation happen in the real world, and I tend to do it the more fun way around of guessing the right cable based on past experience, derating it appropriately, and then checking.
My answer would be that you can't stick a cable that can carry 15.53 amps (if such a cable actually existed with that as it's maximum CCC) in front of a 16 amp protective device. ln needs to be lower or equal to lz so we are back to 16 amps.

 

[westward10]

You aren't ignoring it as such, your calculation suggests Iz is lower than In but you can still use the equation and select a larger Iz it will still follow that Iz is greater or equal to In.

 

[davesparks]

My understanding is that his calculation ends up with a number which is the minimum he could use, taking into consideration the protective device and the rating factors. (Appendix 4 section 5.1.1)

The number has has is the number you use to select a cable size from the tables. It is just a number which exists within the calculation, nothing else, it doesn't exist in the real world, it doesn't represent any actual current.

My answer would be that you can't stick a cable that can carry 15.53 amps (if such a cable actually existed with that as it's maximum CCC) in front of a 16 amp protective device.

That is not what is happening here, the cable can carry the 16A at 25 degrees. What you are doing is selecting a cable from its rating at 30 degrees for an application which is at 25 degrees.

 

[timhoward]

That is not what is happening here, the cable can carry the 16A at 25 degrees. What you are doing is selecting a cable from its rating at 30 degrees for an application which is at 25 degrees.

I think the light bulb has finally come on, thank you for your patience!
So this method essentially comes up with a number based on the ln so that the standard (30 degree in this case) tables can be used.

That also explains why the number drops when the actual CCC of the cable increases.
I think I'll stick to looking up the cable and derating it.

 

[davesparks]

I think I'll stick to looking up the cable and derating it.

The trouble with that is you could end up doing multiple calculations before getting to the size of cable you need.

Not so much of a problem now we all have a calculator to hand on our phones, but when these methods were developed it would have been a long-hand process

 

[pc1966]

I think I'll stick to looking up the cable and derating it.

In most cases a glance at the "standard circuits" in the OSG Table 7.1(ii) will answer most cable-selecting needs, both the cable CCC needed (via combination of size in column 3 and acceptable installation methods in column 4) as well ad VD limits (via columns 5-8). But there will always be the odd case that is not covered there and then you need to compute it (for example if more grouping that mentioned in section 7.2.1).

Also, obviously enough, for exam questions to establish that you can compute it as needed!

The OSG tables are quicker and less error prone for most domestic circuits, but they are not the regs as such.