Fuse selection

[Albert]

It is (I think) a bit complicated to explain, but if someone has the time and the patience, it would be much appreciated, in the books I have I can see partial answers and how to calculate things but it is not always clear and makes sense:
1)Fuse breaking time 0.4 sec and 5 sec, when would you use one or the other.
2)As far I understand the 0.4 sec and 5 sec is for overload, there is an other time factor which is used when calculating the protective conductor's size. finding Ip (thermal constrain), and with it the time factor from the graph can calculate the protective conductor. How do I find the time from the graph if for example using fuse BS 88-2, cartridge, 16A radial circuit, 20m cable, flat twin & Earth 2.5mm2, metod 3. assuming that the circuit Zs is 0.32 Ohm, the Ip will be about 720A, looking in the graph the the Ip value does not fit in the 16A range, so what should I do?
3) Zs for fuses and circuit Zs (Ze+R1+R2), what is the difference.

thanks, Albert

 

[Big_Spark]

Albert, just a quick partial answer at the moment as I am snowed under (just grabbing a cuppa and a five minute breather)

The 0.4 and 5 second disconnection times apply to the nominal voltage of a circuit. If the supply voltage is 110Vac then 5 second rule applies, if the circuit is rated at 230V, then the 0.4 second rule applies.

This information can be found in Table 41A (page 44) of BS7671 and applies to all TN systems, this encompasses TN-S, TN-C and TN-C-S.

Regulation 413-02-09 states:

The maximum disconnection times of Table 41A shall apply to all circuits supplying socket-outlets and other final circuits which supply portable equipment intended for manual moving during use, or hand-held class I equipment.

This does not apply to a final circuit supplying an item of stationary equipment connected by means of a plug and socket outlet where precautions are taken to prevent the use of the socket outlet for supplying hand-held equipment, nor to reduced low voltage circuits as described in regulation 471-15

 

[Albert]

Albert, just a quick partial answer at the moment as I am snowed under (just grabbing a cuppa and a five minute breather)

The 0.4 and 5 second disconnection times apply to the nominal voltage of a circuit. If the supply voltage is 110Vac then 5 second rule applies, if the circuit is rated at 230V, then the 0.4 second rule applies.

This information can be found in Table 41A (page 44) of BS7671 and applies to all TN systems, this encompasses TN-S, TN-C and TN-C-S.
etc...[/b][/i]

You are a star, i'm not surprised that you don't have the time for a 'cuppa', if you are answering all the questions...

Found the tables, and regulation 413-02-09, the second paragraph is the unclear one, as it should talk about the exceptions, I looked at 471-15 page 71, it did not make it clearer. I assume that table 41A is not linear as the time for 120V is 0.8 sec, logically for 110V it should be about 1 sec (110V is not in the table) or less, just trying to understand. If I will put in simple words what it says is that: in a normal household (show me one...) we do not use other than 0.4 sec, this includes lighting & sockets, what about the bath and external sockets (I know that RCD is needed). found 413-02-12 this one mentions the 5 sec, does it mean that sockets that are used for "manual movement during use..." are drilling machine or soldering iron, hoover etc. or it has a differenr meaning? When I put in a socket I almost never know what it will be used for.
Albert

 

[Big_Spark]

Albert, your dilemna over the rule is the same as most people, and getting a straight answer from the IEE on this is like getting Tony Blair to be honest..or any politician for that matter

I think what you need to assume is that standard sockets to BS1363 should all comply with the 0.4s rule as you are unable to know how they will be used in future, and you will certainly comply with the regs.

The only time you need to look at it differently is when your putting in a commando or lewden socket that will supply a piece of static equipment. You could also apply this rule to 2A and 5A socket outlets due to the nature of their use.

However I look at it like this, if everything at 230V complies with a 0.4s disconnection time, then you meet or exceed the regulations, you've covered your proverbial and the client is safer.

Regarding the calculation for protective conductor size, i think you may be refering to the adiabatic equation, you will find this under Regulation 543-01-03 on page 113.

 

[Albert]

Albert, your dilemna over the rule is the same as most people, and getting a straight answer from the IEE on this is like getting Tony Blair to be honest..or any politician for that matter

I think what you need to assume is that standard sockets to BS1363 should all comply with the 0.4s rule as you are unable to know how they will be used in future, and you will certainly comply with the regs.

.

Thanks for this one, i'm familiar with the equation to calculate the csa of the protective conductor, but if you allow me to go back to it, could you please explain using the example in my question? i looked in 543-01-03 (before), and it does not explain what happens in the case that the Ip is out of the range like in the example case 720A, what would be the t , when K is 115, and S(cpc cas) is the one I need to find. in table 54G page 115, it says that we can use the same size for the cpc as for the live if we do not wish to calculate, than we have a problem as we have to use non standard wires or single wires.
Albert

 

[Big_Spark]

I think your misunderstanding something here.

Have you looked at graph 3.3B on page 198?

For a 16A BS88-2.1 the psc at 0.4s is 85A

Give me sometime to eat and I will work this out for you from the circuit characteristics you gave earlier.

 

[Albert]

I think your misunderstanding something here.

finally we agree, i don't understand something. Bon-appetit, and thanks again

Albert

 

[Albert]

FWL_Engineer , if you continue to argue with the others you will not have time for this one... so tel them to leave you alone...

So you had your cuppa and you had something to eat.. yesterday
H E L P!!!
Albert

 

[Big_Spark]

FWL_Engineer , if you continue to argue with the others you will not have time for this one... so tel them to leave you alone...

So you had your cuppa and you had something to eat.. yesterday
H E L P!!!
Albert

Sorry Al, couple of my Business Partners decided we should celebrate St Pats day, so about 9pm I went to the Pub...twice in two days..shocking!!

I have a bit of time on my hands today..phew...so let me addess this and get back to you.

 

[Big_Spark]

Albert, the equation you are looking for is this.

t=K2S2/I2

so for your example.

115sqd x 1.5sqd / 720sqd becomes

13225 x 2.25 / 518400 = 0.057 seconds.

If you want to bone up on this particular problem, look under section 434 in the Regs, it's on page 58.

 

[Big_Spark]

Hey Albert, did this solve your problem?

 

[Albert]

Sorry just came in. I will look at it and come back to you.
in page 58 it seems to me that it relates to the live conductor, my problem is how to find the time for the protective conductor.

Ip=U0 / Zs, Ip =Prospective short current
Uo=230V, Zs for the example lets say 0.5 ohm
Ip=460A

I am trying to find the CSA in sq mm of the protective conductor:
S=(I2*t)-2/K (page 113)
t=value from page198, how do I select it?
there is a small table on the upper left and than the graphs, if i look on the graph X direction it gives the prospective current values, now if I look for 460A it will not meet the 16A curve. do you understand now my problem? it is possible that I am the one who makes it complicated and it is simple.
[/list]

 

[Big_Spark]

Albert, Your looking for something that isn't there.

In your example, 460A is so far along the graph that the disconnection would be almost instantaneous.

The graphs drop off at 0.1 seconds, so as long as your PFC does not show on the graph, and it is below tha maximum fault current of the device, the circuit will comply with the disconnection time.

If you actually look at the graph on page 198, you will see that a fault current of about 115A will cause the fuse to "blow" in 0.1 seconds, well within the 0.4 seconds specified in the regs. If you draw a line along the 0.4 line to were it intersects the 16A graph, you will see they intercept at about 85A, exactly what it states in the table on the upper right of the graph.

As you can see from this, a fault current of 420A cannot be maintained on the circuit long enough to be a problem, a mere 85A will operate the device in 0.4 seconds, the 420A will operate the device in about 0.0043 seconds, so well within spec.

 

[Albert]

Many thanks
Now I can tell you the reson I asked (obviously I like to understand and not only use the data like a monkey), but as it was explained in the class, it did not make sense, and did not pass the 'click ' point in my mind, and I am sure that if I will ask someone in the group to explain, they will not have a clue concerning the 'why' and if they would have a problem to solve that will be different than the example in the class, they will not be able to deal with it. This is the problem when a teacher reads the lesson from his note book without understanding.
Well done, today is my school day and I am going there filling much better (it drove me made for a while, I am sure you know this kind of filling).
Thanks again
Albert

 

[Albert]

What is the difference between the fuse Zs and the circuit Zs (Ze+R1+R2)?, why or when do we use the fuse Zs.
Albert