Studies Help Please

[diGreez]

Hello everyone,

I am working on my C&G 2357 unit 304 task A and one of the questions is:

Determine the minimum cross-sectional-area(csa) for each of the following:
a) earthing conductor, where the incoming live conductor are 25mm2.

I am up and down the regs and web to find something on this, and all I found is this formula:[TABLE="width: 68%, align: center"]
[TR]
[TD="class: Booktext, width: 18%"]

S =​

[/TD]
[TD="class: Booktext, width: 28%"]

Ö(Ia²t)​

[/TD]
[TD="class: Booktext, width: 54%"] [/TD]
[/TR]
[TR]
[TD="class: Booktext, width: 18%"]

​

[/TD]
[TD="class: Booktext, width: 28%"]

k​

[/TD]
[/TR]
[/TABLE]
can someone please direct me right way on this please...

Thanks

 

[telectrix]

the 0.8 is a correction value. it is nothing but a number. and start up current is usually handled by fitting a type C breaker, sometimes even a D type. trial and error is the general practice. start with a C and if it trips, swap it for a D. bearing in mind the max. Zs values will be lower. this is all part of the design process.

 

[Knobhead]

You need to look on Google for "power factor". It is the angle the current lags behind the voltage in an inductive circuit. A well designed motor will have a PF of say .9. As the mechanical load drops the PF gets worse, but this isn't apparent on the ammeter as the current falls in relation to the mechanical loading

 

[diGreez]

so if I am taking "crabtree 16A 3-pole Type C" MCB for this machine - it's fine!?

 

[Baker1988]

so if I am taking "crabtree 16A 3-pole Type C" MCB for this machine - it's fine!?

Motors can take about 5 times there operating current during start up that's why you need a type c or d MCB as these are designed to take the initial inrush or current and won't trip I would say a c would work fine but as said above the zs would drop a type d max zs is a lot lower then a type b

 

[telectrix]

so if I am taking "crabtree 16A 3-pole Type C" MCB for this machine - it's fine!?

the In of the MCB is selected to protect the cable. with a 16A MCB, you would be looking at 1.5mm or 2.5mm cable depending on the ccc of the cable at whatever installation method is used. if 1.5mm cable, i would be tempted to use a 10A type C MCB.

 

[diGreez]

my books.... horrible.... pulls the figures from nowhere explanations and links to reg's is old it said tables that does not exist after 2011 amendment...

 

[diGreez]

am I on the right way:
Ib = 7.216A
In = 10A (if I select mcb of 10A type C)
ccc = 10 / (1.03*0.65*0.88*1) = 16.97A
then looking at table 4D1A method B I need minimum 2.5mm cable as it is 3-phase? did I worked it out correctly?

 

[Richard Burns]

Yes really you have it right as far as I can tell from the information you have given but you are using unknown correction factors.


Ib < In < Iz


You have calculated Ib as 7.216A and Tel has said that using a power factor of 0.8 this gives you a design current of 9A (Ib)


You have selected a BSEN60898 10A type C to cope with start up current. (In)
So In is greater than Ib OK.


Now Iz the current carrying capacity of the cable needs to be above In, which is 10A.


Now I do not know how the cable is run and what other factors that need to be taken into account such as ambient temperature, grouping or insulation which may be the other factors you have used. These total 0.58 to give a current carrying capacity of 16.97A but if one of the factors is the installation method then this is already accounted for in the tables.


However from your description you have single core cables in trunking so you are looking at method B on table 4D1A. Looking on the table for the smallest 3/4 core cable that has a CCC of >16.97A is indeed 2.5mm2.


Well done

 

[diGreez]

Thanks for confirmation!

 

[diGreez]

hello, this one is pulling me apart:
If = Uo / Zs for three-phase would you do 400/0.48= 833.3 * 2=1666.6 or still 230 / 0.48 = 479.2 * 2 = 958.4 ?

 

[Richard Burns]

Uo is the line voltage to earth which will be 230V for single and three phase
U is the line to line voltage 400V.

 

[diGreez]

so if I found that live conductor is 1.5mm and I need to find minimum cpc my calculations:
R=24.20(osg table I1)*15.1m*1.02(osg table I2) / 1000 = 0.37ohm
Zs = Ze+(R1+R2)=0.008+(0.37)=0.378ohm
If=Uo/Zs=400/0.378=1058.2ohm so because it is 3phase I need to double it? 1058.2*2=2116.4ohm?
S=(sqroot(sqI) * t) / k = (sqroot(sq2116.4) * 0.1) / 115 = 5.82mm2

is this right? it looks awfully big... or does it really needs to be 6mm cpc cable for this install?

 

[Richard Burns]

It is a nightmare reading those calculations, try and spread them out a bit.

If you are using the adiabatic equation you need to know the characteristics of the breaker you are using.
If you are using a BSEN60898 10A type C MCB and look at the graph for those breakers a fault current over 100A is off the chart, so you could say (though this would give a high csa) 100A fault current will cause a trip in 0.1s and use these values in the adiabatic equation.

Uo/Zs being used to calculate the PFC would be 230/0.378 and you would multiply by the rule of thumb value of 2 to get a prospective fault current of 1216 Amps. Therefore the breaker would trip far faster than 0.1 seconds, but unless you haev the I2t values for the breaker from the manufacturer you cannnot tell how fast or how the current limiting effect operates.

So using the 100A value of 0.1s trip time would give you a minimum cpc size of 0.28 mm2.
Minimum cpc allowable is 1 mm2.

 

[diGreez]

Thanks A LOT again! You just saved me because I need to hand in these calculations tomorrow morning, and I was ripping my self in to peaces...
my studies book gave an example on BS 3036 fuse with fault current of 163A and nothing else... where no one uses 3036 in the new installs as usually with them they will burn the wires before they break the circuit. a lot of use...