type and ratings of BS EN 60898 circut breakers

[GLENNSPARK]

select protection device according to the cable size. the rating if the device must be less than the current capacity of the cable ( allowing for derating ), e.g. if you have a 2.5mm circuit with a ccc of 23A, you would select a 16A or 20A MCB. on circuits with high start up currents, e,g. motors, you would select a type c or d. then you work out your Zs for the circuit and make sure it's below the max. allowede for the chosen MCB. if it's not, start again.

yes Tel..
and Max. Zs for a C curve is half that for a B.....Max. Zs for a D curve is half that for a C....

 

[GLENNSPARK]

yes, but the ccc of the cable must be more than the MCB rating.

equal or greater to after any correction factors have been applied...

 

[telectrix]

must be st. pedant's day today. :banghead:

 

[GLENNSPARK]

must be st. pedant's day today. :banghead:

sorry Tel...lol...
i couldn`t resist...lol..
call me the patron saint eh...

 

[gutterball]

try this it may help

Cable Calculation For ______________________________________In line with BS 7671
2008 17th Edition.
Known Information - Earthing System TN-S / TN-C-S, Ze =
Length of circuit =
STEP 1. Find the Design Current (Ib) = KW = __________watts = __________amps
V __________volts
For 3 phase (Ib) = _____KW____
V x 1.732 x PF
STEP 2. The Protective Device Rating (In) should be equal to or greater than (Ib), so
(In) = ________amps
STEP 3. Apply any correction factors. If no factors need to be applied then (Iz) = (It).
Cg = ________ (Iz) = ______(In)_______
Ci = ________ Cf x Cg x Ci x Ca
Ca = ________
Cf (BS 3036 factor) = 0.725 (Iz) = ________________ = ________________
STEP 4. Select cable size fromBS 7671 App 4. (Iz/t) > (In) > (Ib).
(Iz/t) = ________amps
From table __________ Reference Method __________ Cable size = _______mm
STEP 5. Volt Drop.
Calculate circuit voltage drop, for a lighting circuit, the maximum permitted
Is 3% of the supply voltage = 6.9V for Uo of 230V, and 12V for U of 400V. For all other
circuits the maximum permitted is 5% of the supply voltage = 11.5V for Uo of 230V and
20V for U of 400V.
Actual volt drop = VD = Length of circuit x design current x mV/A/m
1000
= __________V
= __________________________________________
1000
STEP 6. Shock protection
Maximum permitted Zs from BS 7671 tables 41.2, 41.3, and 41.4
OCPD type and rating __________ maximum Zs = ______ ohms x 80% = ______ohms
To calculate actual Zs = Ze + (R1+R2)
= Ze + (mohm/m x length x 1.2) = ______ohms
STEP 7. Thermal constraints.
To find actual fault current (If) = Uo / Zs = ______amps
Then use adiabatic equation to determine minimum CPC size - S = (Sqroot of (I
squared x T)) / K = ________mm, so nearest available size = ________mm