“Selection of a suitable cable size and Circuit Protective Device for a 230 V, three kW, industrial process heater”
Keywords used: Ambient temperature, current rating, current-carrying capacity (C.C.C.), voltage drop, grouping factor, back-up protection, conduction, radiation, diversity factor, discrimination, adiabatic process, resistivity, cross-sectional area (C.S.A.), prospective short-circuit current, fusing factor, fusing current, breaking capacity, disconnecting time, circuit protective conductor (C.P.C.), fault-loop impedance, breaking capacity, etc.
Circuit Design
Specifications: 230V, 3 kW, industrial process heater Problem: This new circuit is to be run in steel trunking where there are 3 existing circuits. Ambient temperature is 35 oC and
the length of cable is 40 metres. Calculate the correct size of cable to safely supply the above load (Table A52-D, installation method: B1).
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Fig. 1 Electric circuit
 In = nominal fuse/MCB current rating in Amperes (Amps)
 Iz = current-carrying capacity (C.C.C.) of conductors (Amps)
where Iz = It  Ca  Cg  Ci
It = the tabulated current for a single circuit at an ambient temp. of 35 oC
Ca = the correction factor for ambient temperature
Cg = the correction factor for grouping
Ci = the correction factor for thermal insulation
 Ib = design load current (Amps)
 Im = short-circuit current (Amps)
Step 1
 Calculate the design load current, Ib
Power, P (watts) = U (volts)  Ib (Amperes)  cos  (assume  1)
 Ib = P  U
= 3000 watts  230V = 13 Amps
Step 2
 Decide nominal current rating of fuse/MCB (In = 16 Amps), since In has to be equal to or greater than Ib.
 The circuit is adequately protected if Ib  In  Iz  It (433.1)
conductors Iz
Fuse/MCB
In
0 V
230 V
Ib
3 kW
heater
Switch
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Step 3
 Calculate current-carrying capacity of conductors, Iz
a g
n
z C C
I
I

 =
6 ) 5 9 ) (0. 4 (0.
16

= 26.2 Amps
Ca = Temperature correction factor (Table A52-G1)
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Cg = Grouping factor (Table A52H)
Answer: 4 mm2 ( = 2.25 mm) of cross-sectional area of copper conductor is required (Table A52-F1) where It = 32 A.
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Step 4
 Check to see if the voltage drop in the cable carrying 13 Amps
is equal to or less than 4% (for final circuits in domestic installations @ 230V) or 2.5% for three-phase circuits commercial/industrial installations where there exists sub-main
and final circuit cables.
Maximum voltage drop allowed =
100
2 2.5 30
= 5.75 V
 The tables give values of voltage drop in cables. The values are in millivolts for a current of one Ampere for a run of one metre, i.e. voltage drop = mV/A/m (Table A52J-1)
Actual voltage drop = 1000
11
 4  0 13 5.72 V (acceptable)
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Conclusion to Part 1:
40 metres of PVC insulated cable, with a minimum crosssectional area of 4 mm2, will safely carry the load current Ib of 13 Amps under the specified conditions. The final connection to the heater will be made in metallic flexible conduit containing heat resistant (glass fibre) cable of similar cross-sectional area and separate circuit protective conductor (C.P.C.).
Circuit Protective Device:
Step 5
 Every cable must be protected against overload currents (no fault present) which is a function of the load
 Protected against prospective short-circuit currents (caused by a fault) which is a function of the supply
 Thermal or melting energy is released when a circuit is interrupted under short-circuit current conditions, whether by a fuse or MCB, and the melting process is adiabatic. This means that as a result of a short circuit, the rise in conductor temperature results in an increase in resistance which leads to
an increased loss of energy and increased heating, R i W
tm
 
2 dt J
where W = melting energy
i = instantaneous current
R = instantaneous resistance of that part of element which melts on short circuit
t = time
tm = melting time
R is assumed to vary in the same manner with i and t for all short circuits and the quantity dt i
tm
0
2 is approximately constant for the pre-arcing time of the protective device. This quantity is often called the pre-arcing I2t. It is this quantity which determines the amount of excess energy passing through the circuit before the
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circuit is broken and it is particular important in the protection of semiconductor circuits and the reduction of overheating in power circuits.
 The protective device breaking the circuit must absorb this energy and so limits the current which the device can safely interrupt.
 The protective device must be capable of interrupting a short circuit before the “admissible thermal stress” of the wiring is exceeded within a specified time (max. 5 secs) the lowest value of short-circuit current, i.e. at the furthest point of the circuit
 The breaking capacity of a protective device is the maximum current that can be interrupted safely without mechanical damage to itself or the circuit conductors it protects, at a given voltage and power factor, and is measured in kA Choosing the Circuit Protective Device:
 The protective device must be checked for both overload and short-circuit current operation
 Three suitable circuit protective devices are currently available:  16 Amp Type C MCB  NEOZED D01 16 Amp fuse
 16 Amp H.B.C. fuse  A 16 Amp Type C MCB is recommended
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Overload Rating
 The device must satisfy two conditions (ETCI: 433.1):
1) Ib  In  Iz  It where IB is the design load = 13 Amps
In is the MCB rating = 16 Amps
Iz of the cable = 26.2 A
It = 32 Amps
2) I2  1.45 x current-carrying capacity of the cable, Iz
I2 = the minimum operating current of the device and is obtained from the manufacture’s data (thermal tripping = 1.13 – 1.45 In)
 For a 16 Amp Type C MCB, a typical value would be 20 Amps  I2  1.45  26.2 Amps
 20 A  38 A
Here, both conditions are fulfilled.
Short-circuit rating
 A 6 kA (6000 A), 16 Amp Type C MCB will be sufficient and its magnetic tripping = 5 – 10 In caused by Im as shown in Table 1.
Table 1
MCB Type Instantaneous Tripping
B 3 – 5 In
C 5 – 10 In
D 10 – 20 In
 The device should be checked against the fault loop impedance, ZL, requirements on the disconnecting time, i.e.  0.4 seconds for 230V A.C. circuits with ratings not exceeding 35A, i.e. ZL 
0.9 .
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  
0.9
230
L
m Z
V
I 255.6 Amps
where Im = 5 – 10 In, i.e. the type C MCB will trip instantly because Im is greater than 5 – 10 In or 80 A – 160 A.
 The protective device will have a breaking capacity greater than the prospective short-circuit current, and the current will be disconnected almost instantly to prevent overheating.
 To calculate the time, t, in which a given prospective shortcircuit current will raise the temperature of the copper conductors to the limiting final temperature (160 oC) of p.v.c., the following expression (adiabatic equation) is used: 2
2 2
I
k S
t 
where t = PVC melting time in seconds before it reaches its final limiting temperature of 160 oC.
k = a factor taking into account various criteria of the conductor used such as conductivity, temperature coefficient, specific heat capacity and type of insulation used.
S = minimum cross-sectional area of the conductor in mm2
I = short-circuit current in Amps
When using the adiabatic equation, it is assumed that all of the energy dissipated in the conductor remains in it in the form of heat ( 0.1 sec.) because the faulty circuit is opened so quickly.
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 The factor k is determined from the formula:
 


 




 


i
f i
e
c
B
Q B
k

 

l 1 og
2 ) 0 (
20
where Qc = volumetric heat capacity of the conductor (J/oC mm3)
B = reciprocal of temperature coefficient of resistivity at 0 oC for the conductor (0 oC) 20 = electrical resistivity of the conductor at 20 oC ( m)
i = initial temperature of conductor (oC)
f = final temperature of conductor (oC)
 For copper material: B = 234.5 oC
Qc = 3.45  10-3 J/oC mm3
20 = 17.241  10-9 m
20
2 ) 0 (

Q B  c
= 226 and 



 





i
f i
e B 
 
l 1 og = 0.50882
Thus, k = 115
 24 3.
2 .6 55
1 4 15
2
2 2
2
2 2


 
I
k S
t seconds
The cable is therefore protected as the protective device trips quicker ( 0.1 second) than the time it takes for the cable to reach its limiting temperature. Also, for a short circuit of duration  0.1 second (based on the adiabatic equation where it assumes no heat loss), where current asymmetry is important, the energy let-through (I2t) of the circuit protective device should be less than the value of k2S2 for the cable i.e. I2t  k2S2, i.e. 255.62 × 0.015  1152 ×42 (from graph).
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 980 A2s  211,600 A2s
Therefore, the cable is protected if the protective device trips quicker than the time it takes the cable to reach 160oC and k2S2
for the cable is greater than I2t for the protective device.
Conclusion to Part 2: A suitable circuit protective device would therefore be a 16 Amp Type C MCB, this final sub-circuit being additionally protected by a 40 Amp, 30 mA (In), type AC or A RCD. RCD’s are used to prevent earth fault currents from rising above their sensitivity (In), thus greatly reducing the shock and fire risk.
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Advantages of MCB’s
 Automatically trips out under fault conditions
 Cannot manually be held on under fault conditions
 Enables the supply to be restored immediately after fault is cleared
 Enables faulty circuit to be easily identified
 May be used as a switch to control the circuit
 Does not age in service
 Gives accurate protection and is tamper proof
 Tolerates transient overloads
 May be used as an isolator
 Uniform size regardless of the current rating
 Can be mounted in attractive assemblies
Disadvantages of MCB’s
 More expensive than fuses
Advantages of H.B.C. Fuses
 Fusing factor as low as 1.25
 Reliable and consistent
 Accurate discrimination, e.g. as between large currents of short duration and large fault currents (≤ 80 kA) and ideal for back-up protection
 Does not deteriorate with age
 Fast speed of operation, i.e. can clear a heavy fault current in 0.01 seconds while the
mechanism of a circuit breaker could take 0.1 seconds to interrupt the circuit Disadvantages of H.B.C. Fuses
 They are expensive