Sample Response
Nominal voltage = 400 V Power = 50 kW, power factor = 0.8 Length = 100 m Cable type = copper, XLPE insulation, 3-core Installation method = buried in ground Ambient temperature = 25 °C, soil thermal resistivity = 2.5 K.m/W The cable size calculation as per EN50343 consists of four steps: Determine the minimum cable size based on continuous current carrying capacity Determine the minimum cable size based on voltage drop Determine the minimum cable size based on short circuit temperature rise Determine the minimum cable size based on earth fault loop impedance Step 1: Continuous current carrying capacity The first step is to calculate the load current (I) using the formula: I=3VcosϕP where P is the power, V is the nominal voltage, and cosϕ is the power factor. Substituting the values, we get: I=3×400×0.850=0.091A The next step is to find the base current rating (Ib) of the cable from Table B.52.17 of EN50343 for copper, XLPE insulated, 3-core cables buried in ground. The base current rating is the maximum current that a cable can carry continuously under standard conditions of installation and operation. For example, for a 16 mm2 cable, the base current rating is 111 A. The final step is to apply correction factors (Ci) to account for different installation and environmental conditions that affect the current carrying capacity of the cable. The correction factors are obtained from Tables B.52.2 to B.52.16 of EN50343 for various parameters such as ambient temperature, soil thermal resistivity, depth of laying, grouping of cables, etc. For example, for an ambient temperature of 25 °C, the correction factor is 0.94. The corrected current rating (Iz) of the cable is calculated by multiplying the base current rating by all the correction factors: Iz=Ib×C1×C2×C3×... The minimum cable size is then selected such that the corrected current rating is greater than or equal to the load current: Iz≥I For this example, assuming that there are no other cables in close proximity and that the depth of laying is 0.7 m, we can use Table B.52.17 and Table B.52.4 to find that: For a 16 mm2 cable, Ib = 111 A and C1 = 0.94 For a 25 mm2 cable, Ib = 139 A and C1 = 0.96 Therefore, For a 16 mm2 cable, Iz = 111 x 0.94 = 104 A < I For a 25 mm2 cable, Iz = 139 x 0.96 = 133 A > I Hence, based on continuous current carrying capacity, the minimum cable size is 25 mm2. Step 2: Voltage drop The second step is to calculate the voltage drop (VD) across the cable using the formula: VD=A.mmV×I×L where mV/A.m is the voltage drop per ampere per metre of the cable, I is the load current, and L is the length of the cable. The value of mV/A.m can be obtained from Table B.52.18 of EN50343 for copper, XLPE insulated, 3-core cables. For example, for a 25 mm2 cable, mV/A.m = 1.8. Substituting the values, we get: VD=1.8×91×100=16.38V The maximum permissible voltage drop depends on the type and sensitivity of the load, and is usually specified by the relevant standards or codes of practice. For example, BS7671 recommends a maximum voltage drop of 5% for lighting circuits and 4% for other circuits. For this example, assuming a maximum voltage drop of 4%, we get: VDmax=0.04×V=0.04×400=16V Since the calculated voltage drop is greater than the maximum permissible voltage drop, the cable size needs to be increased to reduce the voltage drop. The next cable size from Table B.52.17 is 35 mm2, which has a base current rating of 167 A and a mV/A.m value of 1.4. Repeating the calculations, we get: For continuous current carrying capacity, Iz = 167 x 0.97 = 162 A > I For voltage drop, VD = 1.4 x 91 x 100 = 12.74 V < VDmax Hence, based on voltage drop, the minimum cable size is 35 mm2. Step 3: Short circuit temperature rise The third step is to calculate the maximum short circuit current (Isc) that can flow through the cable in case of a fault using the formula: Isc=ZU where U is the system voltage and Z is the impedance of the circuit. The impedance of the circuit depends on various factors such as the source impedance, the cable impedance, and the load impedance. For simplicity, we can assume that the source impedance is negligible and that the load impedance is purely resistive. The cable impedance consists of two components: resistance and reactance. The resistance of the cable depends on its material, cross-sectional area, and temperature. The reactance of the cable depends on its type, configuration, and frequency. The values of resistance and reactance can be obtained from Tables B.52.19 and B.52.20 of EN50343 for copper, XLPE insulated, 3-core cables. For example, for a 35 mm2 cable at 90 °C (the maximum operating temperature), the resistance is 0.524 ohm/km and the reactance is 0.08 ohm/km. Therefore, for a length of 100 m, the cable impedance is: Zc=Rc2+Xc2=(0.524×0.1)2+(0.08×0.1)2=0.055Ω The load impedance is given by: ZL=IV=91400=4.396Ω Hence, the total impedance of the circuit is: Z=Zc+ZL=0.055+4.396=4.451Ω Substituting in the formula for Isc, we get: Isc=4.451400=89.86kA The next step is to find the adiabatic short circuit current (Ia) of the cable from Table B.52.21 of EN50343 for copper conductors. The adiabatic short circuit current is the maximum current that a cable can withstand for a given duration without exceeding its final conductor temperature. The final conductor temperature depends on various factors such as the initial conductor temperature, the material properties of the conductor and insulation, and the duration and frequency of the fault. For example, for a copper conductor with XLPE insulation and an initial temperature of 90 °C, the final conductor temperature is 250 °C for a fault duration of less than or equal to 5 seconds. For a 35 mm2 cable, Ia = 6.9 kA for t = 5 s. Therefore, For t = 5 s, Ia = 6.9 kA < Isc For t = 1 s, Ia = 10 kA < Isc For t = 0.5 s, Ia = 14.1 kA < Isc For t = 0.1 s, Ia = 31.6 kA < Isc Since the adiabatic short circuit current is less than the maximum short circuit current for all durations, the cable size needs to be increased to withstand the fault. The next cable size from Table B.52.17 is 50 mm2, which has a base current rating of 198 A and a mV/A.m value of 1.1. Repeating the calculations, we get: For continuous current carrying capacity, Iz = 198 x 0.98 = 194 A > I For voltage drop, VD = 1.1 x 91 x 100 = 10 V < VDmax For short circuit temperature rise, Zc = sqrt((0.387 x 0.1)^2 + (0.08 x 0.1)^2) = 0.041 ohm Isc = 400 / (0.041 + 4.396) = 90.27 kA For t = 5 s, Ia = 8.8 kA < Isc For t = 1 s, Ia = 12.7 kA < Isc For t = 0.5 s, Ia = 17.9 kA < Isc For t = 0.1 s, Ia = 40 kA < Isc Hence, based on short circuit temperature rise, the minimum cable size is still not sufficient. The next cable size from Table B.52.17 is 70 mm2, which has a base current rating of 239 A and a mV/A.m value of 0.9. Repeating the calculations, we get: For continuous current carrying capacity, Iz = 239 x 0.99 = 237 A > I For voltage drop, VD = 0.9 x 91 x 100 = 8.19 V < VDmax For short circuit temperature rise, Zc = sqrt((0.268 x 0.1)^2 + (0.08 x 0.1)^2) = 0.029 ohm Isc = 400 / (0.029 + 4.396) = 90.77 kA For t = 5 s, Ia = 11 kA > Isc For t = 1 s, Ia = 15.8 kA > Isc For t = 0.5 s, Ia = 22.4 kA > Isc For t = 0.1 s, Ia = 50 kA > Isc Therefore, based on short circuit temperature rise, the minimum cable size is finally sufficient. Step 4: Earth fault loop impedance The fourth and final step is to calculate the earth fault loop impedance (Zs) of the circuit using the formula: Zs=Ze+R1+R2 where Ze is the external impedance of the source, R1 is the phase conductor resistance of the cable, and R2 is the protective conductor resistance of the cable. The value of Ze depends on the type and location of the source, and can be obtained from the supply authority or measured on site. The values of R1 and R2 can be obtained from Table B.52.19 of EN50343 for copper conductors at different temperatures. For example, for a copper conductor at 90 °C, R1 and R2 are equal to the resistance per kilometre divided by two. For a single-core cable with separate protective conductor or a multi-core cable with reduced neutral conductor cross-section, R2 may be different from R1. For this example, assuming that Ze is equal to 0.35 ohm and that R2 is equal to R1 for simplicity, we can use Table B.52.19 to find that: For a 70 mm2 cable at an initial temperature of 90 °C, R1 and R2 are equal to 0.268 ohm/km divided by two, which is 0.134 ohm/100 m Therefore, for a length of 100 m, Zs = 0.35 + 0.134 + 0.134 = 0.618 ohm The maximum permissible earth fault loop impedance depends on the type and rating of the protective device, and is usually specified by the relevant standards or codes of practice. For example, BS7671 provides tables of maximum Zs values for different types of circuit breakers and fuses. For this example, assuming that the protective device is a type B circuit breaker with a rating of 125 A, we can use Table 41.3 of BS7671 to find that: The maximum Zs value is 0.8 ohm Since the calculated Zs value is less than the maximum Zs value, the cable size is sufficient. Therefore, based on all four steps, the minimum cable size for this example is 70 mm2.
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