[ V_d = \frac\sqrt3 \times L \times I_b \times (R \cos\phi + X \sin\phi)1000 \text (for 3-phase) ]
From IEC table – 2.5 mm² copper carries 24 A. Derating: ( C_t = 0.94 ) (35°C). ( I_z = 24 \times 0.94 = 22.56 A ) – adequate.
Where ( I_z ) = derated cable ampacity. Load: 5 kW heater, 230V AC, power factor = 1. ( I_b = 5000 / 230 = 21.74 A ) how to calculate cable sizes
Increase to 4 mm² (( R_ac ≈ 5.62 \text Ω/km )). Drop = ( 2 \times 40 \times 21.74 \times 5.62 / 1000 = 9.77V ) → 4.25% (still over). Increase to 6 mm² (( R_ac ≈ 3.71 )). Drop = ( 2 \times 40 \times 21.74 \times 3.71 / 1000 = 6.45V ) → 2.8% (acceptable).
If calculated ( V_d ) exceeds limit, until drop is acceptable. Voltage drop almost always dictates final size for long runs. Step 4: Short-Circuit Temperature Rise (Thermal Withstand) Ensure cable can survive a fault until protection operates. [ V_d = \frac\sqrt3 \times L \times I_b
Voltage drop. For 2.5 mm² copper, ( R_ac ≈ 8.91 \text Ω/km ). Single phase drop = ( \frac2 \times 40 \times 21.74 \times 8.911000 = 15.5V ). 15.5V / 230V = 6.7% → exceeds 3% limit (6.9V max).
[ I_z \geq I_b ]
[ I_b \leq I_n \leq I_z ]