Total cost over ( N ) years at interest rate ( r ): [ PV_loss = E_loss \times \textelectricity price \times \frac(1+r)^N - 1r(1+r)^N ] Optimum conductor size is where incremental capital cost equals incremental loss savings. Following IEC/BS 7671 approach: Step 1: Determine Design Current ( I_b ) [ I_b = \fracP\sqrt3 \times V_L \times \cos\phi \ \text(three-phase) \quad \textor \quad \fracPV_P \times \cos\phi \ \text(single-phase) ] Add margin for future expansion (typically 20%). Step 2: Select Protective Device Rating ( I_n ) Choose standard rating (e.g., 16, 20, 25, 32, 40, 50, 63A) such that ( I_n \ge I_b ). Step 3: Determine Installation Method and Correction Factors From standards tables (e.g., IEC 60364-5-52 Table B.52.1 for installation methods: clipped direct, trunking, buried, etc.). Apply ( k_temp ), ( k_group ), ( k_soil ), etc. Step 4: Calculate Required Tabulated Ampacity ( I_tab ) [ I_tab \ge \fracI_nk_total ] Where ( k_total ) = product of all correction factors. Step 5: Select Cable Size from Manufacturer Table Pick smallest standard size (mm²) with base rating ≥ ( I_tab ). Step 6: Check Voltage Drop If drop exceeds limit, increase size until compliant. Step 7: Check Short-Circuit Withstand Compute ( S_min ) from adiabatic equation. If selected size < ( S_min ), increase size. Step 8: Check Earth Fault Loop Impedance (for protection against indirect contact) Ensure disconnection time < 0.4s (for final circuits) or 5s (distribution). This may require larger CPC (circuit protective conductor) or larger line conductor. 7. Worked Example System : 400V three-phase, 50 kW motor, ( \cos\phi = 0.85 ), efficiency 0.94, length 120m, PVC insulated copper cable, clipped direct to wall (Reference Method C), ambient 40°C, grouped with 3 other circuits (4 cables touching), prospective short-circuit 10 kA, permissible voltage drop 3% (12V), disconnection time 0.4 sec.
: ( I_b = \frac50000\sqrt3 \times 400 \times 0.85 \times 0.94 = \frac50000553.6 \approx 90.3 A )
: Select ( I_n = 100A ) (circuit breaker). how to calculate cable size
: The protective device (circuit breaker or fuse) must satisfy: [ I_b \leq I_n \leq I_z ] Where ( I_n ) = nominal rating of protective device, and ( I_z ) = cable’s corrected ampacity. Additionally, for overload protection: [ I_2 \leq 1.45 I_z ] (Where ( I_2 ) is the current ensuring operation of the protective device, typically 1.3–1.45 ( I_n ) for circuit breakers). 3. Voltage Drop Constraint Excessive voltage drop causes poor equipment performance, increased current, and reduced efficiency. Standards (e.g., IEC 60364, BS 7671, NEC) limit total voltage drop from supply to load to typically 3–5% for lighting and 5–8% for other loads. 3.1 Voltage Drop Formula (AC, single-phase and three-phase) For a single-phase circuit: [ V_d = 2 \times I_b \times (R \cos\phi + X \sin\phi) \times L ]
: Check earth fault loop impedance (not shown in detail here) – likely passes with 70 mm². Total cost over ( N ) years at
[ I_z = I_ref \times k_temp \times k_group \times k_soil \times k_depth \times k_therm ]
: Voltage drop. For 50 mm², AC resistance at 70°C ≈ 0.494 Ω/km, reactance ≈ 0.088 Ω/km. Three-phase drop: [ V_d = \sqrt3 \times 90.3 \times (0.494 \times 0.85 + 0.088 \times \sin(\cos^-10.85)) \times 0.120 ] sinϕ = 0.527. R term = 0.494×0.85 = 0.4199 X term = 0.088×0.527 = 0.0464 Sum = 0.4663 Ω/km per phase. ( V_d = 1.732 \times 90.3 \times 0.4663 \times 0.120 = 8.74V ) Percent = 8.74/400×100 = 2.18% < 3% OK. Step 3: Determine Installation Method and Correction Factors
For a three-phase circuit (line-to-line): [ V_d = \sqrt3 \times I_b \times (R \cos\phi + X \sin\phi) \times L ]