- Asked 4/20
- 2075 Baisakh · 4 marks
Cable size MDB to SDB-1 (60 kW) and SDB-2 (75 kW)
Short answer
SDB-1 (60 kW, 80 m): 3½ core 35 mm² Cu (drop 6.10 V). SDB-2 (75 kW, 125 m): 3½ core 50 mm² Cu (drop8.34 V). Both within2.5% limit.
Determine the size of cable used from MDB to SDB-1 and SDB-2 for the following conditions. The total power on the SDB-1 and SDB-2 are 60 kW and 75 kW respectively. The distance between MDB to SDB-1 and SDB-2 are 80m and 125m respectively. The supply system is 3-phase, 4 wire with 400V, 50 Hz. The current ratings of Cu conductor laid in ground are given as follow.
| Conductor area (sq.mm) | 1.5 | 2.5 | 4 | 6 | 10 | 16 | 25 | 35 | 50 | 70 | 95 | 120 | 150 |
|---|---|---|---|---|---|---|---|---|---|---|---|---|---|
| 3.5 core cable current rating (A) | 21 | 27 | 31 | 45 | 60 | 77 | 99 | 120 | 145 | 175 | 210 | 240 | 270 |
Each feeder is sized by current rating from the given table and checked for voltage drop.
Assumptions: pf = 0.8 (not given), allowed drop 2.5 % of 400 V = 10 V, ρ (Cu) = 0.0178 Ω·mm²/m.
I = P / (√3 × V_L × cosφ)
ΔV = √3 × I × ρ × L × cosφ / A
MDB to SDB-1 (60 kW, 80 m)
I = 60000 / (√3 × 400 × 0.8) = 108.25 A
By current: 35 mm² (120 A)
ΔV = √3 × 108.25 × 0.0178 × 80 × 0.8 / 35
= 6.10 V (1.53 %) OK
MDB to SDB-2 (75 kW, 125 m)
I = 75000 / (√3 × 400 × 0.8) = 135.32 A
By current: 50 mm² (145 A)
ΔV = √3 × 135.32 × 0.0178 × 125 × 0.8 / 50
= 8.34 V (2.09 %) OK (< 2.5 %)
| Feeder | I (A) | Cable | Rating (A) | ΔV |
|---|---|---|---|---|
| MDB–SDB-1 | 108.25 | 35 mm² | 120 | 6.10 V (1.53 %) |
| MDB–SDB-2 | 135.32 | 50 mm² | 145 | 8.34 V (2.09 %) |
Answer: MDB–SDB-1: 3½ core 35 mm² Cu; MDB–SDB-2: 3½ core 50 mm² Cu. If a stricter 2 % limit is used, SDB-2 needs 70 mm² (drop 5.96 V = 1.49 %).
✓ Answer checked by Claude Opus 5.5 (high effort)
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