--- title: "Cable size MDB to SDB1 (80 kW) and SDB2 (10 kW)" subject: "Industrial Power Distribution and Illumination" chapter: 5 key: "semester_6/industrial_power_distribution_and_illumination:cable-mdb-sdb-80kw-10kw" source: https://www.bctengineer.com/q/semester_6/industrial_power_distribution_and_illumination/cable-mdb-sdb-80kw-10kw site: BCT Engineering --- # Cable size MDB to SDB1 (80 kW) and SDB2 (10 kW) Industrial Power Distribution and Illumination · Chapter 5: Electrification of Industrial Building - Asked in: 2079 Chaitra, 8 marks; 2071 Bhadra, 4 marks (asked 2 times) - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_6/industrial_power_distribution_and_illumination/cable-mdb-sdb-80kw-10kw - Whole chapter: https://www.bctengineer.com/md/semester_6/industrial_power_distribution_and_illumination/chapter-5 ## Question Determine the size of cable used from MDB to SDB1 and SDB2 for following conditions. The total power on the SDB1 and SDB2 are 80 kW and 10 kW respectively. The distance between MDB to SDB1 and SDB2 are 10 m and 100 m respectively. The supply system is 3 phase 4 wire system with 400 volts. Current ratings of Cu conductor laid in ground is given: | 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 | 36 | 45 | 60 | 77 | 99 | 120 | 145 | 175 | 210 | 240 | 270 | ## Short answer For MDB–SDB1 (80 kW, 10 m), line current 144.34 A, a 3½ core 50 mm² Cu cable is selected; 70 mm² is recommended for safety margin. For MDB–SDB2 (10 kW, 100 m), voltage drop limits the size, requiring a 3½ core 6 mm² Cu cable. ## Answer The cable is first chosen by current rating from the given table, then checked for voltage drop; the larger of the two sizes is selected. Assumptions: power factor 0.8 (not given), allowed voltage drop 2.5 % of 400 V = 10 V for the MDB–SDB feeder. **Formulas and assumptions** ```text Line current I = P / (√3 · V_L · cosφ) Voltage drop ΔV = √3 · I · (ρ·L/A) · cosφ (reactance neglected) Min. area A = √3 · I · ρ · L · cosφ / ΔV_allowed ρ (copper) = 0.0178 Ω·mm²/m ``` #### MDB to SDB1 (80 kW, 10 m) ```text I = 80 000 / (√3 × 400 × 0.8) = 144.34 A From table: 50 mm² → 145 A ≥ 144.34 A ΔV = √3 × 144.34 × 0.0178 × 10 × 0.8 / 50 = 0.71 V (0.18 %) < 10 V OK ``` 50 mm² is just enough by current; since it runs at almost 100 % of its rating, 70 mm² (175 A) is the safer practical choice. #### MDB to SDB2 (10 kW, 100 m) ```text I = 10 000 / (√3 × 400 × 0.8) = 18.04 A From table: 1.5 mm² → 21 A ≥ 18.04 A ΔV (1.5 mm²) = √3 × 18.04 × 0.0178 × 100 × 0.8 / 1.5 = 29.67 V (7.42 %) > 10 V NOT OK A_min = √3 × 18.04 × 0.0178 × 100 × 0.8 / 10 = 4.45 mm² → next size 6 mm² (45 A) ΔV (6 mm²) = 7.42 V (1.85 %) OK ``` | Feeder | Current (A) | By current | By voltage drop | Selected | |---|---|---|---|---| | MDB–SDB1 | 144.34 | 50 mm² | 3.56 mm² | 50 mm² (70 mm² for margin) | | MDB–SDB2 | 18.04 | 1.5 mm² | 4.45 mm² | 6 mm² | **Answer: MDB–SDB1: 3½ core 50 mm² Cu (70 mm² recommended for margin); MDB–SDB2: 3½ core 6 mm² Cu.** The short heavy feeder is decided by current rating; the long light feeder is decided by voltage drop. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. How is the feeder cable size decided in this problem? A. By voltage drop only B. By current rating only C. The average of the current and voltage-drop sizes D. By current rating first, then checked for voltage drop; the larger size is selected Answer: D. A cable must carry the current and keep the drop within limit, so both sizes are found and the larger one is taken. (Key point: Selection method) 2. Which formula gives the line current of the SDB load (pf 0.8 assumed)? A. I = P / (3 x V x cosφ) B. I = P / (√3 x V_L x cosφ) C. I = P x cosφ / (√3 x V_L) D. I = P / V_L Answer: B. For a 3-phase load, I = P/(√3·V_L·cosφ); for 80 kW this gives 144.34 A. (Key point: Line current) 3. 50 mm² (145 A) is just enough for 144.34 A. Why is 70 mm² recommended in practice? A. 50 mm² fails the voltage drop check B. 50 mm² would run at almost 100 % of its rating, leaving no margin C. The distance of 10 m is too long for 50 mm² D. 70 mm² has lower resistivity Answer: B. The drop is only 0.71 V, so 50 mm² passes; the choice of 70 mm² is for safety margin since 144.34 A is near 145 A. (Key point: Why 70 mm² for SDB1) 4. For the 10 kW, 100 m feeder (18.04 A), what decides the size? A. Voltage drop, which needs about 4.45 mm², so 6 mm² is chosen B. Current rating, so 1.5 mm² is enough C. Short-circuit current only D. The 70 mm² chosen for SDB1 Answer: A. 1.5 mm² carries 18.04 A but drops 29.67 V (7.42 %) against 10 V allowed, so the next size above 4.45 mm², i.e. 6 mm², is used. (Key point: Long light feeder) ## Close variants set in other exams - [Cable size MDB to SDB1 (70 kW) and SDB2 (90 kW)](https://www.bctengineer.com/question.md?key=semester_6%2Findustrial_power_distribution_and_illumination%3Acable-mdb-sdb-70kw-90kw) — Industrial Power Distribution and Illumination, chapter 5, asked 5× - [Cable size MDB to SDB-1 (60 kW) and SDB-2 (75 kW)](https://www.bctengineer.com/question.md?key=semester_6%2Findustrial_power_distribution_and_illumination%3Acable-mdb-sdb-60kw-75kw) — Industrial Power Distribution and Illumination, chapter 5, asked 4× - [Power/light sub-circuits and cable for 50 kW, 60 kW](https://www.bctengineer.com/question.md?key=semester_6%2Findustrial_power_distribution_and_illumination%3Asub-circuits-cable-50kw-60kw) — Industrial Power Distribution and Illumination, chapter 5, asked 4×