---
title: "Transformer size for given industry (figure)"
subject: "Industrial Power Distribution and Illumination"
chapter: 4
key: "semester_6/industrial_power_distribution_and_illumination:transformer-size-attached-figure"
source: https://www.bctengineer.com/q/semester_6/industrial_power_distribution_and_illumination/transformer-size-attached-figure
site: BCT Engineering
---

# Transformer size for given industry (figure)

Industrial Power Distribution and Illumination · Chapter 4: Distribution Substation of Industrial Plant

- Asked in: 2076 Bhadra, 8 marks; 2075 Bhadra, 8 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/transformer-size-attached-figure
- Whole chapter: https://www.bctengineer.com/md/semester_6/industrial_power_distribution_and_illumination/chapter-4

## Question

Calculate the size of transformer for the given industry (attached figure) considering factor of maximum utilization Ku and factor of simultaneity Ks. [The attached figure is not included in the source scans.]

## Short answer

For an assumed industry with 5 motors (11 kW each) and lighting/heating loads, compute motor kVA = 11/(0.88×0.85)=14.71 kVA. Apply Ku (0.75 motors, 1.0 others), then Ks per board (0.8, 0.9) and MDB (0.9). Demand = 62.39 kVA, with 20% margin = 74.87 kVA, select 100 kVA transformer.

## Answer

The figure is not available, so the method is shown on an assumed industry with two sub-distribution boards; the same steps apply to any given layout.

**Assumed loads**
- SDB-A: 5 motors, each 11 kW, η = 0.88, pf = 0.85
- SDB-B: lighting 8 kVA, two heaters of 10 kVA each

#### Factors used (IEC / Schneider Electric guide)
- **Ku (factor of maximum utilization)**: motors 0.75; lighting and heating 1.0.
- **Ks (factor of simultaneity)** for a board with n circuits: 2–3 → 0.9, 4–5 → 0.8, 6–9 → 0.7, 10 and above → 0.6.

#### Step 1: rated kVA of each motor
```text
kVA = kW / (η × pf) = 11 / (0.88 × 0.85) = 14.71 kVA
```

#### Step 2: utilized power with Ku
| Load | Rated kVA | Ku | Utilized kVA |
|---|---|---|---|
| Motor (each, 5 nos) | 14.71 | 0.75 | 11.03 |
| Total motors | 73.53 | - | 55.15 |
| Lighting | 8 | 1.0 | 8.0 |
| Heaters 2 × 10 | 20 | 1.0 | 20.0 |

#### Step 3: apply Ks at each board
```text
SDB-A = 55.15 × Ks(5 circuits = 0.8)  = 44.12 kVA
SDB-B = (8 + 20) × Ks(3 circuits = 0.9) = 25.20 kVA
MDB   = (44.12 + 25.20) × Ks(2 = 0.9)  = 69.32 × 0.9
      = 62.39 kVA
```

#### Step 4: future expansion and selection
```text
Required = 62.39 × 1.2 (20 % margin) = 74.87 kVA
```
**Answer: select a 100 kVA, 11/0.4 kV transformer** (next standard rating above 74.87 kVA; standard sizes 50, 100, 160, 200, 250 kVA ...).

**Remarks**
- Connected load is 101.5 kVA but maximum demand is only about 62 kVA; sizing on connected load would over-size the transformer and lower its efficiency.
- If the figure gives Ku and Ks values directly, use those instead of the typical values.
- Check that the largest motor's starting kVA does not cause an excessive dip (usually < 10–15 %) on the chosen transformer.

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. How is the rated kVA of each 11 kW motor found (eta = 0.88, pf = 0.85)?
   A. kVA = 11 / (0.88 x 0.85) = 14.71 kVA
   B. kVA = 11 x 0.88 x 0.85 = 8.23 kVA
   C. kVA = 11 / 0.85 = 12.94 kVA
   D. kVA = 11 x 0.75 = 8.25 kVA
   Answer: A. Input kVA is output kW divided by efficiency and power factor. (Key point: Motor kVA)

2. What is the utilized power of the five motors after applying Ku = 0.75?
   A. 73.53 kVA
   B. 14.71 kVA
   C. 55.15 kVA
   D. 101.5 kVA
   Answer: C. Total rated 5 x 14.71 = 73.53 kVA, times 0.75 gives 55.15 kVA. (Key point: Applying Ku)

3. SDB-B has lighting 8 kVA and two 10 kVA heaters (3 circuits). What is its demand?
   A. 28 x 0.9 = 25.2 kVA
   B. 28 x 0.8 = 22.4 kVA
   C. 28 x 0.6 = 16.8 kVA
   D. 28 kVA with no factor
   Answer: A. Ku is 1.0 for lighting and heating, and 3 circuits have Ks = 0.9. (Key point: Applying Ks at boards)

4. The MDB demand is 62.39 kVA. Which transformer is chosen and why?
   A. 50 kVA, since it is the nearest size to 62.39
   B. 160 kVA, since connected load is 101.5 kVA
   C. 100 kVA, since 62.39 x 1.2 = 74.86 kVA and 100 is the next standard size
   D. 75 kVA, exactly the required value
   Answer: C. The 20% margin gives 74.86 kVA; the next standard rating above is 100 kVA. (Key point: Final selection)

## Related in Industrial Power Distribution and Illumination

- [Selecting distribution transformer size using Ku and Ks](https://www.bctengineer.com/question.md?key=semester_6%2Findustrial_power_distribution_and_illumination%3Atransformer-size-ku-ks) — Industrial Power Distribution and Illumination, chapter 4, asked 3×
- [Short note: Factor of maximum utilization Ku](https://www.bctengineer.com/question.md?key=semester_6%2Findustrial_power_distribution_and_illumination%3Ashort-note-ku) — Industrial Power Distribution and Illumination, chapter 1
- [Short note: Factor of simultaneity Ks](https://www.bctengineer.com/question.md?key=semester_6%2Findustrial_power_distribution_and_illumination%3Ashort-note-ks) — Industrial Power Distribution and Illumination, chapter 1
- [Maximum demand and need to calculate it](https://www.bctengineer.com/question.md?key=semester_6%2Findustrial_power_distribution_and_illumination%3Amaximum-demand-need) — Industrial Power Distribution and Illumination, chapter 1
- [Transformer size for four blocks](https://www.bctengineer.com/question.md?key=semester_6%2Findustrial_power_distribution_and_illumination%3Atransformer-size-four-blocks) — Industrial Power Distribution and Illumination, chapter 4