---
title: "Transformer size for four blocks"
subject: "Industrial Power Distribution and Illumination"
chapter: 4
key: "semester_6/industrial_power_distribution_and_illumination:transformer-size-four-blocks"
source: https://www.bctengineer.com/q/semester_6/industrial_power_distribution_and_illumination/transformer-size-four-blocks
site: BCT Engineering
---

# Transformer size for four blocks

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

- Asked in: 2080 Chaitra, 4 marks
- 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-four-blocks
- Whole chapter: https://www.bctengineer.com/md/semester_6/industrial_power_distribution_and_illumination/chapter-4

## Question

There are four blocks in an industry having connected load of 250 kVA, 200 kVA, 150 kVA and 400 kVA and demand factor of 90%, 80%, 75% and 85% respectively. Use a diversity factor of 1.5. Calculate the size of transformer for the industry if transformer available in the market are 50kVA, 100kVA, 250kVA, 400kVA, 600kVA and 750kVA.

## Short answer

Max demand per block: 250×0.9=225, 200×0.8=160, 150×0.75=112.5, 400×0.85=340 kVA. Sum = 837.5 kVA. Plant max demand = 837.5/1.5 = 558.33 kVA. Next available size is 600 kVA. Select 600 kVA transformer.

## Answer

The transformer is sized on the maximum demand of the industry, found by applying demand factor to each block and then the diversity factor for the whole plant.

#### Step 1: maximum demand of each block
```text
Max demand = connected load × demand factor
Block 1 = 250 × 0.90 = 225.0 kVA
Block 2 = 200 × 0.80 = 160.0 kVA
Block 3 = 150 × 0.75 = 112.5 kVA
Block 4 = 400 × 0.85 = 340.0 kVA
Sum of individual max demands = 837.5 kVA
```

#### Step 2: maximum demand of the industry
```text
Diversity factor = sum of individual max demands
                   / max demand of whole system
Max demand = 837.5 / 1.5 = 558.33 kVA
```

#### Step 3: select the transformer
Available sizes: 50, 100, 250, 400, 600, 750 kVA. The smallest size ≥ 558.33 kVA is 600 kVA.

Loading at peak = 558.33 / 600 = 93 %, which is acceptable.

**Answer: maximum demand = 558.33 kVA; select one 600 kVA transformer.** (If a future expansion margin of 20 % were added, 670 kVA would be needed and the 750 kVA unit would be chosen.)

## Check yourself

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

1. How is the maximum demand of each block found?
   A. Connected load x demand factor
   B. Connected load / demand factor
   C. Connected load x diversity factor
   D. Connected load + demand factor
   Answer: A. E.g. block 1 = 250 x 0.90 = 225 kVA. (Key point: Block demand)

2. How is the plant maximum demand obtained from the sum of block demands (837.5 kVA) and diversity factor 1.5?
   A. 837.5 x 1.5 = 1256.25 kVA
   B. 837.5 - 1.5 = 836 kVA
   C. 837.5 / 1.5 = 558.33 kVA
   D. (250+200+150+400)/1.5 = 666.7 kVA
   Answer: C. Diversity factor = sum of individual maximum demands / maximum demand of the whole system. (Key point: Diversity factor use)

3. Which transformer is selected from 50, 100, 250, 400, 600, 750 kVA?
   A. 600 kVA, the smallest size not less than 558.33 kVA
   B. 400 kVA, the nearest below
   C. 750 kVA, since it is the largest
   D. 250 kVA, one for each block
   Answer: A. The demand 558.33 kVA is covered by 600 kVA, loaded about 93%. (Key point: Selection)

## Related in Industrial Power Distribution and Illumination

- [Connected load, MD, demand, diversity, load factor](https://www.bctengineer.com/question.md?key=semester_6%2Findustrial_power_distribution_and_illumination%3Aload-terminology-five-terms) — Industrial Power Distribution and Illumination, chapter 1
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