---
title: "Peak load, load factor and energy from load pattern"
subject: "Transmission and Distribution System Design"
chapter: 6
key: "semester_8/transmission_and_distribution_system_design:consumer-class-load-pattern-transformer"
source: https://www.bctengineer.com/q/semester_8/transmission_and_distribution_system_design/consumer-class-load-pattern-transformer
site: BCT Engineering
---

# Peak load, load factor and energy from load pattern

Transmission and Distribution System Design · Chapter 6: Electrical Load Characteristics and Load Forecast

- Asked in: 2074 Bhadra, 10 marks
- Video: [Solved Problem 1 - Load Curve & Load Duration Curve Made Easy⚡(Average Load + Load Factor Explained)](https://www.youtube.com/watch?v=z005OTiWeqk) — Power Systemz
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_8/transmission_and_distribution_system_design/consumer-class-load-pattern-transformer
- Whole chapter: https://www.bctengineer.com/md/semester_8/transmission_and_distribution_system_design/chapter-6

## Question

The consumer data for particular distribution transformer for a specified year is as shown as follows.

Table-1:

| Consumer class | Class A | Class B | Class C |
|---|---|---|---|
| Consumer number | 80 | 30 | 40 |
| Monthly energy consumption | 30 kWh | 22 kWh | 25 kWh |
| No. of effective days per month | 30 | 24 | 26 |
| Coincidence factor | 1 | 0.9 | 0.95 |
| Power factor | 0.9 | 0.8 | 0.85 |
| Load pattern | Table 2 | Table 2 | Table 2 |

Table-2 (load pattern):

| Time (hrs.) | 0:00-6:00 | 6:00-10:00 | 10:00-14:00 | 14:00-18:00 | 18:00-21:00 | 21:00-24:00 |
|---|---|---|---|---|---|---|
| Class A | 0.1 | 0.4 | 0.2 | 0.2 | 1 | 0.4 |
| Class B | 0.1 | 0.1 | 0.5 | 1 | 0.5 | 0.1 |
| Class C | 0.1 | 0.1 | 1 | 1 | 0.1 | 0.1 |

Determine the following. For each class of consumer: i. peak load ii. Load factor (daily and annual) iii. Contribution factor iv. Annual energy sell. For distribution transformer (Neglect LT losses): i. peak load ii. Load factor iii. Annual energy sells.

## Short answer

Class peaks: A 10.0 kW, B 2.81 kW, C 3.81 kW. Contribution factors: 1.0, 0.5, 0.1. Transformer peak 11.79 kW, annual load factor 0.47, annual energy sales 48,720 kWh.

## Answer

The load pattern (Table 2) gives the load of each class in per unit of its peak. From it we find the equivalent full-load hours per day, then the peak from the daily energy.

**Method:**

$$
\begin{aligned}
h_{eq} &= \sum (\text{p.u. load} \times \text{hours}) \\
\text{Daily energy per consumer} &= \frac{\text{Monthly kWh}}{\text{Effective days}} \\
P_{ind} &= \frac{\text{Daily energy}}{h_{eq}}, \quad P_{class} = N \times P_{ind} \times \text{coincidence factor} \\
LF_{daily} &= \frac{h_{eq}}{24}, \quad LF_{annual} = \frac{\text{Annual energy}}{P_{ind} \times 8760}
\end{aligned}
$$

#### Equivalent hours from load pattern

- Class A: $0.1(6) + 0.4(4) + 0.2(4) + 0.2(4) + 1(3) + 0.4(3) = 8.0$ h
- Class B: $0.1(6) + 0.1(4) + 0.5(4) + 1(4) + 0.5(3) + 0.1(3) = 8.8$ h
- Class C: $0.1(6) + 0.1(4) + 1(4) + 1(4) + 0.1(3) + 0.1(3) = 9.6$ h

#### For each consumer class

| Quantity | Class A | Class B | Class C |
|---|---|---|---|
| Daily energy (kWh) | 30/30 = 1.000 | 22/24 = 0.9167 | 25/26 = 0.9615 |
| Individual peak (kW) | 1/8 = 0.125 | 0.9167/8.8 = 0.1042 | 0.9615/9.6 = 0.1002 |
| (i) Class peak (kW) | 80×0.125×1 = **10.00** | 30×0.1042×0.9 = **2.81** | 40×0.1002×0.95 = **3.81** |
| (ii) Daily LF | 8/24 = **0.333** | 8.8/24 = **0.367** | 9.6/24 = **0.400** |
| (ii) Annual LF | 360/(0.125×8760) = **0.329** | 264/(0.1042×8760) = **0.289** | 300/(0.1002×8760) = **0.342** |
| (iv) Annual energy (kWh) | 80×30×12 = **28,800** | 30×22×12 = **7,920** | 40×25×12 = **12,000** |

The annual LF is lower than the daily LF for B and C because they consume only on effective days.

#### Transformer load curve (kW) = class peak × p.u. pattern

| Time | A | B | C | Total |
|---|---|---|---|---|
| 0–6 | 1.00 | 0.28 | 0.38 | 1.66 |
| 6–10 | 4.00 | 0.28 | 0.38 | 4.66 |
| 10–14 | 2.00 | 1.41 | 3.81 | 7.21 |
| 14–18 | 2.00 | 2.81 | 3.81 | 8.62 |
| 18–21 | 10.00 | 1.41 | 0.38 | **11.79** |
| 21–24 | 4.00 | 0.28 | 0.38 | 4.66 |

The transformer peak occurs at 18:00–21:00.

#### (iii) Contribution factor of each class

$$
c = \frac{\text{Class load at transformer peak}}{\text{Class peak}}
$$

- Class A: $10/10 = 1.0$
- Class B: $1.406/2.8125 = 0.5$
- Class C: $0.381/3.806 = 0.1$

#### For the distribution transformer

1. **Peak load:** $P = 10 + 1.406 + 0.381 = 11.79$ kW. With reactive power $Q = 10\tan(\cos^{-1}0.9) + 1.406\tan(\cos^{-1}0.8) + 0.381\tan(\cos^{-1}0.85) = 6.13$ kVAR, $S = \sqrt{11.79^2 + 6.13^2} = 13.29$ kVA.
2. **Annual energy sales** (LT losses neglected): $28{,}800 + 7{,}920 + 12{,}000 = 48{,}720$ kWh.
3. **Load factor:**

$$
LF = \frac{48{,}720}{11.787 \times 8760} = 0.472
$$

**Answer:** Transformer peak ≈ 11.79 kW (13.29 kVA), annual LF ≈ 0.47, annual energy sales = 48,720 kWh. Class contribution factors are 1.0, 0.5 and 0.1.

## Check yourself

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

1. How is the equivalent full-load hours per day for a class found?
   A. Sum of (p.u. load x hours of interval)
   B. Average of the p.u. values
   C. Peak p.u. x 24
   D. Sum of p.u. values
   Answer: A. h_eq = sum(p.u. x hours); for Class A it is 8.0 h. (Key point: Equivalent hours)

2. How is the individual peak of a consumer found?
   A. Daily energy x h_eq
   B. Daily energy / h_eq (daily energy = monthly kWh / effective days)
   C. Monthly energy / 720
   D. Daily energy / 24
   Answer: B. Class A: 1 kWh/day / 8 h = 0.125 kW. (Key point: Individual peak)

3. How is the transformer peak found, and when does it occur?
   A. By adding the class peaks; at noon
   B. By adding each class's kW at each time block; at 18:00-21:00 giving 11.79 kW
   C. By taking the largest class only; at midnight
   D. By averaging; at 6:00-10:00
   Answer: B. Class load = class peak x p.u. pattern; the total is highest in the evening block at 11.79 kW. (Key point: Transformer peak time)

4. The contribution factor of a class is:
   A. Class load at transformer peak divided by class peak
   B. Class peak divided by transformer peak
   C. Daily LF divided by annual LF
   D. Coincidence factor squared
   Answer: A. For Class B it is 1.406/2.8125 = 0.5 and for C 0.1. (Key point: Contribution factor)

## Close variants set in other exams

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