---
title: "Dependency of energy loss on load factor"
subject: "Transmission and Distribution System Design"
chapter: 6
key: "semester_8/transmission_and_distribution_system_design:energy-loss-load-factor-dependency"
source: https://www.bctengineer.com/q/semester_8/transmission_and_distribution_system_design/energy-loss-load-factor-dependency
site: BCT Engineering
---

# Dependency of energy loss on load factor

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

- Asked in: 2073 Bhadra, 8 marks
- 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/energy-loss-load-factor-dependency
- Whole chapter: https://www.bctengineer.com/md/semester_8/transmission_and_distribution_system_design/chapter-6

## Question

Explain the dependency of energy loss computation in a transmission/distribution line on the load factor with proper mathematical aid.

## Short answer

Energy loss depends on load factor via the loss factor LLF, which ranges between LF² and LF (empirically LLF = 0.3LF + 0.7LF²). Percent energy loss = percent peak loss × (LLF/LF); lower LF reduces energy loss percentage.

## Answer

Energy loss in a line depends on how the load varies with time, not just on the peak. Since copper loss is proportional to the square of current, the energy loss is linked to the load factor (LF) through the **loss of load factor (LLF)**.

#### Definitions

For a line of resistance $R$ (per phase) carrying load $P(t)$ at voltage $V$ and power factor $\cos\phi$:

$$
\begin{aligned}
P_{loss}(t) &= 3 I^2 R = \frac{P(t)^2 R}{V^2 \cos^2\phi} = k\,P(t)^2 \\
E_{loss} &= \int_0^T k P(t)^2 dt = k P_{max}^2 \, T \cdot \underbrace{\frac{1}{T}\int_0^T \left(\frac{P}{P_{max}}\right)^2 dt}_{LLF}
\end{aligned}
$$

So $E_{loss} = P_{loss,peak} \times LLF \times T$, and $LF = \dfrac{1}{T}\int_0^T \dfrac{P}{P_{max}} dt$.

#### Limits of LLF in terms of LF

Consider a two-step load: $P_{max}$ for time $t$ and $P_{min}$ for the rest $(T - t)$.

- **Case 1, $P_{min} = 0$:** $LF = t/T$ and $LLF = t/T$, so $LLF = LF$ (upper limit).
- **Case 2, very short peak ($t \to 0$), off-peak $P_{min}$ for nearly all the time:** $LF \approx P_{min}/P_{max}$ and $LLF \approx (P_{min}/P_{max})^2$, so $LLF = LF^2$ (lower limit).

Hence

$$
LF^2 \le LLF \le LF
$$

Real load curves lie between these, giving the empirical relation (Buller–Woodrow):

$$
LLF = 0.3\,LF + 0.7\,LF^2
$$

#### Dependency of % energy loss on LF

$$
\%E_{loss} = \frac{P_{loss,peak} \times LLF \times T}{P_{max} \times LF \times T} \times 100 = \%P_{loss,peak} \times \frac{LLF}{LF}
$$

With the empirical formula, $\dfrac{LLF}{LF} = 0.3 + 0.7\,LF$.

**Example:** peak loss 8% of peak demand.

| LF | LLF | LLF/LF | % energy loss |
|---|---|---|---|
| 0.3 | 0.153 | 0.51 | 4.08% |
| 0.5 | 0.325 | 0.65 | 5.20% |
| 1.0 | 1.0 | 1.0 | 8.00% |

**Conclusions:**

- For a given peak, a lower LF means fewer kWh lost but also fewer kWh sold; the % energy loss is always ≤ % peak power loss.
- For a given energy delivered, a higher LF lowers the peak, and since loss ∝ peak², the absolute energy loss falls.
- Energy loss can never be found by multiplying peak loss by hours; LLF must be used.
- Constant losses (transformer core loss, corona) do not depend on load, so for them LLF = 1.

## Check yourself

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

1. Energy loss in a line over time T is:
   A. Peak loss x LF x T
   B. Peak loss x LLF x T
   C. Peak loss x T
   D. Peak loss x LF^2 x T
   Answer: B. Loss is proportional to P^2, so E_loss = P_loss,peak x LLF x T. (Key point: Energy loss formula)

2. What are the limits of LLF in terms of LF?
   A. LF <= LLF <= 1
   B. LF^2 <= LLF <= LF
   C. 0 <= LLF <= LF^2
   D. LLF = LF always
   Answer: B. LLF = LF when off-peak is zero, and LLF = LF^2 for a very short peak on a flat load. (Key point: Limits of LLF)

3. How does % energy loss relate to % peak loss?
   A. Multiply by LF
   B. Multiply by LLF/LF, which is 0.3 + 0.7 LF and is at most 1
   C. Multiply by LF/LLF
   D. They are always equal
   Answer: B. %E_loss = %P_loss,peak x LLF/LF, so it is no more than the % peak loss. (Key point: Percent energy loss)

4. With peak loss 8% and LF = 0.5, % energy loss is:
   A. 8.00%
   B. 4.08%
   C. 5.20%
   D. 2.60%
   Answer: C. LLF = 0.325, LLF/LF = 0.65, so 8 x 0.65 = 5.2%. (Key point: Example value)

## Related in Transmission and Distribution System Design

- [Define loss of load factor](https://www.bctengineer.com/question.md?key=semester_8%2Ftransmission_and_distribution_system_design%3Adefine-loss-of-load-factor) — Transmission and Distribution System Design, chapter 6, asked 2×
- [T/F: % power loss affected by load factor](https://www.bctengineer.com/question.md?key=semester_8%2Ftransmission_and_distribution_system_design%3Atf-power-loss-affected-load-factor) — Transmission and Distribution System Design, chapter 6
- [Peak load, load factor and energy from load pattern](https://www.bctengineer.com/question.md?key=semester_8%2Ftransmission_and_distribution_system_design%3Aconsumer-class-load-pattern-transformer) — Transmission and Distribution System Design, chapter 6, asked 2×
- [Advantages of higher load factor](https://www.bctengineer.com/question.md?key=semester_8%2Ftransmission_and_distribution_system_design%3Ahigher-load-factor-advantages) — Transmission and Distribution System Design, chapter 6
- [Percentage energy loss of 25 kVA transformer](https://www.bctengineer.com/question.md?key=semester_8%2Ftransmission_and_distribution_system_design%3Atransformer-energy-loss-25kva) — Transmission and Distribution System Design, chapter 7
- [Energy loss of 100 kVA transformer in third year](https://www.bctengineer.com/question.md?key=semester_8%2Ftransmission_and_distribution_system_design%3Atransformer-energy-loss-third-year) — Transmission and Distribution System Design, chapter 7

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