---
title: "K1 and K2 of LLF from load pattern"
subject: "Transmission and Distribution System Design"
chapter: 6
key: "semester_8/transmission_and_distribution_system_design:llf-k1-k2-load-pattern"
source: https://www.bctengineer.com/q/semester_8/transmission_and_distribution_system_design/llf-k1-k2-load-pattern
site: BCT Engineering
---

# K1 and K2 of LLF from load pattern

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

- Asked in: 2071 Bhadra, 10 marks; 2071 Magh, 6 marks (asked 2 times)
- 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/llf-k1-k2-load-pattern
- Whole chapter: https://www.bctengineer.com/md/semester_8/transmission_and_distribution_system_design/chapter-6

## Question

Using equation LLF = K1×LF + K2×LF², compute the appropriate value of K1 and K2 of system having following 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 |
|---|---|---|---|---|---|---|
| Demand (MW) | 0 | 4 | 2 | 2 | 10 | 4 |

## Short answer

Compute LF = 74/(10×24) = 0.3083 and LLF = 444/(10²×24) = 0.185. Then K1 = (LLF − LF²)/(LF − LF²) = 0.4217, K2 = 1 − K1 = 0.5783. The result is K1 ≈ 0.42, K2 ≈ 0.58.

## Answer

The **loss of load factor (LLF)** is the ratio of the average power loss to the power loss at peak load over a period. Since loss is proportional to the square of load (at constant voltage and power factor):

$$
LLF = \frac{\sum P_i^2 t_i}{P_{max}^2 \, T}, \qquad LF = \frac{\sum P_i t_i}{P_{max} \, T}
$$

The empirical relation is $LLF = K_1 \, LF + K_2 \, LF^2$ with $K_1 + K_2 = 1$ (so that LLF = 1 when LF = 1). Two equations are thus available to find $K_1$ and $K_2$.

#### Step 1: Tabulate the load pattern

| Interval | $t_i$ (h) | $P_i$ (MW) | $P_i t_i$ (MWh) | $P_i^2 t_i$ |
|---|---|---|---|---|
| 0–6 | 6 | 0 | 0 | 0 |
| 6–10 | 4 | 4 | 16 | 64 |
| 10–14 | 4 | 2 | 8 | 16 |
| 14–18 | 4 | 2 | 8 | 16 |
| 18–21 | 3 | 10 | 30 | 300 |
| 21–24 | 3 | 4 | 12 | 48 |
| Total | 24 | | 74 | 444 |

Peak demand $P_{max} = 10$ MW, $T = 24$ h.

#### Step 2: Load factor

$$
\begin{aligned}
LF &= \frac{74}{10 \times 24} = 0.30833 \\
LF^2 &= 0.09507
\end{aligned}
$$

#### Step 3: Loss of load factor

$$
LLF = \frac{444}{10^2 \times 24} = \frac{444}{2400} = 0.185
$$

#### Step 4: Solve for K1 and K2

Put $K_2 = 1 - K_1$:

$$
\begin{aligned}
LLF &= K_1 LF + (1 - K_1) LF^2 \\
K_1 &= \frac{LLF - LF^2}{LF - LF^2} \\
&= \frac{0.185 - 0.09507}{0.30833 - 0.09507} = \frac{0.08993}{0.21327} \\
&= 0.4217 \\
K_2 &= 1 - 0.4217 = 0.5783
\end{aligned}
$$

**Check:** $0.4217 \times 0.30833 + 0.5783 \times 0.09507 = 0.1300 + 0.0550 = 0.185$ ✓

**Answer: $K_1 \approx 0.42$, $K_2 \approx 0.58$**, i.e. $LLF = 0.42\,LF + 0.58\,LF^2$ for this system.

The commonly used Buller–Woodrow values are $K_1 = 0.3$, $K_2 = 0.7$; this load pattern (a sharp evening peak with zero load at night) gives a somewhat larger $K_1$. The value of LLF (0.185) lies between $LF^2$ (0.095) and $LF$ (0.308), as it must.

## Check yourself

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

1. What condition on K1 and K2 is used so the two unknowns can be found?
   A. K1 = K2
   B. K1 + K2 = 1, so that LLF = 1 when LF = 1
   C. K1 x K2 = 1
   D. K1 - K2 = 1
   Answer: B. At LF = 1 the load is flat and LLF must equal 1, giving K1 + K2 = 1. (Key point: Constraint K1 + K2)

2. For the given pattern, what is the load factor?
   A. 0.185
   B. 0.095
   C. 0.308
   D. 0.500
   Answer: C. Energy = 74 MWh, peak = 10 MW, so LF = 74/(10 x 24) = 0.308. (Key point: Load factor)

3. How is LLF computed and what is its value here?
   A. Sum(P^2 t)/(Pmax^2 T) = 444/2400 = 0.185
   B. Sum(P t)/(Pmax T) = 0.308
   C. Sum(P^2 t)/(Pmax T) = 1.85
   D. (LF)^2 = 0.095
   Answer: A. Loss is proportional to P^2: 444 MW^2 h over 100 x 24 gives 0.185. (Key point: LLF)

4. Solving LLF = K1 LF + (1-K1) LF^2 gives:
   A. K1 = 0.58, K2 = 0.42
   B. K1 = 0.30, K2 = 0.70
   C. K1 = 0.42, K2 = 0.58
   D. K1 = 0.185, K2 = 0.815
   Answer: C. K1 = (0.185-0.0951)/(0.3083-0.0951) = 0.42, so K2 = 0.58. (Key point: K1 and K2)

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