K1 and K2 of LLF from load pattern

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

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.