--- title: "Relation between LLF and LF" subject: "Transmission and Distribution System Design" chapter: 6 key: "semester_8/transmission_and_distribution_system_design:llf-lf-relation-derivation" source: https://www.bctengineer.com/q/semester_8/transmission_and_distribution_system_design/llf-lf-relation-derivation site: BCT Engineering --- # Relation between LLF and LF Transmission and Distribution System Design · Chapter 6: Electrical Load Characteristics and Load Forecast - Asked in: 2080 Chaitra, 6 marks; 2075 Bhadra, 6 marks; 2070 Magh, 6 marks (asked 3 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-lf-relation-derivation - Whole chapter: https://www.bctengineer.com/md/semester_8/transmission_and_distribution_system_design/chapter-6 ## Question Derive the relation between LLF and LF considering a suitable load curve. ## Short answer Using a two-step load curve, LF = (Pm t + P1(T-t))/(Pm T), LLF = (Pm^2 t + P1^2(T-t))/(Pm^2 T). LLF lies between LF^2 and LF; the empirical relation is LLF = k·LF + (1−k)·LF^2, with typical k=0.3 for distribution feeders. ## Answer **Load factor (LF)** is the ratio of average load to peak load. **Loss load factor (LLF)** is the ratio of average power loss to the loss at peak load. Because loss is proportional to $I^2$, LLF always lies between $LF^2$ and $LF$. The usual relation is $$ LLF = k\,LF + (1-k)\,LF^2, \qquad k \approx 0.2\text{–}0.3 $$ #### Definitions For a period $T$ with peak load $P_m$, and assuming constant voltage and power factor: $$ LF = \frac{\frac{1}{T}\int_0^T P\,dt}{P_m}, \qquad LLF = \frac{\frac{1}{T}\int_0^T P^2\,dt}{P_m^2} $$ since loss $\propto I^2 \propto P^2$. #### Suitable load curve Take a two-step daily load curve. The load is at peak $P_m$ for time $t$, and at a lower load $P_1$ for the rest of the time $(T - t)$. ```text P Pm |-----+ | | P1 | +--------------------+ | | +-----+--------------------+--> time 0 t T ``` Then $$ LF = \frac{P_mt + P_1(T-t)}{P_mT}, \qquad LLF = \frac{P_m^2t + P_1^2(T-t)}{P_m^2T} $$ #### Limiting cases **Case 1:** the off-peak load is zero ($P_1 \to 0$, a short sharp peak). $$ LF = \frac{t}{T}, \qquad LLF = \frac{t}{T} \;\Rightarrow\; LLF = LF $$ **Case 2:** the peak lasts a very short time ($t \to 0$, a nearly flat load at $P_1$). $$ LF = \frac{P_1}{P_m}, \qquad LLF = \frac{P_1^2}{P_m^2} \;\Rightarrow\; LLF = LF^2 $$ Every real load curve lies between these two extremes, so $$ LF^2 \le LLF \le LF $$ #### General relation LLF is expressed as a weighted combination of the two limits: $$ LLF = k\,LF + (1-k)\,LF^2 $$ Here $k$ is found from measured load curves of the system. For typical distribution feeders $k = 0.3$ (Buller and Woodrow), giving $LLF = 0.3LF + 0.7LF^2$. Some utilities use $k = 0.2$ for urban feeders, giving $LLF = 0.2LF + 0.8LF^2$. **Example:** for $LF = 0.5$, $LLF = 0.3(0.5) + 0.7(0.25) = 0.325$. The energy loss per year is then $P_{loss,peak}\times0.325\times8760$ kWh. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. Loss load factor (LLF) is defined as: A. Average load divided by peak load B. Average power loss divided by the power loss at peak load C. Peak loss divided by average loss D. Energy sold divided by energy lost Answer: B. LLF = average loss / loss at peak. Since loss is proportional to P^2, LLF = average of P^2 over Pm^2. (Key point: Definition of LLF) 2. In the two-step load curve, when does LLF equal LF? A. When the peak lasts a very short time over a flat load B. When the load is constant C. When the off-peak load P1 is zero D. When the power factor is zero Answer: C. With P1 = 0, LF = t/T and LLF = t/T, so LLF = LF (upper limit). (Key point: Limiting case LLF = LF) 3. When does LLF approach LF^2? A. When off-peak load is zero B. When the peak lasts very briefly (t tends to 0) and the load is nearly flat at P1 C. When the load factor is 1 D. When voltage varies widely Answer: B. For t tending to 0, LF = P1/Pm and LLF = (P1/Pm)^2 = LF^2 (lower limit). (Key point: Limiting case LLF = LF^2) 4. Which relation and range follow from the two limits? A. LLF = k LF + (1-k) LF^2, with LF^2 <= LLF <= LF B. LLF = k LF^2 + (1-k) LF^3, with LLF >= LF C. LLF = LF always D. LLF = k LF + (1-k) LF^2, with LLF >= LF Answer: A. Real curves lie between the two extremes, so LLF is a weighted mix of LF and LF^2, typically with k = 0.2 to 0.3. 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