⚙️ System Parameters
Aluminium ≈ 0.0282
Typical value
OPTIMUM AREA
--
mm²
EQUIVALENT DIAMETER
--
mm
LINE CURRENT
--
A
MIN. ANNUAL COST
--
₹/year
📐 Kelvin's Law
Annual Cost = Annual Fixed Cost + Annual Energy Loss Cost
C(A) = K₁A + K₂/A
At the economic conductor size:
Annual Fixed Cost = Annual Energy Loss Cost
Therefore,
Aopt = √(K₂ / K₁)
C(A) = K₁A + K₂/A
At the economic conductor size:
Annual Fixed Cost = Annual Energy Loss Cost
Therefore,
Aopt = √(K₂ / K₁)
Kelvin's law states that the most economical conductor size is obtained when the annual cost of the energy lost in the conductor is equal to the annual fixed charges on the conductor.
📊 Annual Cost vs Conductor Cross-Sectional Area
Conductor fixed cost
Energy loss cost
Total annual cost
Optimum point
🔌 Physical Meaning of Kelvin's Law
Conductor
Size changes according to optimum cross-section
📈 Larger Conductor
Increasing conductor area increases the initial investment and therefore increases annual fixed charges.
🔥 Smaller Conductor
Smaller area gives higher resistance. Therefore, I²R losses and annual energy-loss cost increase.
⚖️ Economic Size
The optimum size is where the combined annual cost reaches its minimum.
💰 Cost Breakdown at Optimum Size
📋 Comparison of Different Conductor Sizes
| Area (mm²) |
Resistance (Ω) |
Annual Fixed Cost (₹) |
Loss Cost (₹) |
Total Annual Cost (₹) |
Status |
|---|
🧮 Calculation Details
Three-phase line current:
I = P / (√3 × V × cosφ)
Conductor resistance:
R = ρL / (A × n)
Three-phase copper/aluminium loss:
Ploss = 3I²R
Annual energy loss cost:
Closs = Ploss × operating hours × energy tariff
Conductor capital cost:
Ccapital = 3 × n × A × L × conductor cost
Annual fixed charge:
Cfixed = Ccapital × fixed-charge rate
I = P / (√3 × V × cosφ)
Conductor resistance:
R = ρL / (A × n)
Three-phase copper/aluminium loss:
Ploss = 3I²R
Annual energy loss cost:
Closs = Ploss × operating hours × energy tariff
Conductor capital cost:
Ccapital = 3 × n × A × L × conductor cost
Annual fixed charge:
Cfixed = Ccapital × fixed-charge rate