Multiples of Power Loss
4 3. 5 Current Resistance
3 2. 5
2 1. 5
1
1 1. 1 1. 2
1. 3 1. 4 1. 5 1. 6 1. 7
Multiples of Current and Resistance
1. 8
1. 9
2
To minimize copper power loss is to minimize circuit resistance and current. As is evident by the current squared term in the resistive power loss equation (Ploss = I2R), reducing conductor current results in dramatic savings. This equation reveals that two half-loaded feeders each have one-fourth the copper losses of one fully loaded feeder, given all the feeders are the same length and use the same conductor size. This suggests that copper losses could be halved by building dou-ble-circuit distribution primary; however, capital costs would drastically increase.
Therefore, the recommended design practice to economically decrease conductor losses through current reduction is to increase primary voltage. Apparent power (kVA) in a conductor is proportional to voltage and current (kVA = kV x I); doubling primary operating voltage will cut the conductor current in half for the same feeder power flow. Hence, by the resistive power loss equation, the resulting copper loss is 25 percent that of the original voltage using the same feeder conductor and length (see Figure 3). Increased costs for higher voltage class insulation are relatively small compared to total feeder costs. Additionally, the rated feeder power capacity also doubles when doubling the voltage. It is beyond the scope of this article to delve into volt-
Larger overhead conductor often brings with it the added expense of stronger poles, cross-arms, or shorter span lengths (more poles per mile). The additional cost makes it hard to justify larger distribution conductor on the basis of lowering power losses. More capacity and less voltage drop are better drivers for larger conductor.
Many utilities have pockets of 4160 voltage systems in older parts of town surrounded by higher voltage feeders. At this lower voltage, more conductor current flows for the same power delivered, resulting in higher I2R losses. Conversely, converting old 4160-volt feeders to higher voltage is capital-intensive and often not economically justifiable unless the line is already in poor condition and needs major improvements. If parts of the 4160-volt primary are in relatively good condition, installing multiple step-down power transformers at the periphery of the 4160-volt area will reduce copper losses by injecting load current at more points (i.e., reducing overall conductor current and the distance traveled by the current to serve the load).
age drop and power flow equations and calculations; however, simple power flow programs show that doubling operating voltage will not only reduce power loss but will also enable more customers to be served on longer feeders with less power loss and less percentage voltage drop.
This results in lower overall capital and operating costs.
The second variable in the resistive power equation, conductor resistance, is inversely proportional to its cross-sectional area. Hence, larger diameter conductors produce less loss than smaller conductors of the same material for the same current. Voltage drop will also be less with larger diameter conductors; however, construction costs are higher.
100
Percentage of Power Loss (%)
90
80
70
60
50
40
30
20 10 0
4 6 8 10 12
14 16 18 20 22 24 26 28 30 32 34
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