# Optimal DC Cable Selection in PV Designs: Page 6 of 7

## Inside this Article

I =106.92 A × 1.56 = 166.8 A

Thermal and bundling correction factors are 1.00 since there are no more than two current-carrying conductors per conduit. This results in a required cable ampacity of 166.8 A.

2. Select size of cable based on required ampacity: Per Table 310.16 of the 2008 *NEC*, the cable selected is 4/0 AWG, THWN-2 aluminum, and per *NEC* Chapter 9, Table 8, it has a cross-sectional area of 107.2 mm^{2}. Based on insulation type alone, this cable has a permitted ampacity of 205 A at 90°C. This must be further derated for the 75°C rating of the inverter terminals, resulting in a maximum ampacity for the cable of 180 A, which is greater than the required ampacity above.

3. Obtain cable cost data: The THWN-2 aluminum cable specified can be procured for a unit cost of $10.45/m, and we have assumed the cost to install it is $1.00/m.

4. Calculate the shape factor: We use PVSyst to calculate the shape factor *K*. For TMY3 weather data at San Jose Airport, we find that* K* = 0.19.

5. Calculate the time value of the losses: Assuming 25 years of service, an inflation rate of 2% per year and an annual discount rate of 5%, the annuity return factor γ25 is $17.53.

These data are summarized in Table 1 (below).

**THE OPTIMALLY SIZED CABLE**** **

Using the same aluminum cable and assuming an average local utility tariff for delivered solar power of 13¢/kWh, what will be the optimal size for this cable? Inserting the values for *T _{c}* ,

*S*and

_{A}*T*into Equation 14, we find:

Consulting *NEC* Chapter 9, Table 8, we find that the closest conductor to this size is 250-kcmil. Therefore, this is the optimally sized aluminum cable for this installation.** **

**ASSESSING VOLTAGE DROP**** **

From the above example, we conclude that we should upsize the aluminum conductors for this installation from the 4/0 AWG size required for ampacity to the optimal economic size of 250-kcmil. The key point to recognize here is that this result is independent of the length of the conductors. Here we demonstrate this surprising result.

First, assume the length of the conductor run is just 50 feet. Our optimal cable of 250-kcmil aluminum has a resistance of 0.100 ohms per 1,000 feet, so at 50 feet and* I _{mp}* of 98.76 A, the voltage drop will be about 0.494 V. Since

*V*for the string is 357.6 V, this is about 0.14% voltage drop.

_{mp}Now instead assume the conductor run is 1,000 feet. Using the same sized aluminum conductor of 250-kcmil, the voltage loss is 9.88 V, or about 2.8% voltage drop. Contrary to our intuition to increase the cable size in the case of the 1,000- foot run, we gain nothing by doing so because the ratio of *D _{A}*

_{}to

*E*in Equation 3 remains about the same. Increasing the conductor size to reduce voltage drop increases its cost (since cost is proportional to cross-sectional area) and decreases its energy losses (which in the denominator are inversely proportional to cross-sectional area). The result is no net gain, and we would have the added expense for the larger cable if we did so.

_{A}
A useful cost-analysis tool is a single value expressed in the present that can represent the changing costs of a value over the full 25-year life of a solar power plant. We designate this single value as γ is calculated based on an initial value of $1.00, we can multiply it by a value representing economic losses or costs in the first year to yield the total cost of that same value over the expected life of the power plant. This greatly simplifies expected-life economic calculations. (Click on table below to enlarge)._{25} |