Okay, F.Fletcher, let's break down how to determine the proper lead/jumper/bypass wire size for your recloser installation. There's no single magic formula, but a process and some key considerations.

Understanding the Requirements

The wire connecting the recloser terminals to the overhead conductors (the "lead," "jumper," or "bypass" wire as you call it) needs to:1. Carry the load

It must be able to safely conduct the maximum anticipated current during normal operation and during temporary fault conditions.2.

Withstand short-circuit currents

During a fault, the current can be many times higher than normal, and the wire needs to survive long enough for the recloser to trip and clear the fault. It needs to handle the thermal stresses of that high current.3.

Minimize voltage drop

A too-small wire will cause excessive voltage drop, which can affect the performance of downstream equipment.4.

Provide Mechanical Strength

Have adequate tensile strength and resistance to environmental factors (wind, ice, corrosion)5.

Be compatible with connectors

The wire must be appropriately sized for the connectors being used on the recloser terminals and the overhead conductor.6.

Meet applicable codes and standards

NEC, local utility requirements, and other relevant codes must be followed.

Steps to Determine Wire Size

Here's a systematic approach:1. Determine Maximum Continuous Load Current

You mentioned a 200-amp system. This is your starting point. However, consider future growth. Will the load be increased later? Design for the maximum possible load that the system will see, not just the present load. Let's assume you stick with 200A for this example.2.

Calculate Minimum Ampacity based on Continuous Load

NEC Article 310.15 states that conductors shall be protected against overcurrent in accordance with their ampacities specified in Sec. 310.16 through 310.
21. Multiply your maximum continuous load current (200A) by a safety factor (typically 1.25) as required by the NEC for continuous loads. A continuous load is defined as a load that runs for 3 hours or more. If your load is not continuous, this factor may not apply. Consult the NEC or a qualified electrical engineer. 200A 1.25 = 250A (Minimum ampacity requirement)3.

Determine Short-Circuit Current and Duration

This is crucial and often requires a power system study. Your utility or an electrical engineering consultant will need to perform a short-circuit study to determine the maximum symmetrical fault current (Isc) at the recloser location and the clearing time (how long it takes the recloser to interrupt the fault). The recloser's time-current curve (TCC) is critical here. The short-circuit current will depend on the system impedance (source impedance, transformer impedance, conductor impedance, etc.). The closer the recloser is to the source, the higher the short-circuit current will generally be. Let's assume for this example that the short-circuit current (Isc) is 10,000 amps and the clearing time is 0.1 seconds (6 cycles). This is just an example; you MUST get a proper study done.4.

Select a Conductor Material and Insulation

Common materials are copper and aluminum. Copper has higher conductivity (smaller size for the same ampacity) but is more expensive. Aluminum is lighter and less costly. Given this is a high-reliability application, copper is often preferred, but consider the economics. Insulation must be rated for the voltage (15.5 kV) and the environment (temperature, exposure to sunlight, moisture, etc.). Cross-linked polyethylene (XLPE) is a common choice for medium-voltage cables.
5.

Determine Minimum Ampacity based on NEC Table 310.16 Use NEC Table 310.16 (Ampacities of Insulated Conductors Rated Up to and Including 2000 Volts) to find the conductor size required to carry the adjusted current from step 2 (250A). Remember to apply any applicable correction factors for ambient temperature or the number of conductors in a raceway (although these conductors are likely to be run in free air). For example, using the 75°C column in NEC Table 310.16, a copper conductor with THHN/THWN-2 insulation would need to be at least 250 kcmil.Remember to choose the temperature rating of the connector you use, If the connector only supports 75°C then you must use the 75°C rating.6. Check Short-Circuit Withstand Capability

This is where things get more complex. You need to verify that the conductor you selected in Step 5 can withstand the short-circuit current for the clearing time determined in Step 3. Use the following formula to check this

Calculate required conductor size (circular mils)

CM = (Isc sqrt(t)) / K CM = Required conductor size in circular mils Isc = Symmetrical short-circuit current in amps t = Clearing time in seconds K = Constant based on conductor material and insulation: Copper, Thermoplastic Insulation: K = 30 Copper, Thermosetting Insulation: K = 20 Aluminum, Thermoplastic Insulation: K = 12 Aluminum, Thermosetting Insulation: K = 8

Example Calculation

Using the values from our examples above and assuming Copper with Thermosetting Insulation. CM = (10,000A sqrt(0.1s)) / 20 CM = (10,000 0.316) / 20 CM = 158 Based on this calculation, the minimum circular mil is 158. Since we needed 250kcmil (250,000CM) for our ampacity, we are already over our value.
7.

Check Voltage Drop

Calculate the voltage drop under normal load conditions for the proposed conductor size. Voltage drop should be minimized to maintain proper voltage levels at downstream loads. Use a voltage drop calculator or consult engineering tables. A typical rule of thumb is to limit voltage drop to 3% or less on feeders. Since this is likely a short distance, voltage drop is unlikely to be a limiting factor, but always check.8.

Consider Mechanical Factors and Connector Compatibility

The conductor must have sufficient tensile strength to withstand the mechanical stresses of installation and operation (wind loading, ice loading, etc.). Consult with the utility's mechanical standards for overhead line construction. Make sure the conductor size is compatible with the connectors used on the recloser terminals and the overhead conductor. Use properly rated and installed connectors.9.

Consult with the Utility and G&W Electric

This is critical. Your utility likely has specific standards for recloser installations, including approved conductor types and sizes. Consult with them and follow their guidelines. Contact G&W Electric directly to confirm any specific requirements or recommendations for conductor sizes and connector types for their Viper-S recloser.

Example using assumptions Load

200A continuous

Voltage

15.5kV

Short Circuit Current

10,000A

Clearing time

0.1 seconds

Conductor Material

Copper

Insulation

XLPE (Thermosetting)

NEC Table 310.16 Ampacity Requirement

250kcmil based on our assumptions

Important Notes and Cautions



Short-Circuit Study is Essential

The assumed short-circuit current is just an example. You must have a qualified engineer perform a power system study to determine the actual short-circuit current at the recloser location.

Coordinate with the Utility

The utility's standards and practices are paramount. They have the final say on what is acceptable.

Safety First

Working with medium-voltage equipment is dangerous. Only qualified personnel should perform the installation and maintenance.

Documentation

Keep thorough records of all calculations, conductor sizes, connector types, and utility approvals.

Professional Engineering

If you are not a qualified electrical engineer with experience in power system protection, hire one. This is not a task for amateurs.

In Summary

Determining the correct lead/jumper/bypass wire size for a recloser installation requires a careful analysis of load current, short-circuit current, voltage drop, mechanical strength, and applicable codes and standards. A short-circuit study is essential, and you must coordinate closely with the utility and G&W Electric. While I've provided a step-by-step process, consulting with qualified professionals is strongly recommended to ensure a safe and reliable installation. Flag for review