# Insulation Resistance Calculations of Airfield Lighting Circuits

## Presentation on theme: "Insulation Resistance Calculations of Airfield Lighting Circuits"— Presentation transcript:

Insulation Resistance Calculations of Airfield Lighting Circuits
35th annual Hershey conference Insulation Resistance Calculations of Airfield Lighting Circuits Joseph Vigilante, PE

Presentation Objective
To develop a formula to calculate insulation resistance for an airfield lighting circuit and provide theoretical and real-life examples.

Presentation Agenda Cable insulation resistance (IR) background
FAA IR guideline recommendations Formula development Airfield lighting circuit calculations Summary Open Q&A

Anatomy of insulation current flow
Capacitance charging currents, C Absorption current, RA Conduction current, RL Circuit model DC VOLTAGE SOURCE RA C RL S

Anatomy of insulation current flow
CURRENT - MICROAMPERES 100 90 80 70 60 50 40 30 25 20 15 10 9 8 7 6 5 4 3 2.5 2 1.5 1 0.1 0.15 0.2 0.25 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0 SECONDS CAPACITANCE CHARGING CURRENT

Anatomy of insulation current flow
CURRENT - MICROAMPERES 100 90 80 70 60 50 40 30 25 20 15 10 9 8 7 6 5 4 3 2.5 2 1.5 1 0.1 0.15 0.2 0.25 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0 SECONDS CAPACITANCE CHARGING CURRENT ABSORPTION CURRENT

Anatomy of insulation current flow
CURRENT - MICROAMPERES 100 90 80 70 60 50 40 30 25 20 15 10 9 8 7 6 5 4 3 2.5 2 1.5 1 0.1 0.15 0.2 0.25 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0 SECONDS CAPACITANCE CHARGING CURRENT ABSORPTION CURRENT CONDUCTION CURRENT

Anatomy of insulation current flow
CURRENT - MICROAMPERES 100 90 80 70 60 50 40 30 25 20 15 10 9 8 7 6 5 4 3 2.5 2 1.5 1 0.1 0.15 0.2 0.25 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0 SECONDS CAPACITANCE CHARGING CURRENT TOTAL CURRENT ABSORPTION CURRENT CONDUCTION CURRENT

Types of insulation resistance testing
Short-time/spot reading TIME 60 sec MEGOHMS VALUE READ AND RECORDED TIME (MONTHS) MEGOHMS VALUES CHARTED

Types of insulation resistance testing
Time-resistance method TIME 10 Min MEGOHMS INSULATION SUSPECT INSULATION PROBABLY OK

Testing airfield lighting circuits
AC 150/ F, Design & Installation Details for Airport Visual Aids - Chapter 12, Equipment & Material, Section 13, Testing 50MΩ Non-grounded series circuits FAA-C-1391, Installation & Splicing of Underground Cable

Resistance values for maintenance
AC 150/ B, Maintenance of Airport Visual Aid Facilities Suggested minimum values 10,000 ft. or less - 50MΩ 10,000 ft. – 20,000 ft. - 40MΩ 20,000+ ft. - 30MΩ

FAA-C-1391 Installation & Splicing of Underground Cables
Spot Test Take readings no less than 1 minute after readings stabilized Cable IR values = 50MΩ, 40MΩ & 30MΩ Loop IR reduced due to parallel summation Cable IR never less than above values

Insulation resistance formula
Constant current series lighting circuit Provides for parallel summation of circuit components 3 components L-824 cable L-823 cable connectors L-830 isolation transformers RT RN RC I V

FAA minimum IR values L-824 cable
AC 150/5345-7E, Specification for L-824 Underground Electrical Cable for Airport Lighting Circuits Table 1, Test #9 > ICEA S , Section Corresponding to IR Constant 50,000 MΩ - k-ft. at 15.6°C

FAA minimum IR values L-823 cable connectors
AC 150/ D, FAA Specification for L Plug and Receptacle, Cable Connectors Section 5.1 – Type I Connectors 75,000 MΩ

FAA minimum IR values L-830 isolation transformers
AC 150/ C, Specification for Series to Series Isolation Transformers for Airport Lighting Systems Table 3 Insulation Resistance 7,500 MΩ

Base formula 1/IR = 1/RC + 1/RN + 1/RT RT RN RC I V IR

Cable equation The insulation resistance of cable for a certain length can be calculated by the following formula: Where: RC = Insulation resistance in Megohms of cable K = Specific IR in Megohms - k ft at 60˚ F of insulation D = Outer diameter of insulation d = Outer diameter of bare copper wire L = Length of airfield cable in feet Value of K for EPR insulation = 50,000 Megohms 𝑅𝐶=𝐾 ∗𝐿𝑜𝑔 𝐷 𝑑 ∗ 𝐿

Cable equation D= 9.18 mm d= 4.58 mm D d CONDUCTOR JACKET INSULATION
OD

Cable equation 𝑅𝐶=𝐾 ∗𝐿𝑜𝑔 𝐷 𝑑 ∗ 1000 𝐿
𝑅𝐶=𝐾 ∗𝐿𝑜𝑔 𝐷 𝑑 ∗ 𝐿 RC = 50,000 MΩ-k ft * Log ( 9.18/4.58) * (1,000/L) RC = 15,098,860 MΩ / L

Connector equation The insulation resistance of L-823 connector splices can be calculated by the following formula: RN = Rc/Nc Where: RN = Insulation resistance in Megohms of all connectors Rc = Insulation resistance of L-823 connector splice Nc = Quantity of L-823 connector splices RN = 75,000MΩ/Nc

Isolation transformer equation
The insulation resistance of L-830 isolation transformers can be calculated by the following formula: RT = Rt/Nt Where: RT = Insulation resistance in Megohms of all transformers Rt = Insulation resistance of L-830 isolation transformers Nt = Quantity of L-830 isolation transformers RT = 7,500MΩ/Nt

IR Base formula 1 𝐼𝑅 = 𝐿 15,098,860 + 𝑁𝑐 75,000 + 𝑁𝑡 7,500 MΩ⁻¹
1/IR = 1/RC + 1/RN + 1/RT 1 𝐼𝑅 = 𝐿 15,098, 𝑁𝑐 75,000 + 𝑁𝑡 7,500 MΩ⁻¹ RT RN RC I V IR

Developing IR calculation formula
L-823 Connector L-830 Isolation Transformer L-824 Series Lighting Cable Supply Return Section 1 Section 2 Section 3

Developing IR calculation formula
L-823 Connector L-830 Isolation Transformer L-824 Series Lighting Cable Section 1 Section 2 Insulation Resistance to Earth Earth Ground Section 3

Developing IR calculation formula
L-823 Connector Supply L-824 Series Lighting Cable Insulation Resistance to Earth Earth Ground 1 𝐼𝑅𝑠𝑒𝑐𝑡𝑖𝑜𝑛 1 = 𝐿 15,098, 𝑁𝑐 75,000 MΩ⁻¹

Developing IR calculation formula
1 𝐼𝑅𝑠𝑒𝑐𝑡𝑖𝑜𝑛 2 = 𝐿 15,098, 𝑁𝑐 75,000 + 𝑁𝑡 7,500 MΩ⁻¹ L-823 Connector L-830 Isolation Transformer L-824 Series Lighting Cable Insulation Resistance to Earth Earth Ground

Developing IR calculation formula
1 𝐼𝑅𝑠𝑒𝑐𝑡𝑖𝑜𝑛 3 = 𝐿 15,098, 𝑁𝑐 75,000 MΩ⁻¹ Earth Ground Insulation Resistance to Earth L-823 Connector Return L-824 Series Lighting Cable

Developing IR calculation formula
1 𝐼𝑅𝑡𝑜𝑡𝑎𝑙 = 1 𝐼𝑅𝑠𝑒𝑐𝑡𝑖𝑜𝑛 𝐼𝑅𝑠𝑒𝑐𝑡𝑖𝑜𝑛 𝐼𝑅𝑠𝑒𝑐𝑡𝑖𝑜𝑛3 𝐼𝑅𝑡𝑜𝑡𝑎𝑙= 1 1 𝐼𝑅𝑠𝑒𝑐𝑡𝑖𝑜𝑛 𝐼𝑅𝑠𝑒𝑐𝑡𝑖𝑜𝑛 𝐼𝑅𝑠𝑒𝑐𝑡𝑖𝑜𝑛 3 MΩ IR total IRsection 1 IRsection 2 IRsection 3

Calculation example Vault SECTION 1 SECTION 2 SECTION 3 CCR Handhole

Calculation example Cable = 5,000 feet Section 1: Section 2:
Connectors = 2 Isolation XFMRs = 0 Section 2: Cable = 20,500 feet Connectors = 224 Isolation XFMRs = 112 Section 3:

Calculation Example Section 1: Cable = 5,000 feet Connectors = 2 Isolation XFMRs = 0 Vault CCR Handhole Handhole 1 𝐼𝑅𝑠𝑒𝑐𝑡𝑖𝑜𝑛 1 = ,098, ,000 = 𝑀Ω⁻¹

Calculation Example Section 2: Cable = 20,500 feet Connectors = 224 Isolation XFMRs = 112 1 𝐼𝑅𝑠𝑒𝑐𝑡𝑖𝑜𝑛 2 = 20,500 15,098, , ,500 = 𝑀Ω⁻¹

Calculation Example Section 3: Cable = 5,000 feet Connectors = 2 Isolation XFMRs = 0 Vault CCR Handhole Handhole 1 𝐼𝑅𝑠𝑒𝑐𝑡𝑖𝑜𝑛 3 = ,098, ,000 = 𝑀Ω⁻¹

Calculation Example 1 𝐼𝑅𝑠𝑒𝑐𝑡𝑖𝑜𝑛 1 = ,098, ,000 = 𝑀Ω⁻¹ 1 𝐼𝑅𝑠𝑒𝑐𝑡𝑖𝑜𝑛 2 = 20,500 15,098, , ,500 = 𝑀Ω⁻¹ 1 𝐼𝑅𝑠𝑒𝑐𝑡𝑖𝑜𝑛 3 = ,098, ,000 = 𝑀Ω⁻¹ IR total = 50 MΩ Circuit length is 30,500 ft, so the circuit IR is well over the recommended minimum value.

Modifying a circuit Vault SECTION 1 SECTION 2 SECTION 3 CCR Handhole

Modifying a circuit Each circuit modification warrants a reevaluation of the IR for that circuit Assume: Segment 1: Cable = 5,000 feet Connectors = 2 Isolation XFMRs = 0 Segment 2: Cable = 20,500 feet Connectors = 430 Isolation XFMRs = 215 Segment 3: Total circuit: IRsection1 = 2,795 MΩ IRsection3 = 2,795 MΩ IRsection2: RC = 50,000 * Log ( 9.18/4.58) * (1,000/L) = 15,098,860/20,500 = 755 MΩ RN = 75,000/Nc = 75,000/430 = 174 MΩ RT = 7,500/Nt = 7,500/215 = 35 MΩ IRsection2 = 1/(1/ /35 + 1/174) IRsection2 = 28 MΩ IR = 1/ ( 1/2, /28 + 1/2,795 ) IR = 27.4 MΩ

Modifying a circuit Total circuit IR = 27.4 MΩ
Circuit Length = 30,500 feet Recommended value for +20k feet circuit = 30 MΩ If you remove the transformer component, the circuit IR = 128 MΩ

Case study example: Measuring IR of a TDZ Circuit
Section 1: Cable = 1,615 feet Connectors = 5 Isolation XFMRs = 0 Section 2: Cable = 13,200 feet Connectors = 365 Isolation XFMRs = 180 Section 3: Total circuit IR = 33.2 MΩ Circuit Length = 16,430 feet 10k-ft – 20k-ft = 40 MΩ W/O Transformers; IR = 164.3 Measured value = MΩ

Case study example: Measuring IR of a REL Circuit
Section 1: Cable = 1,540 feet Connectors = 5 Isolation XFMRs = 0 Section 2: Cable = 27,777 feet Connectors = 180 Isolation XFMRs = 90 Section 3: Total circuit IR = 61.6 MΩ Circuit Length = 30,857 feet + 20k-ft = 30 MΩ Measured value = 998 MΩ

Summary Good engineering practice to perform circuit load and insulation resistance calculations Best practice – establish field test baseline and track results Standards provide recommended values - reductions are allowed Check & verify transformers, connectors and cable types and sizes Minimum allowable component values – higher factory test values High initial field value does not necessarily indicate a good circuit

Questions