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Insulation Resistance Calculations of Airfield Lighting Circuits 35 TH ANNUAL HERSHEY CONFERENCE Joseph Vigilante, PE

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

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

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

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Anatomy of insulation current flow CURRENT - MICROAMPERES SECONDS CAPACITANCE CHARGING CURRENT

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Anatomy of insulation current flow CURRENT - MICROAMPERES SECONDS CAPACITANCE CHARGING CURRENT ABSORPTION CURRENT

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Anatomy of insulation current flow CURRENT - MICROAMPERES SECONDS CAPACITANCE CHARGING CURRENT ABSORPTION CURRENT CONDUCTION CURRENT

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Anatomy of insulation current flow CURRENT - MICROAMPERES SECONDS CAPACITANCE CHARGING CURRENT TOTAL CURRENT ABSORPTION CURRENT CONDUCTION CURRENT

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Types of insulation resistance testing Short-time/spot reading 0 TIME 60 sec MEGOHMS VALUE READ AND RECORDED TIME (MONTHS) MEGOHMS VALUES CHARTED

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Types of insulation resistance testing Time-resistance method 0 TIME 10 Min MEGOHMS INSULATION SUSPECT INSULATION PROBABLY OK

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

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

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

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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 RTRT RTRT RNRN RNRN RCRC RCRC I V

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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 M - k-ft. at 15.6°C

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

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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 M

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

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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

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

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Cable equation

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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 = 75,000M/Nc

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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 = 7,500M/Nt

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Base formula RTRT RTRT RNRN RNRN RCRC RCRC I V M¹M¹ IR 1/IR = 1/RC + 1/RN + 1/RT

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

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

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Developing IR calculation formula L-823 Connector Insulation Resistance to Earth Earth Ground Supply L-824 Series Lighting Cable M¹M¹

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Developing IR calculation formula L-823 Connector L-830 Isolation Transformer L-824 Series Lighting Cable Insulation Resistance to Earth Earth Ground M¹M¹

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Developing IR calculation formula L-824 Series Lighting Cable Insulation Resistance to Earth Earth Ground L-823 Connector Return M¹M¹

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Developing IR calculation formula M IRsection 3 IR total IRsection 2 IRsection 1

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Calculation example Vault CCR SECTION 1 SECTION 2 SECTION 3 Handhole

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

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

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Calculation Example Section 2: Cable = 20,500 feet Connectors = 224 Isolation XFMRs = 112

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Calculation Example Vault CCR Handhole Section 3: Cable = 5,000 feet Connectors = 2 Isolation XFMRs = 0

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Calculation Example IR total = 50 M Circuit length is 30,500 ft, so the circuit IR is well over the recommended minimum value.

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

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Modifying a 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: Cable = 5,000 feet Connectors = 2 Isolation XFMRs = 0 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 M IRsection2 = 1/(1/ /35 + 1/174) IRsection2 = 28 MΩ IR = 1/ ( 1/2, /28 + 1/2,795 ) IR = 27.4 MΩ Each circuit modification warrants a reevaluation of the IR for that circuit

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

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

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

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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

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Questions

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