Insulation Resistance Calculations of Airfield Lighting Circuits

35TH ANNUAL HERSHEY CONFERENCE

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 VOLTAGESOURCE

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 1.5 2 2.5 3 4 5 6 7 8 9 10

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 1.5 2 2.5 3 4 5 6 7 8 9 10

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 1.5 2 2.5 3 4 5 6 7 8 9 10

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 1.5 2 2.5 3 4 5 6 7 8 9 10

SECONDS

CAPACITANCE CHARGING CURRENT

TOTAL CURRENT

ABSORPTION CURRENT

CONDUCTION CURRENT

Types of insulation resistance testing

Short-time/spot reading

0 TIME 60 sec

ME

GO

HM

S

VALUE READ AND RECORDED

TIME (MONTHS)

ME

GO

HM

S

VALUES CHARTED

Types of insulation resistance testing

Time-resistance method

0 TIME 10 Min

ME

GO

HM

S

INSULATION SUSPECT

INSULATION PROBABLY OK

Testing airfield lighting circuits

• AC 150/5340-30F, 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 150/5340-26B, 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-96-659, Section 7.11.2

– Corresponding to IR Constant 50,000 MΩ - k-ft. at 15.6°C

FAA minimum IR values

• L-823 cable connectors– AC 150/5345-26D, FAA

Specification for L-823 Plug and Receptacle, Cable Connectors

– Section 5.1 – Type I Connectors 75,000 MΩ

FAA minimum IR values

• L-830 isolation transformers– AC 150/5345-47C, 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

VIR

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(𝐷𝑑 )∗( 1000𝐿 )

Cable equation

JACKET

INSULATION

CONDUCTOR

OD

D dD= 9.18 mmd= 4.58 mm

Cable equation

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/NcWhere:• 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/NtWhere:• 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

Base formula

RT RN RC

I

V

1𝐼𝑅

=𝐿

15,098,860+

𝑁𝑐75,000

+𝑁𝑡7,500 MΩ⁻¹

IR

1/IR = 1/RC + 1/RN + 1/RT

Developing IR calculation formula

L-823 Connector L-830 Isolation Transformer

L-824 Series Lighting Cable

Supply

Return

Section 1 Section 2

Section 3

Section 1 Section 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

Developing IR calculation formula

L-823 Connector

Insulation Resistance to Earth

Earth Ground

Supply

L-824 Series Lighting Cable

1𝐼𝑅 𝑠𝑒𝑐𝑡𝑖𝑜𝑛1

=𝐿

15,098,860+

𝑁𝑐75,000

MΩ⁻¹

Developing IR calculation formula

L-823 ConnectorL-830 Isolation Transformer

L-824 Series Lighting Cable

Insulation Resistance to Earth

Earth Ground

MΩ⁻¹1

𝐼𝑅 𝑠𝑒𝑐𝑡𝑖𝑜𝑛2=

𝐿15,098,860

+𝑁𝑐75,000

+𝑁𝑡7,500

Developing IR calculation formula

L-824 Series Lighting Cable

Insulation Resistance to Earth

Earth Ground

L-823 Connector

Return

1𝐼𝑅𝑠𝑒𝑐𝑡𝑖𝑜𝑛3

=𝐿

15,098,860+

𝑁𝑐75,000MΩ⁻¹

Developing IR calculation formula

MΩ

1𝐼𝑅𝑡𝑜𝑡𝑎𝑙

=1

𝐼𝑅𝑠𝑒𝑐𝑡𝑖𝑜𝑛1+

1𝐼𝑅𝑠𝑒𝑐𝑡𝑖𝑜𝑛2

+1

𝐼𝑅𝑠𝑒𝑐𝑡𝑖𝑜𝑛3

𝐼𝑅𝑡𝑜𝑡𝑎𝑙=1

1𝐼𝑅𝑠𝑒𝑐𝑡𝑖𝑜𝑛1

+1

𝐼𝑅𝑠𝑒𝑐𝑡𝑖𝑜𝑛2+

1𝐼𝑅𝑠𝑒𝑐𝑡𝑖𝑜𝑛3

IRsection 3IR total IRsection 2IRsection 1

Calculation example

VaultCCR

SECTION 1SECTION 2SECTION 3

Handhole

Handhole

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

Calculation Example

VaultCCR

Handhole

Handhole

Section 1:Cable = 5,000 feet

Connectors = 2

Isolation XFMRs = 0

=

Calculation Example

= .

Section 2:Cable = 20,500 feet

Connectors = 224

Isolation XFMRs = 112

Calculation Example

VaultCCR

Handhole

Handhole

=

Section 3:Cable = 5,000 feet

Connectors = 2

Isolation XFMRs = 0

Calculation Example

IR total = 50 MΩ

Circuit length is 30,500 ft, so the circuit IR is well over the recommended minimum value.

1𝐼𝑅𝑠𝑒𝑐𝑡𝑖𝑜𝑛2

=20,500

15,098,860+22475,000

+1127,500

=.0192777𝑀Ω⁻ ¹

1𝐼𝑅𝑠𝑒𝑐𝑡𝑖𝑜𝑛1

=5000

15,098,860+

275,000

=.0003578𝑀Ω⁻ ¹

1𝐼𝑅𝑠𝑒𝑐𝑡𝑖𝑜𝑛3

=5000

15,098,860+

275,000

=.0003578𝑀Ω⁻ ¹

Modifying a circuit

VaultCCR

SECTION 1SECTION 2SECTION 3

Handhole

Handhole

Modifying a circuit

Assume:Segment 1:

Cable = 5,000 feetConnectors = 2Isolation XFMRs = 0

Segment 2:Cable = 20,500 feetConnectors = 430Isolation XFMRs = 215

Segment 3:Cable = 5,000 feetConnectors = 2Isolation XFMRs = 0

Total circuit:IRsection1 = 2,795 MΩIRsection3 = 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 MΩRT = 7,500/Nt = 7,500/215 = 35 MΩ

IRsection2 = 1/(1/755 + 1/35 + 1/174)IRsection2 = 28 MΩ

IR = 1/ ( 1/2,795 + 1/28 + 1/2,795 ) IR = 27.4 MΩ

Each circuit modification warrants a reevaluation of the IR for that circuit

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

Total circuit IR = 33.2 MΩCircuit Length = 16,430 feet10k-ft – 20k-ft = 40 MΩW/O Transformers; IR = 164.3

Measured value = 58.10 MΩ

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

Case study example: Measuring IR of a REL Circuit

Total circuit IR = 61.6 MΩCircuit Length = 30,857 feet+ 20k-ft = 30 MΩ

Measured value = 998 MΩ

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

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