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

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


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


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


ABSORPTION CURRENT


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


ABSORPTION CURRENT

CONDUCTION CURRENT


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


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


• 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


• 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


L-823 Connector L-830 Isolation Transformer


Insulation Resistance to Earth

Earth Ground


L-823 Connector


Earth Ground

Supply


1𝐼𝑅 𝑠𝑒𝑐𝑡𝑖𝑜𝑛1

=𝐿

15,098,860+

𝑁𝑐75,000

MΩ⁻¹


L-823 ConnectorL-830 Isolation Transformer



Earth Ground

MΩ⁻¹1

𝐼𝑅 𝑠𝑒𝑐𝑡𝑖𝑜𝑛2=

𝐿15,098,860

+𝑁𝑐75,000

+𝑁𝑡7,500




Earth Ground

L-823 Connector

Return

1𝐼𝑅𝑠𝑒𝑐𝑡𝑖𝑜𝑛3

=𝐿

15,098,860+

𝑁𝑐75,000MΩ⁻¹


MΩ

1𝐼𝑅𝑡𝑜𝑡𝑎𝑙

=1

𝐼𝑅𝑠𝑒𝑐𝑡𝑖𝑜𝑛1+


+1

𝐼𝑅𝑠𝑒𝑐𝑡𝑖𝑜𝑛3

𝐼𝑅𝑡𝑜𝑡𝑎𝑙=1


+1

𝐼𝑅𝑠𝑒𝑐𝑡𝑖𝑜𝑛2+


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


Connectors = 2

Isolation XFMRs = 0

=

Calculation Example

= .


Connectors = 224


Calculation Example

VaultCCR

Handhole

Handhole

=


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.


=20,500

15,098,860+22475,000

+1127,500

=.0192777𝑀Ω⁻ ¹


=5000

15,098,860+

275,000

=.0003578𝑀Ω⁻ ¹


=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


Connectors = 5

Isolation XFMRs = 0


Connectors = 365



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


Connectors = 5

Isolation XFMRs = 0


Connectors = 180



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

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

Documents

insulation resistance

megohms slide

cable equation slide

rt slide

v ir slide

v slide

epr insulation

cable ir values