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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