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
S
DC VOLTAGE
SOURCE
C
RL
RA
Anatomy of insulation current flow
100
90
80
CAPACITANCE CHARGING CURRENT
70
60
CURRENT - MICROAMPERES
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
SECONDS
2
2.5 3
4
5
6
7
8 9
10
Anatomy of insulation current flow
100
90
80
CAPACITANCE CHARGING CURRENT
70
60
CURRENT - MICROAMPERES
50
40
30
25
20
15
10
9
8
7
6
5
4
3
2.5
2
ABSORPTION
CURRENT
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
SECONDS
2
2.5 3
4
5
6
7
8 9
10
Anatomy of insulation current flow
100
90
80
CAPACITANCE CHARGING CURRENT
70
60
CURRENT - MICROAMPERES
50
40
30
25
20
15
10
9
8
7
6
5
4
3
2.5
2
ABSORPTION
CURRENT
1.5
CONDUCTION
CURRENT
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
SECONDS
2
2.5 3
4
5
6
7
8 9
10
Anatomy of insulation current flow
100
90
80
CAPACITANCE CHARGING CURRENT
70
60
CURRENT - MICROAMPERES
50
40
TOTAL
CURRENT
30
25
20
15
10
9
8
7
6
5
4
3
2.5
2
ABSORPTION
CURRENT
1.5
CONDUCTION
CURRENT
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
SECONDS
2
2.5 3
4
5
6
7
8 9
10
Types of insulation resistance testing
Short-time/spot reading
MEGOHMS
VALUE READ AND
RECORDED
MEGOHMS
VALUES CHARTED
0
TIME
60 sec
TIME (MONTHS)
Types of insulation resistance testing
Time-resistance method
MEGOHMS
INSULATION
PROBABLY OK
INSULATION
SUSPECT
0
TIME
10 Min
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
I
V
RT
RN
RC
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-96659, 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
I
IR
V
RT
RN
RC
Cable equation
The insulation resistance of cable for a certain length can
be calculated by the following formula:
∗
∗
1000
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
Cable equation
CONDUCTOR
JACKET
INSULATION
D
d
D= 9.18 mm
d= 4.58 mm
OD
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/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
Base formula
1/IR = 1/RC + 1/RN + 1/RT
MΩ⁻¹
I
IR
V
RT
RN
RC
Developing IR calculation formula
Section 1
L-823 Connector
Supply
Return
Section 3
L-824 Series Lighting Cable
L-830 Isolation
Transformer
Section 2
Developing IR calculation formula
Section 1
L-830 Isolation
Transformer
L-823 Connector
Insulation
Resistance to
Earth
Earth Ground
Section 3
L-824 Series Lighting Cable
Section 2
Developing IR calculation formula
L-823 Connector
Supply
L-824 Series Lighting Cable
Insulation
Resistance to
Earth
Earth Ground
MΩ⁻¹
Developing IR calculation formula
1
2
15,098,860
75,000
L-830 Isolation
Transformer
Insulation
Resistance to
Earth
Earth Ground
L-824 Series Lighting Cable
7,500
L-823 Connector
MΩ⁻¹
Developing IR calculation formula
MΩ⁻¹
Insulation
Resistance to
Earth
Earth Ground
L-823 Connector
Return
L-824 Series Lighting Cable
Developing IR calculation formula
MΩ
IR total
IRsection 1
IRsection 2
IRsection 3
Calculation example
Vault
SECTION 1
SECTION 2
SECTION 3
CCR
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
Section 1:
Vault
Cable = 5,000 feet
Connectors = 2
Isolation XFMRs = 0
CCR
Handhole
Handhole
,
,
,
= .0003578 Ω⁻¹
Calculation Example
Section 2:
Cable = 20,500 feet
Connectors = 224
Isolation XFMRs = 112
,
,
,
,
,
= .0192777 Ω⁻¹
Calculation Example
Section 3:
Vault
Cable = 5,000 feet
Connectors = 2
Isolation XFMRs = 0
CCR
Handhole
Handhole
,
,
,
= .0003578 Ω⁻¹
Calculation Example
1
1
5000
15,098,860
2
75,000
.0003578 Ω⁻¹
2
20,500
15,098,860
224
75,000
112
7,500
3
5000
15,098,860
2
75,000
.0003578 Ω⁻¹
1
1
.0192777 Ω⁻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
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:
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 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
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
Total circuit IR = 33.2 MΩ
Circuit Length = 16,430 feet
10k-ft – 20k-ft = 40 MΩ
W/O Transformers; IR = 164.3
Section 2:
Cable = 13,200 feet
Connectors = 365
Isolation XFMRs = 180
Section 3:
Cable = 1,615 feet
Connectors = 5
Isolation XFMRs = 0
Measured value = 58.10 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:
Cable = 1,540 feet
Connectors = 5
Isolation XFMRs = 0
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