Insulation Resistance Calculations of Airfield Lighting Circuits 35 TH ANNUAL HERSHEY CONFERENCE...
-
Upload
haylee-garber -
Category
Documents
-
view
256 -
download
3
Transcript of Insulation Resistance Calculations of Airfield Lighting Circuits 35 TH ANNUAL HERSHEY CONFERENCE...
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