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VOLTAGE DROP CALCULATION FOR
LIGHTING AND CONVENIENCE SOCKET
CIRCUITS
By: V. IDOS, REE
JGC Philippines, Inc.
43RD IIEE ANNUAL NATIONAL CONVENTION
SMX CONVENTION CENTER, PASAY CITY, METRO MANILA
CONTENTS
 Effect of Voltage Variation
 Definition of Terms
 Philippine Electrical Code Provisions
 Formula for Voltage Drop Calculation
 Sample Calculations
 VPCM Calculation Software
 Q&A
EFFECTS OF VOLTAGE VARIATION
INDUCTION MOTORS
Reference: IEEE Std. 141-1993
EFFECTS OF VOLTAGE VARIATION
LIGHTING FIXTURES
Reference: IEEE Std. 141-1993
FLUORESCENT LAMPS Light output is directly
proportional to the applied
voltage for magnetic ballast.
Light output for electronic
ballast may be more or less
dependent on input voltage.
High Intensity Discharge
(HID) Lamps
Mercury lamps with typical
reactor ballast will have a
12% change in light output
for a 5% change in voltage
input
HID lamps may extinguish if
input voltage drops below
75% of rated voltage.
EFFECTS OF VOLTAGE VARIATION
RESISTANCE HEATING DEVICES
o Heat output of resistance heaters
varies approximately as the square
of the impressed voltage.
o A 10% drop in voltage will cause a
drop of approximately 19% in heat
output.
Reference: IEEE Std. 141-1993
EFFECTS OF VOLTAGE VARIATION
CAPACITORS
o The reactive power of the capacitor
varies as the square of the
impressed voltage.
o A 10% drop in voltage will cause a
drop of approximately 19% in
reactive power output.
Reference: IEEE Std. 141-1993
TO SUMMARIZE:
o The voltage impressed in an
equipment has an effect on the
performance of a device or
equipment.
o Excessive voltage drop can cause
heating of equipment. Thus,
degradation of insulation is
possible.
Reference: IEEE Std. 141-1993
DEFINITION OF TERMS
o System Voltage- the root-mean-
square phase-to-phase voltage of a
portion of an ac electric system.
o Nominal System Voltage – the
voltage by which a portion of the
system is designated and to which
certain operating characteristics
of the system are related.
Reference: IEEE Std. 141-1993
DEFINITION OF TERMS
o Voltage Drop- is the amount of
voltage loss that occurs through
all or part of the circuit due to
impedance.
% VD = VD/VN x 100%
Where:
VD – Voltage drop (volts)
VN – nominal system voltage (volts)
SOURCE: Novec
 According to PEC (Article 2.15.1.2 (A)
FPN No. 2)
“Conductors for feeders as defined in Article 1.1
sized to prevent a voltage drop exceeding 3
percent at the farthest outlet of power, heating
and lighting loads, or combinations of such loads,
and where the maximum total voltage drop on
both feeders and branch circuit to the farthest
outlet does not exceed 5 percent, provide
reasonable efficiency of operation.”
PHILIPPINE ELECTRICAL CODE
PROVISIONS
PHILIPPINE ELECTRICAL CODE
PROVISIONS
 According to PEC (Article 2.10.2.2 (A)
FPN No. 4)
“Conductors for branch circuits as defined in
Article 1.1 sized to prevent a voltage drop
exceeding 3 percent at the farthest outlet of
power, heating and lighting loads, or combinations
of such loads, and where the maximum total
voltage drop on both feeders and branch circuit
to the farthest outlet does not exceed 5 percent,
provide reasonable efficiency of operation.”
SOURCE
Dist. Panel
Farthest
Load
FEEDER
BRANCH
CIRCUIT
% VD(FDR) ≤ 3%
% VD (BRANCH) ≤ 3%
% VD(FDR) +
% VD (BRANCH) ≤ 5%
PHILIPPINE ELECTRICAL CODE
PROVISIONS
[Art. 2.15.1.2 (A)]
[Art. 2.10.2.2 (A)]
[Art. 2.10.2.2 (A)
& 2.15.1.2 (A)]
FORMULA FOR VOLTAGE DROP
CALCULATION
Phasor Diagram of Voltage Relations
For Voltage Drop Calculations
Reference: IEEE Std. 141-1993
FORMULA FOR VOLTAGE DROP
CALCULATION
Approximate Formula for Voltage Drop:
Reference: IEEE Std. 141-1993
VD = KI (R cos ф + X sin ф )
Where:
VD – Voltage drop (volts)
K – multiplying constant (2 for single phase, 𝟑 𝐟𝐨𝐫 𝐭𝐡𝐫𝐞𝐞 𝐩𝐡𝐚𝐬𝐞)
I – current flowing in conductor (amperes)
R – line resistance of conductor (ohms)
X – line reactance of conductor (ohms)
ф – angle whose cosine is the load power factor
cos ф – load power factor in decimals
sin ф – load reactive factor in decimals
Equation 1
FORMULA FOR VOLTAGE DROP
CALCULATION
o Formula for resistance of a copper conductor with
reference to cross-sectional area and length of a
conductor:
Reference: IEEE Std. 141-1993
R = ρL/ A
Where:
R – Resistance (ohms)
ρ – material resistivity (ohm-m)
For copper ρCu@20C = 1.7241 x 10-8 ohm-m @ 20°C
For aluminum ρAl@20C = 2.65 x 10-8 ohm-m @ 20°C
L – length of the conductor (m)
A – cross-sectional area of the conductor (m2)
Equation 2
FORMULA FOR VOLTAGE DROP
CALCULATION
o Substituting equation 3 to equation 2, the resistance of a
conductor with reference to ambient temperature,
length and cross-sectional area is:
o For correction of resistivity according to ambient
temperature
ρT-AMB = ρ20°C [ 1 + α (TAMB – 20°C)] Equation 3
R = ρ20°C [ 1 + α (TAMB – 20°C)]L/A
Where:
α –temperature coefficient of resistivity (/°C)
αCu = 3.93 x 10-3 /°C for copper
TAMB – ambient temperature (°C)
Equation 4
FORMULA FOR VOLTAGE DROP
CALCULATION
Reference: IEEE Std. 141-1993
o For lighting and small power, the power factor is high
(more than 0.90). Hence, we can consider a pf = 1.0, and
ф = 0 degrees.
o Solving voltage drop for single-phase lighting loads
using equations 1 and 4:
VD = KI (R cos ф + X sin ф )
VD = 2I ρ20°C [ 1 + α (TAMB – 20°C)]L/A
2 0
1
ρ20°C [ 1 + α (TAMB – 20°C)]L/A
Eq. 5
FORMULA FOR VOLTAGE DROP
CALCULATION
Reference: IEEE Std. 141-1993
o Solving single phase voltage drop for using copper
conductor, and ambient temperature of 40°C:
VD = 2I ρCu@20°C [ 1 + αCu (TAMB – 20°C)]L/A
1.7241 x 10-8 ohm-m
Equation 6
3.93 x 10-3 /°C
40 °C
VD = 3.719x10-8 I L/A
SAMPLE CALCULATION NO. 1
Given:
System Voltage, VN = 230V, 1 phase
Allowable Voltage Drop (%), %VD = 3%
Allowable Voltage Drop (V), VD = 6.9 V
Problem:
Two- 250W flood lights will be installed 300m from
the lighting panel. What will be the minimum
copper conductor size required to meet the
allowable voltage drop at ambient temperature of
40°C and 1.30A load for each flood light?
Solution:
Lighting
Panel
(230V,1ph)
FLOOD LIGHT
2 x (250W,1.3Amp)
L=300m
A = ? for
VD ≤ 6.9V
%VD ≤ 3%
USING EQUATION 6:
I = 2 x 1.3 Amp = 2.6 Amp
A = 3.719x10-8 I L/ VD
SUBSTITUTING:
A = 3.719x10-8 (2.6 Amp) (300m)/ 6.9V
A = 4.2x10-6 m2
CONVERT TO mm2:
A = 4.2x10-6 m2 x (1000mm/1m)2
A = 4.2 mm2
USE NEXT SIZE, SAY 6mm2
CALCULATE % VD USING 6 mm2:
% VD = [3.719x10-8 I L/A]/VN x 100%
% VD = [3.719x10-8 (2.6Amp) (300m)/(6 x10-6m2)]/230V x 100%
% VD = 2.10 %
% VD = [3.719x10-8 I L/A]/VN x 100%
SAMPLE CALCULATION NO. 2
Problem:
Two- additional 250W flood lights, 50m apart, will be
connected to the junction box of the flood lights in
Sample Calculation No. 1. The first flood light is
200m from the junction box. What will be the
minimum copper conductor size required to meet the
allowable voltage drop assuming all conductor sizes
are same?
Solution:
Lighting
Panel
(230V,1ph)
2 x 1.3Amp
L=300m
VD ≤ 6.9V
%VD ≤ 3%
VDTOTAL ≤ 6.9 V
L=200m
L=50m
1.3Amp
1.3Amp
1
2
3
VDTOTAL = VD1 + VD2 +VD3
VD1 = 3.719x10-8 (2.6+1.3+1.3 Amp)(300m)/A1
VD1 = 5.80164x10-5 /A1
VD2 = 3.719x10-8 (1.3+1.3 Amp)(200m)/A2
VD2 = 1.93388x10-5 /A2
VD3 = 3.719x10-8 (1.3 Amp)(50m)/A3
VD3 = 2.41735x10-6 /A2
A= A1 = A2 = A3
VDTOTAL = (5.80164x10-5 /A) + (1.93388x10-5 /A)
+ (2.41735x10-6 /A)
A = 1.156 x 10-5 m2 (1000mm/1m)2
VDTOTAL = (5.80164x10-5 /A) + (1.93388x10-5 /A) + (2.41735x10-6 /A)
A = 1.156 x 10-5 m2 (1000mm/1m)2
A = 11.56 mm2
USE NEXT SIZE, SAY 16mm2
VDTOTAL = (5.80164x10-5 + 1.93388x10-5 + 2.41735x10-6)(1/16mm2)(1000mm/1m)2
VDTOTAL = 4.99 V
% VDTOTAL = 4.99 V/230 V x 100%
% VDTOTAL = 2.17 %
Imagine what it is like
when doing voltage
drop calculation for
multiple circuits and
hundreds of lighting
fixtures?
Photo credit to: Vectorstock.com
IN JGC, WE DEVELOP OUR IN-HOUSE
SOFTWARE:
V- Voltage Drop Calculator
P- Panel Schedule
C- Cable Schedule
M- Material Take-Off (MTO)
Lighting VPCM
VPCM (JGC’S IN-HOUSE SOFTWARE)
FILL UP THE SETTINGS
• System
Information
• Criteria
• Set up load
information
• Cable
information
• Cable gland
information
• Panel Board
Information
• Location
information
SELECT THE MODULE
Cable
Load
Load
Current
10 Fixtures
SAMPLE OUTPUT:
Lighting Block Diagram (CAD file)
PANEL SCHEDULE MODULE:
PANEL SCHEDULE MODULE:
OUTPUT IN EXCEL FILE
CABLE SCHEDULE MODULE:
CABLE SCHEDULE MODULE:
OUTPUT IN EXCEL FILE
MATERIAL TAKE-OFF
OUTPUT IN EXCEL FILE
ADVANTAGES OF THE VPCM:
• Easy voltage drop calculation
• Easy development of lighting block
diagram
• Easy production of panel board
schedule
• Create your cable schedule
• Optimize cable size (reduce cost)
• Can automatically select size of your
cable
• Reduce your man-hour cost in design
QUESTIONS AND ANSWERS
THANK YOU VERY MUCH!

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Voltage Drop Calculation.pdf

  • 1. VOLTAGE DROP CALCULATION FOR LIGHTING AND CONVENIENCE SOCKET CIRCUITS By: V. IDOS, REE JGC Philippines, Inc. 43RD IIEE ANNUAL NATIONAL CONVENTION SMX CONVENTION CENTER, PASAY CITY, METRO MANILA
  • 2. CONTENTS  Effect of Voltage Variation  Definition of Terms  Philippine Electrical Code Provisions  Formula for Voltage Drop Calculation  Sample Calculations  VPCM Calculation Software  Q&A
  • 3. EFFECTS OF VOLTAGE VARIATION INDUCTION MOTORS Reference: IEEE Std. 141-1993
  • 4. EFFECTS OF VOLTAGE VARIATION LIGHTING FIXTURES Reference: IEEE Std. 141-1993 FLUORESCENT LAMPS Light output is directly proportional to the applied voltage for magnetic ballast. Light output for electronic ballast may be more or less dependent on input voltage. High Intensity Discharge (HID) Lamps Mercury lamps with typical reactor ballast will have a 12% change in light output for a 5% change in voltage input HID lamps may extinguish if input voltage drops below 75% of rated voltage.
  • 5. EFFECTS OF VOLTAGE VARIATION RESISTANCE HEATING DEVICES o Heat output of resistance heaters varies approximately as the square of the impressed voltage. o A 10% drop in voltage will cause a drop of approximately 19% in heat output. Reference: IEEE Std. 141-1993
  • 6. EFFECTS OF VOLTAGE VARIATION CAPACITORS o The reactive power of the capacitor varies as the square of the impressed voltage. o A 10% drop in voltage will cause a drop of approximately 19% in reactive power output. Reference: IEEE Std. 141-1993
  • 7. TO SUMMARIZE: o The voltage impressed in an equipment has an effect on the performance of a device or equipment. o Excessive voltage drop can cause heating of equipment. Thus, degradation of insulation is possible. Reference: IEEE Std. 141-1993
  • 8. DEFINITION OF TERMS o System Voltage- the root-mean- square phase-to-phase voltage of a portion of an ac electric system. o Nominal System Voltage – the voltage by which a portion of the system is designated and to which certain operating characteristics of the system are related. Reference: IEEE Std. 141-1993
  • 9. DEFINITION OF TERMS o Voltage Drop- is the amount of voltage loss that occurs through all or part of the circuit due to impedance. % VD = VD/VN x 100% Where: VD – Voltage drop (volts) VN – nominal system voltage (volts) SOURCE: Novec
  • 10.  According to PEC (Article 2.15.1.2 (A) FPN No. 2) “Conductors for feeders as defined in Article 1.1 sized to prevent a voltage drop exceeding 3 percent at the farthest outlet of power, heating and lighting loads, or combinations of such loads, and where the maximum total voltage drop on both feeders and branch circuit to the farthest outlet does not exceed 5 percent, provide reasonable efficiency of operation.” PHILIPPINE ELECTRICAL CODE PROVISIONS
  • 11. PHILIPPINE ELECTRICAL CODE PROVISIONS  According to PEC (Article 2.10.2.2 (A) FPN No. 4) “Conductors for branch circuits as defined in Article 1.1 sized to prevent a voltage drop exceeding 3 percent at the farthest outlet of power, heating and lighting loads, or combinations of such loads, and where the maximum total voltage drop on both feeders and branch circuit to the farthest outlet does not exceed 5 percent, provide reasonable efficiency of operation.”
  • 12. SOURCE Dist. Panel Farthest Load FEEDER BRANCH CIRCUIT % VD(FDR) ≤ 3% % VD (BRANCH) ≤ 3% % VD(FDR) + % VD (BRANCH) ≤ 5% PHILIPPINE ELECTRICAL CODE PROVISIONS [Art. 2.15.1.2 (A)] [Art. 2.10.2.2 (A)] [Art. 2.10.2.2 (A) & 2.15.1.2 (A)]
  • 13. FORMULA FOR VOLTAGE DROP CALCULATION Phasor Diagram of Voltage Relations For Voltage Drop Calculations Reference: IEEE Std. 141-1993
  • 14. FORMULA FOR VOLTAGE DROP CALCULATION Approximate Formula for Voltage Drop: Reference: IEEE Std. 141-1993 VD = KI (R cos ф + X sin ф ) Where: VD – Voltage drop (volts) K – multiplying constant (2 for single phase, 𝟑 𝐟𝐨𝐫 𝐭𝐡𝐫𝐞𝐞 𝐩𝐡𝐚𝐬𝐞) I – current flowing in conductor (amperes) R – line resistance of conductor (ohms) X – line reactance of conductor (ohms) ф – angle whose cosine is the load power factor cos ф – load power factor in decimals sin ф – load reactive factor in decimals Equation 1
  • 15. FORMULA FOR VOLTAGE DROP CALCULATION o Formula for resistance of a copper conductor with reference to cross-sectional area and length of a conductor: Reference: IEEE Std. 141-1993 R = ρL/ A Where: R – Resistance (ohms) ρ – material resistivity (ohm-m) For copper ρCu@20C = 1.7241 x 10-8 ohm-m @ 20°C For aluminum ρAl@20C = 2.65 x 10-8 ohm-m @ 20°C L – length of the conductor (m) A – cross-sectional area of the conductor (m2) Equation 2
  • 16. FORMULA FOR VOLTAGE DROP CALCULATION o Substituting equation 3 to equation 2, the resistance of a conductor with reference to ambient temperature, length and cross-sectional area is: o For correction of resistivity according to ambient temperature ρT-AMB = ρ20°C [ 1 + α (TAMB – 20°C)] Equation 3 R = ρ20°C [ 1 + α (TAMB – 20°C)]L/A Where: α –temperature coefficient of resistivity (/°C) αCu = 3.93 x 10-3 /°C for copper TAMB – ambient temperature (°C) Equation 4
  • 17. FORMULA FOR VOLTAGE DROP CALCULATION Reference: IEEE Std. 141-1993 o For lighting and small power, the power factor is high (more than 0.90). Hence, we can consider a pf = 1.0, and ф = 0 degrees. o Solving voltage drop for single-phase lighting loads using equations 1 and 4: VD = KI (R cos ф + X sin ф ) VD = 2I ρ20°C [ 1 + α (TAMB – 20°C)]L/A 2 0 1 ρ20°C [ 1 + α (TAMB – 20°C)]L/A Eq. 5
  • 18. FORMULA FOR VOLTAGE DROP CALCULATION Reference: IEEE Std. 141-1993 o Solving single phase voltage drop for using copper conductor, and ambient temperature of 40°C: VD = 2I ρCu@20°C [ 1 + αCu (TAMB – 20°C)]L/A 1.7241 x 10-8 ohm-m Equation 6 3.93 x 10-3 /°C 40 °C VD = 3.719x10-8 I L/A
  • 19. SAMPLE CALCULATION NO. 1 Given: System Voltage, VN = 230V, 1 phase Allowable Voltage Drop (%), %VD = 3% Allowable Voltage Drop (V), VD = 6.9 V Problem: Two- 250W flood lights will be installed 300m from the lighting panel. What will be the minimum copper conductor size required to meet the allowable voltage drop at ambient temperature of 40°C and 1.30A load for each flood light?
  • 20. Solution: Lighting Panel (230V,1ph) FLOOD LIGHT 2 x (250W,1.3Amp) L=300m A = ? for VD ≤ 6.9V %VD ≤ 3% USING EQUATION 6: I = 2 x 1.3 Amp = 2.6 Amp A = 3.719x10-8 I L/ VD SUBSTITUTING: A = 3.719x10-8 (2.6 Amp) (300m)/ 6.9V A = 4.2x10-6 m2 CONVERT TO mm2: A = 4.2x10-6 m2 x (1000mm/1m)2 A = 4.2 mm2 USE NEXT SIZE, SAY 6mm2 CALCULATE % VD USING 6 mm2: % VD = [3.719x10-8 I L/A]/VN x 100% % VD = [3.719x10-8 (2.6Amp) (300m)/(6 x10-6m2)]/230V x 100% % VD = 2.10 % % VD = [3.719x10-8 I L/A]/VN x 100%
  • 21. SAMPLE CALCULATION NO. 2 Problem: Two- additional 250W flood lights, 50m apart, will be connected to the junction box of the flood lights in Sample Calculation No. 1. The first flood light is 200m from the junction box. What will be the minimum copper conductor size required to meet the allowable voltage drop assuming all conductor sizes are same?
  • 22. Solution: Lighting Panel (230V,1ph) 2 x 1.3Amp L=300m VD ≤ 6.9V %VD ≤ 3% VDTOTAL ≤ 6.9 V L=200m L=50m 1.3Amp 1.3Amp 1 2 3 VDTOTAL = VD1 + VD2 +VD3 VD1 = 3.719x10-8 (2.6+1.3+1.3 Amp)(300m)/A1 VD1 = 5.80164x10-5 /A1 VD2 = 3.719x10-8 (1.3+1.3 Amp)(200m)/A2 VD2 = 1.93388x10-5 /A2 VD3 = 3.719x10-8 (1.3 Amp)(50m)/A3 VD3 = 2.41735x10-6 /A2 A= A1 = A2 = A3 VDTOTAL = (5.80164x10-5 /A) + (1.93388x10-5 /A) + (2.41735x10-6 /A) A = 1.156 x 10-5 m2 (1000mm/1m)2
  • 23. VDTOTAL = (5.80164x10-5 /A) + (1.93388x10-5 /A) + (2.41735x10-6 /A) A = 1.156 x 10-5 m2 (1000mm/1m)2 A = 11.56 mm2 USE NEXT SIZE, SAY 16mm2 VDTOTAL = (5.80164x10-5 + 1.93388x10-5 + 2.41735x10-6)(1/16mm2)(1000mm/1m)2 VDTOTAL = 4.99 V % VDTOTAL = 4.99 V/230 V x 100% % VDTOTAL = 2.17 %
  • 24. Imagine what it is like when doing voltage drop calculation for multiple circuits and hundreds of lighting fixtures? Photo credit to: Vectorstock.com
  • 25. IN JGC, WE DEVELOP OUR IN-HOUSE SOFTWARE: V- Voltage Drop Calculator P- Panel Schedule C- Cable Schedule M- Material Take-Off (MTO) Lighting VPCM
  • 27. FILL UP THE SETTINGS • System Information • Criteria • Set up load information • Cable information • Cable gland information • Panel Board Information • Location information
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  • 33. SAMPLE OUTPUT: Lighting Block Diagram (CAD file)
  • 39. ADVANTAGES OF THE VPCM: • Easy voltage drop calculation • Easy development of lighting block diagram • Easy production of panel board schedule • Create your cable schedule • Optimize cable size (reduce cost) • Can automatically select size of your cable • Reduce your man-hour cost in design
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  • 44. THANK YOU VERY MUCH!