Voltage Drop Calculation.pdf

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

LIGHTING FIXTURES
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.

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.

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.

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.

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.

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

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%
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

CALCULATION
Approximate Formula for Voltage Drop:
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

CALCULATION
o Formula for resistance of a copper conductor with
reference to cross-sectional area and length of a
conductor:
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

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

CALCULATION
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

CALCULATION
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

SAMPLE OUTPUT:
Lighting Block Diagram (CAD file)

PANEL SCHEDULE MODULE:
OUTPUT IN EXCEL FILE

CABLE SCHEDULE MODULE:

MATERIAL TAKE-OFF

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

Voltage Drop Calculation.pdf

Related slideshows

More Related Content

Voltage Drop Calculation.pdf