Distribution System Engineering
(8024510-3)
Dr. Hani A. Aldhubaib
hadhubaib@uqu.edu.sa
Lecture 8
Voltage Drop Calculations
2
Introduction
▪ One of the main goals of electric utilities is to deliver electric power to their
customer within acceptable voltage range.
▪ According to ANSI C84.1* standard, electric supply systems should be designed
and operated to limit the maximum voltage unbalance to 3% when measured at
the electric utility revenuer meter under a no-load condition.
▪ What is voltage drop VD?
The difference between the sending-end and the receiving-end voltage of the line.
*ANSI C84.1 is the American National Standard for Electric Power Systems and Equipment – Voltage Ratings.
3
Voltage Levels Standard
ANSI C84.1 Standard defines two voltage ranges:
▪ Range A (normal steady-state voltages):
Electric systems should be designed to operate within the limits specified by this
range. Operation at voltage levels outside these limits should be uncommon.
▪ Range B (emergency voltages):
This range includes voltage levels above and below those specified by range A.
Operation within voltage limits in this range should be corrected and improved
within minimum time to meet those specified by range A.
4
Voltage Drop Calculations
Part 1: Voltage-drop calculations for allocated loads
5
Voltage Drop Calculation Methods
Two general methods are used to calculate the voltage drop along a feeder:
A. Approximate method.
B. Kdrop Factor method.
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Voltage Drop Calculation Methods
A. Approximate method
VD = VS − VL  Re ( Z  I ) , where Z is the total impedance in Ω.
VD
%VD =
100
VL − N
7
Voltage Drop Calculation Methods
Example 8-1:
A single-phase lateral provides service to three distribution transformers. Using the approximate
method, calculate the voltage drop between N1 and N2, given that:
the impedance of the line is z = 0.3+j0.6 Ω/mi, the current flowing through the line segment is
I12 = 43.0093 − 25.8419 A, and the voltage at node N1 is V1= 24000 V.
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Voltage Drop Calculation Methods
Example 8-1:
VD  Re ( Z  I )
= Re[(0.282 + j 0.57)  (43.0093 − 25.8419)]
= 21.6 V
z = 0.3+j0.6 Ω/mi;
V1= 24000 V.
I12 = 43.0093 − 25.8419 A;
1 mile = 5280 feet
5000 ft =
VD
21.6
%VD =
100 =
100
VL − N
2400
= 0.9%
5000
= 0.95 mi
5280
VD = VS − VL  Re ( Z  I )
Z12 = (0.3 + j 0.6)  0.95 = 0.282 + j 0.57 Ω
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Voltage Drop Calculation Methods
B. Kdrop Factor method
▪ The voltage drop can be calculated using the concept of Kdrop factor.
▪ The Kdrop can be determined in two ways:
o Analytical evaluation of Kdrop factor.
o Using K constant curves.
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Voltage Drop Calculation Methods
B. Kdrop Factor method:
Analytical evaluation of Kdrop factor.
▪ The Kdrop can be determined by computing the percent voltage drop down a
line that is one mile long and serving a balanced three phase load of 1 kVA.
VD = Re ( z  I ) , where z is the impedance in Ω/mile.
K drop
VD
=
100
VL − N
%VD = K drop  kVA  l , where l is the length of the feeder in mile.
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Voltage Drop Calculation Methods
B. Kdrop Factor method:
K constant curves.
▪ The Kdrop can be determined using the K
constant curve. Most standard conductors
have K constant curves that determine the
Kdrop factor.
The K constant curves for copper
conductors, assuming a lagging load power
factor of 0.9 is shown in this slide.
%VD = K drop  kVA  l
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Voltage Drop Calculation Methods
Example 8-2:
A distribution feeder has an impedance of z = 0.306+j0.6272 Ω/mi. Calculate the %VD
using the approximate method and the Kdrop factor method; given that a load power factor
of 0.9 lagging, 7500 kVA load, and line to line voltage of 12.47 kV are assumed. The total
length of the feeder is 1.5 miles.
7500 kVA
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Voltage Drop Calculation Methods
Example 8-2:
 I = 347.24 − 25.84 A
7500 kVA Z = (0.306 + j 0.6272)  1.5 = 0.459 + j 0.9408 Ω
VD  Re ( Z  I )
z = 0.306+j0.6272 Ω/mi; pf = 0.9 lag;
S = 7500 kVA; VL-L =12.47 kV;
l = 1.5 miles;
θ = cos-1(0.9)=25.84̊
Approximate method:
VD = VS − VL  Re ( Z  I )
S
7500
I=
=
= 347.24 A
3 VL − L
3 12.47
= Re[(0.459 + j 0.9408)  (347.24 − 25.84)]
= 285.83V
%VD =
VD
285.83
100 =
100
VL − N
12470 / 3
= 3.97%
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Voltage Drop Calculation Methods
Example 8-2:
 I = 0.0463 − 25.84 A
7500 kVA
VD = Re ( z  I )
= Re[(0.306 + j 0.6272)  (0.0463 − 25.84)] Ω
z = 0.306+j0.6272 Ω/mi; pf = 0.9 lag;
S = 7500 kVA; VL-L =12.47 kV;
l = 1.5 miles;
θ = cos-1(0.9)=25.84̊
Kdrop Factor method:
VD = Re ( z  I )
S
1
I=
=
= 0.0463 A
3 VL − L
3 12.47
= 0.025408 V
K drop
VD
0.025408
=
100 =
100 = 0.00035291
VL − N
12470 / 3
%VD = K drop  kVA  l
= 0.00035291 7500 1.5 = 3.97%
15
Voltage Drop Calculation Methods
Example 8-3:
Calculate the %VD of a three-phase #4 copper conductor feeder feeding a 500 kVA
load at 0.9 power factor lagging. The line-to-line voltage is 4.16 kV, and the length of
the feeder is 1.2 miles.
500 kVA
%VD = K drop  kVA  l
16
%VD = K drop  kVA  l
%VD = 0.01 500 1.2 = 6%
17
Voltage Drop Calculation Methods
Example 8-4:
Determine the percent voltage drop from N0 to N3; given that The Kdrop factor for the
line segments is 0.00035291 %/(kVA-mi2).
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Voltage Drop Calculation Methods
Example 8-4:
Total kVA from N1 to N2:
kVA1-2: 750+500 = 1250 kVA.
%VD1-2= 0.00035291×1250×0.75 = 0.33%.
Kdrop = 0.00035291
%VD = K drop  kVA  l
Total kVA from N2 to N3:
kVA2-3: 500 kVA.
%VD2-3= 0.00035291×500×0.5 = 0.0882%.
Total kVA from N0 to N1:
kVA0-1: 300+750+500 = 1550 kVA.
%VD0-3= 0.821+0.33+0.0882 = 1.2392%.
%VD0-1= 0.00035291×1550×1.5 = 0.821%.
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