This document provides information about a course on electrical transmission and distribution systems. It discusses various topics that will be covered in the course, including approximate models for analyzing voltage drop and line impedance on distribution lines. It also discusses 'K' factors that can be used to calculate voltage drop and rise percentages. Other topics covered include uniformly distributed loads on distribution laterals and calculating the total power loss on a distribution line. The document provides examples and equations for calculating various parameters related to distribution system analysis.

1. ECNG 3013
Electrical Transmission and
Distribution Systems
Course Instructor: Prof. Chandrabhan Sharma
Email: chandrabhan.sharma@sta.uwi.edu
Phone: Ext. 3141
Office: Rm. 221 B

2. Approximate Models for Analysis
Assumption: Balanced 3Φ transposed system.
1.Voltage Drop:
(ZI)
VS = VL + (R + jX)I
= VL + RI + jXI … (1)

3. Figure below gives the phasor representation
The voltage drop down the line is defined as the difference
between the magnitudes of VS and VL
∴drop = |VS| - |VL| ... (2)
V
Since δ is small for a distribution line.
 V ≈ Re(ZI) … (3)
drop
Equation (3) would be used as a definition for voltage drop.

4. Example: For the feeder given before, the impedance of the first line segment is:
Also I12 = 43.0093 ∠-25.8419 A
Z12 = (0.2841 + j0.5682) Ω
VN1 = 2400 ∠0 V
V2 = 2400 ∠0 – (0.2841 + j0.5682)(43.0093 ∠-25.8419)
The exact voltage at N2 is:
∠-0.4015
∴V = 2378.40982378.4098 V 21.5902 V
drop (exact) = 2400 – =
Vdrop = Re[(0.2841 + j0.5682)(43.0093 ∠-25.8419)]
Computing Vdrop according to Eq. 3
∴ = 21.64926 V
Error = (21.5902 -21.6426) 100
21.5902
= -0.27% (small)

5. Line Impedance → Approximate Model
Assumption → Line fully transposed → +ve
sequence only needs to be determined.
Given: the equation for +ve Sequence
Z1 = r + j0.12134ln(Deq/GMR) Ω/mile …(4)
where r = conductor resistance/ mile
3
Deq = equivalent spacing = Dab.Dbc.Dca (ft)
GMR = conductor geometric mean radius
(from tables)

6. A 3Φ line segment has a T configuration as given before. The spacings between
the conductors are:
Dab = 2.5ft ; Dbc = 4.5 ft and Dca = 7.0 ft
The conductors are ACSR with r = 0.306Ω/ mile and GMR = 0.0244ft (obtained
from tables)
Find the Z1 for the line.
Solution:
Equivalent Spacing = (2.5)(4.5)(7) = 4.2868 ft
3
From equation (4)
Z1 = 0.306 + j0.12134ln(4.2863/0.0244) Ω/mile
= 0.306 + j0.6272 Ω/mile
‘K’ Factors
Another method of obtaining voltage drop is by use of a ‘k’ factor. There are two
types of k factors:
(a) one for voltage drop and
(b) another for voltage rise calculations.

7. Kdrop Factor ∆ % Phase Voltage drop/ kVA.mile …(5)
(km)
This gives the % voltage drop per mile (km) of conductor with a 1kVA load. A
p.f. of 0.9 is assumed!
Example:
For the T line given before, calculate the K drop factor assuming p.f. = 0.9lag
and nominal LL voltage is 12.47kV.
Solution:
Previously Z = (0.306 + j0.6272) Ω/mile
I = 1kVA/ (√3.kVLL) ∠–cos-1 0.9
For load 1kVA at 0.9 pf.
= 1/ (√3 x 12.47) ∠-25.84 = 0.046299 ∠-25.84 A
∴ = Re[ZI] = Re{[0.306 + j0.6272][0.046299 ∠-25.84]}
Vdrop
= 0.025408 V
VLN = 12.47/ √3 = 7199.6 V
Kdrop factor = (0.025408/7199.6)100
(km)
Kdrop = 0.00035291 % / kVA.mile

8. Most utilities have standard conductors, spacings and voltages.
Therefore a simple spreadsheet would be developed to
compute the Kdrop factors for the utility.
How do we use Kdrop?
Assume for the configuration given before it is required to find
the voltage drop if a load of 7500kVA is fed 1.5 miles away?
Vdrop = Kdrop. kVA. Mile
= (0.00035291)(7500)(1.5) = 3.9702%
Suppose the utility’s standard for voltage drop is 3% max. Then
the max. load which can be fed is given by:
kVAload = 3.0%/(0.00035291 x 1.5) = 5667.2 kVA

9. Given Kdrop = 0.00035291. Determine the percentage voltage drop from N0 to
N3.
Solution: Total kVA flowing in segment N0 to N1:
kVA01 = (300 + 750 + 500) = 1550 kVA
% voltage drop from N0 to N1 = (0.0003591)(1550)(1.5) = 0.8205%
∴ 12 = 750 + 500 = 1250 kVA
kVA
Voltage drop from N1 to N2:
∴ Vdrop12 = (0.00035291)(1250)(0.75) = 0.3308%
Vdrop23 = (0.00035291)(500)(0.5) = 0.0882%
Total % voltage drop from N0 to N3
∴ Vdroptotal = (0.8205 + 0.3308 + 0.0882) = 1.2396%
(approximate value)

10. Krise Factor
In this case the load is a shunt capacitor.
N .B. | When a leading current flows through an inductive reactance (Z line) there
is a voltage rise across the reactance NOT a drop!
|
Observe the phasor diagram below:
∴ rise = |Re(ZIcap)| = X|Icap|
V … (6)
∴Krise = Voltage rise % … (7)
kVAr . mile
where the load is a 1 kVAr capacitor

11. For the example before:
(1)Calculate the Krise factor
(2) Determine the rating of a 3Φ capacitor bank to limit the voltage drop to
2.5% in the last example.
Solution:
Z/mile = (0.306 + j0.6272)Ω
Icap = 1kVAr ∠90
The current taken by a 1kVAr bank:
∠90 = 0.046299 ∠90
√3 kVLL
= 1
√3 x 12.47
∴Voltage rise per kVAr mile is:
= |Re(0.306 + j0.6272)(0.046299 ∠90 )| = 0.029037 V
Vrise = Re|Z.Icap|
But VLN = 7199.6V
∴
Krise factor = (0.029037/7199.6)100 = 4.0331 x 10 -4 % rise/ kVAr.mile

12. From the last example the % voltage drop for a 7500kVA load at
1.5 mile was 3.9702%.
To limit this to 2.5%, the required voltage rise need is
Vrise = 3.9702 – 2.5 = 1.4702%
∴Required rating of shunt capacitor is
kVAr = Vrise = 1.4702
Krise . mile (4.0331x10-4)(1.5)
= 2430.18 kVAr

13. Uniformly Distributed Loads
Distribution laterals with transformers of the
same rating can be assessed as a uniformly
distributed load.
Generalized Line with n uniformly distributed loads.

14. Voltage Drop
Given n uniformly spaced loads dx apart.
The loads are treated as constant current loads di.
Let L = length of feeder
Z = r + jx = impedance of line in Ω / mile
dx = length of each section
di = load at each node
n = # of nodes & no. of line sections
IT = total current into the feeder

15. ∴ Load currents are given by: di = IT /n … (8)
Voltage drop of first line segment is given by:
Vdrop1 = Re[ Z.dx.(n.di)] … (9)
Similarly,
Vdrop2 = Re[ Z.dx.{(n-1).di}] etc … (10)

16. ∴ Voltage drop is given by:
Total
Vdroptotal = Vdrop1 + Vdrop2 + … + Vdropn
= Re { Z.dx.di.[n + (n-1) + (n-2) + … + 1] } … (11)
Equation (11) can be reduced by using the expansion:
n
[ ∑ m]
m =1 1 + 2 + 3 + … + n = n(n+1)/2 … (12)
Sub into (11)
→ Vdroptotal = Re{ Z.dx.di [ n(n+1)/2] } … (13)
But dx = L/ n … (14)
and di = IT/ n … (15)
Sub (14) and (15) into (13) yields

17. VdropT = Re[ Z. L/n. IT /n (n(n+1)/2) ]
= Re{ Z. L. IT. ½. (n+1)/n}
VdropT = Re{ ½ . Z. IT (1 + 1/n)} … (16)
where Z T = Z.L
This is the general equation for an evenly
distributed load.
As n→∞ then
Vdrop total = Re[ ½ . Z T . IT] …

18. The load (distributed) can be represented using either of the two models:
Model # 1
Load at
midpoint
OR
Model # 2
½ load at last
node
where Vdrop total = Re [ ½ ZT. IT]

19. Power Loss
Using the line model shown previously:
Model # 1 gives the 3Φ Power Loss as:
Ploss = 3.|IT|2 R/2 = 3/2|IT|2.R … (18)
where R = total resistance of line
Model # 2 gives the 3Φ Power Loss as:
Ploss = 3.|IT/2|2.R = ¾.|IT|2.R … (19)
Which one is correct? → Neither!

20. Correct Model
Recall before:
L
Power loss to node 1 => Ploss1 = 3 (r.dx).|n.di|2 … (20)
Similarly Ploss2 = 3 (r.dx).[|(n-1).di|]2 … (21)
Etc.
∴P
loss Total = 3(r.dx). |di| [n + (n-1) + (n-3) + … + 1 ] … (22)
2 2 2 2 2
where n2 + (n-1)2 + (n-2)2 + …+ 12 = n (n+1) (2n+1) … (23)
6
∴ loss Total = 3 ( r . L/n )(|IT/n|)2 [ n (n+1) (2n+1) ]
P … (24)
6

21. Simplifying (24)
Ploss Tot = 3. R. |IT|2 [ (n+1) (2n+1) ]
6n2
Ploss Tot = 3R|IT|2 [ 2n2 + 3n + 1 ]
6n2
Ploss Tot = 3R|IT|2 [ 1 + 1 + 1 ] … (25)
3 2n 6n2
where R = r . L = total resistance / phase
As n → ∞ → Ploss total = 3R|IT|2[ 1/3 ]
= 3. R/3 |IT|2 … (26)
Model # 1
for Power
Loss/ phase
Model # 2
for Power
Loss/ phase

22. The Exact Lumped Load Model
Need to Find ‘k’ and ‘c’
Vdrop Total = Re[ kZIT + (1-k).Z.c.IT] … (27)
where Z = Total line impedance in ohms
k = as given above
c = fraction of current at end of line s.t.
IT = IX + cIT
But (from before) Vdrop Total = Re[ ½. Z. IT ] … (28)
Equating (27) and (28)
Re [ ½ ZIT ] = Re [ kZIT + (1 –k)ZcIT ] … (29)
Or ½ ZIT = kZIT + (1-k) ZcIT … (30)

23. Dividing across by ZIT gives
½ = k + c( 1-k ) … (31)
Solving for k:
 0. 5 − c  … (32)
k = 
 1− c 
Applying the procedure to the power loss model yields:
Ploss Total = 3[ kR|IT|2 + (1-k)R(c.|IT|)2 ]… (33)
But before Ploss Total = 3[ 1/3 R |IT|2 ] … (34)
Equating (33) and (34)
3[ 1/3R|IT|2] = 3[kR|IT|2 + (1-k)R (c|IT|2) ]
1/3 = [ k (1-c2) + c2 ] … (35)
Solving equation (32) and (35) for k and c
Sub. for k from (32) into (35)
 0.5 − c 
 
 1− c 
1/3 = [ ( 1 – c 2 ) + c2 ] … (36)
Solving for c → c = 1/3 … (37)
Sub into (32) → k = ¼ … (38)

24. The exact model for the distribution line is: