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 C

2. Lumping Loads in Geometric
Configurations
• Another method for approximate calculations
of voltage drop and power loss for an area.
• Helps in the determination of max. load that
can be served in a specific area at a given
voltage level and conductor size.
• Area can be represented as either:
i. Rectangle
ii. Triangle or
iii. Trapezoid.

3. Rectangular Representation
Let: D = load Diversity (kVA/ mile2)
PF = assumed lagging power factor
Z = line impedance in Ω/ mile
l = length of area
w = width of area
kVLL = line to line voltage
Assume all loads are modeled as constant
current loads.

4. Constant load Diversity Rectangular area.
• Area l x w served by a primary feeder n → m.
• Area assumed to have a constant load density with 3Φ laterals
uniformly tapped off the primary main.
Q? → Determine the total voltage Drop and
total 3Φ power loss from n → m?

5. IT = D. l. w ∠ -cos-1 (p.f.)
Total Current Entering area = IT
… (1)
√3 kVLL
For an incremental section x miles from node n, the incremental current (di)
serving this incremental section is given by:
di = IT/ l A/mile … (2)
∴ Current is incremental segment (dx) is given by
i = IT – (x.di)
∴ = IT – (x.IT/l) = IT (1 –x/l) … (3)
Voltage drop in the incremental segment is
dV = Re(Z.i.dx)
∴ = Re[ Z.IT (1 – x/l). dx] … (4)
l l
Total voltagedV = Re[ ZIthe(1 − x )dx ]
drop down feeder:
V
drop ∫
=
0
T ∫
0
l
∴
Vdrop = Re[ Z.IT. ½ .l] = Re[ ½ . Z. IT] … (5)
where Z= Zl
*** Note that this equation (5) is the same as that obtained before for uniformly
distributed along a feeder. The only difference being the manner in which the
total current (I ) was determined.

6. ∴ The total load of a rectangular area can be modeled at the
centroid of a rectangle as given below:
The voltage drop computed to the load point (l/2) will
represent the total voltage drop from node n to node m.

7. 3Φ Power Loss
The power loss per unit length dx from a point x
dp = 3.|i|2.r.dx
= 3[ |IT|2. (1-x/l)2.r.dx]
= 3.r.IT2 [ 1-2x/l + x2/l2 ]dx … (6)
∴ total 3Φ power loss down the primary main is
l l
x x2
Ploss = ∫ dp = 3rI ∫(1 −2 + 2 ) dx
2
T
l l
0 0 … (7)
= 3 [1/3 R. IT2] … (8)
where R = r.l
This is similar to what was obtained before.

8. Example:
It is proposed to serve a rectangular area of dimension
10,000 ft by 6000 ft wide. The load density of the area is
2500 kVA/ mile2 at p.f. 0.9 lag. The primary feeder uses
336, 400 26/7 ACSR on a T pole structure given before.
a)What is the min. standard nominal voltage level that
can be used to serve this area without exceeding a
voltage drop of 3%? The choices of voltage levels are
4.16kV and 12.47kV.
b)What is the expected total 3Φ power loss.

9. From tables:
z = 0.306 + j0.6272 Ω/ mile.
Dimension of area = (10,000 x 6000) ft
= (1.8939 x 1.1364) miles
Area = 2.1522 miles2
∴The total Load of the area = D.A kVA
= (2500)(2.1522) = 5380.6 kVA
Total impedance of line = z.l = (0.306 + j0.6272)(1.8939)
= (0.5795 + j1.1897) Ω
IT = 5380.6 kVA ∠ -cos-1 (0.9) = 746.7∠ -25.84 A
For Voltage = 4.16kV
√3 x 4.16
∴ Total voltage drop along the primary main:
= Re [ ½ (0.5795 + j1.1879)(746.7 ∠ -25.84)]
Vdrop = Re[ ½ . Z . IT]
= 388.1 V

10. VLN = 4160/√3 = 2401.8 V
∴% Voltage drop = Vdrop = 388.1 = 16.16%
VLN 2401.8
∴Nominal voltage of 4160 does not meet criteria of V < 3%.
drop
∠-cos-1 (0.9)
Nominal Voltage = 12.47 kV
IT = kVA = 5380.6
= 249.1 ∠ -25.84 A
√3 VLL √3 x 12.47
∴
= Re [ ½ (0.5795 + j1.1879)(249.1 ∠-25.84) ]
Vdrop = Re [ ½. Z. I]
= 129.5 V
∴ VLN = 7199.6 V
But
∴Vdrop% = 129.5/ 7199.6 = 1.8%
1 .R.I T2 1 x0.5795 x(249.1) 2
12.47 kV is adequate to serve the load.
Ploss3 Φ = 3[ 3 ]kW = 3[ 3 ]
1000 1000

11. What was accomplished:
1)Development of formulas and techniques for
computing Vdrop and Ploss for line segments with
uniformly distributed loads.
2)As above but for geometric area (rectangle) with
constant load densities.
3)Techniques useful for making quick, approximate
calculations.
4)Most times all that is needed is a “ballpark” value for
giving a guide as to how the feeder/ area is
performing.