This document defines various terms related to electrical load including demand, maximum demand, average demand, diversified demand, maximum diversified demand, maximum non-coincident demand, demand factor, utilization factor, load factor, diversity factor, and load diversity. It then provides examples of how to calculate these values for distribution transformers and feeders using load data from customers and diversity factors. Specifically, it shows calculating the non-coincident and maximum diversified demands for three distribution transformers using customer energy consumption data and a load forecasting equation.

1. ECNG 3013
Electrical Transmission and
Distribution Systems
Nature of Loads
Course Instructor: Prof. Chandrabhan Sharma
Email: chandrabhan.sharma@sta.uwi.edu
Phone: Ext. 83141
Office: Rm. 221
A

2. NATURE OF LOADS ON DISTRIBUTION
SYSTEMS
Definitions:
The load is constantly changing.
To describe this changing load,
the following terms are defined.

3. a) Demand:
• Load averaged over a specific time.
• Load can be kW, kVAr, KVA or A.
• Time interval important (usually 15 minutes)
a) Maximum Demand:
• Largest demand over time period.
• Must state demand interval, period and units e.g. 15 mins. max kW demand
for week = 150kW } ( Let’s call this bullet ‘P’)
a) Average Demand :
• Average of the demands over a specific period (day, week, month).
• ‘P’
a) Diversified Demand:
• Sum of demands imposed by a group of loads over a particular period. plus ‘P’
a) Maximum Diversified Demand:
• Max of the sum of the Demands imposed by a group of loads over a particular
period. plus ‘P’

4. f) Maximum Non-coincident Demand:
• For a group of loads, the sum of the individual max. demands without any
restriction that they occur at the same time. plus ‘P’
g) Demand Factor:
• Ratio of maximum Demand to total connected load.
• Utilization Factor:
= Max. Demand / Rated Capacity
i) Load Factor:
= Avg. Demand of any individual (or group) customers
Max. Demand for the same period
j) Diversity Factor:
= Max. non-coincident Demand
Max. diversified Demand
k) Load Diversity:
= (Max. non-coincident Demand) – (Max. diversified Demand)

5. a) Demand
•To define Load, the Demand Curve is broken into equal time intervals (15 min.).
•In each interval, the average demand is determined (red)
•This average value is termed the 15 min. kW Demand.

6. 24 HOUR DEMAND CURVE
(15 min. period)

7. (b) Max. Demand
• For the 24H/15min. Demand curve, each bar
represents the 15 min. Avg. Demand.
• Customer has 3 periods where demand >
6kW.
• The largest of these is the 15-minute max. kW
demand.
• For this customer it is 6.18kW at 13.15

8. (c) Average Demand
During the 24 hour period, energy will be consumed (kWH)
The energy in each 15 min. period is
kWH = (15 min. kW Avg. Demand) x ¼ hr.
∴ Total Energy/Day = ∑(kWH) = 58.96 kWH (say)
24
∴ Average Demand = Total Energy = 58.96
Hours 24
= 2.46kW

9. (d) Load Factor
• Load Factor = Avg. Demand
Max. Demand
L.F. gives an indication of how “well” the
utility’s facilities are being used.

10. For customer given before:
L.F. = Avg. 15 min. kW Demand = 2.46
Max. 15 min. kW Demand 6.18
= 0.4

11. Distribution Transformer Loading
• Distribution Transformers provide service to
several customers.
• Each customers has a unique 24 Hour Demand
Curve.
• Peaks and Valleys as well as max. demands
differ for each customer.
• Max. demands and load factors differ.
• How to determine transformer loading?

12. Allocating Load to Distribution
Transformers
Four methods can be used:
1.Application of Diversity Factors
2.Load Survey
3.Transformer Load Management
4.Metered Feeder Max. Demand

13. Assume 4 Customers Connected to a Transformer
Given: Individual Customer Load Characteristics
Cust # 1 Cust # 2 Cust # 3 Cust # 4
Energy Usage (kWH) 58.57 36.46 95.64 42.75
Max. kW Demand 6.18 6.82 4.93 7.05
Time of Max. kW Demand 13.15 11.30 6.45 20.30
Load Factor 0.40 0.22 0.81 0.25
Avg. kW Demand 2.44 1.52 3.98 1.78
And Given also the 4 (24 hr/15 min) demand curves for the
customers.

14. Diversified Demand
• The sum of the four individual 15 min. kW Demand over the 24
hour period gives the diversified Demand for the transformer.
• Note as you sum the loads, the resulting curve starts to smooth
out.

15. Maximum Diversified Demand
•For the transformer, the 15 min kW demand exceeds
16kW twice.
•The greater of this is the 15 min max. diversified kW
demand of the transformer.
•It is 16.16kW and occurs at 17:30.
•Note that this does not occur at any of the individual
load’s maximum demand and it is not the sum of the max.
demands of each load.

16. Load Duration Curve for a Transformer
• Curve shows that the transformer operates with
15 min kW Demand of 8kW or greater 22% of
time.
• Curve gives an indication whether transformer
needs to be replaced due to overload conditions.

17. Max. Noncoincident Demand
For the transformer, the 15min max. non-
coincident kW demand for the day is sum of
individual customer 15min max. kW Demands.
From table:
∴ Max noncoincident demand = 6.18 + 6.82 + 4.93+7.05
= 24.98kW

18. Diversity Factor
D.F. = (Max. noncoincident Demand)/(Max. Diversified Demand)
= 24.98/16.16 = 1.5458
As the number of customers increases the D.F. increases and then tends
to a constant.

19. Therefore, if one knows the individual max.
demands of customers, then the max. diversified
demand on the transformer can be calculated.
However this presupposes that the D.F. curve
(shown on last slide) is known for the utility.
This can only be obtained from load surveys.

20. DEMAND FACTOR
• This can be defined for an individual
customer. (ONLY)
• Demand Factor = (Max Demand)/(Total
Connected Load)

21. For the customer #1 given before:
≈
∴ 15 min kW Max. Demand = 6.18kW
Total connected load = sum of ratings of all devices
35kW (say)
Demand Factor = 6.18/35 = 0.1766 = 17.7%
∴
Gives an indication of % of electrical devices
(Load) connected when max. demand occurs.

22. UTILISATION FACTOR
• Let the rating of the transformer serving the 4 loads =15kVA.
• Max. diversified demand = 16.16kW
• Let p.f. = 0.9
• Max. 15 min KVA diversified demand = 16.16/0.9 =
17.96kVA
• UF = (Max KVA Demand)/(transformer KVA rating) =
17.96/15= 1.197
• Gives an indication of how well the capacity of the device
(transformer) is being utilized.

23. LOAD DIVERSITY: (L.D.)
∆
L.D. (For Transformer)
(non-coincident max. demand)- (max. diversified Demand)
= 24.98 – 16.16 = 8.82kW

24. FEEDER LOAD
A feeder Load would be “smoother” than a transformer load as
the feeder load would comprise several transformer loads.
The most comprehensive model of a feeder will include every
distribution transformer.
In this case the use of the Diversity Factors become very
important where :
Max. diversified demand
= (Max noncoincident Demand)/DFn
where n = # of customers

25. Load Surveys
Utilities usually perform load surveys to determine the relationship between energy
consumption in kWh and max. kW demand.
At end of survey max demand vs. kWh for each customer is plotted on a common graph.
Linear regression is then used to determine the best fit straight line.
Assuming : From regression analysis the curve found is
Max Demand (kW) = 0.1058 + (0.005014)(kWh)

26. Example #1
A single phase lateral feeder provides service to three distribution transformers as shown.
The energy in kWh consumed by each customer during a month is known. A load survey
has been conducted for customers in this class and it has been found that the customer 15
min max. kW demand is given by the equation
kWdemand max = 0.2 + (0.008)*kWh)
The energy consumed by each customer is given (survey):
1523, 1645, 1984, 1590 and 1456 kWh for Tf # 1
1235, 1587, 1698, 1745, 2015 and 1765 kWh for Tf # 2
2098, 1856, 2058, 2265, 2135, 1985 and 2103 for Tf # 3
1)Determine for each transformer the 15 min noncoincident max. kW demand and using a
table of diversity factors, determine the 15 min max. diversity kW demand. [Given DF 5 =
2.2; DF6 = 2.3; DF7 = 2.4]
2)Determine the 15 min noncoincident max. kW demand and the 15 min max. diversified
kW demand for each of the line segments.
[Given DF13 = 2.74; DF18 = 2.86]

27. Solution: (1) First apply equation kW = 0.2 + (0.008)(kWh) to obtain max. demand for
each load. (Customer)
Transformer # T1
Customer # #1 #2 #3 #4 #5
kWh 1523 1645 1984 1590 1456
kW 12.4 13.4 16.1 12.9 11.9
Transformer # T2
Customer # #6 #7 #8 #9 #10 #11
kWh 1235 1587 1698 1745 2015 1765
kW 10.1 12.9 13.8 14.2 16.3 14.3
Transformer # T3
Customer # #12 #13 #14 #15 #16 #17 #18
kWh 2098 1856 2058 2256 2135 1985 2103
kW 17.0 15.1 16.7 18.3 17.3 16.1 17.0

28. T1: Noncoincident Max. Demand = 12.4 + 13.4 + 16.1 + 12.9 + 11.9
∴ = 66.7 kW
Max. Diversified Demand = (Noncoincident Demand)/ DF5
= 66.7/2.2
∴ = 30.3 kW
T2: Noncoincident Max. Demand = 10.1 + 12.9 + 13.8 + 14.2 + 16.3 + 14.3
= 81.6 kW
∴
Max. Diversified Demand = 81.6/ DF6 = 81.6/2.3
= 35.5 kW
T3: Noncoincident Max. Demand = 17.0 + 15.1 + 16.7 + 18.3 + 17.3 + 16.1 + 17.0
= 117.5 kW
Max. Diversified Demand = 117.5/ DF7 = 117.5/2.4
= 48.9 kW

29. Not Asked in Question
Assuming a power factor of 0.9, the max. kVA diversified demand on each
transformer is:
T1 = 30.3/ 0.9 = 33.6 kVA
T2 = 35.5/ 0.9 = 39.4 kVA
T3 = 48.9/0.9 = 54.3 kVA
Let transformer ratings be 25, 37.5 and 50kVA respectively.
Only T1 would have significant kVA demand
greater than rating (135%)

30. 2) Segment N1 to N2
Max. Noncoincident demand:
= ∑ customers max. demands
18
∴
= 66.7 + 81.6 + 117.5 = 265.5 kW
∴
∴
∴ Given DF18 = 2.86
∴Max. diversified demand = 265.5/ 2.86
= 92.8 kW

31. Segment N2 to N3
This segment sees only T2 & T3 (13 customers)
Max. noncoincident demand = 81.6 + 117.5
= 199.1 kW
DF13 = 2.74 (given)
Max. diversified demand = 199.1/ 2.74 = 72.6 kW

32. Segment N3 to N4
Sees T3 and 7 customers.
Max. Diversified demand = 117.4/ 2.4
= 48.9 kW
• Note: Max. diversified demand of line segments and
transformers do not occur at same time refer to
transformers max diversified demand and compare
to line diversified demand. [ Except for last segment
(N3→N4) and T3]

33. Transformer Load Management
• Used by utilities to determine the loading on
distribution transformers.
• Based on kWh supplied by the transformer
during a peak loading month.
• Need to develop the straight line relationship
between the max. diversified demand of a
distribution transformer to the total kWh
supplied by the transformer for a specific
month. (Assumption → Customers are of
similar type)

34. • Data obtained from load survey.
• In this case, the data is logged for similar
transformers (loads are similar
• From energy supplied ( Σ sum monthly
Bills of all customers connected to T/F),
the max. diversified demand obtained.
• Decision can be taken on whether T/F is
overloaded or needs changing.

35. Metered Feeder Max. Demand
• Usually a table of Diversity Factors is not
available.
• Also for load management, database of
customers to transformers also required. This
is not always available.
• Allocating loads to transformers based on
metered substation (ss) feeder consumption
requires the least amount of data.

36. The KVA ratings of distribution transformers
connected to a load is always known.
From this an “allocation factor ” is determined.
Allocation Factor = AF
∴ = Metered Demand (in S.S.)*
kVATotal (connected)
where kVAT = Σ kVA ratings of every T/F
connected.

37. Load allocation is done by:
Transformer Demand = ( A.F.) (kVA Transformer)
Note: Metered Demand could be either kW, kVA.

38. Example 2:
For the example before, assume metered
demand on feeder is 92.8kW, allocate the
loadings on each transformer based on kVA
rating. (i.e. load for N1→ N2)

39. Solution:
Total kVA = (25 + 37.5 + 50) = 112.5 kVA
∴ A F = 92.8/112.5 = 0.8249 kW/kVA
∴ Allocation of loads to transformer:
T1: kW1 = (0.8249)(25) = 20.62kW
T2: kW2 = (0.8249)(37.5) = 30.93kW
T3: kW3 = (0.8249)(50) = 41.24kW
Compared Before:
Transformer Max. Diversified Using Allocation Factor
Demand kW kW
T1 30.3 20.6
T2 35.5 30.9
T3 48.9 41.2

40. Voltage Drop Calculations Using
Allocated Loads
Assumption →
(1) Loads are constant real/ reactive power
(2) Source voltage = 1.0 p.u.

41. Example #3:
For the example # 1, given that N1 = 2400V calculate the secondary voltages on
T1, T2 and T3 using diversity factors. You are given:
Power Factor = 0.9 lag
Z cable/ line = (0.3 + j0.6) Ω/mile
T1: 25kVA, 2400/240V ; Z = 1.8 /40 %
Ratings of Transformers:
T2: 37.5kVA, 2400/240V ; Z = 1.9 /45 %
T3: 50kVA, 2400/240V ; Z = 2.0 /50 %
The lengths of the line segments are given in feet on the diagram above.

42. Solution Ex #3:
From before, using the diversity factors, the loadings on the transformer and line
were found in KW. Applying the 0.9 p.f. we obtain:
Segment N1-N2 P12 = 92.8 kW → S12 = (92.8 + j45) kVA
Segment N2-N3 P23 = 72.6 kW → S23 = (72.6 + j35.2) kVA
Segment N3-N4 P34 = 48.9 kW → S34 = (48.9 + j23.7) kVA
Transformer T1 PT1 = 30.3 kW → ST1 = (30.3 + j14.7) kVA
Transformer T2 PT2 = 35.5 kW → ST2 = (35.5 + j17.2) kVA
Transformer T3 PT3 = 48.9kW → ST3= (48.9 + j23.7) kVA

43. Converting the transformer p.u. impedances to
ohms referred to high voltage side:
For T1:
Zb = (2.4)2 / 0.025) = 230.4Ω
ZT1 = (0.018 /40 )(230.4)
= (3.18 + j2.67)Ω
Similarly,
T2 = (2.06 + j2.06) Ω
T = (1.48 + j1.77) Ω

44. Line segment impedances:
N1→N2 → Z12 = (0.3 + j0.6)(5000/5280) = (0.2841 + j0.5682)Ω
N2→N3 → Z23 = (0.3 + j0.6)(500/5280) = (0.0284 + j0.0568)Ω
∴
N3→N4 →∴34 = (0.3 + j0.6)(750/5280) = (0.0426 + j0.0852)Ω
Z
∴
Calculating Current in segments (P = VI*)
I = (P/V)*
I12 = [(92.8 + j45.0)/2.4 ∠0]* = 43.0 /-25.84 A
V2 = V1 – I12Z12
= (2400 /0) – (43.0 /-25.84)(0.2841 + j0.5682)
= 2378.4 /-0.4 V

45. Current flowing into T1:
IT1 = [(30.3 + j14.7)/2.378 /-0.4]*
= 14.16 /-26.24 A
Secondary Voltage referred to primary
VT1 = V2 – ZT1 IT1
= 2378.4 /-0.4 – (3.18 + j2.67)(14.16 /-26.24)
= 2321.5 /-0.8 V
Secondary Voltage is: (Turns ratio = 10)

46. Current flowing in Section N2-N3
= [(72.6 + j35.2)/(2.378 /-0.4)]*
I23 = [(kW + jkVAr) / kV ]*
= 33.9 /-26.24 A
∴ Voltage at N3:
= 2378.4 /-0.4 – (0.0284 + j0.0568)(33.9 /-26.24)
V3 = V2 – Z23.I23
= 2376.7 /-0.4 V

47. Current flowing into T2:
IT2 = [(35.5 + j17.2)/ 2.3767 /-0.4]*
= 16.58 /-26.27 A
Primary Voltage = VT2 = V3 – ZT2 IT2
= 2367.7 /-0.4 – (2.06 + j2.06)(16.58 /-26.27)
= 2331.1 /-0.8 V
Secondary Voltage on T2
= 233.11 /-0.8 V

48. Current flowing in line Section N3 - N4:
I34 = [(49.3 +j23.7) / 2.3767/-0.4]*
= 22.9 /-26.27 A
Voltage at N4:
V4 = V3 – Z34 I34
= 2376.7 /-0.4 – (0.0426 + j0.0852)(22.9 /-26.27)
= 2375.0 /-0.5 V
But current into T3 ≡ current in N3 - N4
IT3 = 22.9 /-26.27 A

49. VT3 = V4 – ZT3 IT3
= 2375.0 /-0.5 – (1.48 + j1.77)(22.9 /-26.27)
= 2326.9 /-1.0 V
Secondary Voltage VT3 = 232.7 /-1.0 V
% Voltage drop to transformer T3 :
| V1 | − | VT 3 |  2400 − 2326.90 
Vdrop = =  = 3.08%
| V1 |  2400 

50. Load Allocation Based on Transformer
Ratings
When only the ratings of the distribution
transformers are known, the feeder can be
allocated based upon the metered demand
and the transformer kVA ratings.

51. Example:
For the system given before, assume the voltage
at N1 is 2400V. Derive the secondary voltages
on the three transformers, allocating the loads
based on the transformer ratings.
Assume that the metered kW demand at N1 is
92.8kW.

52. Solution:
Given that the impedances of transformers and line
∴
segments are as before and power factor is 0.9 lag.
KVA demand at N1 from metered demand is given by:
= 103.2 /25.84 kVA
S12 = 92.8/0.9 kVA at cos-1(0.9)
= 103.2 /25.84 / (25 + 37.5 + 50)
Calculation of Allocation Factor
= 0.9175 /25.84

53. Allocating loads to each transformer:
ST1 = (A.F.) (kVAT1)
= (0.9175 /25.84)(25)
= (20.6 + j10.0) kVA
ST2 = (0.9175 /25.84)(37.5)
= (31.0 + j15) kVA
ST3 = (0.9175 /25.84)(50)
= (41.3 + j20) kVA

54. Line power flows are:
S12 = ST1 + ST2 + ST3
= (92.9 + j45) kVA
S23 = ST2 + ST3
= (72.3 + j35) kVA
S34 = ST3
= (41.3 + j20) kVA

55. Using these and the line flows to the transformers as
before we can calculate the voltages on each
transformer:
V2 = 2378.4 /-0.4 V VLT1 = 234.0 /-0.4 V
V3 = 2376.7 /-0.4 V VLT2 = 233.7 /-0.8 V
V4 = 2375.3 /-0.4 V VLT3 = 233.5 /-0.9 V

56. % Voltage Drop:
| V1 | − | VT 3 |  2400 − 2335 
Vdrop = =  = 2.72%
| V1 |  2400 

57. Standard Voltage Ratings
The American National Standards Institute (ANSI)
standard ANSI C84.1-1995 for “Electric Power Systems
and Equipment Voltage Ratings (60 Hertz)” provides
the following definitions for system voltage terms.1
• System Voltage: the root mean square (rms) phasor
voltage of a portion of an alternating-current electric
system that is bounded by transformers or utilization
equipment.
• 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. Each nominal system voltage pertains to a
portion of the system bounded by transformers or
utilization equipment.

58. • Maximum System Voltage: the highest system voltage that occurs under
normal operating conditions, and the highest system voltage for which
equipment and other components are designed for satisfactory continuous
operation without derating of any kind.
• Service Voltage: the voltage at the point where the electrical system of the
supplier and the electrical system of the user are connected.
• Utilization Voltage: the voltage at the line terminals of utilization
equipment.
• Nominal Utilization Voltage: the voltage rating of certain utilization
equipment used on the system.
The ANSI standard specifies two voltage ranges. An oversimplification of the
voltage ranges is
• Range A: Electric supply systems shall be so designated and operated such
that most service voltages will be within the limits specified for Range A. The
occurrence of voltages outside of these limits should be infrequent.
• Range B: Voltages above and below Range A. When these voltages occur,
corrective measures shall be undertaken within a reasonable time to improve
voltages to meet Range A.

59. For a normal three-wire 120/240 volt service to a user, Range A
and Range B voltages are:
• Range A
- Nominal utilization voltage = 115 V
- Maximum utilization and service voltage = 126 V
- Minimum service voltage = 114 V
- Minimum utilization voltage = 110 V
• Range B
- Nominal utilization voltage = 115 V
- Maximum utilization and service voltage = 127V
- Minimum service voltage = 110 V
- Minimum utilization voltage = 107 V

60. These ANSI standards give the distribution engineer a range of
normal steady-state voltages (Range A) and a range of
emergency steady-state voltages (Range B) that must be
supplied to all users.
In addition to the acceptable voltage magnitude ranges, the
ANSI standard recommends that the “electric supply systems
should be designed and operated to limit the maximum voltage
unbalance to 3% when measured at the electric-utility revenue
meter under a no-load condition.” Voltage unbalance is defined
as:
Voltageunbalance = Max. deviation from average voltage . 100%
Average voltage