2 The Nature of Loads

The modeling and analysis of a power system depend upon the “load”. What is load?
The answer to that question depends upon what type of analysis is desired. For exam-
ple, the steady-state analysis (power-flow study) of an interconnected transmission
system will require a different definition of load than that used in the analysis of a
secondary in a distribution feeder. The problem is that the “load” on a power system
is constantly changing. The closer you are to the customer, the more pronounced will
be the ever-changing load. There is no such thing as a “steady-state” load. To come
to grips with load, it is first necessary to look at the “load” of an individual customer.

2.1 DEFINITIONS
The load that an individual customer or a group of customers presents to the distribu-
tion system is constantly changing. Every time a light bulb or an electrical appliance
is switched on or off the load seen by the distribution feeder changes. To describe the
changing load, the following terms are defined:

1. Demand
• Load averaged over a specific period
• Load can be kW, kvar, kVA, A
• Must include the time interval
• Example: The 15-minute kW demand is 100 kW
2. Maximum Demand
• Greatest of all demands which occur over a specific period
• Must include demand interval, period, and units
• Example: The 15-minute maximum kW demand for the week was
150 kW
3. Average Demand
• The average of the demands over a specified period (day, week, month,
etc.)
• Must include demand interval, period, and units
• Example: The 15-minute average kW demand for the month was 350 kW
4. Diversified Demand
• Sum of demands imposed by a group of loads over a particular period
• Must include demand interval, period, and units
• Example: The 15-minute diversified kW demand in the period ending at
9:30 was 200 kW

DOI: 10.1201/9781003261094-2 9
10 Distribution System Modeling and Analysis

5. Maximum Diversified Demand

• Maximum of the sum of the demands imposed by a group of loads over
a particular period
• Must include demand interval, period, and units
• Example: The 15-minute maximum diversified kW demand for the week
was 500 kW
6. Maximum Non-coincident Demand
• For a group of loads, the sum of the individual maximum demands with-
out any restriction that they occur at the same time
• Must include demand interval, period, and units
• Example: The maximum non-coincident 15-minute kW demand for the
week was 700 kW
7. Demand Factor
• Ratio of maximum demand to connected load
8. Utilization Factor
• Ratio of the maximum demand to rated capacity
9. Load Factor
• Ratio of the average demand of any individual load or a group of loads
over a period to the maximum demand over the same period
10. Diversity Factor
• Ratio of the “maximum non-coincident demand” to the “maximum
diversified demand”
11. Load Diversity
• Difference between “maximum non-coincident demand” and the “maxi-
mum diversified demand”

2.2 INDIVIDUAL CUSTOMER LOAD

Figure 2.1 illustrates how the instantaneous kW load of a customer changes for two
15-minute intervals.

FIGURE 2.1 Customer demand curve.

The Nature of Loads 11

FIGURE 2.2 24-hour demand curve for Customer #1.

2.2.1 Demand
To define the load, the demand curve is broken into equal time intervals. In Figure
2.1, the selected time interval is 15 minutes. In each interval, the average value of the
demand is determined. In Figure 2.1, the straight lines represent the average load in a
time interval. The shorter the time interval, the more accurate will be the value of the
load. This process is very similar to numerical integration. The average value of the
load in an interval is defined as the “15-minute kW demand”.
The 24-hour, 15-minute kW demand curve for a customer is shown in Figure 2.2.
This curve is developed from a spreadsheet that gives the 15-minute kW demand for
a period of 24 hours.

2.2.2 Maximum Demand
The demand curve shown in Figure 2.2 represents a typical residential customer. Each
bar represents the “15-minute kW demand”. Note that during the 24-hour period,
there is a great variation in the demand. This customer has three periods in which
the kW demand exceeds 6.0 kW. The greatest of these is the “15-minute maximum
kW demand”. For this customer, the “15-minute maximum kW demand” occurs at
13:15 and has a value of 6.18 kW.

2.2.3 Average Demand
During the 24-hour period, energy (kWh) will be consumed. The energy in kWh used
during each 15-minute time interval is computed by

1
kWh  15  minute kW demand    hour (2.1)
4

The total energy consumed during the day is then the summation of all of the
15-minute interval consumptions. From the spreadsheet, the total energy consumed
12 Distribution System Modeling and Analysis

during the period by Customer #1 is 58.96 kWh. The “15-minute average kW

demand” is computed by

Total Energy 58.96

=
kWaverage = = 2.46 kW (2.2)
Hours 24

2.2.4 Load Factor
“Load factor” is a term that is often referred to when describing a load. It is defined
as the ratio of the average demand to the maximum demand. In many ways, load fac-
tor gives an indication of how well the utility’s facilities are being utilized. From the
utility’s standpoint, the optimal load factor would be 1.00 since the system has to be
designed to handle the maximum demand. Sometimes utility companies will encour-
age industrial customers to improve their load factor. One method of encouragement
is to penalize the customer on the electric bill for having a low load factor.
For Customer #1 in Figure 2.2, the load factor is computed to be:

kWaverage 2.46
=
Load Factor = = 0.40 (2.3)
kWmaximum 6.18

2.3 DISTRIBUTION TRANSFORMER LOADING

A distribution transformer will provide service to one or more customers. Each cus-
tomer will have a demand curve like that of Figure 2.2. However, the peaks and val-
leys and maximum demands will be different for each customer. Figures 2.3, 2.4, and
2.5 give the demand curves for the three additional customers connected to the same
distribution transformer.
The load curves for the four customers show that each customer has its unique
loading characteristic. The customers’ individual maximum kW demand occurs at
different times of the day. Customer #3 is the only customer who will have a high
load factor. A summary of individual loads is given in Table 2.1.

FIGURE 2.3 24-hour demand curve for Customer #2.

The Nature of Loads 13

FIGURE 2.4 24-hour demand curve for Customer #3.

FIGURE 2.5 24-hour demand curve for Customer #4.

TABLE 2.1
Individual Customer Load Characteristics
Cust. #1 Cust. #2 Cust. #3 Cust. #4

Energy Usage (kWh) 58.57 36.46 95.64 42.75

Maximum kW Demand 6.18 6.82 4.93 7.05
Time of Max. kW Demand 13:15 11:30 6:45 20:30
Average kW Demand 2.44 1.52 3.98 1.78
Load Factor 0.40 0.22 0.81 0.25

These four customers demonstrate that there is great diversity between their loads.

2.3.1 Diversified Demand
It is assumed that the same distribution transformer serves the four customers dis-
cussed previously. The sum of the four 15 kW demands for each time interval is the
“diversified demand” for the group in that time interval, and in this case, the distri-
bution transformer. The 15-minute diversified kW demand of the transformer for the
14 Distribution System Modeling and Analysis

FIGURE 2.6 Transformer diversified demand curve.

day is shown in Figure 2.6. Note in this figure how the demand curve is beginning
to smooth out. There are not as many significant changes as seen by some of the
individual customer curves.

2.3.2 Maximum Diversified Demand

The transformer demand curve of Figure 2.6 demonstrates how the combined cus-
tomer loads begin to smooth out the extreme changes of the individual loads. For the
transformer, the 15-minute kW demand exceeds 16 kW twice. The greater of these is
the “15-minute maximum diversified kW demand” of the transformer. It occurs at
17:30 and has a value of 16.16 kW. Note that this maximum demand does not occur
at the same time as any one of the individual demands nor is this maximum demand
the sum of the individual maximum demands.

2.3.3 Load Duration Curve

A “load duration curve” can be developed for the transformer serving the four cus-
tomers. Sorting in descending order, the kW demand of the transformer develops the
load duration curve shown in Figure 2.7.
The load duration curve plots the 15-minute kW demand versus the percent of
time; the transformer operates at or above the specific kW demand. For example, the
load duration curve shows the transformer operates with a 15-minute kW demand of
12 kW or greater than 22% of the time. This curve can be used to determine whether
a transformer needs to be replaced due to an overloading condition.

2.3.4 Maximum Non-coincident Demand

The “15-minute maximum non-coincident kW demand” for the day is the sum of
the individual customer’s 15-minute maximum kW demands. For the transformer in
question, the sum of the individual maximums is

kWmaximum non  coincident demand  6.18  6.82  4.93  7.05  24.98 kW (2.4)
The Nature of Loads 15

FIGURE 2.7 Transformer load duration curve.

2.3.5 Diversity Factor
By definition, diversity factor is the ratio of the maximum non-coincident demand
of a group of customers to the maximum diversified demand of the group. With
reference to the transformer serving four customers, the diversity factor for the four
customers would be

kWmaximum non  coincident demand 24.98

Diversity Factor    1.5458 (2.5)
kWmaximum diversified demand 16.15

The idea behind the diversity factor is that when the maximum demands of the
customers are known, then the maximum diversified demand of a group of custom-
ers can be computed. There will be a different value of the diversity factor for
different numbers of customers. The previously computed value would apply to
four customers. If there are five customers, then a load survey would have to be set
up to determine the diversity factor for five customers. This process would have to
be repeated for all practical numbers of customers. Table 2.2 is an example of the
diversity factors for the number of customers ranging from one up to 70. The table
was developed from a different database than the four customers that have been
discussed previously.
A graph of the diversity factors is shown in Figure 2.8.
Note in Table 2.2 and Figure 2.8 that the value of the diversity factor has basically
leveled out when the number of customers has reached 70. This is an important
observation because it means, at least for the system from which these diversity fac-
tors were determined, that the diversity factor will remain constant at 3.20 from 70
customers and up. In other words, as viewed from the substation, the maximum
diversified demand of a feeder can be predicted by computing the total non-­
coincident maximum demand of all of the customers served by the feeder and divid-
ing by 3.2.
16 Distribution System Modeling and Analysis

TABLE 2.2
Diversity Factors
N DF N DF N DF N DF N DF N DF N DF

1 1.0 11 2.67 21 2.90 31 3.05 41 3.13 51 3.15 61 3.18

2 1.60 12 2.70 22 2.92 32 3.06 42 3.13 52 3.15 62 3.18
3 1.80 13 2.74 23 2.94 33 3.08 43 3.14 53 3.16 63 3.18
4 2.10 14 2.78 24 2.96 34 3.09 44 3.14 54 3.16 64 3.19
5 2.20 15 2.80 25 2.98 35 3.10 45 3.14 55 3.16 65 3.19
6 2.30 16 2.82 26 3.00 36 3.10 46 3.14 56 3.17 66 3.19
7 2.40 17 2.84 27 3.01 37 3.11 47 3.15 57 3.17 67 3.19
8 2.55 18 2.86 28 3.02 38 3.12 48 3.15 58 3.17 68 3.19
9 2.60 19 2.88 29 3.04 39 3.12 49 3.15 59 3.18 69 3.20
10 2.65 20 2.90 30 3.05 40 3.13 50 3.15 60 3.18 70 3.20

Note: DF = diversity factors

FIGURE 2.8 Diversity factors.

2.3.6 Demand Factor
The demand factor can be defined for an individual customer. For example, the
15-minute maximum kW demand of Customer #1 was found to be 6.18 kW. In order
to determine the demand factor, the total connected load of the customer needs to be
known. The total connected load will be the sum of the ratings of all of the electri-
cal devices at the customer’s location. Assume that this total comes to 35 kW, the
demand factor is computed to be

kWmaximum demand 6.18

=
Demand Factor = = 0.1766 (2.6)
kWtotal connected load 35
The Nature of Loads 17

The demand factor gives an indication of the percentage of electrical devices that
are on when the maximum demand occurs. The demand factor can be computed for
an individual customer but not for a distribution transformer or the total feeder.

2.3.7 Utilization Factor
The utilization factor gives an indication of how well the capacity of an electrical
device is being utilized. For example, the transformer serving the four loads is rated
15 kVA. Using the 16.16 kW maximum diversified demand and assuming a power
factor of 0.9, the 15-minute maximum kVA demand on the transformer is computed
by dividing the 16.16 kW maximum kW demand by the power factor and would be
17.96 kVA. The utilization factor is computed to be

kVAmaximum demand 17.96

=
Utilization Factor = = 1.197 (2.7)
kVAtransformer rating 15

2.3.8 Load Diversity
Load diversity is defined as the difference between the non-coincident maximum
demand and the maximum diversified demand. For the transformer in question, the
load diversity is computed to be

Load Diversity  KWnon  conincident demand  KWmaximum diversified demand

(2.8)
 24.97  16.16  8.81 KW

2.4 FEEDER LOAD
The load that a feeder serves will display a smoothed-out demand curve as shown in
Figure 2.9.

FIGURE 2.9 Feeder demand curve.

18 Distribution System Modeling and Analysis

The feeder demand curve does not display any of the abrupt changes in demand
of an individual customer demand curve or the semi-abrupt changes in the demand
curve of a transformer. The simple explanation for this is that with several hundred
customers served by the feeder, the odds are good that as one customer is turning off
a light bulb another customer will be turning a light bulb on. The feeder load, there-
fore, does not experience a jump as would be seen in the individual customer’s
demand curve.

2.4.1 Load Allocation
In the analysis of a distribution feeder “load,” data will have to be specified. The
data provided will depend upon how detailed the feeder is to be modeled and the
availability of customer load data. The most comprehensive model of a feeder will
represent every distribution transformer. When this is the case, the load allocated to
each transformer needs to be determined.

2.4.1.1 Application of Diversity Factors

The definition of the diversity factor is the ratio of the maximum non-coincident
demand to the maximum diversified demand. A table of diversity factors is shown in
Table 2.2. When such a table is available, then it is possible to determine the maxi-
mum diversified demand of a group of customers such as those served by a distribu-
tion transformer. That is, the maximum diversified demand can be computed by

kWmaximum non  coincident demand

kWmaximum diversified demand  (2.9)
DFnumber of customers

This maximum diversified demand becomes the allocated “load” for the
transformer.

2.4.1.2 Load Survey
Many times, the maximum demand of individual customers will be known either
from metering or from a knowledge of the energy (kWh) consumed by the customer.
Some utility companies will perform a load survey of similar customers in order to
determine the relationship between the energy consumption in kWh and the maxi-
mum kW demand. Such a load survey requires the installation of a demand meter at
each customer’s location. The meter can be the same type as is used to develop the
demand curves previously discussed, or it can be a simple meter that only records the
maximum demand during the period. At the end of the survey period, the maximum
demand vs. kWh for each customer can be plotted on a common graph. Linear regres-
sion is used to determine the equation of a straight line that gives the kW demand
as a function of kWh. The plot of points for 15 customers, along with the resulting
equation derived from a linear regression algorithm, is shown in Figure 2.10.
The straight-line equation derived is

kWmaximum demand  0.1058  0.005014  kWh (2.10)

The Nature of Loads 19

FIGURE 2.10 kW demand vs. kWh for residential customers.

Knowing the maximum demand for each customer is the first step in developing a
table of diversity factors, as shown in Table 2.2. The next step is to perform a load
survey where the maximum diversified demand of groups of customers is metered.
This will involve selecting a series of locations where demand meters can be placed
that will record the maximum demand for groups of customers ranging from at least
two to 70. At each meter location, the maximum demand of all downstream custom-
ers must also be known. With that data, the diversity factor can be computed for the
given number of downstream customers.

Example 2.1: A single-phase lateral provides service to three

distribution transformers, as shown in Figure 2.11.

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-minute maximum kW demand is given by the equation

kWdemand  0.2  0.008  kWh

The kWh consumed by Customer #1 is 1,523 kWh. The 15-minute maximum kW

demand for Customer #1 is then computed as
20 Distribution System Modeling and Analysis

FIGURE 2.11 Single-phase lateral.

kW1  0.2  0.008  1523  12.4 kW

The results of this calculation for the remainder of the customers are summarized
next by transformer.

Transformer T1
Customer #1 #2 #3 #4 #5

kWh 1,523 1,645 1,984 1,590 1,456

kW 12.4 13.4 16.1 12.9 11.9

Transformer T2
Customer #6 #7 #8 #9 #10 #11

kWh 1,235 1,587 1,698 1,745 2,015 1,765

kW 10.1 12.9 13.8 14.2 16.3 14.3

Transformer T3
Customer #12 #13 #14 #15 #16 #17 #18

kWh 2,098 1,856 2,058 2,265 2,135 1,985 2,103

kW 17.0 15.1 16.7 18.3 17.3 16.1 17.0

1. Determine for each transformer the 15-minute non-coincident maximum

kW demand, and using the Table of Diversity Factors in Table 2.2, deter-
mine the 15-minute maximum diversified kW demand.

kWmaximum non  coincident demand  12.4  13.4  16.1 12.9  11.8  66.6 kW
T1 : kWmaximum non  concideent demand
kWmaximum diversified demand   30.3 kW
DF5

kWmaximum non  coincident demand  10.1 12.9  13.8  14.2  16.3  14.3
 81.6 kW
T2 :
kWmaximum non  cooncident demand
kWmaximum diversified demand   35.4 kW
DF6
The Nature of Loads 21

kWmaximum non  coincident demand  17.0  15.0  16.7  18.3  17.3  16.1 17.0
 117.4 kW
T3 :
kWmaximum non  concident demand
kWmaximum diversified demand   48.9 kW
DF7

Based upon the 15-minute maximum kW diversified demand on each

transformer and an assumed power factor of 0.9, the 15-minute maximum
kVA diversified demand on each transformer would be as follows:

30.2
=
kVAT1 maximum diversified demand = 33.6 kVA
.9
35.5
= = 39.4 kVA
kVAT 2 maximum diversified demand
.9
48.9
= = 54.4 kVA
kVAT 3 maximum diversified demand
.9

The kVA ratings selected for the three transformers would be 25 kVA, 37.5 kV,
and 50 kVA respectively. With those selections, only Transformer T1 would
experience a significant maximum kVA demand greater than its rating (135%).
2. Determine the 15-minute non-coincident maximum kW demand and
15-minute maximum diversified kW demand for each of the line segments.

Segment N1 to N2: The maximum non-coincident kW demand is the sum

of the maximum demands of all 18 customers.

kWmaximum non  coincident demand  66.6  81.6  117.4  265.6 kW

The maximum diversified kW demand is computed by using the diversity

factor for 18 customers.

265.5
kWmaximum diversified =
demand = 92.8 KW
2.86

Segment N2 to N3: This line segment “sees” 13 customers. The non-coin-

cident maximum demand is the sum of Customers #6 through #18. The
diversity factor for 13 (2.74) is used to compute the maximum diversified
kW demand.

kWmaximum non  coincident demand  81.6  117.4  199.0 kW

199.0
kWmaximum diversified demand   72.6 kW
2.74

Segment N3 to N4: This line segment “sees” the same non-coincident

demand and diversified demand as that of Transformer T3.

kWmaximum non  coincident demand  117.4  117.4 kW

199.0
kWmaximum diversified demand   48.9 kW
2 .4
22 Distribution System Modeling and Analysis

Example 2.1 demonstrates that Kirchhoff’s Current Law (KCL) is not obeyed
when the maximum diversified demands are used as the “load” flowing through the
line segments and through the transformers. For example, at node N1, the maximum
diversified demand flowing down the line segment N1–N2 is 92.8 kW and the maxi-
mum diversified demand flowing through Transformer T1 is 30.3 kW. KCL would
then predict that the maximum diversified demand flowing down line segment N2–
N3 would be the difference of these or 62.5 kW. However, the calculations for the
maximum diversified demand in that segment were computed to be 72.6 kW. The
explanation for this is that the maximum diversified demands for the line segments
and transformers don’t necessarily occur at the same time. At the time that the line
segment N2–N3 is experiencing its maximum diversified demand, line segment N1–
N2 and Transformer T1 are not at their maximum values. All that can be said is that
at the time segment N2–N3 is experiencing its maximum diversified demand, the
difference between the actual demand on the line segment N1–N2 and the demand of
Transformer T1 will be 72.6 kW. There will be an infinite number of combinations of
line flow down N1–N2 and through Transformer T1 that will produce the maximum
diversified demand of 72.6 kW on line N2–N3.

2.4.1.3 Transformer Load Management

A transformer load management program is used by utilities to determine the load-
ing on distribution transformers based upon a knowledge of the kWh supplied by the
transformer during a peak loading month. The program is primarily used to deter-
mine when a distribution transformer needs to be changed out due to a projected
overloading condition. The results of the program can also be used to allocate loads
to transformers for feeder analysis purposes.
The transformer load management program relates the maximum diversified
demand of a distribution transformer to the total kWh supplied by the transformer
during a specific month. The usual relationship is the equation of a straight line. Such
an equation is determined from a load survey. This type of load survey meters the
maximum demand on the transformer in addition to the total energy in kWh of all of
the customers connected to the transformer. With the information available from sev-
eral sample transformers, a curve similar to that shown in Figure 2.10 can be devel-
oped, and the constants of the straight-line equation can then be computed. This
method has the advantage because the utility will have in the billing database the
kWh consumed by each customer every month. As long as the utility knows which
customers are connected to each transformer, by using the developed equation, the
maximum diversified demand (allocated load) on each transformer on a feeder can be
determined for each billing period.

2.4.1.4 Metered Feeder Maximum Demand

The major disadvantage of allocating load using the diversity factors is that most
utilities will not have a table of diversity factors. The process of developing such a
table is generally not cost beneficial. The major disadvantage of the transformer load
management method is that a database is required that specifies which transformers
serve which customers. Again, this database is not always available.
Allocating load based upon the metered readings in the substation requires the
least amount of data. Most feeders will have metering in the substation that will at
The Nature of Loads 23

minimum give either the total three-phase maximum diversified kW or kVA demand
and/or the maximum current per phase during a month. The kVA ratings of all distri-
bution transformers are always known as feeders. The metered readings can be allo-
cated to each transformer based upon the transformer rating. An “allocation factor”
(AF) can be determined based upon the metered three-phase kW or kVA demand and
the total connected distribution transformer kVA.

kVAmetered demand (2.11)

AF =
kVAtotal kVA rating

where kVAtotal kVA rating = Sum of the kVA ratings of all distribution transformers.
The allocated load per transformer is then determined by

kVAtransformer demand  AF  kVAtransformer rating (2.12)

The transformer demand will be either kW or kVA depending upon the metered
quantity.
When the kW or kVA is metered by phase, then the load can be allocated by phase
where it will be necessary to know the phasing of each distribution transformer.
When the maximum current per phase is metered, the load allocated to each dis-
tribution transformer can be done by assuming nominal voltage at the substation and
then computing the resulting kVA. The load allocation will now follow the same
procedure as outlined earlier.
If there is no metered information on the reactive power or power factor of the
feeder, a power factor will have to be assumed for each transformer load.
Modern substations will have microprocessor-based metering that will provide
kW, kvar, kVA, power factor, and current per phase. With this data, the reactive power
can also be allocated. Since the metered data at the substation will include losses, an
iterative process will have to be followed so that the allocated load plus losses will
equal the metered readings.

Example 2.2: Assume that the metered maximum diversified kW

demand for the system of Example 2.1 is 92.9 kW. Allocate this load
according to the kVA ratings of the three transformers.

kVAtotal  25  37.5  50  112.5 kVA

92.9
=
AF = 0.8258
112.5

The allocated kW for each transformer becomes

T1: kW1 = 0.8258 · 25 = 20.64 kW,

T2: kW2 = 0.8258 · 37.5 = 30.97 kW,
T3: kW3 = 0.8258 · 50 = 41.29 kW
24 Distribution System Modeling and Analysis

2.4.1.5 What Method to Use?

Four different methods have been presented for allocating load to distribution
transformers:

• Application of diversity factors

• Load survey
• Transformer load management
• Metered feeder maximum demand

Which method to use depends upon the purpose of the analysis. If the purpose of the
analysis is to determine as closely as possible the maximum demand on a distribu-
tion transformer, then either the diversity factor or the transformer load manage-
ment method can be used. Neither of these methods should be employed when the
analysis of the total feeder is to be performed. The problem is that using either of
those methods will result in a much larger maximum diversified demand at the sub-
station than actually exists. When the total feeder is to be analyzed, the only method
that gives good results is that of allocating load based upon the kVA ratings of the
transformers.

2.4.2 Voltage Drop Calculations Using Allocated Loads

The voltage drops down line segments, and through distribution, transformers are of
interest to the distribution engineer. Four different methods of allocating loads have
been presented. The various voltage drops can be computed using the loads allocated
by the three methods. For these studies, it is assumed that the allocated loads will be
modeled as constant real power and reactive power.

2.4.2.1 Application of Diversity Factors

The loads allocated to a line segment or a distribution transformer using diversity
factors are a function of the total number of customers “downstream” from the line
segment or distribution transformer. The application of the diversity factors was dem-
onstrated in Example 2.1. With a knowledge of the allocated loads flowing in the line
segments and through the transformers and the impedances, the voltage drops can be
computed. The assumption is that the allocated loads will be constant real power and
reactive power. In order to avoid an iterative solution, the voltage at the source end
is assumed and the voltage drops calculated from that point to the last transformer.
Example 2.3 demonstrates how the method of load allocation using diversity fac-
tors is applied. The same system and allocated loads from Example 2.1 are used in
Example 2.3.

2.4.2.2 Load Allocation Based Upon 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.
This method was discussed in Section 2.3.3. Example 2.4 demonstrates this
method.
The Nature of Loads 25

Example 2.3: For the system of Example 2.1, assume the voltage
at N1 is 2,400 volts; compute the secondary voltages on the three
transformers using the diversity factors.

The system of Example 2.1, including segment distances, is shown in Figure 2.12.

FIGURE 2.12 Single-phase lateral with distances.

Assume that the power factor of the loads is 0.9 lagging.

The impedance of the lines are z = 0.3 + j0.6 Ω/mile.
The ratings of the transformers are

T1: 25 kVA, 2400–240 volts, Z = 1.8/40%,

T2: 37.5 kVA, 2400–240 volts, Z = 1.9/45%,
T3: 50 kVA, 2400–240 volts, Z = 2.0/50%

From Example 2.1, the maximum diversified kW demands were computed.

Using the 0.9 lagging power factor, the maximum diversified kW and kVA
demands for the line segments and transformers are as follows:

Segment N1–N2: P12 = 92.9 kW S12 = 92.9 + j45.0 kVA

Segment N2–N3: P23 = 72.6 kW S23 = 72.6 + j35.2 kVA
Segment N3–N4: P34 = 49.0 kW S34 = 49.0 + 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 = 49.0 kW ST3 = 49.0 + j23.7 kVA

Convert transformer impedances to ohms referred to the high voltage side.

kV 2  1000 2.42  1000

T1: Zbase1    230.4 
kVA1 25

ZT1   0.018 /40   230.4  3.18  j 2.67 

26 Distribution System Modeling and Analysis

kV 2  1000 2.42  1000

T2: Zbase2    153.6 
kVA2 37.5

ZT 2   0.019 /45  153.6  2.06  j 2.06 

kV 2  1000 2.42  1000

T3: Zbase3    115.2 
kVA3 50

ZT 3   0.02 /50   115.2  1.48  j1.77 

Compute the line impedances.

5000
N1 – N2: Z12  0.3  j 0.6    0.2841 j 0.5682 
5280

500
N2 – N3: Z23  0.3  j 0.6    0.0284  j 0.0568 
5280

750
N3 – N4: Z34  0.3  j 0.6    0.0426  j 0.0852 
5280

Calculate the current flowing in segment N1 – N2.

 
 kW  jkvar   92.9  j 45.0 
I12       43.0 / 25.84 A
 kV   2.4 / 0 

Calculate the voltage at N2.

V2  V1  Z12·I12

V2  2400 / 0   0.2841 j 0.5682  43.0 / 25.84  2378.4 / 0.4 V

Calculate the current flowing into T1.

 
 kW  jkvar   30.3  j14.7 
IT1       14.16 / 26.84 A
 kV   2.378 / 0.4 

Calculate the secondary voltage referred to the high side.

VT1  V2  ZT1·IT1

VT1  2378.4 / 0.4   3.18  j 2.67   14.16 / 26.24  2321.5 / 0.8 V

The Nature of Loads 27

Compute the secondary voltage by dividing by the turns ratio of 10.

2321.5 / 0.8
VlowT1   232.15 / 0.8 V
10

Calculate the current flowing in line section N2–N3.

 
 kW  jkvar   72.6  j 35.2 
I23       33.9 / 25.24 A
 kV   2.378 / 0.4 

Calculate the voltage at N3.

V3  V2  Z23·I23

V2  2378.4 / 0.4   0.0284  j 0.0568   33.9 / 26.24  2376.7 / 0.4 V

Calculate the current flowing into T2.

 
 kW  jkvar   35.5  j17.2 
IT 2       16.58 / 26.27 A
 kV   2.3767 / 0.4 

Calculate the secondary voltage referred to the high side.

VT 2  V3  ZT 2·IT 2

VT 2  2376.7 / 0.4   2.06  j 2.06   16.58 / 26.27  2331.1 / 0.8 V

Compute the secondary voltage by dividing by the turns ratio of 10.

2331.1 / 0.8
VlowT 2   233.1 / 0.8 V
10

Calculate the current flowing in line section N3–N4.

 
 kW  jkvar   49.0  j 23.7 
I34       22.9 / 25.27 A
 kV   2.3767 / 0.4 

Calculate the voltage at N4.

V4  V3  Z34  I34

V4  2376.7 / 0.4   0.0426  j 0.0852  22.9 / 26.27  2375.0 / 0.5 V

The current flowing into T3 is the same as the current from N3 to N4.
28 Distribution System Modeling and Analysis

 
 kW  jkvar   51.0  j 24.7 
IT 3       23.91 / 26.30 A
 kV   2.375.0 / 0.5 

Calculate the secondary voltage referred to the high side.

VT 3  V4  ZT 3  IT 3

VT 3  2375.0 / 0.5  1.48  j1.77   23.91 / 26.30  2326.1 / 1.0 V

Compute the secondary voltage by dividing by the turns ratio of 10.

2326.1 / 1.0
VlowT 3   232.6 / 1.0 V
10

Calculate the percent voltage drop to the secondary of Transformer T3. Use the
secondary voltage referred to the high side.

V1  | VT 3 | 2400  2326.1
Vdrop   100   100  3.0789%
| V1 | 2400

Example 2.4: For the system of Example 2.1, assume the voltage
at N1 is 2,400 volts, compute the secondary voltages on the three
transformers allocating the loads based upon the transformer ratings.
Assume that the metered kW demand at N1 is 92.9 kW.

The impedances of the line segments and transformers are the same as in Example
2.3.
Assume the load power factor is 0.9 lagging and compute the kVA demand at
N1 from the metered demand.

92.9
S12  / cos1  0.9   92.9  j 45.0  103.2 / 25.84 kVA
0 .9

Calculate the AF.

103.2 / 25.84
AF   0.9175 / 25.84
25  37.5  50

Allocate the loads to each transformer.

ST1  AF  kVAT1   0.9175 / 25.84   25  20.6  j10.0 kVA

ST 2  AF  kVAT 2   0.9175 / 25.84   37.5  31.0  j15.0 kVA

The Nature of Loads 29

ST 3  AF  kVAT1   0.9175 / 25.84   50  41.3  j 20.0 kVA

Calculate the line flows.

S12  ST1  ST 2  ST 3  92.9  j 45.0 kVA

S23  ST 2  ST 3  72.3  j 35.0 kVA

S34  ST 3  41.3  j 20.0 kVA

Using these values of line flows and flows into transformers, the procedure for
computing the transformer secondary voltages is exactly the same as in Example
2.3. When this procedure is followed, the node and secondary transformer volt-
ages are as follows:

V2  2378.1/  0.4 V VlowT1  234.0 /  0.6 V

V3  2376.4 /  0.4 V VlowT 2  233.7 /  0.6 V

V4  2374.9 /  0.5 V VlowT 3  233.5 /  0.9 V

The percent voltage drop for this case is

V1  | VT 3 | 2400  2334.8
Vdrop   100   100  2.7179%
｜V1 ｜ 2400

2.5 INDIVIDUAL CUSTOMER LOADS WITH ELECTRIC VEHICLES

Electric vehicles (EVs) are quickly becoming popular and are a considerable load on
the grid compared with the remaining connected load for residential and commercial
customers. Figure 2.2 showed a 24-hour demand curve for Customer #1. This section
will show the effect of adding a Level 2 EV charger to this customer.
Chapter 9 explains EV chargers in detail and how they behave as a load from a
circuit perspective, but in this section, they will be covered at a high level. Figure
2.13 shows the 24-hour demand curve for Customer #1 with the addition of a Level
2 EV charger.
Level 2 chargers commonly consume 9.6 kW of real power while charging. For
Customer #1, the EV charger begins charging at 18:15 and charges for one hour until
19:15. To show the drastic difference of this load with and without the EV charger,
Figure 2.2 is shown again in Figure 2.14 using the same scale as Figure 2.13.
This shifts the occurrence of the maximum kW demand from 13:15 to 19:00 and,
in addition, changes the maximum kW demand from 6.18 kW to 15.73 kW. This is
more than double the original maximum kW demand. In addition, this short charging
period increased the energy consumed by 9.6 kWh and the average kW demand by
0.44 kW. The most significant impact, however, is on the load factor. The load factor
30 Distribution System Modeling and Analysis

FIGURE 2.13 24-Hour demand curve for Customer #1 with EV charger.

FIGURE 2.14 24-hour demand curve for Customer #1 without EV charger.

was severely reduced with the addition of the EV charger from 0.4 to 0.18. Lastly, the
loading on the distribution transformer is a concern. It was previously calculated that
the maximum diversified demand for Customers #1 through #4 was 16.16 kW, and
those four customers were connected to a 15 kVA transformer. During the period of
maximum diversified demand, the kVA load on the transformer was calculated to be
The Nature of Loads 31

17.96, which produced a transformer utilization factor of 1.197. This is slightly over-
loaded but tolerable for short periods of time. EV chargers that use smart charging
assume a power factor of 0.95 lagging, so the total kVA load on the transformer for
the new maximum diversified demand is

16.16 9.6
kVAmaximum demand    28.06 kVA.
0.9 0.95

This gives a new transformer utilization factor of

kVAmaximum demand 28.06

=
Utilization Factor = = 1.871,
kVAtransformer rating 15

which is an unacceptable kVA load on the transformer, even for short periods of time.
This would require the distribution engineer to intervene by replacing this trans-
former. In addition, it will be shown in Chapter 9 that the high current drawn by these
EV chargers has a significant impact on the customer’s service voltage, which also
must be considered.
The impact that EV chargers have on the grid is significant; however, the cost of
an “electrical gallon” or an eGallon as defined by the Department of Energy, is still
much cheaper than a gallon of petroleum-based gasoline [1]. Due to the differential
in cost of petroleum-based energy and electrical energy, and the continued electrifi-
cation of traditionally non-electrified devices, the trend of EVs is likely to continue
to grow. This will have to be addressed by distribution engineers to keep the grid
stable and reliable and to keep voltages within limits.

2.6 SUMMARY
This chapter has demonstrated the nature of the loads on a distribution feeder. There
is a great diversity between individual customer demands, but as the demand is moni-
tored on line segments working back toward the substation, the effect of the diver-
sity between demands becomes very slight. It was shown that the effect of diversity
between customer demands must be taken into account when the demand on a dis-
tribution transformer is computed. The effect of diversity for short laterals can be
taken into account in determining the maximum flow on the lateral. For the diversity
factors of Table 2.2, it was shown that when the number of customers exceeds 70, the
effect of diversity has pretty much disappeared. This is evidenced by the fact that the
diversity factor has become almost constant as the number of customers approached
70. It must be understood that the number 70 will apply only to the diversity factors
of Table 2.2. If a utility is going to use diversity factors, then that utility must perform
a comprehensive load survey to develop the table of diversity factors that apply to
that particular system.
Examples 2.3 and 2.4 show that the final node and transformer voltages are
approximately the same. There is very little difference between the voltages when the
32 Distribution System Modeling and Analysis

loads were allocated using the diversity factors and when the loads were allocated
based upon the transformer kVA ratings.
Section 2.5 showed that EV chargers add a significant load to the grid. The load
due to just one EV charger is substantial enough to cause overloading of transform-
ers, which could require replacement of the device.

PROBLEMS
2.1 The following are the 15-minute kW demands for four customers
between the hours of 17:00 and 21:00. A 25 kVA single-phase trans-
former serves the four customers.

Time Cust #1 Cust #2 Cust #3 Cust #4

KW KW KW kW
17:00 8.81 4.96 11.04 1.44
17:15 2.12 3.16 7.04 1.62
17:30 9.48 7.08 7.68 2.46
17:45 7.16 5.08 6.08 0.84
18:00 6.04 3.12 4.32 1.12
18:15 9.88 6.56 5.12 2.24
18:30 4.68 6.88 6.56 1.12
18:45 5.12 3.84 8.48 2.24
19:00 10.44 4.44 4.12 1.12
19:15 3.72 8.52 3.68 0.96
19:30 8.72 4.52 0.32 2.56
19:45 10.84 2.92 3.04 1.28
20:00 6.96 2.08 2.72 1.92
20:15 6.62 1.48 3.24 1.12
20:30 7.04 2.33 4.16 1.76
20:45 6.69 1.89 4.96 2.72
21:00 1.88 1.64 4.32 2.41

(a) For each of the customers, determine

(1) maximum 15-minute kW demand,
(2) average 15-minute kW demand,
(3) total kWh usage in the time period, and
(4) load factor.
(b) For the 25 kVA transformer, determine
(1) maximum 15-minute diversified demand,
(2) maximum 15-minute non-coincident demand,
(3) utilization factor (assume unity power factor),
(4) diversity factor, and
(5) load diversity.
(c) Plot the load duration curve for the transformer.
2.2 Two transformers each serving four customers are shown in Figure 2.15:
The following table gives the time interval and kVA demand of the four
customer demands during the peak load period of the year. Assume a
power factor of 0.9 lagging.
The Nature of Loads 33

FIGURE 2.15 System for Problem 2.2.

Time #1 #2 #3 #4 #5 #6 #7 #8

3:00–3:30 10 0 10 5 15 10 50 30
3:30–4:00 20 25 15 20 25 20 30 40
4:00–4:30 5 30 30 15 10 30 10 10
4:30–5:00 0 10 20 10 13 40 25 50
5:00–5:30 15 5 5 25 30 30 15 5
5:30–6:00 15 15 10 10 5 20 30 25
6:00–6:30 5 25 25 15 10 10 30 25
6:30–7:00 10 50 15 30 15 5 10 30

(a) For each transformer, determine the following:

(1) 30-minute maximum kVA demand,
(2) non-coincident maximum kVA demand,
(3) load factor,
(4) diversity factor,
(5) suggested transformer rating (50, 75, 100, 167),
(6) utilization factor, and
(7) energy (kWh) during the four-hour period.
(b) Determine the maximum diversified 30-minute kVA demand at the
“tap”
2.3 Two single-phase transformers serving 12 customers are shown in
Figure 2.16.

FIGURE 2.16 Circuit for Problem 2.3.

34 Distribution System Modeling and Analysis

The 15-minute kW demands for the 12 customers between the hours

of 5:00 p.m. and 9:00 p.m. are given on the next page. Assume a load
power factor of 0.95 lagging. The impedance of the lines are z = 0.306 +
j0.6272 Ω/mile. The voltage at node N1 is 2,500/0 volts.
Transformer ratings:

T1: 25 kVA 2400–240 volts Zpu = 0.018/40

T2: 37.5 kVA 2400–240 volts Zpu = 0.020/50

(a) Determine the maximum kW demand for each customer

(b) Determine the average kW demand for each customer
(c) Determine the kWH consumed by each customer in this time period
(d) Determine the load factor for each customer
(e) Determine the maximum diversified demand for each transformer
(f) Determine the maximum non-coincident demand for each
transformer
(g) Determine the utilization factor (assume 1.0 power factor) for each
transformer
(h) Determine the diversity factor of the load for each transformer
(i) Determine the maximum diversified demand at Node N1
(j) Compute the secondary voltage for each transformer taking diversity
into account

Transformer #1– 25 kVA

Time #1 #2 #3 #4 #5
KW KW KW KW kW
05:00 2.13 0.19 4.11 8.68 0.39
05:15 2.09 0.52 4.11 9.26 0.36
05:30 2.15 0.24 4.24 8.55 0.43
05:45 2.52 1.80 4.04 9.09 0.33
06:00 3.25 0.69 4.22 9.34 0.46
06:15 3.26 0.24 4.27 8.22 0.34
06:30 3.22 0.54 4.29 9.57 0.44
06:45 2.27 5.34 4.93 8.45 0.36
07:00 2.24 5.81 3.72 10.29 0.38
07:15 2.20 5.22 3.64 11.26 0.39
07:30 2.08 2.12 3.35 9.25 5.66
07:45 2.13 0.86 2.89 10.21 6.37
08:00 2.12 0.39 2.55 10.41 4.17
08:15 2.08 0.29 3.00 8.31 0.85
08:30 2.10 2.57 2.76 9.09 1.67
08:45 3.81 0.37 2.53 9.58 1.30
09:00 2.04 0.21 2.40 7.88 2.70
The Nature of Loads 35

Transformer #2 – 37.5 kVA

Time #6 #7 #8 #9 #10 #11 #12

KW KW KW KW KW KW kW
05:00 0.87 2.75 0.63 8.73 0.48 9.62 2.55
05:15 0.91 5.35 1.62 0.19 0.40 7.98 1.72
05:30 1.56 13.39 0.19 5.72 0.70 8.72 2.25
05:45 0.97 13.38 0.05 3.28 0.42 8.82 2.38
06:00 0.76 13.23 1.51 1.26 3.01 7.47 1.73
06:15 1.10 13.48 0.05 7.99 4.92 11.60 2.42
06:30 0.79 2.94 0.66 0.22 3.58 11.78 2.24
06:45 0.60 2.78 0.52 8.97 6.58 8.83 1.74
07:00 0.60 2.89 1.80 0.11 7.96 9.21 2.18
07:15 0.87 2.75 0.07 7.93 6.80 7.65 1.98
07:30 0.47 2.60 0.16 1.07 7.42 7.78 2.19
07:45 0.72 2.71 0.12 1.35 8.99 6.27 2.63
08:00 1.00 3.04 1.39 6.51 8.98 10.92 1.59
08:15 0.47 1.65 0.46 0.18 7.99 5.60 1.81
08:30 0.44 2.16 0.53 2.24 8.01 7.74 2.13
08:45 0.95 0.88 0.56 0.11 7.75 11.72 1.63
09:00 0.79 1.58 1.36 0.95 8.19 12.23 1.68

2.4 On a different day, the metered 15-minute kW demand at node N1 for

the system of Problem 2.3 is 72.43 kW. Assume a power factor of 0.95
lagging. Allocate the metered demand to each transformer based upon
the transformer kVA rating. Assume the loads are constant current and
compute the secondary voltage for each transformer.
2.5 A single-phase lateral serves four transformers as shown in Figure 2.17.

FIGURE 2.17 System for Problem 2.5.

Assume that each customer’s maximum demand is 15.5 kW + j7.5 kvar.

The impedance of the single-phase lateral is z = 0.4421 + j0.3213 Ω/1000
ft. The four transformers are rated as follows:

T1 and T2: 37.5 kVA, 2,400–240 volts, Z = 0.01 + j0.03 per unit
T3 and T4: 50 kVA, 2,400–240 volts, Z = 0.015 + j0.035 per unit
36 Distribution System Modeling and Analysis

Use the diversity factors found in Table 2.2 to complete the following:

(a) Determine the 15-minute maximum diversified kW and kvar demands on

each transformer.
(a) Determine the 15-minute maximum diversified kW and kvar demands
for each line section.
(b) If the voltage at Node 1 is 2,600/0 volts, determine the voltage at Nodes
2, 3, 4, 5, 6, 7, 8, 9. In calculating the voltages, take into account diversity
using the answers from (1) and (b) above.
(c) Use the 15-minute maximum diversified demands at the lateral tap (sec-
tion 1–2) from part (b) above. Divide these maximum demands by 18
(number of customers) and assign that as the “instantaneous load” for
each customer. Now calculate the voltages at all of the nodes listed in
part (c) using the instantaneous loads.
(d) Repeat part (d) above except assume the loads are “constant current”.
To do this, take the current flowing from Node 1 to Node 2 from part
(d) above, divide by 18 (number of customers), and assign that as the
“instantaneous constant current load” for each customer. Again, calcu-
late all of the voltages.
(e) Take the maximum diversified demand from Node 1 to Node 2 and “allo-
cate” that out to each of the four transformers based upon their kVA rat-
ings. To do this, take the maximum diversified demand and divide it by
175 (total kVA of the four transformers). Now multiply each transformer
kVA rating by that number to give how much of the total diversified
demand is being served by each transformer. Again, calculate all of the
voltages.
(f) Compute the percent differences in the voltages for parts (d), (e), and (f)
at each of the nodes using the part (c) answer as the base.

REFERENCES
1. “eGallon”. Accessed on Dec. 8, 2021. [Online]. Available: https://www.energy.gov/
maps/egallon.