ECE 611 Power Distribution
Systems
Load Characteristics
Rasoul Narimani
California State University Northridge
Department of Electrical and Computer Engineering
Rasoul.narimani@csun.edu
Announcements
 First Quiz next Friday (09/16/22)
 Next Session we will write code in MATLAB.
1
Radial feeders
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Primary “main” feeder
Laterals
a, b, c phases
3, V, 1-phase connections
Voltage regulators
Delta-Wye transformers
Shunt capacitor banks
Transformers
Secondaries
Protection devices
Loads
[Kersting’s, Fig. 1.4]
Electrical Loads
Resistive Load
Any load that consists of a heating element is generally referred to as a
resistive load.
Inductive Load
An inductive load is quite different and will use a magnetic field to do all
of the work.
Capacitive Load
The current wave leads the voltage wave in a capacitive load. There are
no standalone capacitive loads – no one load is capacitive, such as a
lightbulb or a motor. Capacitors are used in large power circuits to control
power usage.
The majority of modern power circuits will use a combination of
resistive, inductive, and capacitive loads.
Electrical Loads
Constant power: The current will increase if the voltage decreases.
Example: motor, regulated driver to an LED lamp that tries to deliver the
same DC power to the LED array independent of the incoming voltage.
A constant power load is designed to
dynamically adjust the load current
inversely with the load voltage so that the
load power is constant, P = VI.
Electrical Loads
Constant current: The power will decrease with the same ratio if the
voltage decreases.
Example: Airport lighting it is the system of transformer and lamps that
is constant current, not the lamps themselves.
Constant current load dynamically adjusts
its resistance as the load voltage varies,
such that the load current remains
constant, I = V/R.
Electrical Loads
Constant impedance: If the voltage decreases the current will
decrease with the same ratio also and then the power at the
square of the voltage.
Example: Electric resistance heater.
There is ZIP load models in which
consist of constant-impedance,
constant-current, and constant-power
loads.
Electrical Loads
 Distribution systems obviously exist to supply electricity to end users,
so loads and their characteristics are important.
 Utilities supply a broad range of loads, from rural areas with load
densities of 10 kVA/mi2 to urban areas with 300 MVA/mi2.
 A utility may feed houses with a 10- to 20-kVA peak load on the same
circuit as an industrial customer peaking at 5 MW.
 The electrical load on a feeder is the sum of all individual customer
loads. And the electrical load of a customer is the sum of the load
drawn by the customer’s individual appliances.
Electrical Loads
 Load is constantly changing. There (actually) is no steady-state.
 But we can look at averages and maximums over specified time
periods.
 Steady-state analysis such as power-flow study are based on the
“snapshot” view of the system loads.
 Planning also depends on the statistics of loads.
Individual Customer Load-Demand
 Demand : load (kW, kVA, …) averaged over a time period
 For example, the 15min kW demand (D15min) is 4.6 kW at
6:15.
Customer #1 demand curve [Fig. 2.1]
Individual customer load – max/avg demand
 Maximum demand over a time interval: the 15min max kW
demand (MaxD15min) over the day is 6.19kW and it occurs at
13:15,
 Average demand over a time interval: the 15min average kW
demand (AvgD15min) over the day is 2.5kW
Customer #1’s 15min demand [Fig. 2.2]
Individual Customer Load – Load Factor
 Load factor (LF) = AvgD/MaxD (always ≤ 1) for a
given period and interval.
 With the 15min demand, the LF is 2.5/6.19 = 0.40
 Ideally we want to have almost unity LF.
 The smaller the LF is, the more likely that facilities are
under-utilized.
Load Factor
 The loads of certain customer classes tend to vary in
similar patterns.
 Commercial loads are highest from 8 a.m. to 6 p.m.
Residential loads peak in the evening.
 Weather significantly changes loading levels. On hot
summer days, air conditioning increases the demand and
reduces the diversity among loads.
 At the transformer level, load factors of 0.4 to 0.6 are
typical.
Individual Customer Load - Demand Factor
 Defined for an individual customer.
 Total connected load is the sum of ratings for all electric
devices that are connected to this customer
 The demand factor (always ≤ 1) indicates the percentage of
electrical devices that are on when the maximum demand
occurs.
Distribution Transformer Loading
 Diverse loads are aggregated in various points of a
distribution system, e.g., at feeder head or transformer.
 A distribution transformer typically provides service to a
couple of individual customers.
 Diversified demand and max diversified demand can be
defined for the aggregated load accordingly.
Diversified Demand
A transformer with four feeders, each peaking at 100 A, will peak at less
than 400 A because of diversity between feeders. The coincident factor
between four feeders is normally higher than coincident factors at the
individual customer level. Expect coincident factors to be above 0.9.
Each feeder is already highly diversified, so not much more is gained by
grouping more customers together if the sets of customers are similar.
If some feeders are mainly residential and others are commercial, the
peak load of the feeders together can be significantly lower than the sum
of the peaks.
Diversified Demand
 A transformer supplies customers #1-4 (on the right)
 the demand curve is beginning to smooth out.
 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.
Load Duration Curve
 Sort the diversified demand in descending order (and each bar will be
15min/24h =1.04%), used for evaluating transformer stress.
 For example, 22% of the time the transformer operates with a 15min
demand ≥ 12kW
Max Demand for Aggregated Loads
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Max diversified demand: max for the aggregated load.
15min max diversified kW demand is 16.16kW at 17:30.
Max noncoincident demand: sum of the max demand.
15min max noncoincident kW demand is 6.18+…+7.05=24.98kW.
 Their diff is the load diversity = 24.98-16.16 = 8.82kW.
Diversity Factor
 Diversity factor (DF) (≥ 1) is the ratio of the two max demand
 The more number of customers served, the larger the DF is
Utilization Factor
 Similar to the demand factor but defined for a transformer
 Assuming a power factor of 0.9, the max kVA demand is
16.16/0.9 = 17.96kVA
 It indicates how well the capacity of an equipment is being
utilized
 Diff from the demand factor, utilization factor could be >1
Feeder Load
 Smoother demand curve
 Similar metrics as for transformer (diverse demand, diversity
factor...)
Application of Diversity Factor
 This max diversified demand value calculated above is the
allocated “load”
How to Get the Max Demand for Each
Customer?
 Metering data showing
temporal demand curve.
 If only the knowledge of the
energy (kWh) is available,
utilities can perform a load
survey to determine the
relationship between kWh
and max kW demand.
Transformer Load Management
 If we know which customers are
connected to which transformer,
we can estimate the xfmr max.
diversified demand to check for
overloaded transformers.
 Prevent transformer failures
 Prevent transformer fires
 Better usage of capital
(money)
[Ian Dobson, ISU]
Example
 Analyze a single-phase lateral
 Given: monthly energy usage for each customer (kWh)
 Compute: max div demand for each Xformer and line segment
The energy in kWh consumed by each customer during a
month is known.
Solution
 Load survey shows
 For customer #1 of T1
 For T1:
Solution
Solution
The 15min maximum kVA diversified demand on each transformer would be
The kVA ratings selected for the three transformers would be 25, 37.5, and
50kVA, respectively.
With those selections, only transformer T1 would experience a significant
maximum kVA demand greater than its rating (135%).
Example & Solution
b) Determine the 15min noncoincident maximum kW demand and 15min
maximum diversified kW demand for each of the line segments.
Segment N1 to N2:
Segment N2 to N3:
Segment N3 to N4:
Example & Solution
 The allocated load results do not obey KCL!
The maximum diversified demand flowing down the line segment N1–N2 is 92.9kW,
and the maximum diversified demand flowing through transformer T1 is 30.3kW. KCL
would then predict that the maximum diversified demand flowing down line segment
N2–N3 would be the difference of these or 62.6kW. However, the calculations for the
maximum diversified demand in that segment were computed to be 72.6kW.
A transformer load management program is used by utilities to determine
the loading on distribution transformers in order to determine when a
distribution transformer needs to be changed out due to a projected
overloading condition.
Voltage Drop Calculations
 With known allocated loads flowing in line segments/transformers and
the impedances, the voltage drops can be computed by assuming
constant real/reactive power loads.
Example
Assume the voltage at N1 is 2400 V and compute the secondary voltages on
the three transformers using the DFs. The segment distances are shown in
below.
Assume that the power factor of the loads is 0.9 lagging. The impedance of the
lines are z = 0.3 + j0.6Ω/mile. The ratings of the transformers are as follows:
Example-cont
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:
Segment N1–N2: P12 = 92.9kW S12 = 92.9 + j45.0kVA
Segment N2–N3: P23 = 72.6kW S23 = 72.6 + j35.2kVA
Segment N3–N4: P34 = 49.0kW S34 = 49.0 + j23.7kVA
Transformer T1: PT1 = 30.3kW ST1 = 30.3 + j14.7kVA
Transformer T2: PT2 = 35.5kW ST2 = 35.5 + j17.2kVA
Transformer T3: PT3 = 49.0kW ST3 = 49.0 + j23.7kVA
Example-cont
Convert transformer impedances to Ohms referred to the high voltage side
Compute the line impedances:
Example-cont
Calculate the current flowing in segment N1–N2:
Calculate the voltage at N2:
Calculate the current flowing into T1:
Calculate the secondary voltage referred to the high side:
Example-cont
Compute the secondary voltage by dividing by the turns ratio of 10:
Calculate the current flowing in line segment N2–N3:
Calculate the voltage at N3:
Calculate the current flowing into T2:
Example-cont
Calculate the secondary voltage referred to the high side:
Compute the secondary voltage by dividing by the turns ratio of 10:
Calculate the current flowing in line section N3–N4:
Calculate the voltage at N4:
Example-cont
The current flowing into T3 is the same as the current from N3 to N4:
Calculate the secondary voltage referred to the high side:
Compute the secondary voltage by dividing by the turns ratio of 10:
Calculate the percent voltage drop to the secondary of transformer T3. Use
the secondary voltage referred to the high side:
Alternative Approach for Allocated Load
 The major disadvantage of allocating load using the Diversity Factors
(DFs) is that most utilities will not have a table of DFs.
 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.
 We need a method that requires the least amount of data.
Alternative Approach for Allocated Load
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
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 distribution transformers are always known for a
feeder.
The metered readings can be allocated to each transformer based upon
the transformer rating.
Alternative Approach for Allocated Load
An “allocation factor” (AF) can be determined based upon the metered
three-phase kW or kVA demand and the total connected distribution
transformer kVA:
Metered_demand can be either kW or kVA.
is the sum of the kVA ratings of all distribution transformers.
The allocated load per transformer is then determined by
Example
Assume that the metered maximum diversified kW demand
for the system of previous example is 92.9kW. Allocate this
load according to the kVA ratings of the three transformers.
For each xformer the allocated load
T1: kW1 = 0.8258 · 25 = 20.64 kW
T2: kW2 = 0.8258 · 37.5 = 30.97 kW
T3: kW1 = 0.8258 · 50 = 41.29 kW
Voltage Drop Calculations
Accordingly, voltage drop can be calculated based on AF.
Lets revisit the previous example. From the metered demand
For each xformer
Example-cont
Calculate the line flows:
Using these values of line flows and flows into transformers, the procedure
for computing the transformer secondary voltages is exactly the same. The
node and secondary transformer voltages are:
The percent voltage drop for this case is:
Comparisons of the Two Approaches
Example (DF Approach)
Example (AF Approach)
Minimal difference between the two approaches.
However, this feeder analysis framework still has limits in terms of its
scalability and generalizability for unbalanced systems
Questions?
Rasoul Narimani
Rasoul.narimani@csun.edu
Questions