1. ECE498HZ: Power Distribution System Analysis
January 26, 2016
ECE498HZ: Power Distribution System Analysis
Lecture 3: Electrical Loads
ECE Main Slide
Hao Zhu
Dept. of Electrical & Computer Engineering
University of Illinois, Urbana-Champaign

2. Announcements
Tamer Rousan from Ameren Illinois will give a talk on Thursday, presenting some Ameren activities on power distribution systems
Max Liu will teach next Tuesday
Homework 1 posted, due on Feb 4

3. Radial feeders 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]

4. 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

5. 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]

6. 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 itoccurs 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]

7. 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

8. 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

9. 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

10. Diversified demand
Suppose a transformer supplies customers #1-4 (on the right)
15min diversified kW demand is 9.0kW at 4:00

11. 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

12. Max demand for aggregated loads
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 = = 8.82kW

13. 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

14. 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

15. Feeder load Smoother demand curve
Similar metrics as for transformer (diverse demand, diversity factor...)

16. Application of diversity factor
This max diversified demand value calculated in (2.9) is the allocated “load”

17. How to get the max demand for every 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

18. 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]

19. Progress so far
[Ian Dobson, ISU]

20. Basic feeder analysis Example 2.1: analyze a single-phase lateral
Given: monthly energy usage for each customer (kWh)
Compute: max div demand for each Xformer and line segment

21. Example 2.1 Load survey shows For customer #1 of T1
For T1, max noncoin = = 66.6 kW
What size should the transformers be? (Max kVA demand)
The allocated load results do not obey KCL!

22. 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 2.3: voltage at N1 = 2400V; PF 0.9 lagging; line z = j0.6 Ω/mile; T1: 25 kVA, 2400/240V, Z = 1.8/40%

23. Example 2.3 Equivalent xformer impedance Line impedance
From Ex 2.1: N1–N2: P12 = 92.9kW S12 = j45.0 kVA
Hence the current flow
Voltage at N2

24. Example 2.3 - cont Current flowing into T1
Secondary voltage referred to the high-side
Secondary voltage with turns ratio = 10

25. Alternative approach for allocated load
Difficult to obtain the DF table + xformer/customer connection
What if we only know
The metered peak demand for the whole feeder
The apparent power rating of all transformers connected to the feeder
The allocated load per transformer (either kW or kVA)

26. Example 2.2
Assume the metered max div kW demand for the lateral is 92.9 kW
For each xformer the allocated load
T1: kW1 = · 25 = kW
T2: kW2 = · 37.5 = kW
T3: kW1 = · 50 = kW

27. Voltage drop calculations
Accordingly, voltage drop can be calculated based on AF
Example 2.4: from the metered demand
For each xformer
For each line segment

28. Comparisons of the two approaches
Example 2.3
Example 2.4
Minimal difference between the two approaches
However, this feeder analysis framework still has limits in terms of its scalability and generalizability for unbalanced systems