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 haozhu@illinois.edu

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

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]

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 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]

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

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

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 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 in (2.9) is the allocated “load”

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

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]

Progress so far [Ian Dobson, ISU]

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

Example 2.1 Load survey shows For customer #1 of T1 For T1, max noncoin = 12.4 + 13.4 + 16.1 + 12.9 + 11.9 = 66.6 kW What size should the transformers be? (Max kVA demand) The allocated load results do not obey KCL!

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 = 0.3 + j0.6 Ω/mile; T1: 25 kVA, 2400/240V, Z = 1.8/40%

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

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

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)

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 = 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 Example 2.4: from the metered demand For each xformer For each line segment

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