Distribution System Modeling And Analysis - The Nature Of Loads

The nature of loads

Contents 1

Definitions

2

Individual Customer Load

3

Distribution Transformer Loading

4

Relationship Between Load and Loss Factors

5

Feeder Load

What is load? The answer to that question depends upon what type of an analysis is desired. For example, the steady-state analysis (powerflow study) of a transmission system will require a different definition of load than that used in the analysis of a secondary in a distribution feeder.

What is load? 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.

Definitions - Demand Load averaged over a specific period of time. Load can be kW, KVAR, kVA, or A.

Must include the time interval.

Example: The 15-minute kW demand is 100 kW.

Definitions – Maximum Demand

Greatest of all demands that occur during a specific time.

Must include demand interval, period, and units.

Example: The 15-minute Maximum kW demand for the week was 100 kW.

Definitions – 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.

Definitions – 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.

Definitions – Max. Diversified Demand Maximum of the sum of demands imposed by a group of loads over a particular period.

Must include demand interval, period, and units.

Example: The 15-minute max. diversified kW demand for a week was 500 kW.

Definitions – Max. Noncoincident Demand For a group of loads, the sum of the individual maximum demands without any restriction that they occur at the same time. Must include demand interval, period, and units.

Example: The maximum noncoincident 15minute kW demand for a week was 700 kW.

Definitions Demand Factor  Ratio of maximum demand to connected load.

Utilization Factor  Ratio of maximum demand to rated capacity.

Diversity Factor  Ratio of the maximum noncoincident demand to the maximum diversified demand.

Definitions Load Factor  Ratio of the average demand of any individual customer or group of customers over a period to the maximum demand over the same period.

Load Diversity  Difference between maximum noncoincident demand and the maximum diversified demand.

Individual Customer Load – Demand Interval Demand Interval  It is the period over which the load is averaged. This selected Δt period may be 15 min, 30 min, 1 hr, or even longer. Of course, there may be situations where the 15- and 30-min demands are identical.

Individual Customer Load - Demand Demand  Load averaged over a specific period of time.  The demand of an installation or system is the load at the receiving terminals averaged over a specified interval of time.  In order to define the load, demand curve is broken into equal time.  For example, in Figure 1 the selected time interval is 15 minutes.  The straight lines represent the average load in a time interval.

Individual Customer Load - Demand

15 Minute kW Demand

 The shorter the time interval, the more accurate will be the value of the load.  The average value of the load in an interval is defined as the 15-minute kW demand. 6.0

Instantaneous

5.0 4.0 3.0 2.0 1.0 6:15

6:30

6:45

Time of Day

Fig. 1 Customer demand curve

Individual Customer Load - Demand  The 24-hour 15-minute kW demand value for a customer is shown in Figure 2. kW Demand

15 10 5 0 00:15 02:45 05:15 07:45 10:15 12:45 15:15 17:45 20:15 22:45 Tim e of Day

Fig. 2 24-hour demand curve.

Individual Customer Load – Max. Demand Maximum Demand  Greatest of all demands that occur during a specific time.  During the 24-hour period (Fig. 2), there is a great variation in the demand.  The greatest of these is the 15-minute maximum kW demand.  For this customer the 15-mimute maximum kW demand occurs at 11:45 and has a value of 12.68 kW.

Individual Customer Load – Average Demand Average Demand  The average of the demands over a specified period. Total energy Average Demand = Hours  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 − min kW demand ) • hour 4

Individual Customer Load – Average Demand  The total energy consumed during the day is the summation of all of the 15-minute interval consumptions.  If the total energy consumed during the period by customer is 58.96 kWh, then the 15-minute average kW demand is computed by Total energy 58.96 Average Demand = = = 2.46 kW Hours 24

Individual Customer Load – Load Factor Load Factor  Ratio of the average demand of any individual customer or group of customers over a period to the maximum demand over the same period.  The ratio of the average load (or average demand) over a designated period of time to the peak load (or maximum demand) occurring on that period.

Individual Customer Load – Load Factor  Therefore, the load factor FLD is FLD =

average load Average 15 − min kW demnad = peak load Maximum 15 − min kW demnad

 In Figure 2, the load factor can be found by average load 2.46 FLD = = = 0.194 peak load 12.68

 Load factor gives an indication of how well the utility’s facilities are being utilized.

Individual Customer Load – Load Factor  From the utility’s standpoint, the optimal load factor would be 1.0, since the system has to be designed to handle the maximum demand.  Sometimes utility companies will encourage industrial customers to improve their load factors. One method of encouragement is to penalize the customer on the electric bill for having a low power factor.

Individual Customer Load

– Load Factor

 By energy, the load factor can be expressed as average load × T FLD = peak load × T where T = time, in days (24), weeks (168), months (730), or years (8760).

 For the annual factor, it can be expressed as Total annual energy Annual load factor = Annual peak load × 8760

Distribution Transformer Loading A distribution transformer will provide service to one or more customers. Each customer will have a demand curve similar to Figure 2. For example, there are four customers connected to the same distribution transformer. The load curves for the four customers show that each customer has his unique loading characteristic.

Distribution Transformer Loading 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

Energy 2.44 Time

1.52

3.98

1.78

Average Demand Maximum Demand

0.22

0.81

0.25

Load Factor

0.40

Diversified Demand Diversified Demand)

Demand

(Coincident

 It is the demand of the composite group, as a whole, of somewhat unrelated loads over a specified period of time.  It is assumed that one distribution transformer serves four customers discussed previously. The sum of the four 15-minute kW demands for each time interval is the diversified demand for the group in that time interval.

Diversified Demand 15-mimute Max. Diversity Demand

Customer #4 #3 Customer #2 #1

Maximum demand = 5.7 kW

Maximum demand Maximum demand==11.5 8.5kW kW

Time Day Timeofof ofDay Day Time of Day Time

Fig. 3 24-hour demand curve

23:00

615 10 15 10 410 5 2 55 0 00 000:: :11155 0:15 5 222:: :11155 2:00 5 444:: :11155 3:45 5 666:: :11155 5:30 88:: 7:15 1155 1100 9:00 ::11 55 1122 10:45 ::11 55 1 1 12:30 44:: 1155 116 14:15 6::1 155 16:00 1188 ::11 55 17:45 2200 ::11 55 19:30 2222 ::11 55 21:15

kW kW Demand kWDemand Demand kW Demand

Maximum demand = 13.1 kW

Maximum Diversified Demand The importance of the maximum diversified demand is the maximum sum of the contributions of the individual demands to the diversified demand over a specific time interval. 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.

Load Duration Curve A load duration curve can be developed for the transformer serving the four customers. Sorting in descending order, the kW demand of the transformer develops the load duration curve. The load duration curve plots the 15-minute kW demand versus the percent of time. The curve can be used to determine whether a transformer needs to be replaced due to an over-loading condition.

Load Duration Curve  For example, the load duration curve shows the transformer operates with a 15-mimute kW demand of 20 kW or greater 15% of the time.

Fig. 4 Transformer load duration curve.

Demand factor The demand factor can be defined for an individual customer. The definition is  The ratio of the maximum demand to the total connected load. Therefore, the demand factor (DF) can be expressed as maximum demand DF = total connceted demand

 The demand factor is usually less than 1.0.  It is an indicator of the simultaneous operation of the total connected load.

Demand factor For example the 15-minute maximum kW demand of Customer #1 was found to be 6.18 kW.  The total connected load will be the sum of the ratings of all of the electrical devices at the customer’s location.  Assume that this total comes to 35 kW, then 

Maximum Demand 6.18 Demand Factor = = = 0.1766 Total Connected Load 35

Connected Load The sum of the continuous ratings of the load-consuming apparatus connected to the system. Or, the sum of the ratings of the electricity consuming apparatus connected to a generating system. That is, the electric load (in watts), if all apparatus and equipment connected to the system are energized simultaneously.

Noncoincident demand Noncoincident demand  The demands of a group loads are with no restrictions on the interval.

Maximum Noncoincident Demand  The 15-minute maximum noncoincident kW demand for the day is the sum of the individual customer 15-minute maximum kW demands.  For the transformer, the sum of the individual maximum is

Maximum Noncoincident Demand = 13.1 + 8.5 + 11.5 + 5.7 = 38.8 kW

Diversity Factor It is the ratio of the sum of the individual maximum demands of the various subdivisions of a system to the maximum demand of the whole system. That is, diversity factor is the ratio of the maximum noncoincident demand of a group of customers to the maximum diversified demand of the group.

Diversity Factor Therefore, the diversity factor (FD) is Sum of Individual Maximum Demands FD = Coincident Maximum Demand Maximum Noncoincident Demands = Maximum Coincident Demand D1 + D2 + D3 +  + Dn FD = Dg n

∑ Di

= i =1 Dg

Di = maximum demand of load i, disregarding time of occurrence. Dg = D1+2+3+…+n = coincident maximum demand of group of n loads.

Diversity Factor From the definition of demand factor, we can obtain Maximum Demand = Total Connceted Demand ( TCD ) × DF then, n

∑ Di

n

∑ TCDi × DFi

FD = i =1 = i =1 Dg

Dg

Diversity Factor  The diversity factor can be equal to or greater than 1.  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 customers can be computed.  There will be a different value of the diversity factor for different numbers of customer.

Diversity Factor  Table 1 developed from a database is an example of the diversity factors for the number of customers ranging from one to 70. N 1 2 3 4 5 6 7 8 9 10

DF 1.0 1.60 1.80 2.10 2.20 2.30 2.40 2.55 2.60 2.65

N 11 12 13 14 15 16 17 18 19 20

DF 2.67 2.70 2.74 2.78 2.80 2.82 2.84 2.86 2.88 2.90

N 21 22 23 24 25 26 27 28 29 30

DF 2.90 2.92 2.94 2.96 2.98 3.00 3.01 3.02 3.04 3.05

N 31 32 33 34 35 36 37 38 39 40

DF 3.05 3.06 3.08 3.09 3.10 3.10 3.11 3.12 3.12 3.13

N 41 42 43 44 45 46 47 48 49 50

DF 3.13 3.13 3.14 3.14 3.14 3.14 3.15 3.15 3.15 3.15

Table 1 Diversity Factor

N 51 52 53 54 55 56 57 58 59 60

DF 3.15 3.15 3.16 3.16 3.16 3.17 3.17 3.17 3.18 3.18

N 61 62 63 64 65 66 67 68 69 70

DF 3.18 3.18 3.18 3.19 3.19 3.19 3.19 3.19 3.20 3.20

Diversity Factor  A graph of the diversity factors is shown in Figure 8. 3.5 Diversity Factors

3 2.5 2 1.5 1 0.5 0 1

5

9 13 17 21 25 29 33 37 41 45 49 53 57 61 65 69 Number of Customers

Fig. 5 Diversity Factor

Diversity Factor  Note in Table 2.2 and Figure 2.8 that the value of the diversity factor basically leveled out when the number of customers reached 70.  This is an important observation because it means that as viewed from the substation, the maximum diversified demand of a feeder can be predicted by computing the total noncoincident maximum demand of all of the customers served by the feeder and dividing by 3.2.

Utilization Factor It is the ratio of the maximum demand of a system to the rated capacity of the system. Therefore, maximum demand (coincident) Fu = rated system capacity

The utilization factor gives an indication of how well the capacity of an electrical device is being utilized. For transformer, it can be expressed as

Utilization Factor maximum kVA demand Fu of transformer = transformer kVA rating  For example The transformer serving four loads is rated 15 kVA. Using the 16.16 kW maximum diversified demand and assuming a power factor of 0.9. Find the utilization factor.

Utilization Factor Maximum kW demand Transformer kVA rating = Power Factor 16.16 = = 17.96 0.9 Utilization factor =

Maximum kVA demand 17.96 = = 1.197 Transformer kVA rating 15

Load Diversity Load diversity is defined as the difference between the noncoincident maximum demand and the maximum diversified demand. For the transformer, the load diversity is represented as Load Diversity = Max. noncoincident demand - Max. diversified demand

Load Diversity It is the difference between the sum of the peaks of two or more individual loads and the peak of the combined load. Then, the load diversity (LD) is LD = Max. noncoincident demand - Max. diversified demand  n  =  ∑ Di  − Dg  i =1 

Coincidence Factor  It is the ratio of the maximum coincident total demand of a group of consumers to the sum of the maximum power demands of individual consumers comprising the group both taken at the same point of supply for the same time.  The coincidence factor (FC) is coincident maximum demand sum of individual maximum demands

FC = =

Dg n

∑ Di

i =1

1 = FD

Contribution Factor It is defined as the contribution factor of the ith load to the group maximum demand. Therefore, Dg = c1 D1 + c 2 D2 + c 3 D3 +  + c n Dn

where ci is called contribution factor. class demand at time of system ( i .e., group ) peak ci = class noncoincident maximum demand n

FC =

c1 D1 + c2 D2 + c3 D3 +  + cn Dn n

∑ Di

i =1

∑ ci Di

= i =1 n ∑ Di i =1

Contribution Factor Special case Case 1 ： D1 = D 2 = D3 =  = D n n

FC =

D ∑ ci i =1

nD

, then

n

∑ ci

= i =1 n

That is, the coincident factor is equal to the average contribution factor.

Contribution Factor Case 2 ： c1 = c2 = c3 =  = cn , then n

FC =

c× ∑ Di i =1 n

=c

∑ Di

i =1

That is, the coincident factor is equal to the contribution factor.

Example 1 Problem 2.3

Example 2 There are six residential customers connected to a distribution transformer. The connected load is 9 kW for each house, and the demand factor and diversity factor for the group of six houses have been decided as 0.65 and 1.10, respectively. Determine the diversified demand of the group of six houses on the distribution transformer.

Example 3 Assume that example 2 has a system peak of 3000kW per phase and a copper loss of 0.5 percent at the system peak. Determine the following ：  The copper loss of the feeder in kilowatts per phase.  The total copper losses of the feeder in kilowatts per three-phase.

Example 4 Assume that annual peak load of a primary feeder is 2000 kW, at which the power loss, i.e., total copper loss, or ∑ I2R , is 80 kW per three-phase. Assuming an annual loss factor of 0.15, determine:  The average annual power loss.  The total annual energy loss due to the copper losses of the feeder circuits.

Example 5  Assume that there are two primary feeders supplied by one transformer. One of the feeders supplies an industrial load which occurs primarily between 8 am and 11 pm, with a peak of 2000kW at 2 pm. The other one feeds residential loads which occur mainly between 6 am and 12 pm, with a peak of 2000kW at 9 pm. Determine the following: (System peak load is 3000kW at 7 pm.)  The diversity factor of the load connected to the transformer.  The load diversity of the load connected to transformer.  The coincidence factor of the load connected to transformer.

Example 5 System peak load

3000

Residential load peak

Industrial load peak

2000

1000

12

2

4

6

8

10

A.M.

12

2

4

6

8

10

12

Noon

Transmission Line Distribution Transformer

Primary Feeder Industrial load

Reserved for Future loads

Residential load

Example 6 Use the data shown in Table 2. Note that the peak occurs at 5 P.M. Determine the following:  The class contribution factors for each of the three load classes.  The diversity factor for the primary feeder.  The diversified maximum demand of the load group.  The coincidence factor of the load group.

Time 1 2 3 4 5 6 7 8 9 10 11 12 noon 1 2 3 4 5 6 7 8 9 10 11 12 A.M.

Street Lighting 100 100 100 100 100 100 100 0 0 0 0 0 0 0 0 0 0 100 100 100 100 100 100 100

Load , kW Residential 200 200 200 200 200 200 300 400 500 500 500 500 500 500 500 500 600 700 800 1000 1000 800 600 300

Commercial 200 200 200 200 200 200 200 300 500 1000 1000 1000 1000 1200 1200 1200 1200 800 400 400 400 200 200 200

Example 7 Assume a substation supplied an annual peak load of 3500 kW. The total annual energy supplied to the primary feeder circuits is 10,000,000 kWh. The peak demand occurs in July or August and is due to air-conditioning load.  Find the annual average power demand.  Find the annual load factor.

Example 8  Use the data given in Example 7 and suppose that a new load of 100 kW with 100 percent annual load factor is supplied from the substation. The investment cost, or capacity cost, of the power system upstream, i.e., toward the generator, from the substation is $3.00/kW per month. Assume that the energy delivered to these primary feeders is $0.03/kWh.  Find the new annual load factor on the substation.  Find the total annual cost to the utility to serve this load.

Relationship Between Load and Loss Factors

Assume that the primary feeder shown in Figure 6 is connected to a variable load. PLS1

Fig. 6 The primary feeder.

P1

Figure 7 shows an arbitrary and idealized load curve.

Relationship Between Load and Loss Factors

Fig. 7 Idealized load curve.

Relationship Between Load and Loss Factors

Assume that the off-peak loss is PLS,1 at some off-peak load P1 and that the peak loss is PLS,2 at some off-peak load P2.  The load factor is Pav Pav FLD = = Pmax P2

 From Figure 7, we can obtain P2 × t + P1 × ( T − t ) Pav = T

Relationship Between Load and Loss Factors  Substituting this equation into the previous one. P2 × t + P1 × ( T − t ) FLD = P2 × T

or t P1 ( T − t ) FLD = + × T P2 T

Relationship Between Load and Loss Factors  The loss factor is FLS =

PLS ,av PLS ,max

=

PLS ,av PLS ,2

where PLS,av = average power loss. PLS,max = maximum power loss. PLS,2 = peak loss at peak load.

 From Figure 9, we also can obtain PLS ,av =

PLS ,2 × t + PLS ,1 × ( T − t ) T

Relationship Between Load and Loss Factors  Then, the loss factor can be expressed as FLS =

PLS ,2 × t + PLS ,1 × ( T − t ) PLS ,2 × T

where PLS,1 = off-peak loss at off-peak load. t = peak load duration. T-t = off-peak load duration.

 The copper losses are the function of the associated loads. PLS ,1 = k × P12

and

PLS ,2 = k × P22

Relationship Between Load and Loss Factors  Thus, the loss factor can be expressed as

( k × P22 ) × t + (k × P12 ) × ( T − t ) FLS = (k × P22 ) × T or

t  P1  FLS = +   T  P2 

2

(T − t ) T

Relationship Between Load and Loss Factors t P1 ( T − t ) FLD = + × T P2 T

t  P1  FLS = +   T  P2 

2

(T − t ) T

The load factor can be related to loss factor for three different cases:  Case 1: Off-peak load is zero, i.e. PLS ,1 = 0, since P1=0. t FLD = FLS = T  That is, the load factor is equal to the loss factor and they are equal to the t/T constant.

Relationship Between Load and Loss Factors  Case 2: Very short lasting peak, that is t → 0 . Then, T −t → 1.0 T Therefore, FLS → ( FLD ) 2

 That is, the value of the loss factor approaches the value of the load factor squared.

Relationship Between Load and Loss Factors  Case 3: Load is steady. That is, t → T . It means the difference between peak load and off-peak load is negligible. Thus,

t P1 ( T − t ) T P1 ( T − T ) FLD = + × = + × =1 T P2 T T P2 T 2

FLS =

2

t  P1  ( T − t ) T  P1  ( T − T ) +   = +   =1 T  P2  T T  P2  T

∴ FLS → FLD

Relationship Between Load and Loss Factors  That is, the value of the loss factor approaches the value of the load factor.

Therefore, in general, the relationship between loss factor and load factor can be shown as 2 FLD < FLS < FLD

An approximate formula to relate the loss factor to the load factor as 2 FLS = 0.3FLD + 0.7 FLD 2 FLS = 0.2 FLD + 0.8 FLD (T. P.C)

Feeder Load  The load that a feeder serves will display a smoothed demand curve as shown in Figure 8.  The feeder demand curve does not display any of the abrupt changes in demand of an individual customer demand curve.  The simple explanation for this is that the feeder serves with several hundred customers, and one customer is turning off a light bulb, then another customer will be turning a light bulb on.

Feeder Load

15000 10000 5000

Time of Day

Fig. 8 Feeder demand curve.

22:15

20:15

18:15

16:15

14:15

12:15

10:15

8:15

6:15

4:15

2:15

0 0:15

kW Demand

Feeder Total

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. Then, the load allocated to each transformer needs to be determined.

Application of Diversity Factors The definition of the diversity factor (DF) is the ratio of the maximum noncoincident demand to the maximum diversified demand. When diversity factor is available, then it is possible to determine the maximum diversified demand of a group of customers such as those served by a distribution transformer.

Application of Diversity Factors That is, the maximum diversified demand can be computed by: Maximum diversified Demand =

Maximum Noncoincident Demand DFn

This maximum diversified demand becomes the allocated load for the transformer.

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 maximum kW demand.

Load Survey At the end of the survey period the maximum demand vs. kWh for each customer can be plotted on a common graph. Linear regression is used to determine the equation of a straight line that gives the kW demand as a function of kWh. For example, the straight-line equation can be expressed as

Load Survey 15 Minute Maximum kW Demand (kW)

10.5

12

10

8 kWi kW1i

6

4

2

1.9

0 400 500

600

800

1000

1200

1400

kWhi Energy (kWh)

1600

1800

2000 2000

Max. kW demand = 0.1058 + 0.005014 • kWh

Load Survey Knowing the maximum demand for each customer is the first step in developing a table of diversity factors as shown in Table 2. The next step is to perform a load survey where the maximum diversified demand of group of customer is metered. This will involve selecting a series of location where demand meters can be placed.

Load Survey The meters will record the maximum demand for groups of customers ranging from at least 2 to 70. At each meter location the maximum demand of all downstream customer must also be known. With that data, the diversity factor can be computered for the given number of downstream customers.

Load Survey The first step  Knowing the maximum demand for each customer.  The results can use to find “Maximum Noncoincident demand”.

The second step  The maximum diversified demand of groups of customer is metered.  The results are used to obtain “Maximum diversified demand”

Load Survey The third step  By the previous data, the diversity factor can be computered.

Maximum Noncoincident Demand DFn = Maximum diversified Demand

Example 9 A single-phase lateral provides service to three distribution transformer as shown in Figure 9. N1

N3

N2

T1

1

2

3

4

5

N4

T2

6

7

8

9

T3

10 11 12 13 14 15 16 17 18

Fig. 9 Single-phase lateral.

Example 9  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-mimute maximum kW demand is given by the equation kWdemand = 0.2 + 0.008 • kWh

Example 9 T1 Customer

T2 Customer

#1

#2

#3

#4

#5

kWh

1523

1645

1984

1590

1456

kW

12.4

13.4

16.1

12.9

11.9

#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

T3 Customer

#12

#13

#14

#15

#16

#17

#18

kWh

2098

1856

2058

2265

2135

1985

2103

kW

17.0

15.1

16.7

18.3

17.3

16.1

17.0

Example 9  Determine for each transformer the 15mimute noncoincident maximum kW demand and, using the Table 2 (Diversity Factor), determine the 15-mimute maximum diversified kW demand.  Determine the 15-mimute noncoincident maximum kW demand and 15-minute maximum diversified kW demand for each of the line segments.

Example 9 Discussing  This Example demonstrates that Kirchhoff’s currentKCL law is (KCL) is notorobeyed obeyed not? when the maximum diversified demands are used as the load flowing through the line segments and through theWhy? transformers.  At node N1 the maximum diversified demand flowing down the line segment N1-N2 is 92.8 kW, and the maximum diversified demand flowing through transformer T1 is 30.3 kW.

Example 9  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 is that the maximum diversified demands for the line segments and transformers don’t necessarily occur at the same time.

Example 9  At the time that 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 line segment N1-N2 and the demand of transformer T1 will be 72.6 kW, not 62.5 kW.

Transformer Load Management 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.

Transformer Load Management 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. A transformer load management program is used by utilities to determine the loading on distribution transformers.

Transformer Load Management The program is primarily used to determine 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. Because the utility will have in the billing database the kWh consumed by each customer every month.

Transformer Load Management 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.

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 effective. The major disadvantage of the transformer load management method is that a database is required that specifies which transformer serve which customer.

Metered Feeder Maximum Demand 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.

Metered Feeder Maximum Demand 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. An “allocation factor” (AF) can be determined based upon the metered threephase kW or kVA demand and the total connected distribution transformer kVA

Metered Feeder Maximum Demand Metered Demand AF = kVAtotal Where metered demand can be either kW or kVA, and kVAtotal = sum of the kVA ratings of all distribution transformers.

 The allocated load per transformer is then determined by Transformer demand = AF • kVA transformer

 The transformer demand will be either kW or kVA depending upon the metered quantity.

Metered Feeder Maximum Demand

When the kW or kVA is metered by phase, 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 distribution transformer can be done by assuming nominal voltage at the substation and then computing the resulting kVA.

Example 10 Assume that the metered maximum diversified kW demand for the system of Example 9 is 92.8 kW. Allocate this load according to the kVA ratings of the three transformers. kVA total = 25 + 37.5 + 50 = 112.5 Metered demand AF = kVAtotal 92.8 = = 0.8249 kW/kVA 112.5

Example 10  The allocated kW for each transformer becomes T1 ： kW1 = 0.8249 ⋅ 25 = 20.62

kW

T2 ： kW2 = 0.8249 ⋅ 37.5 = 30.93 kW T3 ： kW3 = 0.8249 ⋅ 50 = 41.24

kW

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.

What Method to Use? If the purpose is to determine as closely as possible the maximum demand on a distribution transformer, then either the diversity factor or the transformer load management method can be used. Neither of these methods should be employed when the analysis of the total feeder is to be performed.

What Method to Use? The problem is that using those methods will result in a much larger maximum diversified demand at the substation 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, that is, allocation factor.

Voltage-Drop Calculation Using Allocation Loads

The various voltage drops will be computed using the loads allocated by two of the methods in the following examples. (Diversity factor and allocation factor) For these studies it is assumed that the allocated loads will be modeled as constant real power and reactive power.

Application of Diversity Factor The loads allocated to a line segment or a distribution transformer using diversity factors are a function of the total number of customers down stream from the line segment or distribution transformer. 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.

Example 11 For the system of Example 9, assume the voltage at N1 is 2400 volts. Compute the secondary voltages on the three transformers and calculate the percent voltage drop to the secondary of transformer T3 using the diversity factor.  Assume that the power factor of the loads is 0.9 lagging.  The impedance of the lines are: z = 0.3 + j 0.6 Ω / mile

Example 11  The ratings of the transformers are T1 : 25kVA, 2400-240 volts, Z = 1.8∠40% T2 : 37.5kVA, 2400-240 volts, Z = 1.9∠45%  Z = 2 . 0 ∠ 50 % T3 : 50kVA, 2400-240 volts, N1

5000’

N2

N3

500’

T1

1

2

3

4

5

750’

T2

6

7

8

9

N4

T3

10 11 12 13 14 15 16 17 18

Example 12 For the system of Example 9, assume the voltage at N1 is 2400 volts and 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 11. Assume the load power factor is 0.9 lagging.