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ECE 6606PD Distribution Systems Engineering ‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐ ‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐

Assignment 1 Distribution System Load Characteristics 1-a)Average Power=

total annual energy 8760 7 10 = =1141kW 8760

1-b)Annual Load Factor= F =

Average demand

LD

Peak demand 1141 = =0.326 3500 2-a) Transformer kW=

1800+2000+2000

1.15 To Calculate Transformer S, calculate Q

=5043.48 kW

-1

first Q=Ptancos (power factor) -1

Q1 =1800tancos (0.95)=591.63 kVAR Similarily Q2 =1239.49 kVAR Q3 =968.64 kVAR Transformer kVAR=

Q1 +Q2 +Q3 1.15 2

2

=2434.578 kVAR 2

2

Transformer kVA= P +Q = 5043.48 +2434.578 =5600.34 kVA 2-b)Load diversity=(1800+2000+2000)-(5043.48)=756.52 kW 2-c) First transformer's emergency rating =3125×1.25=3906.25 kVA (Not Suitable) Second transformer's emergency rating =4687×1.25=5858.75 kVA (Suitable) Choose second transformer

ECE6606PD

Distribution Systems Engineering

Assignment # 2 Distribution System Load Forecasting 1. Explain briefly the main difference between the extrapolation, the simulation, and the econometric methods for load forecast. Your answer should address the following: a. Their use for distribution system load forecast. b. The data requirements. c. The accuracy of their results. [30 marks] Lecture 2, Section 4 2. For the demand data given in Table 1, forecast the nearest three future points and calculate the MAE the RMSE over the historical data points sets using the following forecasting techniques: a. 4-order weighted moving average (WMA) b. ARIMA (2,0,0) [70 marks] Table 1 Hourly demand data variation Hour Power (kW) Hour Power (kW) Hour Power (kW) Hour Power (kW)

1 28969.26 7 22187.94 13 29886.12 19 37448.58

2 26392.32 8 22463.73 14 30487.05 20 36800.67

3 24254.58 9 22850.31 15 30160.56 21 35871.45

4 22707.33 10 24301.62 16 30874.11 22 34722.09

5 21787.5 11 26440.89 17 33837.27 23 32497.41

6 21642.96 12 28536.93 18 37946.61 24 28453.05

Solution: i) 4-order weighted moving average (WMA) The mathematical expression for the 4-order moving average model can be re-written as follows 4

Yˆ i 



Wm Y i m  W1 Y i 1W2 Y i 2W3 Y i

3W4 Y i 4 m1

i 4

The chosen weights are 0.4, 0.3, 0.2, and 0.1, respectively. Table 2 and Fig. 1 present the generated results when using the 4-order weighted moving average. These results reveal that the MAE is 2314.41 kW and the RMSE is 2848.89 kW. Table 2 Load forecasting using 4-order weighted moving average Hour (i) 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27

Actual 28969.26 26392.32 24254.58 22707.33 21787.5 21642.96 22187.94 22463.73 22850.31 24301.62 26440.89 28536.93 29886.12 30487.05 30160.56 30874.11 33837.27 37946.61 37448.58 36800.67 35871.45 34722.09 32497.41 28453.05 28453.05 28453.05

2

Power (kW) Forecasted

24534.696 23017.347 22160.358 21996.297 22149.216 22481.127 23287.281 24683.277 26492.394 28233.867 29512.131 30041.256 30483.834 31877.958 34520.703 36218.28 36927.891 36673.158 35755.263 34269.948 31662.006 29888.826 28857.486

e(i)

2747.20 1374.39 27.58 467.43 701.09 1820.49 3153.61 3853.65 3393.73 2253.18 648.43 832.85 3353.44 6068.65 2927.88 582.39 1056.44 1951.07 3257.85 5816.90

{e(i)}^2

7547085.86 1888939.63 760.77 218493.61 491532.80 3314194.76 9945249.72 14850641.44 11517376.16 5076833.63 420460.17 693645.79 11245533.01 36828537.10 8572463.73 339178.11 1116067.59 3806666.34 10613606.17 33836302.34

40 35 30 D e 25 m 20 an d 15 ( 10 5 0 1

2

3

4

5

6

7

8

9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 Time Intervals (hours) Actual

Forecasted

Fig. 1 Load forecasting using 4-order weighted moving average

ii) ARIMA (2,0,0) The

forecasting model for the ARIMA (2,0,0) Yˆ i a b1 Y i 1b2 Y

mathematically

by,

i 3, 4, 5,......

i 2

where a , b , 1 and

is expressed

b2 represent the model coefficients that can be estimated using the Least

Square Method as follows, ⎡a ⎤ ⎢ ⎥ T T A  b   . 1  . Y 1 ⎢ ⎥ ⎢ b2 ⎥ ⎣ ⎦ Y Y 1 ⎤ ⎥ 2 Y 2 ⎥ ⎡1 ⎢1 ⎥ Y ⎢ ⎥   3 ⎢ ⎢



⎣1 Y (n 1)



Y (n 2)⎦

Y  3 ⎡ ⎢ ⎥⎤ Y 4 ⎢ ⎥ ⎥ Y ⎢ ⎢ ⎥  n⎦  ⎣ There for this problem the model coefficients are ⎡a ⎤ ⎡3009.50⎤ ⎢ ⎥ ⎢ ⎥ A  b1 ⎢

⎥



⎢

1.726

⎥

⎢ b2 ⎥ ⎣ ⎦

⎢⎣- 0.8318⎥⎦

3

Table 3 and Fig. 2 present the generated results when using the ARIMA (2,0,0). These results reveal that the MAE is 796.71 kW and the RMSE is 1134.88 kW. 45 40 35 D e m an d (

30 25 20 15 10 5 0 1

2

3

4

5

6

7

8

9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 Time Intervals (hours) Actual

Forecasted

Fig. 2 Load forecasting using ARIMA (2,0,0) Table 3 Load forecasting using ARIMA (2,0,0) Hour (i) 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27

Actual 28969.26 26392.32 24254.58 22707.33 21787.5 21642.96 22187.94 22463.73 22850.31 24301.62 26440.89 28536.93 29886.12 30487.05 30160.56 30874.11 33837.27 37946.61 37448.58 36800.67 35871.45 34722.09 32497.41 28453.05 28453.05 28453.05

4

Power (kW) Forecasted

24466 22920 22028 21727 22243 23303 23326 23764 25947 28433 30271 30856 30771 29708 31211 35732 40360 36082 35378 34313 33102 30218 25088 28452 28452

e(i)

211.42 212.67 240.50 84.04 55.06 839.27 475.69 537.62 493.89 103.93 384.88 368.95 610.44 1166.11 2626.27 2214.61 2911.42 718.67 493.45 409.09 604.59 1764.95

{e(i)}^2

44698.42 45228.53 57840.25 7062.72 3031.60 704374.13 226280.98 289035.26 243927.33 10801.44 148132.61 136124.10 372636.99 1359812.53 6897294.11 4904497.45 8476366.42 516486.57 243492.90 167354.63 365529.07 3115048.50

Distribution Systems Engineering ECE 6606PD ‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐ ‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐

Assignment 3

Distribution System Planning Q1) Lecture 3, Sections 1, 3 and 4 Q2) it is required to calculate the cost function between different sections

 

Since it is a transportation problem, then we assume a separate cable is extended from each substation to each load point, no tapping‐of is allowed (tapping of is considered only in a transhipment problem) Also since there is no limit to the power that could be supplied by each substation, each substation is assumed to supply the full load of a load point if associated with the least cost function for this load point

For straight distance: cost function is 1 pu. For straight river crossing: cost function is 10 pu. Thus the following table can be constructed Load center 1 2 3 4 5 6 7 8 9

Shortest distance from1 (4‐1) 2 (4‐1‐2) 5 (4‐7‐8‐9‐6‐3) 0 1 (4‐5) 4 (4‐7‐8‐9‐6) 1 (4‐7) 2 (4‐7‐8) 3 (4‐7‐8‐9)

Shortest distance from4 (9‐8‐7‐4‐1) 3 (9‐8‐5‐2) 2 (9‐6‐3) 3 (9‐8‐7‐4) 2 (9‐8‐5) 1 (9‐6) 2 (9‐8‐7) 1 (9‐8) 0

The minimum distance table is constructed for each load point Load point 1 2 3 4 5 6 7 8 9

Closest substation SS1 SS1 SS2 SS1 SS1 SS2 SS1 SS2 SS2

Cost function 1 2 2 0 1 1 1 1 0

Total cost function is 9 SS1 is used to supply load points: 1, 2, 4, 5, and 7 SS2 is used to supply load points 3, 6, 8 and 9 Rating of SS1=2+4+1+5+2=14 MVA Rating of SS2=4+6+5+5=20 MVA Results are illustrated in following table Substation Fed load points Total size

SS1 12457 14 MVA

SS2 3689 20 MVA

ECE6606PD

Distribution Systems Engineering

Assignment # 4 Solution Distribution System Automation & Demand Side Management 1. Discuss the main factors influencing efficient and reliable load supply to customers. Lecture 4, Section 1 [30 marks] 2. Explain briefly the main basic areas required to be developed in order to implement distribution system automation. Lecture 4, Section 6 [30 marks] 3. Discuss the different requirements of successful load management and its possible impacts Lecture 4, Section 8 [40 marks]

Assignment # 5 Solution Sub-Transmission Lines and Non-Technical Distribution Substations Design Factors 1. Discuss briefly the different types of sub-transmission circuit’s configurations. Your answer should address the following: a. The one line diagram of the electric circuit. b. The reliability of the configuration. c. The relative cost of each configuration. d. The main drawbacks of each configuration. Lecture 5, Section 2.1

2. Discuss briefly how the most optimal substation locations (sites) are determined and the different factors affecting the selection process. Lecture 5, Section 4

Assignment # 6 Solution Distribution Substation Design Aspects 1. Discuss briefly the different types of substation bus configurations. Your answer should address the following: a. The one line diagram of the electric circuit. b. The possible operating voltages for each configuration. c. The main drawbacks of each configuration. d. The main advantages of each configuration Lecture 6, Section 3

2. A distribution substation services a square area with the substation at the center of the square. Assume that the substation is served by four three phase four wire 13.2/22.9 KV grounded-wye primary feeders. The feeder mains are made of either #2 AWG copper or #1/0 ACSR conductors. The three phase open wire overhead lines have a geometric mean spacing of 37 in 2 between phase conductors. Assume a lagging-load power factor of 0.9 and a 1000 KVA/mi uniformly distributed load density. From tables, the conductor ampacity for #2 AWG copper is 230A. (a) Consider thermally loaded feeder mains: Since, the thermally limited case is considered, therefore, the feeder conductor is equal to its current carrying capacity, i.e. I = Imax = 230 A. (i) Maximum load per feeder. S feeder  3 x VLL 3

x I max

x 22.9 x 230

9122.7 KVA (ii) Substation size. Ssubstation 4 x S feeder 4 x 9122.57 36490.8 KVA

1

(iii) Substation spacing, both ways. S feeder Afeeder x D 2

9122.7 l x 1000 4 l 4 3.02 mi Substation Spacing, both Ways 2 l4 6.04 mi (iv) Total percent voltage drop from the feed point to the end of the main. From Fig. 1, k for #2 AWG copper at 22.9 kV is 0.00025 2 3  x k x D x l

%VD 4

4

23

 x 0.0025 x 1000 x (3.02) 3 4.59 %

3

(b) Consider voltage drop-limited feeders which have 3% voltage drop and find: (i) Substation spacing, both ways. 2 %VD 3  x k x D x l 4

4

23

3

 x 0.00025 x 1000 x l

l4

3 2.621 mi

3 4

Substation Spacing, both Ways 2 l4 5.242 mi (ii) Maximum load per feeder. 2 S feeder Afeede x D x 1000 r l 4 2 S feeder (2.621) x 1000 6869.641 KVA (iii) Substation size. Ssubstation 4 x S feeder 4 x 6869.641 27.47856 MVA

(iv) Ampere loading of the main in per unit of conductor ampacity.

I

S feeder 3 x VLL

36869.941 x 22.9 

173.196 A

2

I pu

 I

I

max

173.19 0.753  6 230

Fig. 1 K constant curves for copper conductors with 0.9 lagging power factor

3

Distribution Systems Engineering ECE 6606PD -------------------------------------------------------------------------------------------------------------------------

Assignment 7

Primary Distribution Systems Design and Operation 1) (40 marks) Explain briefly the diferent primary

distribution system configurations. Your answer should address the following: a) The single line diagram of each configuration. b) The main advantage of each configuration. c) The disadvantages of each configuration. d) The degree of reliability of each configuration. Lecture 7, Section 3 2) (60 Marks) - First, determine the k factor for the main and the

lateral From Figure 2, The K factor for the main = 0.0004 and for the lateral = 0.001 -

Then calculate the maximum diversified demand

-

Determine the peak load for each substation as follows

-

For design A: The voltage drop will be the summation of the voltage drops of the shown lines

-

For design B: The voltage drop will be summation of the voltage drop of the shown three lines

Total Voltage Drop 1.6 1.6 0.0625 3.2625% Thus , design A is better than Design B

ECE6606PD

Distribution Systems Engineering

Assignment # 8 Solution Secondary Distribution Systems Design, Services and Metering 1. Explain briefly the different secondary distribution system configurations. Your answer should address the following: a. The single line diagram of each configuration. b. The main advantage of each configuration. c. The disadvantages of each configuration. d. The degree of reliability of each configuration. Lecture 8, Section 1.2

2. Discuss the main components of the secondary distribution system. Your answer should consider the following issues: a. The secondary system voltage level. b. The design consideration of the secondary system. c. The degree of reliability of each component in the secondary system. Lecture 8, Section 1.1

Assignment # 9 Solution Voltage Drops and Power Loss Calculations 2

1. Figure #1 shows a square-shaped service area (A = 4 mi ) with a uniformly distributed load density 2

of D kVA/mi and 2 mi of #4/0 AWG copper overhead main from a to b. There are many closely spaced primary laterals, which are not shown in the square-shaped service area of the figure. In this voltage-drop study, use the precalculated voltage-drop curves of Figure #2 when applicable. Use the nominal primary voltage of 19,920/34,500 V for a three-phase four-wire wye-grounded 2

system. Assume that at peak loading the load density is 1000 kVA/mi and the lumped load is 2000 2

kVA, and that at off-peak loading the load density is 333 kVA/mi and the lumped load is still 2000 kVA, The lumped load is of a small industrial plant working three shifts a day. The substation bus voltages are 1.025 pu V of 19,920 base volts at peak load and 1.000 pu V during off-peak load. The transformer located between buses c and d has a three-phase rating of 2000 kVA and a deltarated high voltage of 34,500 V and grounded-wye- rated low voltage of 277/480 V. It has 0 + j0.05 per unit impedance based on the transformer ratings. It is tapped up to raise the low voltage 5.0 percent relative to the high voltage, i.e., the equivalent turns ratio in use is (19,920/277) x 0.95. Use the given information and data for peak loading and determine the following: a- The percent voltage drop from the substation to point a, from a to b, from b to c, and from c to d on the main. b- The per unit voltages at the points a, b, c, and d on the main. c- The line-to-neutral voltages at the points a, b, c, and d.

Figure #1, Question #1

Figure #2, Question #1 Solution: a- The kVA rating of the square-shaped service area is Ssquare = Dsquare x Area = 1000 x 4 = 4000 kVA. The total kVA load on the main feeder is Stotal = Ssquare + Slumped = 4000 + 2000 = 6000 kVA. Using Fig. 2 the K constant for # 4/0 Copper is 0.000065 and that for # 4 Copper is 0.00017. Then, the %VD from the substation to bus A is %VDOA = K x Stotal x l = 0.000065 x 6000 x 1 = 0.4 % . The %VD from bus A to bus B is

l

%VDAB = (K x Ssquare x

2

) + (K x Slumped x l)

= (0.000065 x 4000 x 1) + (0.000065 x 2000 x 2) = 0.52 % . The %VD from bus B to bus C is %VDBC =K x Slumped x l =0.00017 x 2000 x 2 = 0.68 % . The %VD from bus O to bus C is %VDOC =%VDOA +%VDAB +%VDBC = 0.4 + 0.52 + 0.68 = 1.6 % . The per unit voltage at bus C (similarly at the lumped load terminal) is: Vc = Vo - VDOC = 1.025 -0.016 = 1.009 pu To find the %VD from bus C to bus D: The lumped load current is I

Slumped



3 x VC ,LL

2000 kVA

3 x 1.009 x 34.5 kV



The base load current is S lumped

Ibase 

3 x VC ,LL

33.17 A



2000 kVA

33.47 A

3 x 34.5 kV

The per unit lumped load current is 33.17 I  0.99 pu I pu  I 33.47 base

Since pf = 0.9, therefore = 25.84and sin = 0.4359. and the low voltage side has been tapped up 5%

Then the %VD from bus C to bus D is %VDC = )

I (R cos  X sin Vbase

0.05 

0.99 x 0.05 x 0.4359 1

0.0 5

= -2.84 % .

b- The per unit values for the voltages at buses A, B, C, and D on the main: VA = V0 – V0A = 1.025 – 0.004 = 1.021 pu VB = VA – VAB = 1.021 – 0.0052 = 1.0158 pu B

VC = VB – VBC = 1.0158 – 0.0068 = 1.009 B

pu VD = VC – VCD = 1.009 – (-0.0284) = 1.0374 pu c- The line-to-neutral voltages at buses A, B, C, and D on the main: VA = 19,920 x 1.021 = 20,338.32 V VB = 19,920 x 1.0158 = 20,234.73 V B

VC = 19,920 x 1.009 = 20,099.28 V VD = 277 x 1.0374 = 287.36 V

2. Figure 3 show a single-line representation of a three-phase, 69 kV network. Substation 1 supplies substations 2 and 3. Substations 2 and 3 are connected via a tie line. Calculate: a. The voltage difference between substations 2 and 3 when the tie line is open. b. The line currents when the tie line is connected. c. The total power loss when the tie line is connected.

Figure #3, Question #2 Solution:

I L1 112.5  j 54.49 125 25.84 A

For the 125 A, PF = 0.9 load;

For the 195 A, PF = 0.85 load; I 165.75  j 102.72 195 31.78 A L2 a. when the tie line is open as shown in Figure # 4 we have;

I1 I L1 112.5  j 54.49 125 25.84 A

and

I 2 I L2 165.75  j 102.72 195 31.78 A

Figure # 4, Question #2 j 1.8V1 199.332 j 153.459  V1 112.5  j 54.49x 0.9 

The voltage at substation 2 (V2) can be expressed by; V2 V1 I 1 x Z12 

Similarly, the voltage at substation 3 (V3) can be expressed by; V3 V1 I 2

x Z13 V1 165.75  j 102.72 x

1 

j 1V1 268.47 j 63.03 

Subtract the last two equations from each other, the voltage difference between substations 2 and 3 (V23) can be found as V23 V2 V3 199.332  j 153.459 268.47  j 63.0369.138 j 90.429 113.8 52.6 V

b. when the tie line is connected as shown in Figure # 5 we have;

Figure # 5, Question #2

ECE6606PD

Distribution Systems Engineering

I1 I 3 I L1 112.5  j 54.49 125 25.84 A

and

I 2 I 3 I L2 165.75  j 102.72 195 31.78 A Therefore, I1 112.5 j 54.49  I 3

and

I 2 165.75 j 102.72 I 3

Using kVL, we get; I1 x Z12 I 2

x Z13 I x Z 23 0 3

Substituting by I1 and I2 in the last equation we get;

112.5 j 54.49 I 3 x 0.9 j 1.8 165.75 j 102.72 I 3 x 1 j 1  I x 1 j 20 3

 j 1.8 165.75 j 102.72x 112.5 j 54.49x0.9 1  

Therefore,

I3 

0.9 

j 1

j 1.8 1 j 1 1 j 2  

I 3 7.43 j 18.89 20.368.54 

A

I1 112.5 j 54.49 7.43 j 18.89105.07  j 73.38 128.1634.9 A  I 2 165.75 j 102.72  7.43  j 18.89173.18 j 83.83 192.4025.83 A c. The total power loss when the tie line is connected can be calculated as follows; 2 2 Ploss

ECE6606PD



3 x I

2

x r12

I

2

x r13  I 3

Distribution Systems Engineering x r23



1

x 0.9

3 x

128.16

x1

192.4

2

2

156.64 kW

20.3

2



x1

Distribution Systems Engineering ECE 6606PD -------------------------------------------------------------------------------------------------------------------------

Assignment 10

Application of Capacitors for Distribution Systems 1) (20 marks) Power capacitors are used to improve

distribution system performance. Discuss the benefits for using power capacitor in distribution systems. (Lecture 10, Sections 4.3 & 4.4) (20 marks) Explain the main diference between applying series and shunt capacitors to distribution system and also identify the main functions and the target applications of each capacitor. (Lecture 10, Sections 5 & 6) 2) (60 marks) Assume that a three – phase 400-hp, 50-

Hz, and 4160-V star connected induction motor has a full load efficiency of 85%, a lagging pf of 0.75, and is connected to a feeder. If it is desired to correct the power factor of the load to 0.88 lagging, by connecting three capacitors at the load, determine following: a) The rating of capacitors bank in Kilovars (without any approximations). b) The capacitance of each unit if capacitors are connected in delta (in microfarads). c) The capacitance of each unit if capacitors are connected in star (in microfarads). Solution a) The input power of the induction motor can be found

as:

P  hp)

(400 hp)(0.7457 kW /

350.92 kW

0.85

The reactive power of the motor at the uncorrected power factor is:

Distribution Systems Engineering ECE 6606PD -------------------------------------------------------------------------------------------------------1 -----------------Q P 350.92 tan(cos 0.75)

tan 1

1

Q1 350.920.8819 309.48 k var

The reactive power of the motor at the corrected power factor is 1

Q P tan

350.92 tan(cos 0.88)

2

2

Q2 350.920.5397 189.41k var Thus, the reactive power provided by the capacitor equals

Qc Q1 Q2

309.48 189.41 120.07 k var

b) The capacitor reactance can be calculated as

X c, phase

V 

2 phase

Qc, phase

1 

2 fC

Thus, the capacitance of the capacitor is calculated as

C Where

Qc, phase 2 f  V

2 phase

Q 

Q



120.07

k var

c c, phase

3

3

For delta connected capacitor, Vphase is equal to 4160 V (The voltage across each individual capacitor is the total line voltage) as shown in Figure 1

Figure 1 Substituti ng

120.07 1000 C  3250 41602 7.36F c) In this case, Vphase will be equal to (4160/sqrt(3))

as shown in Figure 2

Figure 2 Substituti ng

C

120.07 1000 4160 2 22.09F 3260 ( ) 3

Comment: Most utilities prefer to install delta connected capacitors as the capacitance required in this case is (1/3) the capacitance required for star connected capacitors. However, it has to be noted that delta connected capacitors are subjected to (sqrt (3)) more voltage than star connected capacitors

Assignment 11 Distribution System Voltage Regulation 1) (30 marks) Voltage regulators are used to improve the

quality of the distribution system performance. Discuss briefly the main components of the automatic voltage regulator and their main functions. Also explain the main function of the line drop compensator (LDC) and the electric circuit of the LDC. [Sections 6 & 7]

2) (70 marks) An industrial customer’s bus is located at

the end of a 3 mile primary line with a resistance of 0.3 Ω/mi and an inductive reactance of 0.8 Ω/mi. The customer’s transformer is rated 5000 kVA with transformer impedance of 0 + j0.05 pu Ω based on the rated kVA. Assume that the industrial load is at the annual peak of 4000 kVA at 80 % lagging power factor. Select a proper three-phase capacitor bank size to be connected to the industrial load 4 kV bus to achieve the following goals: a) Produce a voltage rise of at least 0.02 pu. b) Raise the ON-peak power factor to at least 88 % lagging power factor. Hint: Use multiples of three-phase, 150 kvar capacitor units in sizing the required capacitor bank. - The primary voltage of the transformer is 12.8 kV. -

Solution For the original load (4000 kVA at 80 % lagging power factor), 1

S o 4000 cos (0.8) 400036.87

kVA

After correcting the load power factor to 88 % lagging, the active power component will not be afected. Therefore,

Pn Po

So pfo

Howeve Sn r

4000 0.8 3200 kW 3200 Sn 0.88

pfn

Pn Sn 3636.364 kVA Then, it is required to find the old and new reactive powers

Q 

2

) (P ) 

(S

o

o

Q 

2

o 2

) (P ) 

(S

n

n

2

2400 kVAR 40002 32002 3636.362 32002 1727.17 kVAR

n

Therefore, the additional reactive power required by installing the capacitor bank to raise the ON-peak power factor to at least 88 % lagging power factor is

Qc Qo Qn

2400 1727.17 672.83 kVAR

Using multiples of three-phase 150 kvar capacitor units, then the required reactive power of the capacitor bank is 750 kVAR (this value will raise the power factor to 88.88 %) Considering the transformer primary voltage as 12.8 kV, the transformer impedance can be calculated as follows;

X X tr

tr tr

(base) 0.05

( pu) X

3 2

(12.810 ) 5000 10

1.6384

3

The resultant voltage rise from installing the 750 kVAR capacitor bank is

VR( pu)



Q X L

1000 (kVB,LL )



750 (0.8

2

3 1.6384) 2 pu 0.018486 1000 (12.8)

This is less than the required 2% Thus, 750 KVAR don’t meet the design specifications

Increase the capacitor size to 900 KVAR

VR( pu) 

Q 900 (0.8 3 1.6384) 2 X L  0.0222 pu 1000 (12.8) 2 1000 (kVB,LL )

Check for the new power factor 1 Q 1 Q Qc 1 2400 900 pf cos(tan ( n )) cos(tan ( o ) cos(tan ( ) n Pn 0.905 Pn pfn 3200

Assignment 12 Distribution Service Reliability 1) (50 marks) Improvement of electric power delivery

reliability is an important task that is carried out by the electric utility due to the high cost of customer outages. Explain briefly the main reliability indices that are commonly used in measuring distribution system reliability. Also explain the widely used reliability practices in distribution systems. Your answer should address this subject for the following customer’s type: Residential, light load commercial, commercial, industrial, and agricultural. [Sections 3 & 5]

2) (50 marks) Each component in the system has its

internal risk of failure along with the external factors. Discuss briefly how the diferent existing components in the distribution system can afect the system reliability. [Section 6.1]

Notes -

One typed page per essay-type question Submission Due Date: April, 5, 2013