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ECE 6606PD Distribution Systems Engineering

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Assignment

1

Distribution

System

Load

Characteristics

1-a)Average Power= total annual₈₇₆₀ energy 107

=₈₇₆₀=1141kW

1-b)Annual Load Factor= F = Average demand

LD Peak demand = 1141 =0.326 3500 2-a) Transformer kW= 1800+2000+2000 =5043.48 kW 1.15

To Calculate Transformer S, calculate Q first Q=Ptancos-1 (power factor)

Q =1800tancos-1 (0.95)=591.63 kVAR Similarily

Q₂ =1239.49 kVAR Q₃ =968.64 kVAR

Transformer kVAR= Q1 +Q2 +Q3 _{=2434.578 kVAR}

1.15

Transformer kVA= P2 +Q2 = 5043.482 +2434.5782 =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)

Table 1 Hourly demand data variation

[70 marks] Hour 1 2 3 4 5 6 Power (kW) 28969.26 26392.32 24254.58 22707.33 21787.5 21642.96 Hour 7 8 9 10 11 12 Power (kW) 22187.94 22463.73 22850.31 24301.62 26440.89 28536.93 Hour 13 14 15 16 17 18 Power (kW) 29886.12 30487.05 30160.56 30874.11 33837.27 37946.61 Hour 19 20 21 22 23 24 Power (kW) 37448.58 36800.67 35871.45 34722.09 32497.41 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







W_m 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) Power (kW) {e(i)}^2

Actual Forecasted e(i)

1 28969.26 2 26392.32 3 24254.58 4 22707.33 5 21787.5 24534.696 2747.20 7547085.86 6 21642.96 23017.347 1374.39 1888939.63 7 22187.94 22160.358 27.58 760.77 8 22463.73 21996.297 467.43 218493.61 9 22850.31 22149.216 701.09 491532.80 10 24301.62 22481.127 1820.49 3314194.76 11 26440.89 23287.281 3153.61 9945249.72 12 28536.93 24683.277 3853.65 14850641.44 13 29886.12 26492.394 3393.73 11517376.16 14 30487.05 28233.867 2253.18 5076833.63 15 30160.56 29512.131 648.43 420460.17 16 30874.11 30041.256 832.85 693645.79 17 33837.27 30483.834 3353.44 11245533.01 18 37946.61 31877.958 6068.65 36828537.10 19 37448.58 34520.703 2927.88 8572463.73 20 36800.67 36218.28 582.39 339178.11 21 35871.45 36927.891 1056.44 1116067.59 22 34722.09 36673.158 1951.07 3806666.34 23 32497.41 35755.263 3257.85 10613606.17 24 28453.05 34269.948 5816.90 33836302.34 25 28453.05 31662.006 26 28453.05 29888.826 27 28857.486 2

40 35 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

ii) ARIMA

(2,0,0)

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

The forecasting model for the ARIMA (2,0,0) is expressed mathematically by,

Yˆ



i



a b₁ Y



i 1



b₂ Y



i 2



i 3, 4, 5,... where a , b1 ,

and

b₂ represent the model coefficients that can be estimated using the Least Square Method as follows,

⎡a ⎤ A ⎢b ⎥



T .



1.T Y ⎢1 _⎥ ⎢⎣b2 ⎥⎦ ⎡1 ⎢ ⎢1 ⎢ ⎢ Y



2



Y



3



Y



1



⎤ Y



2



⎥ ⎥ ⎥ ⎣1 Y (n 1) Y (n _2)⎦ ⎡Y



3



⎤ ⎢ ⎥ Y _⎢⎢Y



_⎥4



⎥ ⎢

_

n



⎥

There for this problem the model coefficients are

⎡a ⎤ _{⎡3009.50⎤} ⎢ ⎥ ⎢ ⎥ A _⎢b₁ ⎥ ⎢ ⎦ D e m an d ( ⎥ ⎣

1.726

⎥ _⎢⎣b

2 ⎥⎦ ⎢_{⎣- 0.8318⎥⎦}

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 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) Power (kW) {e(i)}^2

Actual Forecasted e(i)

1 28969.26 2 26392.32 3 24254.58 24466 211.42 44698.42 4 22707.33 22920 212.67 45228.53 5 21787.5 22028 240.50 57840.25 6 21642.96 21727 84.04 7062.72 7 22187.94 22243 55.06 3031.60 8 22463.73 23303 839.27 704374.13 9 22850.31 23326 475.69 226280.98 10 24301.62 23764 537.62 289035.26 11 26440.89 25947 493.89 243927.33 12 28536.93 28433 103.93 10801.44 13 29886.12 30271 384.88 148132.61 14 30487.05 30856 368.95 136124.10 15 30160.56 30771 610.44 372636.99 16 30874.11 29708 1166.11 1359812.53 17 33837.27 31211 2626.27 6897294.11 18 37946.61 35732 2214.61 4904497.45 19 37448.58 40360 2911.42 8476366.42 20 36800.67 36082 718.67 516486.57 21 35871.45 35378 493.45 243492.90 22 34722.09 34313 409.09 167354.63 23 32497.41 33102 604.59 365529.07 24 28453.05 30218 1764.95 3115048.50 25 28453.05 25088 26 28453.05 28452 27 28452 4 D e m an d (

ECE 6606PD

Distribution Systems Engineering

‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐ ‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐

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

The minimum distance table is constructed for each load point

Load point Closest substation Cost function

1 SS1 1 2 SS1 2 3 SS2 2 4 SS1 0 5 SS1 1 6 SS2 1 7 SS1 1 8 SS2 1 9 SS2 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 SS1 SS2

Fed load points 1 2 4 5 7 3 6 8 9

Distribution Systems Engineering

ECE6606PD

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.

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 between phase conductors. Assume a lagging-load power factor of 0.9 and a 1000 KVA/mi2 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

V_LL

x I _max

 x 22.9 x 230

9122.7 KVA (ii) Substation size.

S_substation 4 x S _feeder 4 x 9122.57 36490.8 KVA

(iii) Substation spacing, both ways.

S _feeder Afeeder x D

9122.7 l 2 _{x 1000}

l₄ 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

%VD _2 x k x D x l 3 4 3 4 2 x 0.0025 x 1000 x (3.02)3 3 4.59 %

(b) Consider voltage drop-limited feeders which have 3% voltage drop and find: (i) Substation spacing, both ways.

%VD _2 x k x D x l 3 4 3 4 3 2 x 0.00025 x 1000 x l 3 3 4 l₄ 2.621 mi

Substation Spacing, both Ways 2 l₄ 5.242 mi (ii) Maximum load per feeder.

2 S _feeder A_feede r x D l₄ 2 x 1000 S _{feeder (2.621) x 1000} 6869.641 KVA (iii) Substation size.

S_substation 4 x S _feeder 4 x 6869.641 27.47856 MVA

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

S _feeder I  6869.941 173.196 A 4 3 x V_LL 3 x 22.9

I _pu  _I I max 173.19 6 230 0.753

---

---ECE 6606PD

Distribution Systems Engineering

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

-

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%

Distribution Systems Engineering

ECE6606PD

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.

Assignment # 9 Solution

Voltage Drops and Power Loss

Calculations

1. Figure #1 shows a square-shaped service area (A = 4 mi2) with a uniformly distributed load density

of D kVA/mi2 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 system. Assume that at peak loading the load density is 1000 kVA/mi2 and the lumped load is 2000 kVA, and that at off-peak loading the load density is 333 kVA/mi2 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 delta-rated high voltage of 34,500 V and grounded-wye- delta-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 #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 % .

%VDAB = (K x Ssquare x l ) + (K x Slumped x l) 2 = (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   2000 kVA

3 x



1.009 x 34.5 kV



33.17 A

The base load current is

I_base  S lumped 3 x V_{C ,LL} 2000 kVA 3 x 34.5 kV 33.47 A

The per unit lumped load current is

I _{pu } I I base

33.17 0.99 pu 33.47

Since pf = 0.9, therefore = 25.84and sin = 0.4359.

and the low voltage side has been tapped up 5%

x V_{C ,LL}

3

Then the %VD from bus C to bus D is %VD = I (R cosX sin ) V_base 0.05  0.99 x 0.05 x 0.4359 1 0.0 5 = -2.84 % . C

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 VBB = VA – VAB = 1.021 – 0.0052 = 1.0158 pu VC = VBB – VBC = 1.0158 – 0.0068 = 1.009 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

VBB = 19,920 x 1.0158 = 20,234.73 V

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:

For the 125 A, PF = 0.9 load;

I _L1 112.5 j 54.49 125 25.84 A

For the 195 A, PF = 0.85 load; I _L2 165.75 j 102.72 195 31.78 A 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

I ₂ I _L2 165.75 j 102.72 195 31.78 A

Figure # 4, Question #2 The voltage at substation 2 (V2) can be expressed by;

V₂ V₁ I₁ x Z₁₂ V₁ 



112.5 j 54.49



x



0.9 

j 1.8



V₁ 



199.332 

j 153.459



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

ECE6606PD

Distribution Systems Engineering

I₁ I ₃ I _L1 112.5 j 54.49 125 25.84 A I ₂ I₃ I _L2 165.75 j 102.72 195 31.78 A and Therefore, I₁ 



112.5 j 54.49



I₃ I ₂ 



165.75 j 102.72



I₃ and Using kVL, we get; I₁ x Z₁₂ I 2 x Z13 I 3 x Z ₂₃ 0

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



112.5 j 54.49



I₃



x



0.9 j 1.8







165.75 j 102.72



I₃



x



1 j 1



I 3 x



1  j 2



0 Therefore,



112.5 j 54.49



x



0.9  j 1.8







165.75 j 102.72



x



1  j 1



I ₃ 

_

0.9  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;

ECE6606PD

Distribution Systems Engineering

3 x



I 2 x r12 _{ I} 2 x _r 13  I3 x r23



3 x





128.16



2 x 0.9 



192.4



2 x 1 



20.3



2 x 1



156.64 kW 1

---

---ECE 6606PD

Distribution Systems Engineering

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

_

(400

hp

)(0.7457

kW

/

hp

)

0.85



350.92 kW

The reactive power of the motor at the uncorrected power

factor is:

---

---ECE 6606PD

Distribution Systems Engineering

Q



P

tan





350.92 tan(cos

1

0.75)

1 1

Q

₁



350.92



0.8819



309.48 k var

The reactive power of the motor at the corrected power

factor is

Q



P

tan





350.92 tan(cos

1

0.88)

2 2

Q

₂



350.92



0.5397



189.41k var

Thus, the reactive power provided by the capacitor equals

Q

_c



Q

₁



Q

₂



309.48



189.41



120.07 k var

b)

The capacitor reactance can be calculated as

X

_c,

phase

V

2

₁



phase



Q

_{c, phase}

₂

_

_fC

Thus, the capacitance of the capacitor is calculated as

C

_

Q

c, phase₂ phase

Where

Q



Q

c



120.07

k var

c, phase

3

3

For delta connected capacitor, V

phase

is equal to 4160 V

(The voltage across each individual capacitor is the

total line voltage) as shown in Figure 1

f

V

2

Substituti

ng

Figure 1

C

_

120.07

₃

_

₂

_

_

₅₀



1000

_

₄₁₆₀

_

2

7.36



F

c)

In this case, V

_phase

will be equal to (4160/sqrt(3))

as shown in Figure 2

Substituti

ng

C

_

120.07



1000

_

22.09



F

3



2





60



(

4160

)

2

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),

S



4000



cos

1

_(0.8)



4000



36.87

kVA

After correcting the load power factor to 88 % lagging, the

active power

P

_n



P

_o



S

_o



pf

_o



4000



0.8



3200 kW

Howeve

r

P

_n



S

_n



pf

_n



3200



S

_n



0.88

S

_n



3636.364 kVA

Then, it is required to find the old and new reactive powers

Q



(S

)

2



(P )

2





2400 kVAR

o o o

Q



(S

)

2



(P )

2





1727.17 kVAR

n n 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

Q

_c



Q

_o



Q

_n



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

( pu)

_

_X

(12.8



10

3

)

2

(base)



0.05





1.6384



tr tr tr

5000



10

3

The resultant voltage rise from installing the 750 kVAR

capacitor bank is

VR( pu)

_

_Q

_

_X

L

3636.36

2

_

₃₂₀₀

2

1000 (kV

_B,LL

)



750



(0.8



3



1.6384)

1000 (12.8)



0.018486 pu

2

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

X

_L

1000 (kV

_B,LL

)



900



_{1000 (12.8)}

(0.8



3



1.6384)

2



0.0222 pu

Check for the new power factor

pf



cos(tan

1

(

Q

n

₎₎

_

_cos(tan

1

(

Q

o



Q

c

₎

_

_cos(tan

1

(

2400



900

)

n

pf

_n



0.905

P

_n

P

_n

3200

2

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