Criteria For The Load Flow Study 1 Collection

Criteria For The Load Flow Study 1. Collection of all data about the Nablus network including the one-line diagram, information about the power stations, transformers, transmission lines and loads. 2. Investigating the problems from which the Nablus network suffers in max. , min load & fault condition Applying method of reactive power compensation to improve the operation (using tap changer transformer, capacitor bank) 3. 4. Performing the economical analysis of the saving achieved by the implementation of reactive power compensation

CH: 2 Existing System The present load in the West Bank is supplied from several points within the IEC network. There are supply points at Ø 22 KV supplying Qalailya and Tulkarm Ø 33 KV feeders from Beisan (in Israel)supplying Jenin , Tubas and Nablus as well as feeders from M. Afraym and feeders from portable substations.

Load Forcast

3. 1: Element Of The Network A. Sources. Generators are one of the essential components of the power systems. Synchronous generators are widely used in power systems. . Nablus are fed from 3 connection point by Israel Electrical Company (IEC), at 33 KV. 1. Asker (odeleh & Almeslekh) →→→ 30 MVA namely 2. Quseen →→→ 18 MVA namely 3. Innab →→→ 5 MVA namely

B. Transformers & load • There are two type of Transformer (power & distribution Transformer ) which is ∆-y connected • So, the source always balanced without looking to the load since the P. T is ∆-y ground connected • The distribution Transformer capacity & its voltage are summarized in table (1)

• There also 3 power Transformer with (10 MVA capacity, 33/6. 6 KV& %Tap=± 12 ; for 17 taps with 1. 5% for each step changer) • Note the impedance of each transformer are determined by the typical value at ETAP program • The loads will be seen later in the table of L. F of transformer

C. Transmission Line • There are two type of conductor 1. O. H lines →→→ACSR 2. cables →→→ cu XLPE which have these rating as in table (2) . In Quseen they use double T. L (3*95 mm 2 ACSR) for the transmission of power since the current is above 400 A in max load

3. 2 : Electrical Problem In The Network 1. High drop voltage 2. 3. Low power factor Low load factor at most of distribution transformer which reduce the efficiency of transformer & so increase the losses (the electrical losses in distribution network must not exceed 8 -10 % from total active power) Exceeding the permissible capacity of each connection point MVA 4.

Remedy 1. • • • High drop voltage & low power factor problem It is important to keep the power factor above 0. 92 on the distribution transformer so as to minimize the electrical losses in the network & do not paying penalties For max load the voltage of buses must rise to (1. 05 Vnom<= Vbus <=1. 1 Vnom) as we can to decrease the current & so to decrease the losses The first step for this improvement is done by using the taps , if not enough it can solve by adding capacitor banks

2. Low load factor problem • • • Re-arrange the distribution transformer (if it can be) to increase the load factor LF=[. 65 -. 75] give the max. efficiency distribution transformer The engineers choose the transformer in distribution network with load factor [. 45 -. 55] expressed to the growth of the load by years 3. Capacity problem • For the fourth problem there is a study to get another connection point or to move the connection to another point as in Qussen

Taps • Almost all transformers provide taps on windings to adjust the ratio of transformation by changing taps down when we need to raise the voltages up and vice versa • There are two types of transformer taps: 1. Tap changing without loads (fixed tap) changer on either side or both sides of transformers 2. Tap changing under load (LTC)

capacitor banks • Shunt capacitor banks is very important method of controlling voltage at the buses at both transmission and distribution levels along lines or at substation and load. • Essentially capacitor is a means of supplying mega-vars (MVAR) at the point of installation. • Capacitor banks may be permanently connected, or regulators • Switching may be manually or automatically controlled either by time clock or in response to voltage or reactive - power requirement • capacitor reduces the line current necessary to supply the load and reduce the voltage drop in the line as the power factor is improved

4. 1: Etap Power Station Program • It is a load flow program which can simulink the power system receiving the input data (source , transformer , T. L & loads) as One Line Diagram schematic And results output report that includes bus voltage , branch losses , load factors power factors …etc. • It is also able to do the Fault analysis. . Harmonic analysis. . Transient stability analysis.

4. 2: Simulation For Max. Load Case This step done by the following criteria 1. drawing the one line diagram (source , transformer T. L, buses & loads) 2. entering R&X in Ω or (Ω /any unit of length) & its length. note (Y) value is not important since the T. L is short (L<80 Km) 3. entering the typical value (X/R & %Z) for each transformer 4. entering the rated voltage for each bus 5. entering the actual MVA & P. F for each load 6. entering the source as a swing bus, for load flow studies a swing power grid will take up the slack of the power flows in the system, i. e. , the voltage magnitude and angle of the power grid terminals will remain at the specified operating values ( V & δ are given , P & Q are unknown) 7. run the load flow analysis to get the output result

A: Max. load case results without improvement • The total demand for Qussen Swing bus P= 20. 27 MW Q=14. 846 MVAr S=24. 929 MVA ∆P=1. 06 ∆Q=2. 555 I=436 A ∆P%=1. 06/20. 27=5. 456% pf= 80. 33 lagging • The total demand for Innab Swing bus P= 5. 702 MW Q=3. 741 MVAr S=6. 82 MVA ∆P=. 075 ∆Q=. 307 I=119 A ∆P%=. 075/5. 695=1. 31% pf= 83. 61 lagging B: Qussen-with tap changer improvement Swing bus P= 19. 841 MW Q=14. 318 MVAr S=24. 467 MVA pf= 81. 09 lagging ∆P=. 920 ∆Q=2. 026 I=428 A ∆P%=. 92/19. 841=4. 63% Method of iteration: Newton Raphson method Number of Iterations: 3

L. F for Quseen

L. F for Innab

P. F for Quseen

P. F for Quseen

P. F for Innab

V% for Quseen

V% for Quseen

V% for Innab

Problems In The Network • we notice # of problems: 1. low load factor (L. F<. 45) for the most of transformer 2. high load factor (L. F>1) for some transformer {T 82, T 103 in Quseen} 3. The P. F for all buses are low (P. F<. 92) except {bus. 33, 34 in Quseen & bus 25, 27, 29, 54 in Innab} 4. %V does not lies between(1. 05 Vnom-1. 1 Vnom) for any bus 5. considerable losses in Quseen (∆P%=5. 456)

V% for Quseen with Taps

V% for Quseen with Taps

V% for Quseen with Taps

For this case we notice the following result 1. there is a small increase in the P. F 2. 36% of buses lies between(1. 05 Vnom 1. 1 Vnom) & the other is >95% Vnom 3. P. F of the swing bus increase from 80. 33 to 81. 09 4. The current decrease from 436 A to 428 A 5. the losses decrease in Quseen. 826% from the original case 6. There is a saving in the capacity of. 5 MVA

C: Max. load with capacitor improvement * Capacitor bank are used to solve P. F problem & its penalties, we put these capacitor bank at the load side (0. 4 Kv side) * Qc=pold(tancos-1 p. fold-tancos-1 p. fnew) Where standard capacitor are: 0. 4 Kv→→→ 25, 40, 60, 100 KVAr 6. 6 or 11 Kv→ 3, 6 MVAr

1. Quseen Qc=19. 841*(tancos-1. 8109 -tancos-1. 92) =5. 866 MVAr Qcact=5. 476 MVAr for P. F=. 9202 2. Innab Qc=5. 702(tancos-1. 8361 -tancos-1. 92) =1. 312 MVAr Qcact=1. 3 MVAr for P. F=. 9221 * we use suitable rated capacitor bank for each load to rise its p. f above. 92, so to increase the overall p. f of swing bus

PF for Quseen with cap

PF for Quseen with cap

PF for Inab with cap

4. 2: Simulation For Min. Load Case A : Min. load case results without improvement The total demand for Qussen Swing bus P= 7. 721 MW Q=5. 21 MVAr S=9. 348 MVA pf= 82. 59 lagging ∆P=. 152 ∆Q=. 354 ∆P%=1. 96% I=164 A The total demand for Innab Swing bus P= 2. 262 MW Q=1. 421 MVAr S=2. 672 MVA pf= 84. 69 lagging ∆P=. 012 ∆Q=. 047 ∆P%=. 53% I=47 A

V% for Quseen without tap

V% for Quseen without tap

B: Qussen-with tap changer improvement Swing bus P= 7. 709 MW Q=5. 2 MVAr S=9. 32 MVA pf= 82. 72 lagging ∆P=. 141 ∆Q=. 32 I=163 A ∆P%=1. 82% * At this case half turn of tap changer are used to increase the voltage of the bus (vbus>=vnom) [only Quseen region have under this value Vbus=. 95 -. 98. 5 Vnom]. tap changer have affect to increase the voltage but less affect on p. f.

V% for Quseen with tap

V% for Quseen with tap

C: Min load using capacitor 1. Quseen Qc=7. 709*(tancos-1. 8272 -tanco-1. 92) =1. 952 MVAr Qcact=1. 995 MVAr for P. F=. 9224 2. Innab Qc=2. 262(tancos-1. 8469 -tancos-1. 92) =. 456 MVAr Qcact=. 547 MVAr for P. F=. 9209

PF for Quseen without cap

PF for Quseen without cap

PF for Inab without cap

PF for Inab with cap

PF for Quseen with cap

PF for Quseen with cap

* At min load less capacitor bank are used to rise the p. f at the load & so the overall p. f. the losses in the network are become very low since the currents is reduced. * Some of these capacitor bank are used at max &min which is called fixed capacitor bank. And other capacitor which only used at max or at min are called regulated one. regulated capacitor bank are more expensive than fixed since it need to controller for use.

Changing of the switch gear & connection point simultaneously Ø change the switch gear from 33/6. 6 KV to 33/11 KV except Jumblat region • Jumblat region will kept as it is to exploits the distribution transformer which have two level voltage at primary side (11, 6. 6 KV/. 4 KV) at Jumblat with transformer have only 6. 6/. 4 KV side at other places ( East & West Mojeer aldeenregion ) • This operation will save the price of a new transformer with 11/. 4 KV Ø The change of position of the connection point is from Qussen to sarrah with 3 Km double T. L

§ The change is starting from replacing the distribution transformer of 6. 6/. 4 KV in East & West Mojeer aldeen region to 11, 6. 6/. 4 KV(from Jumblat & East part region from Nablus) § This step also taken some case of the L. F distribution rearrangement which are sumerize at the next table

T# capacity L. F New capacity from 62 630 58. 6 630 Jumblat 63 630 29. 5 400 Jumblat 64 630 69. 1 630 Jumblat 68 400 91. 6 630 Jumblat 72 250 36. 8 250 East part 74 160 47. 3 160 East part 75 250 59. 4 250 East part 76 400 38. 2 400 Jumblat 77 400 29. 2 400 Jumblat 84 84 42. 8 84 Jumblat 90 630 26. 5 630 Jumblat 97 400 44. 9 400 Jumblat 99 630 40. 7 630 Jumblat 100 630 40. 7 630 Jumblat 101 630 52. 2 630 Jumblat 103 300 105. 7 400 Jumblat 104 400 91. 2 400 Jumblat 108 1000 East part

§ The transformer at Jumblat region which exchange are T 25, T 26, T 27, T 28, T 30, T 38, T 42, T 45, T 46, T 47, T 49, T 50, T 51 &T 61 630 KVA→→ T 25, T 26, T 30, T 42, T 45&T 50 400 KVA→→ T 27, T 28, T 39, T 46, T 47, T 49, T 51&T 61 § the transformer of Jumblat are back as its default capacity except 1)T 39(400 KVA) of L. F=18. 3 with T 103(300 KVA) 2)exchange T 47(400 KVA) of L. F=80. 4% with T 25(630 KVA) of L. F=7. 7% The result of this case with taps: Swing bus P= 19. 647 MW Q=14. 172 MVAr S=24. 225 MVA pf= 81. 1 lagging ∆P=0. 727 ∆Q=0. 881 I=424 A ∆P%=0. 727/19. 647=3. 7%

From ETAP result we notice that: 1. the current in The main T. L at Quseen is decreased in widely range from 400 to 35 A approximately due to change of the position of connection point to Sarah. at Sarah the power distribute directly for more branches. The decrease in current will decrease the losses in The main T. L (at the old case) 2. the change from 6. 6 to 11 KV also decrease the current in the branches & so this mean decreasing in the losses 3. the voltage is slowly decreased at Quseen busses region since the supply is exchange 4. the voltage is slowly increased at the region which the switch gear is change

Quseen, old case

w n, ne usee e Q cas

e s old ca

e cas new

Ø The first economical study is using capacitor bank & this study followed by this criteria: ∆∆P=∆Pbefore, cap - ∆Pafter, cap ∆∆P: saving in real power losses ∆Pbefore, cap : real power losses before adding capacitor ∆Pafter, cap : real power losses after adding capacitor Z∆p=∆∆p*T*140 Z∆p : annual saving in real power cost T=8760(0. 124+0. 0001 tmax)^2 T≈3500 hour 140: cost per MWh($/MWh) Kc=C*Qc Kc: cost of capacitor C: cost of capacitor per KVAr($/KVAr) Qc: capacitor KVAr

knowing that the cost of the capacitor are: Fixed Cap=5$/Kvar Regulated Cap=22$/Kvar The type of capacitors used are regulated only since the loads are vary every time , increase by years & may be decrease under min. so the P. F become leading & this will damage the transformer. also the P. F correction range is considerable to use regulated capacitor. Zc=0. 22*Kc Zc: annual capacitor running cost. 22: maintenance & life time of capacitor (depreciation factor) ∆Z=Z∆p-Zc ∆Z: annual saving Saving of penalties=1% from the total bill for every 1% p. f <92% ∆Zt=∆Z+ Saving of penalties ∆Zt: total annual saving S. P. B. P=investment(capacitors initial cost)/ total annual saving S. P. B. P < 2 year →→→project is visible S. P. B. P > 2 year →→→project is not visible

1)Quseen ∆∆P=∆Pbefore, cap-∆Pafter, cap ∆∆P=0. 92 -0. 702=0. 218 Z∆p=∆∆p*T*140 Z∆p=0. 218*3500*140=106820 Kc=C*Qc Kc=5. 476*10^3*22=120472 Zc=0. 22*Kc=26503. 84 ∆Z=Z∆p-Zc=106820 -26503. 84 =80316 Total Bill=19. 841*10^3*3500=69443500 kwh Saving of penalties=1%*69443500 (92 -81. 09)*140*10^-3=1060680 ∆Zt=80316. 16+1060680=1140996(This is the annual saving)

2)Innab ∆∆P=∆Pbefore-∆Pafter ∆∆P=0. 075 -0. 061=0. 014 Z∆p=∆∆p*T*140 Z∆p=0. 014*3500*140=6860 Kc=C*Qc Kc=1. 3*10^3*22=28600 Zc=0. 22*Kc=6292 ∆Z=Z∆p-Zc=6860 -6292 =568 Total Bill=5. 702*10^3*3500=19957000 Saving of penalties=1%*19957000 (92 -83. 61)*140*10^-3 = 234415 ∆Zt=568+234415=234983 (This is the annual saving)

S. P. B. P=total investment/total annual saving =(120472+28600) / (1140996+234983) =. 1083 year =1. 296 month Ø The second economical study is changing of switch gear & the connection point. this step will make saving in power without paying money Annual saving=saving in power*3500 h*140$/MW =Pold-Pnew *3500 h*140 =(19. 841 -19. 647) *3500*140 =95060$/year