1. Name :
Ahmad Sobhi Abu Sadah
Hemzeh Ahmad Qados

2. Criteria For The Load Flow Study
Ch.1: Load Flow Concept
Criteria For The Load Flow Study
Collection of all data about the Nablus network including the one-line diagram, information about the power stations, transformers, transmission lines and loads.
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)
Performing the economical analysis of the saving achieved by the implementation of reactive power compensation

3. 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
22 KV supplying Qalailya and Tulkarm
33 KV feeders from Beisan (in Israel)supplying Jenin ,Tubas and Nablus as well as feeders from M. Afraym and feeders from portable substations.

4. Load Forcast

5. CH.3: Electrical Network Study
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 33KV. 1.Asker (odeleh & Almeslekh) →→→30MVA namely 2.Quseen →→→18MVA namely 3.Innab →→→5MVA namely
CH.3: Electrical Network Study

6. 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)

7. The loads will be seen later in the table of L.F of transformer
There are also 3 power Transformer with (10MVA capacity,33/6.6KV& %Tap=±12 ;for 17 taps with1.5% 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

8. 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*95mm2ACSR) for the transmission of power since the current is above 400A in max load

9. 3.2 : Electrical Problem In The Network
High drop voltage
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

10. Remedy 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
(1.05Vnom<= Vbus <=1.1Vnom) 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

11. Low load factor problem
Re-arrange the distribution transformer (if it can be) to increase the load factor
LF=[ ] give the max. efficiency distribution transformer
The engineers choose the transformer in distribution network with load factor [ ] expressed to the growth of the load by years
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

12. 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:
Tap changing without loads (fixed tap) changer on either side or both sides of transformers
Tap changing under load (LTC)

13. 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

14. 4.1: Etap Power Station Program
CH.4:Load Flow Analysis
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.

15. 4.2: Simulation For Max. Load Case
This step done by the following criteria
drawing the one line diagram (source ,transformer T.L, buses & loads)
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<80Km)
entering the typical value (X/R & %Z) for each transformer
entering the rated voltage for each bus
entering the actual MVA & P.F for each load
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)
run the load flow analysis to get the output result

16. A: Max. load case results without improvement
The total demand for Qussen
Swing bus P= 20.27MW Q=14.846MVAr S=24.929MVA pf= lagging
∆P= ∆Q= I=436A
∆P%=1.06/20.27=5.456%
The total demand for Innab
Swing bus P= 5.702MW Q=3.741MVAr S=6.82MVA pf= lagging
∆P= ∆Q= I=119A
∆P%=.075/5.695=1.31%
B: Qussen-with tap changer improvement
Swing bus P= MW Q=14.318MVAr S=24.467MVA pf= lagging
∆P= ∆Q= I=428A
∆P%=.92/19.841=4.63%
Method of iteration: Newton Raphson method Number of Iterations: 3

17. L.F for Quseen

18. L.F for Innab

19. P.F for Quseen

20. P.F for Quseen

21. P.F for Innab

22. V% for Quseen

23. V% for Quseen

24. V% for Innab

25. Problems In The Network
we notice # of problems:
low load factor (L.F<.45) for the most of transformer
high load factor (L.F>1) for some transformer {T82,T103 in Quseen}
The P.F for all buses are low (P.F<.92) except {bus.33,34 in Quseen & bus 25,27,29,54 in Innab}
%V does not lies between(1.05Vnom-1.1Vnom) for any bus
considerable losses in Quseen (∆P%=5.456)

26. V% for Quseen with Taps

27. V% for Quseen with Taps

28. V% for Quseen with Taps

29. For this case we notice the following result
there is a small increase in the P.F
36% of buses lies between(1.05Vnom-1.1Vnom) & the other is >95% Vnom
P.F of the swing bus increase from to 81.09
The current decrease from 436A to 428 A
the losses decrease in Quseen .826% from the original case
There is a saving in the capacity of .5MVA

30. 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 (0.4Kv side)
* Qc=pold(tancos-1p.fold-tancos-1p.fnew)
Where standard capacitor are:-
0.4Kv→→→25,40,60,100KVAr
6.6or11Kv→3,6MVAr

31. Qc=5.702(tancos-1.8361-tancos-1.92) =1.312MVAr
1.Quseen
Qc=19.841*(tancos tancos-1.92)
=5.866MVAr
Qcact=5.476MVAr for P.F=.9202
2.Innab
Qc=5.702(tancos tancos-1.92)
=1.312MVAr
Qcact=1.3MVAr for P.F=.9221
for each load to we use suitable rated capacitor bank * rise its p.f above .92,so to increase the overall p.f of swing bus

32. PF for Quseen with cap

33. PF for Quseen with cap

34. PF for Inab
with cap

35. 4.2: Simulation For Min. Load Case
A : Min. load case results without improvement
The total demand for Qussen
Swing bus P= 7.721MW Q=5.21MVAr S=9.348MVA pf= lagging
∆P= ∆Q= I=164A
∆P%=1.96%
The total demand for Innab
Swing bus P= 2.262MW Q=1.421MVAr S=2.672MVA pf= lagging
∆P= ∆Q= I=47A
∆P%=.53%

36. V% for Quseen without tap

37. V% for Quseen without tap

38. B: Qussen-with tap changer improvement
Swing bus P= 7.709MW Q=5.2MVAr S=9.32MVA pf= lagging
∆P= ∆Q= I=163A
∆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= Vnom] .tap changer have affect to increase the voltage but less affect on p.f.

39. V% for Quseen with tap

40. V% for Quseen with tap

41. C: Min load using capacitor
1.Quseen
Qc=7.709*(tancos tanco-1.92)
=1.952MVAr
Qcact=1.995 MVAr for P.F=.9224
2.Innab
Qc=2.262(tancos tancos-1.92)
=.456MVAr
Qcact=.547 MVAr for P.F=.9209

42. PF for Quseen without cap

43. PF for Quseen without cap

44. PF for Inab
without cap

45. PF for Inab
with cap

46. PF for Quseen with cap

47. PF for Quseen with cap

48. 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.

49. Ch.5:future project
Changing of the switch gear & connection point simultaneously
change the switch gear from 33/6.6KV to 33/11KV except Jumblat region
Jumblat region will kept as it is to exploits the distribution transformer which have two level voltage at primary side (11, 6.6KV/.4KV) at Jumblat with transformer have only 6.6/.4KV side at other places ( East & West Mojeer aldeenregion )
This operation will save the price of a new transformer with 11/.4KV
The change of position of the connection point is from Qussen to sarrah with 3Km double T.L

50. The change is starting from replacing the distribution transformer of 6.6/.4KV in East & West Mojeer aldeen region to 11,6.6/.4KV(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

51. from
New capacity
L.F
capacity T#
Jumblat
630
58.6 62
400
29.5 63
69.1 64
91.6 68
East part
250
36.8 72
160
47.3 74
59.4 75
38.2 76
29.2 77 84
42.8
26.5 90
44.9 97
40.7 99
100
52.2
101
105.7
300
103
91.2
104
1000
108

52. The result of this case with taps:-
The transformer at Jumblat region which exchange are T25,T26,T27,T28,T30,T38,T42,T45,T46,T47,T49,T50,T51&T61
630KVA→→ T25,T26,T30,T42,T45&T50
400KVA→→ T27,T28,T39,T46,T47,T49, T51&T61
the transformer of Jumblat are back as its default capacity except
1)T39(400KVA) of L.F=18.3 with T103(300KVA)
2)exchange T47(400KVA) of L.F=80.4% with T25(630KVA) of L.F=7.7%
The result of this case with taps:-
Swing bus P= MW Q=14.172MVAr S=24.225MVA pf= 81.1 lagging
∆P= ∆Q= I=424A
∆P%=0.727/19.647=3.7%

53. From ETAP result we notice that:- 1. the current in The main T
From ETAP result we notice that:- 1. the current in The main T.L at Quseen is decreased in widely range from 400 to 35A 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 11KV 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

54. Quseen,old
case

55. Quseen,new
case

56. old case

57. new case

58. CH.6: Econnmical Study
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( tmax)^2 T≈3500hour 140:cost per MWh($/MWh) Kc=C*Qc Kc:cost of capacitor C:cost of capacitor per KVAr($/KVAr) Qc: capacitor KVAr

59. 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 < 2year →→→project is visible S.P.B.P > 2year →→→project is not visible

60. 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=
∆Z=Z∆p-Zc=
=80316
Total Bill=19.841*10^3*3500= kwh
Saving of penalties=1%* ( )*140*10^-3=
∆Zt= = (This is the annual saving)

61. 2)Innab ∆∆P=∆Pbefore-∆Pafter ∆∆P=0. 075-0. 061=0. 014 Z∆p=∆∆p. T
2)Innab ∆∆P=∆Pbefore-∆Pafter ∆∆P= =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= =568 Total Bill=5.702*10^3*3500= Saving of penalties=1%* ( )*140*10^ = ∆Zt= = (This is the annual saving)

62. S.P.B.P=total investment/total annual saving =( ) / ( ) =.1083 year =1.296 month
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*3500h*140$/MW =Pold-Pnew *3500h* =( ) *3500* =95060$/year

63. Thank You