PowerPoint Presentation

Improvement the electrical distribution networkBenefits and advantages of improving the electrical distribution networksReduction of power losses.increasing of voltage levels .correction of power factor.increasing the capability of the distribution transformer.

Methods of improvement of distribution electrical networks1. swing buses 2.transformer taps3. capacitor banks (compensation)Tubas Electrical Distribution Network

TUBAS ELECTRICAL NETWORK is provided by Israel Electrical Company (IEC) with two connection point

Electrical Supply :SourcesTyaseerAl zawiahCapacity 15 MVA 5 MVAVoltages 33 KV 33KVRated C.B 300 A 150 AElements Of The Network :The network consists of 151 distribution transformer (33/0.4Y KV). The transformers range from 50KVA to 630 KVA the following table shows them in details: Distribution Transformers

Number of transformersRating (KVA)3501610018160462503540033630

Overhead lines

The conductors used in the network are ACSR with different diameters as the following table:

Cable NameCross sectional area (mm2)R (/Km)X (/Km)Nominal Capacity (A)Ostrich1500.190.28350Cochin1100.250.29300Lenghorn700.390.31180Aprpcot500.810.29130Underground cables

The under ground cables used in the network are XLPE Cu as shown :

Diameter (mm2)R (/Km)X (/Km)950.410.121 Problems in The Network :The P.F is less than 0.92% , this cause penalties and power losses.There is a voltage drop.There is power losses.Over loaded transformerOver loaded connection point

Analysis of the network In first stage of the analysis of tubas network we have to take the maximum load in daily load curve.Then applied it on ETAB we started the study of this case after we applied the data needed Like load consumption of power and other data.

Maximum load case We have to summarize the results, total generation, demand, loading, percentage of losses, and the total power factor.

The swing current = 326 AMWMVARMVA% PFSwing Bus(es):16.7557.47418.34691.33 lag.Generators:0.000.000.000.00Total Demand:16.7557.47418.34691.33 lag.Total Motor Load:9.3684.14810.24591.44 lag.Total Static Load:6.7602.2457.12394.9 lag.Apparent Losses:0.6271.081Bus #rated(kv)operating(kv)operating %Bus1790.4000.36791.8Bus1800.4000.37593.7Bus1860.4000.37493.5Bus1870.4000.37794.2Bus1890.4000.37994.6Bus1900.4000.37894.6Bus1910.4000.37794.2Bus1960.4000.37694.0Bus1970.4000.37192.8Bus1980.4000.37694.0Bus1990.4000.37493.5Bus2000.4000.37192.8Bus2010.4000.36490.9Bus2020.4000.37393.1Bus2070.4000.37794.2Bus2080.4000.37994.8Bus2090.4000.37994.7Bus2100.4000.37593.6The P.F in the network equals 91.33% and this value causes many problems specially paying banalities and this value must be (0.92-0.95)

The voltages of buses are not acceptable and this voltage will be less when it reaches the consumer

the network have over loaded transformer .Over loaded connection point .High losses of power .The maximum load case improvementThe methods we used to do that are:Tab changing in the transformers.Adding capacitors to produce reactive power.Changing and replace transformer.Add another connection point .

Improvement the maximum case using taps changing and power factor improving .

In the first part of project this step is done and the result had been taken The method of tab changing involves changing in the tab ratio on t he transformer but in limiting range which not accede (5%) .The P.F need to be improved to reduce the penalties on municipalities, reduce the current flows in the network which reduces the losses. The power factor after the improving must be in the range (0.92- 0.95) lag

Improvement the maximum case using taps changing and power factor improving .

We use this equation to calculate the reactive power needing for this improvement is:

Qc = P (tan cos-1 (p.f old)- tan cos-1 (p.f new))PF old = 91.33PF new at least = 92%Q=16.755*(tan (24.783) tan (23.074)) = 774 KVAR

Improvement the maximum case using taps changing and power factor improving .

The following table shown the summary .

The swing current = 328 A

MWMVARMVA% PFSwing Bus(es):17.4236.94618.75792.89 lagTotal Demand:17.4236.94618.75792.89 lagTotal Motor Load:9.3684.14810.24591.44 lagTotal Static Load:7.3991.6687.58597.55 lagApparent Losses:0.6561.131Bus numberV rated (KV)Operating %Bus650.497.646Bus680.499.519Bus690.497.426Bus700.497.275Bus730.497.309Bus1790.499.029Bus1800.499.483Bus1810.4100.755Bus1820.497.209Bus1830.497.114Bus1840.496.815Bus1850.497.207Bus1860.499.632Bus1870.4100.218Bus1880.497.264 overloaded transformers This problem was solved by changing transformers locations where the transformers which are large and the load on them small were changed with small highly loaded transformsThen another transformers connected in parallel with the left overloaded transformers this will need to buy new transformers.TransformerSrated oldSavgLF oldSrated newLF newAAUJ1400402.51.00625250+2500.644Serees Western250262.486251.0499454000.4824Tamoon Albatmah160169.231251.0576952500.5415Tamoon Almeshmas250423.18751.69275250+2500.6771Tamoon Alrafeed250316.196251.2647854000.6323Tamoon jalamet Albatmah100125.121.25121600.6256Tamoon first of the town250264.186251.056745160+1600.5885Tamoon National Security 160161.17251.0073282500.4837Aqaba Eastern400439.458751.0986476300.558Aqaba Western400485.90751.2147696300.617Faraa Camp Old Station630854.84251.356893630+4000.6639wadi alfaraa alhafreia250254.461.017844000.4614Wadi alfaraa gas station400409.4651.0236636300.5199Housing250261.9751.04794000.5239Abu Omar400499.616251.2490416300.6344Allan Alsood250281.541251.126165250+2500.4504Almasaeed250459.511.838046300.5835Alhawooz400476.4051.1910136300.6049Althoghra160163.0751.0192192500.4538Almghier Marah Alkaras100114.2766251.1427661600.5713Tayaseer Main250305.656251.2226254000.6113Aljarba Eastern160174.86751.0929222500.5595Merkeh Abu Omar5064.5616251.2912331000.5164 overloaded transformersThe following table shows the transformers which are needed to be bought:

shows the extra transformers left after solving the overloaded transformers problem

Number of transformersKVA66301250Number of transformersKVA1100150 overloaded transformersFlowing table summarizes the analysis results after changing transformers

The swing current = 327 A

MWMVARMVA% PFSwing Bus(es):17.3886.86718.69593.01 lagTotal Demand:17.3886.86718.69593.01 lagTotal Motor Load:9.3944.16310.27591.43 lagTotal Static Load:7.3741.6647.55997.55 lagApparent Losses:0.6201.039Bus numberVratedOperating (%)Bus650.498.353Bus680.499.519Bus690.497.774Bus700.497.286Bus730.497.322Bus1790.4101.288Bus1800.4100.658Bus1810.4100.769Bus1820.497.223Bus1830.497.128Bus1840.496.829Bus1850.497.221Bus1860.4100.719Bus1870.4100.234Bus1880.497.279New connection Point Tubas Electrical Distribution Company (TEDCO) is planning to add new connection point for the company in Zawya area.

This connection point is 5MVA rated.

And circuit breaker is 150A

New connection Point The following table shows the results summary after the new connection point

The swing current = 325 A

MWMVARMVA% PFSwing Bus(es):17.4306.62218.64693.48 lagSwing bus (1):12.8654.92013.7793.4 lagSwing bus (2):4.5651.7024.87293.7lagTotal Demand:17.4306.62218.64693.48 lagTotal Motor Load:9.3944.16310.27591.43 lagTotal Static Load:7.5991.7127.79097.55 lagApparent Losses:0.4370.747BusVratedOperating (%)Bus650.498.484Bus680.499.519Bus690.498.241Bus700.497.853Bus730.497.991Bus1790.4102.013Bus1800.4101.362Bus1810.4101.475Bus1820.497.905Bus1830.497.81Bus1840.497.515Bus1850.497.962Bus1860.4101.609Bus1870.4101.211Bus1880.498.211Improving the network with the new connection point

As before the improvement is done by tap changing and adding capacitor banks.

Now all buses are operating over 100% voltages. This will make the voltages reach to the consumer with fewer losses.

Improving the network with the new connection point

The results of the improving are summarized in the following table

The swing current = 322A

MWMVARMVA% PFSwing Bus(es):17.4546.55818.64593.61 lag.Swing bus (1):12.6654.82013.9793.35 lagSwing bus (2):4.7651.8024.57293.82lagTotal Demand:17.4546.55818.64593.61 lagTotal Motor Load:9.3944.16310.27591.43 lagTotal Static Load:7.6241.6507.80197.74 lagApparent Losses:0.4350.744Bus numberVratedOperating (%)Bus650.4101.454Bus680.4102Bus690.4100.73Bus700.4100.853Bus730.4100.45Bus1790.4102.03Bus1800.4101.38Bus1810.4101.5Bus1820.4100.37Bus1830.4100.28Bus1840.4100.969Bus1850.4100.43Bus1860.4101.63Bus1870.4101.23Bus1880.4100.71We note that :When we improve the not work the losses in the network decrease and the total current decrease.

Losses before improvement = 627 kW. Losses after improvement =435 kW.

Total current in origin case =326 A

Total current after voltage improvement= 322A

Minimum Case

In the minimum load case the load is assumed to be half the maximum loadThe network analysis in this case shows the results in the following table

The swing current =166 A

MWMVARMVA% PFSwing Bus(es):8.3813.4809.67592.36 lagTotal Demand:8.3813.4809.67592.36 lagTotal Motor Load:4.6992.0825.14091.43 lagTotal Static Load:3.5291.1323.70695.22 lagApparent Losses:0.1530.265Bus numberVratedOperating (%)Bus650.40098.454Bus680.40099.760Bus690.40098.284Bus700.40098.666Bus730.40098.682Bus1790.40096.900Bus1800.40097.504Bus1810.40098.063Bus1820.40098.635Bus1830.40098.589Bus1840.40098.367Bus1850.40098.631Bus1860.40097.630Bus1870.40097.785Bus1880.40098.303Minimum Case

Now taking the taps fixed as in the maximum load case the results shows that all the buses have good voltage level and the power factor is in the range so no need to add capacitor banks for this caseso the capacitor banks used in the network are all regulated.

Minimum Case

The following table shows the analysis summary with the taps changed

The swing current = 165 A

MWMVARMVA% PFSwing Bus(es):8.7203.6149.43992.38 lagTotal Demand:8.7203.6149.43992.38 lagTotal Motor Load:4.6992.0825.14091.43 lagTotal Static Load:3.8551.2444.05195.17 lagApparent Losses:0.1660.287Bus numberVratedOperating (%)Bus650.40098.445Bus680.40099.760Bus690.40098.251Bus700.40098.626Bus730.400101.099Bus1790.400101.654Bus1800.400102.194Bus1810.400102.842Bus1820.40098.582Bus1830.40098.534Bus1840.40098.318Bus1850.40098.581Bus1860.400102.434Bus1870.400102.598Bus1880.40098.247Minimum Load Study After The Connection Point And Solving Overloaded Transformers ProblemAfter solving overloaded transformers problem in maximum case as seen before some transformers were changed and new transformers connected in parallel with some of overloaded transformers. Also the new connection point is connected to the network.

Minimum Load Study After The Connection Point And Solving Overloaded Transformers Problem The results for minimum load study in this case are shown in the following table

The swing current = 163 A

MWMVARMVA% PFSwing Bus(es):8.7383.5419.42892.68 lagSwing bus (1):6.1572.5246.65492.52lagSwing bus (2):2.5811.0172.77493.03lagTotal Demand:8.7383.5419.42892.68 lagTotal Motor Load:4.6992.0825.14091.43 lagTotal Static Load:3.9281.2704.12895.15 lagApparent Losses:0.1110.189BusVratedOperating (%)Bus650.40099.003Bus680.40099.760Bus690.40098.819Bus700.40098.920Bus730.400101.456Bus1790.400103.328Bus1800.400103.158Bus1810.400103.214Bus1820.40098.938Bus1830.40098.890Bus1840.40098.675Bus1850.40098.967Bus1860.400103.300Bus1870.400103.105Bus1880.40098.728Final improving as with the fixed tabIt is noticed that the voltages and the power factor in this case are goodso no need to add new capacitor banks to the network in this case therefore all capacitor banks connected are regulated. Also it can be seen that the losses decreased.

Final improving as with the fixed tabThe final results for the minimum load case are summarized in the following table:

The swing current = 164 A

MWMVARMVA% PFSwing Bus(es):8.7553.5489.44792.68 lagSwing bus (1):6.1672.5266.66692.51lagSwing bus (2):2.5881.0182.78193.1lagTotal Demand:8.7553.5489.44792.68 lagTotal Motor Load:4.6992.0825.14091.43 lagTotal Static Load:3.9451.2764.14695.15 lagApparent Losses:0.1110.190BusVratedOperating (%)Bus650.400101.435Bus680.400102.252Bus690.400101.283Bus700.400101.388Bus730.400101.455Bus1790.400103.327Bus1800.400103.157Bus1810.400103.212Bus1820.400101.409Bus1830.400101.360Bus1840.400101.136Bus1850.400101.439Bus1860.400103.299Bus1870.400103.104Bus1880.400101.192When we increase power factor the losses in the network decrease and the total current decrease.

Losses before improvement = 153 KW. Losses after improvement =111KW.

Total current in origin case =166A

Total current after voltage improvement= 164A

Economical studyWhile we are improving the power factor of our network, the amount of reactive power which had been added as inserting capacitors is 845kvarP max=16.755 MWP min=8.381 MWLosses before improvement = 0.627 MWLosses after improvement = 0.435 MWPF before improvement(MAX) = 91.33%PF after improvement(MAX) = 93.61%PF before improvement(MIN)= 92.36%PF after improvement(MIN)= 92.68%

To find the economical operation of the network we must do the following calculation:PAV = (Pmax + Pmin)/2 =(16.755+8.381)/2 = 12.568MWLF=PAV/Pmax = 0.748Total energy per year=P max*LF*total hour per year = 109786 MWHTotal cost per year=total energy*cost (NIS/KWH)= =49404.061 M NIS 62.977392 MILLION NIS/YEAR

Saving in penalties of (PF): Table follow shows relation of PF to the penalties:

Penalties=0.01*(0.92-pf)*total bill =0.01*0.0066*62.977*106 =7620.26 NIS/YEARPFPenalties0.92 or moreNo penaltiesLess than 0.92 to 0.81%of the total bill for every 0.01 of PF less than 0.92Less than 0.8 to 0.71.25%of the total bill for every 0.01 of PF less than 0.92Less than 0.71.5%of the total bill for every 0.01 of PF less than 0.92Losses before improvement = 468.996 KWEnergy = power loss hour/year = 410.8404 104 KWHTotal cost=energy cost = 1848782.232 NIS/YEARLosses after improvement = 325.38 KWEnergy= 285.03288 104 KWHCost of losses= 128.2647 104 NIS/YEARSaving in cost of losses=cost before improvement-cost after improvement =566134 NIS/YEAR

Total capacitor = 905 KVARCost per KVAR with control circuit = 15JD = 90NISTotal cost of capacitors= 81450 NIS

Total cost of transformers = 186200 NISTotal investment cost = 267650 NISTotal saving=saving in penalties+ saving in losses = 3876206 NISS.P.B.P= (investment) / (saving) =0.69 YEAR

Transformer ratedNumber of transformercost ($)6306820025014000If we divide the network to two network depend on the capacity ofconnection point the losses is more than the losses on the first network as followingMaximum case Losses before improvement = 720 kW. Losses after improvement =531 kW.

Minimum caseLosses before improvement = 180 KW. Losses after improvement =151 KW

And S.P.B.P= (investment) / (saving) =1.80 YEAR

We not that :

Monitoring System

The monitoring system designed for this project consists of the following parts:

Measurement devices.The remote terminal unit (RTU).Computer interfaceCurrent Measurement

the supervisor have to know the current in the network

high short circuit currents can cause damages in the system

Then the supervisor can cut the power if the protective devices in the network did not work well.

Current Measurement

In our project we choose the current transformer that converts from 60/5 A this device is MSQ-30Like any other transformer it has :primary windinga magnetic core, and a secondary winding. The alternating current flowing in the primary produces a magnetic field in the core which then induces a current in the secondary winding circuit.

50Current Measurement

The current transformer (C.T) gives 4 volts at 10 A amperes flowing in the primary side, then the output voltage of the current transformerThe signal then amplified and inverted by the op-amp (op amp amplification ratio is 100/22 =4.5 )

the buffer is used to get the signal in its actual shape.The buffer also do the task of current isolation.

Current Measurement

a rectifier circuit is used to take the peak of the voltage

The low pass filter is to remove the high frequencies. The diode is to cut the negative half wave of the voltage signal. The capacitor is to smooth the output DC signal.

Voltage Measurement

Voltage is another important parameter in the network,conventional transformer is used here with ration is 230v:6v RMS And we need a buffer circuit As shown

Voltage Measurement

As in the current measurement it is needed to rectify the voltage output signal

Power Factor Measurement

The power factor is defined as cosine the angle between current and voltage signals. Here the current and voltage signals will be transform to pulses

Power Factor Measurement

then they will be injected to PLL (CD4046)

the output of PLL will be the pulse which its width represents the phase shift between the signals.

Power Factor Measurement

The following figure shows this operation

1 shows the two signals A and B.

2 shows signal V pulses.

3 shows signal I pulses.

4 shows the output of PLL

Power Factor Measurement

A counter in the microcontroller will count the duration of the phase shift signal.

Then the power factor will be cosine the angle.

P.F = COS

Frequency Measurement

other PLL will be used.

The voltage pulse of amplifying circuit is the first input

the second input of the PLL will bea fixed signal with 20ms(i.e. 50Hz)

The output of the PLL will be the difference between the fixed signal and the voltage pulses,

the difference duration will be either added or subtracted from the 50Hz.

Frequency Measurement

If Y>20ms(F