D ntroduction to Distriution Systems 30 Module D ntroduction to Distriution Systems Primry Author: Gerld B. Shele, ow Stte University Emil Address: gshele@istte.edu Prerequisite Competencies: Stedy Stte nlysis of circuits using phsors, three-phse circuit nlysis nd three-phse power reltionships, found in Module B3. Module Ojectives:. dentify sic distriution susttion equipment including trnsformers nd protection equipment.. Perform distriution circuit voltge regultion nd power efficiency clcultions. 3. Perform power fctor correction clcultions for lrge industril lods. 4. dentify sic wiring requirements for residentil nd commercil instlltions including wiring color codes nd grounding requirements. Distriution Overview Energy is consumed t nominl voltge of 0 or 40 volts for most residentil equipment nd 08, 40, 77 pr 480 volts for commercil nd industril customers. The power flows through metering device to determine the mount of rel power consumed y the equipment. The rective power is provided without chrge. The exception to this is if commercil or industril customers operte t low power fctor, they could incur power fctor correction chnge. The distriution system trnsports the complex power from the trnsmission grid to the customer. Distriution systems re typiclly rdil ecuse networked systems, lthough more relile, re more expensive. Since threephse power is more efficient thn single phse, sets of three lines strt from ech distriution susttion. However, close to the end of the rdil feeder, it is often efficient to drop one or two of the three phses. Often only single phse is used in very rurl res to rech the most remote customers. This less expensive design results in uneven loding of ll three phses. The loding is equlized s much s possile y proper ssignment of customers to ech phse. t is the selection of the line cpcities, numer of phses over distnce, the setting of trnsformers, voltge levels nd customer geogrphic distriution nd usge which complictes the engineering tsk of distriution design. D. Equipment Description / Functions The equipment ssocited with the distriution system includes the susttion trnsformers connected to the trnsmission grid (network), the distriution lines from the trnsformers to the customers nd the protection nd control equipment etween the susttion distriution trnsformer nd the customer. The protection equipment includes lightning protectors, circuit rekers (with or without reclosers), disconnects, nd fuses. The control equipment includes voltge regultors, cpcitors, nd demnd side mngement (lod control) equipment. The susttion trnsformers typiclly reduce the trnsmission grid voltge from 38 k to 69, 34.5 or k. Older distriution systems my still use four k s the primry voltge. The distriution lines re rdilly configured, due to cost considertions. However, lrge city distriution my e networked. Especilly if the system is underground through numer of lrge uildings previously connected y stem pipes or col distriution network. All mterils re under copyright of PowerLern. Copyright 000, ll rights reserved.
D ntroduction to Distriution Systems 3 Distriution lines re three-phse on leving the susttion ut re often reduced to two-phses or one-phse for lods distnt from the susttion. The loding of ech phse t the susttion is s equl s possile rring unusul lod usge ptterns y ssigning customers to ech phse sed on historicl usge. The susttion trnsformer reduces ny finl lod imlnce y pproprite design of the secondry windings. The loding of ech phse is distriuted eqully under norml conditions nd norml lod ptterns for the customers connected to the distriution line. The control devices include voltge regultors nd cpcitors. Both devices ttempt to rise the voltge to reduce the losses nd to provide the proper voltge t the customer loction. A voltge regultor is n utotrnsformer with tp positions to rise the voltge to trget vlue. A cpcitor hs timer or voltge-sensing device to connect the device when the voltge is elow trget level. A new device used for lrge commercil or industril customer is sttic vr controller (SC). A SC consists of inductors nd cpcitors tht re switched into use y power electronic circuits to implement desired control. A desired control my e desired voltge or desired hrmonic content of the voltge (see module PQ). Similr devices re considered s FACTS devices. FACT is pneumonic for flexile AC trnsmission system. These devices use microprocessor technology to sense the rel-time conditions of the system nd respond, normlly within qurter of cycle or less, to electroniclly switch components. Components re either inductors or cpcitors. The protection equipment includes lightning protectors, fuses, disconnects nd rekers. Lightning protectors (rrestors) remove high voltge surges y switching the line to ground using sprk gps. Once the sprk gp voltge limit hs een exceeded, mgnetic fields estlished y the flowing current y coils extend the rc to increse the voltge need to sustin the rc, thus extinguishing the rc when the voltge returns to norml. A typicl resistnce versus voltge grph is shown in Figure D.. Note tht this is nonliner device. Resistnce oltge Figure D. Lightning Arrestor mpednce Model A sequence of events, shown s grph of current versus voltge for typicl lightning rrestor is shown in Figure D.. As the voltge increses to the strikeover vlue ( s), the device strts to conduct very quickly, resulting in fst increse in current connected to ground. After sufficient current hs pssed to de-energize the excessive voltge, the current will decrese until the reclosing voltge ( r) is reched when the device stops conducting. Fuses nd circuit rekers re used to isolte the distriution system when fult occurs on the distriution line. A fult is simply the filure of one or more components of the system. Typicl distriution fults include wire on ground fter oject (cr, plne, etc.) hits pole, therey reking pole. Other fults cn e temporry. One typicl temporry fult is tree lim touching the distriution wire. The tree conducts electricity to ground. Another typicl temporry fult occurs when iologicl oject (ird, squirrel, humn, etc.) connects the distriution wire to ground. A fuse is simply piece of metl with lower melting point thn the distriution line wire. When fult occurs on the line, significntly higher current flows. This cuses the metl to melt, n rc results clering the pth of ll metl nd the rc is extinguished since the norml operting voltge cn not sustin the rc. The mount of overcurrent determines how quickly the fuse melts. All mterils re under copyright of PowerLern. Copyright 000, ll rights reserved.
D ntroduction to Distriution Systems 3 s oltge r Current Figure D. oltge Current Chrcteristics of SiC Gpped Arrestor A circuit reker is n utomtic switch, which is commnded to open when rely senses excessive current for defined period. The rely cn e set to reclose the circuit reker in given sequence in the expecttion tht the fult is only temporry. A typicl sequence is to reclose the reker fter two minutes fter the initil opening. Then reclose the reker five minutes fter ech susequent opening. Normlly, only three ttempts re tried to close the circuit reker. A recloser is used when the most likely cuse of the fult is tree lim or niml on the distriution equipment. The fuse selection nd the circuit reker timing hve to e coordinted such tht the lest mount of equipment is removed to isolte fult. D. One-Line Digrm for Distriution System The one-line digrm for distriution system is the sme s for the trnsmission grid except tht the numer of phses nd the connections of ech phse to customer hve to e recorded. Distriution mps for residentil neighorhoods show routing of the distriution lines. They show where ech pole or cle vult is locted (numered). Which phses re connected (continued), which wire (cle) type is used, where the protective nd control equipment is locted, wht control nd protection is present, nd wht settings re used to set or to control ech system re lso shown. The customer lod is viewed t the low voltge trnsformer (0/0 volts) such tht the individul service drops to ech house re not represented. A simplified exmple is shown in Figure D.3. Note tht the loction of ech line is shown geogrphiclly to show the loctions. Most modern digrms would show Glol Positioning Stellite (GPS) coordintes nd simplified street mp. Other utility services, nturl gs, telephone nd cle T might e shown. A commercil distriution system includes distriution lines (cles) nd customer owned equipment within the delivery point susttion (or vult). All mterils re under copyright of PowerLern. Copyright 000, ll rights reserved.
D ntroduction to Distriution Systems 33 TSS:Electric Junction Lincoln Wy Grnd to Duff Note: mp is not to scle North Screen Shows 5 Lincoln Wy Duff to Skunk River Fst Foods West So 5 th Street Shipping Cos South Mrlotn Ave West Frm Access Moile Home Courts Foodstores Figure D.3 Distriution System One-Line Digrm Light industril systems re normlly t higher voltges (e.g. 34.5 nd 69 k) nd re lwys three phses. ndustril distriution systems my e served t even higher voltges (e.g. 69, 38, 30 nd 345 k). Such lods re directly connected to the trnsmission grid (ll three phses). D.3 Lod Types The lod equipment t customer loction includes mny different devices, which my e ctegorized into the following types: resistive, inductive, nd (very infrequently) cpcitive. The ctul lods re primrily resistive for heting, inductive for motors, nd cpcitive for filtering. However, power electronics cn chnge the ctul lod into n lterntive chrcteristic y high speed switching nd signl conditioning. Cpcitors my lso e present for power fctor correction. These re control devices nd not lods from the customer perspective. The customer models re generted from surveys of home pplinces sed on the econometrics of ech neighorhood. Additionlly, lod profiles re ville from chrt records t ech susttion per distriution feeder. Such chrts re normlly ville for ech month of the yer. Digitl recording of feeder lods is more prevlent nd llows more extensive modeling of the customer use ptterns. n ddition, the power fctor of ech feeder is recorded to determine if corrective ction is required. The use of power electronic motor drives nd power supplies re quickly chnging the chrcteristics of lod devices. D.4 Design Guidelines The primry regultor design guide is the voltge trget t the customer loction. The voltge mgnitude is required to e etween 0 nd 5 volts t the customer side of the low voltge trnsformer. Stte commissions set this limit. Additionlly, voltge dips t the customer site re required to e within specified limits in most sttes. The voltge will dip s the inrush currents for strting mchines temporrily lod the system. The stte commissions lso typiclly set reliility trgets for the customer. Such considertions re eyond the scope of this text. The power qulity t the customer site is lso concern s covered in Module PQ. Power qulity now involves mny spects of delivery including frequency, hrmonic content, reliility, etc. This text ssumes tht power qulity only includes hrmonic content. Other delivery specifictions re discussed individully. Any specil lod considertions my hve to e considered y the electric utility, when such lods dversely ffect the distriution system. Such considertions re eyond this text. All mterils re under copyright of PowerLern. Copyright 000, ll rights reserved.
D ntroduction to Distriution Systems 34 D.5 Customer Triffs The customer pys for electricity ccording to the triff pproved y the stte pulic utility commission. An exmple of such triffs would contin the following detils: Residentil Triff: Price for electric usge is $0.05/kWHr. The electricity delivered is nominlly 60 Hertz, 05-30 volts AC, with no more thn one hour of interruption per yer. Light ndustril Triff: Price for electric usge is $0.0/kWHr energy chrge nd $0.5/kW demnd chrge for the pek demnd in ny hlf-hour intervl. The electricity delivered is nominlly 60 Hertz, 05-30 volts AC, with no more thn one dy of interruption per yer. Medium ndustril Triff: Price for electric usge is $0.5/kWHr energy chrge nd $0.0/kW demnd chrge for the pek demnd in ny fifteen-minute intervl. Power fctor correction chrge is ssessed if the power fctor is less thn 0.95. The power fctor correction chrge is $0.50/kAr for correcting demnd elow 0.95 nd $0.70/kAr for correcting demnd elow 0.90. The electricity delivered is nominlly 60 Hertz, 05-30 volts AC, with no more thn eight hours of interruption per yer. Hevy ndustril Triff: Price for electric usge is $0.8/kWHr energy chrge nd $0.5/kW demnd chrge for the pek demnd in ny fifteen-minute intervl. Power fctor correction chrge is ssessed if the power fctor is less thn 0.98. The power fctor correction chrge is $0.50/kAr for correcting demnd elow 0.98 nd $0.70/kAr for correcting demnd elow 0.95. The electricity delivered is nominlly 60 Hertz, 05-30 volts AC, with no more thn one hour of interruption per yer. The ove triffs re only exmples. Prcticl triffs vry gretly. However, the ove detil is exemplry of the detil of most triffs. Note tht ech stte hs different triff structure nd price philosophy for ech customer. t is common for sttes to give preferentil tretment to compnies who relocte into stte from nother stte or country. Such preferentil tretment is termed cross susidies since the totl cost of production, trnsporttion, distriution nd finncing of the utility hs to come from electricity sles. ncome is derived primrily from sles. Thus, other customers must py for the preferentil tretment of others. Such cross susidies re extremely politicl in nture. The mount nd numer of cross susidies is very lrge nd is one of the focl points for de-regultion of electric utilities. This module ssumes tht the customer will e chrged n energy chrge, cpcity chrge, nd power fctor correction chrge. Tle D. lists the ssumed chrges for the exmples in this chpter. Tle D. Pricing Dt Triff Fctor Qulifying fctor Rte Energy Chrge $0.0/kWHr Demnd Chrge $0.05/kAr Power Fctor Correction Pf less thn 0.95 $0.0/kAr Pf less thn 0.9 $0.30/kAr Pf less thn 0.85 $0.40/kAr This module lso ssumes tht the cost nd price estimtes cn e sed on single nlysis for the pek period. The ssumption is tht the pek period is one hour of the 8,760 hours per yer. The costs of this pek period cn e used s n estimte of the yerly cost if functionl reltionship cn e found. Assume, for the exmples, tht the pek period cn e multiplied y utiliztion fctor nd then y the numer of hours in yer to estimte the income nd the costs for the yer. Use utiliztion fctor of.6 for the exmples in this chpter. The mrginl cost for the pek period is n index to use for cost estimtes. The mrginl cost is the cost of dding one more increment of demnd when ll other prmeters sty fixed. All mterils re under copyright of PowerLern. Copyright 000, ll rights reserved.
D ntroduction to Distriution Systems 35 D.6 Distriution System (oltge Drop) Clcultions We will descrie voltge regultion clcultions for rdil system. Single-phse clcultion will e used s distriution feeders re often single phse. For three-phse distriution circuits we must use per-phse equivlent circuit. Power flow equtions (conservtion of energy) re solved y ssuming voltge t the most remote point (the receiving end) of the distriution system such tht the voltge will e within the required rnge. The receiving end voltge is set to the trget; the receiving end current is clculted from the complex power demnd t the receiving end, then the sending end voltge nd current re clculted for the receiving end vlues. This process is itertively used for ech lod until the voltge t the trnsmission susttion is clculted. The lod demnd could e constnt power drin (sink). The lod demnd could e constnt current sink. The lod demnd could e constnt impednce. The type of lod demnd my even e composition of ll three lod types. The recursive steps in the process re roken down into the following. First, ssume voltge t the frthest end (lst customer). Use this voltge to find the current t the customer us. Find the sending end voltge nd current for this distriution segment. Use the voltge t this us to find the current required for the customer t this us. Add to this current the current needed for ny distriution lines leving this us. Now repet the process until the voltge t the sending end us (trnsmission susttion) is found. The procedure for distriution clcultions cn e depicted y the following pseudo-code in Tle D.. This procedure is implemented in MATLAB code for the corresponding simultor. Tle D. oltge Regultion Clcultions. For the lst customer clculte the current for the required voltge:. f impednce lod, then use ohm s lw,. f power lod, then use S= *.. Clculte the sending end voltge nd current using current from. 3. For the next customer clculte the current for the required lod (sending end voltge from.):. f impednce lod, then use ohm s lw,. f power lod, then use S= *. 4. Clculte the totl current, using KCL to find the current required from the next line towrd the source. 5. Clculte the sending end voltge nd current mtrix using current from 4. 6. Repet steps 3 through 5 for ll customers nd lines until the source is reched. 7. Clculte the sending end power from S= *. 0. Clculte the voltge regultion ( s- r)/ r. The voltge regultion is clculted s shown in (D.). The voltge regultion is mesure of the stiffness of the distriution system. delly, the voltge regultion t ll lod uses should e smll. However, recurring prolem is voltge flicker due to motor strt-up inrush currents. The presence of such flicker is n indiction tht the distriution system is not stiff nd tht significnt voltge correction is needed. s R R *00% (D.) s Once ll line currents re known, then the losses cn e completed s the sum of R over ll lines. Then efficiency is PN PLOSS PN where P N is the power from the source. All mterils re under copyright of PowerLern. Copyright 000, ll rights reserved.
D ntroduction to Distriution Systems 36 One line segment is 000 feet. Note tht the equivlent pi model for distriution lines often does not include the impct of mutul coupling etween the phses since the length of such lines re "electriclly short." This is equivlent to sying tht the inclusion of such effects is negligile. Aluminum Conductor Size Copper Neutrl Resistnce (90c) (/000 ft) Tle D.3 Line Dt Rectnce Ampcity Direct Buril Ampcity n Duct (A) Cost ($k/000 ft) (/000 ft) 4 AWG 6-#4.53.035 30 90.5 AWG 0-#4.33.030 70 0.7 AWG 3-#4.6.09 90 40.9 /0 AWG 6-#4..08 0 60. /0 AWG 3-#.6.06 50 80.3 3/0 AWG 6-#.3.04 80 00.5 4/0 AWG 0-#..03 30 30.7 50 kcml 5-#.09.0 360 60.9 300 kcml 8-#0.07.0 400 90 3. 350 kcml 0-#0.06.00 440 30 3.4 400 kcml -#0.04.09 470 330 3.6 Exmple D. (Bus Distriution System Anlysis) Consider the following three-phse distriution system. The usses re numered: source (src),,,3,4, nd 5. The numer of segments is given in ech line (,,3,4,5 respectively). The conductor used is 4AWG. Also, lod levels (3-phse power) re given under ech us in kw nd kars. Buses,,3, nd 5 hve constnt power lods. The lod t us 4 is constnt impednce nd is given s (000+j4000) ohms per phse (Wye connected). Determine the source voltge required to hold us 5 t k (line-to-line). Also, determine the source current, the source power, the totl system losses, the efficiency, nd the percent-regultion. Note gin compenstor tht ll ka quntities given re three-phse. All mterils re under copyright of PowerLern. Copyright 000, ll rights reserved.
D ntroduction to Distriution Systems 37 5 5 4 S 4 4 000 30 6350 30 3 S Z 4 4 5 5 Z 45 4LL 5 0 j 0 3 6350 30 3 0.40 j0.3533 A 6350 30 0.40 j.3533.65 j0.75 00.44.00 635.3 30.00 0.3338 j0.37575 000 j4000 S 4 0.40 j0.3533 0.3338 j0.37575 0.7353 j0.79 635.3 30.00 3 S 3 3 S 4 4 Z 3LL 34 635.3 30.00 0.7353 j0.79. j0.4 006.5.006 40 j5 3 0 3.6859 j.765 6354.4630.006 4 S 3 3 3 Z 3 LL 0.7353 j0.79.6859 j.765.4 j.0055 6354.46 30.006.4 j.0055.59 j0.05 04.88.006 80 j0 3 0 3 3.0674 j3.00506 6359.4430.006 3 S Z LL.4 j.0055 3.0674 j3.00506 5.599 j5.006 6359.44 30.006 5.599 j5.006 0.53 j0.035 0.66.058 0 j30 3 0 3 S 4.6568 j4.5053 6363.3630.058 S SRC P SRC SRC P 5.579 j5.006 4.6568 j4.5053 0.847 j9.558 Z SRC, LL N 3Re S 6363.36 30.058 0.847 j9.558.06 j0.07 046.98.037 6354.46 30.006 6359.44 30.006 6363.36 30.058 6377.97 30.037 SRC SRC 3Re 6377.97 30.037 0.847 j9.558 j85056.09 3Re 5854.066 j85056.09 3Re 5854.066 7756.98 W All mterils re under copyright of PowerLern. Copyright 000, ll rights reserved.
D ntroduction to Distriution Systems 38 Losses : 5 4 3 R R R R R 45 34 3 S.759.73 5.76 9.50 05.935 TOTAL 3 (54.9) P Efficiency : LOSS regultion% reg : % reg LL, SRC LL, SRC LL,5 00% 046.98 000 % reg 00% 0.45% 046.98 PN PLOSS 00% PN 7756.98 76.57 00% 99.0% 7756.98 D.7 Split Distriution Line Anlysis The student should determine how to solve the equtions if the distriution line splits into two sections s shown in Figure D.6. Note tht the clcultions re prolem when the two distriution lines connect t us. The solution is to invent n itertive procedure to djust the customer voltge t us 6 until the voltges t the connection point re (nerly) the sme. An exmple procedure is to updte the ending voltge (customer 5) ssuming tht the lst three solutions cn pproximte qudrtic curve to find the sme solution s found for the first rnch (Customers nd 3). Other power flow techniques, such s Guss-Seidel, re typiclly used. Such techniques re the suject of senior elective or grdute courses. Source 3 4 DL- DL- DL-3 DL-4 Customer DL-5 5 Customer DL-6 Customer 3 6 Customer 4 Customer 5 Figure D.6 Multiple Brnch Distriution System All mterils re under copyright of PowerLern. Copyright 000, ll rights reserved.
D ntroduction to Distriution Systems 39 D.8 Power Fctor Correction Compenstion The customer is required to mintin the power fctor within limits since rective power is not chrged to the customer t this time. The stndrd correction configurtion is to plce cpcitor in prllel to the lod s shown in Figure D.7. The vlue of the cpcitnce is to reduce the impednce s seen y the electric utility to rel component only. Alterntively, this implies tht the rective power component is zero. + s > c d s - S h u n t L o d Figure D.7 Shunt Compenstion The clcultions re shown in (D.) through (D.6). The demnd requirement clcultion (D.) determines the mount of rective power required y the lod equipment. The complex power required (provided) y the compenstion equipment (D.38) determines how much rective demnd requirement cn e provided. The configurtion of the compenstion determines how the requirements re stisfied (D.39 or D.40). The one-port (shunt device/equipment) determines the component size of the compenstion. Sd d d cos jd sin (D.) S d * d d P jq d d (D.) S c * c c P jq c c d (D.3) s (D.4) c d S s S S (D.5) c d S c * c c c jc 0 jq ( jc ) c c * c (D.6) All mterils re under copyright of PowerLern. Copyright 000, ll rights reserved.
D ntroduction to Distriution Systems 40 Exmple D. (Power Fctor Correction) Consider one demnd operting t 500 t 60 Hz with P = 48kW @ pf = 0.60 lgging nd P = 4 kw t pf = 0.96 leding. Add cpcitor in prllel to these nd find the cpcitnce vlue to zero the rective demnd. jq 48 j ka jq 4 j ka P P j Q Q 7 j57 ka 9.8 ka @ pf 0. lgging S P 64 S P 7 S net 784 Q 57 kar c Q net c 500 377 rd sec C Q / c c 605 F Other compenstion configurtions nd clcultions my e required for unique lod conditions. nductors my lso e dded to compenste for excessive cpcitnce t the customer site. Alterntively, utomtic compenstion my e required if the lod chrcteristics chnge quickly. A power electronic device clled sttic AR or voltge compenstor (svc) s mentioned ove (D.) ccomplishes such utomtic compenstion. The power system specilist should e le to derive ll of the equtions needed for compenstion y shunt inductors, series cpcitors, or series inductors. D.9 Blncing Power Feeder Demnds Distriution line lod demnds re "lnced" y dividing the numer of customers etween the three phses such tht the expected demnd (ka) is evenly divided. The mount of demnd per customer hs to e known or estimted s well s the time schedule for the demnd. Then the demnd mount nd schedule is used to determine to which phse the customer should e connected. f the demnd is unknown, then it is typicl to use the rted cpcity of the pole trnsformer s the customer demnd. Consider feeder, which hs loding profile of (A-0, B-60, nd C-00), ll in ka t given point on the line. The next customer closest to the distriution susttion to e dded is expected to demnd 40 ka. This customer should e plced on the B phse to yield resulting loding profile of (A-0, B-00, C-00). Thus this process strts t the frthest point of the feeder nd completes t the distriution susttion. The sme process is used if only two of the phses re present. Oviously, if only one phse is present, there is no decision to e mde. An exmple clcultion would e to divide the lods in Tle D.4 mong the three phses s shown. Note tht the feeder requires only the c phse to extend to the lst customer nd only the phse to extend to the second to lst customer. Thus, one would only find one phse t the end of the circuit, two phses s the circuit is trced to the source. Note tht loding is not equl. Note tht the demnd is only n estimte nd does not contin informtion s to the time of dy when the demnd will occur. This dt is normlly dopted from typicl demnd ptterns for typicl homes on typicl dys. Almost s ccurte s wether forecst! f the distnce etween customers is lrge compred to the cost of line, then the numer of phses extended to the lst customers cn e significntly chnged s shown in Tle D.5. All mterils re under copyright of PowerLern. Copyright 000, ll rights reserved.
D ntroduction to Distriution Systems 4 Note tht phse A does not extend eyond the fifth customer. Phse B does not extend eyond the ninth customer. Thus, the numer of conductors cn e reduced to sve costs. Additionlly, the complexity of the pole top design to crry one conductor is simpler thn for three. Also, less expensive. The economics of instlltion will determine the pproch tken. The student my wish to consider the reliility of the circuit. Would the extension of ll three phses to the lst customer provide service tht is more relile? Under wht conditions would the service e more relile? Under wht conditions would the service e of the sme reliility? Tle D.4 Distriution Customer Dt (Pln A) Customer Demnd Phse A Phse B Phse C (ka) 0 0 35 35 3 40 40 4 0 0 5 0 0 6 35 35 7 5 5 8 5 5 9 30 30 0 05 05 5 5 5 5 3 5 5 4 30 30 Totl 5 05 90 Tle D.5 Distriution Customer Dt (Pln B) Demnd Phse A Phse B Phse C Customer (ka) 0 0 35 35 3 40 40 4 0 0 5 0 0 6 35 35 7 5 5 8 5 5 9 30 30 0 05 05 5 5 5 5 3 5 5 4 30 30 Totl 5 05 90 All mterils re under copyright of PowerLern. Copyright 000, ll rights reserved.
D ntroduction to Distriution Systems 4 D.0 Tolernce of Current Tle D.6 shows the current impct on the humn ody. Note tht the mount of current is mesured in millimps. Wheres, the current in distriution system is normlly mesured in kilo-mps. Tle D.6 Tolernce of Current - Humns Current Effect -5 ma Senstion 0-0 ma nvoluntry muscle contrctions 0-00 ma Pin, rething difficult 00-300 ma entriculr firilltion, possile deth >300mA Respirtory prlysis, urns, unconsciousness, permnent loss of short-term memory, compction of soft tissue (spinl cord, etc.) D. Customer Wiring The distriution trnsformer connected etween the distriution line nd the customers my e configured in numer of wys. The configurtion is sed on selected prctice (preferences) nd economic cost. Once in the customer s site, the wiring is the responsiility of the customer. The Ntionl Electric code requires ll designs to e within 5% of equipment cpility. Figure D.8 shows how single-phse connection cn e trnsformed into two-phse connection. Note tht the key is to ground the middle of the secondry coil to e the reference for the customer. k 0 0 Figure D.8 Distriution to Customer Wiring Trnsformtion Figure D.9 shows typicl distriution ox found in most residentil nd commercil fcilities. The incoming phses () re connected to min circuit reker to isolte the customer from the source if current exceeds mximum (typiclly 00 or 00 mps). The min circuit reker is connected to two () us rs to which individul circuit rekers re ttched. The sic circuit reker will feed one or two rooms nd is limited to 5 or 0 mps. Note tht the lck wire is lwys ssumed hot unless the circuit reker is open. The white is the norml return (ground). The green wire is the sfety ground. The sfety ground protects the user y providing lessor resistnce pth from the pplince to ground s shown in Figure D.0. All mterils re under copyright of PowerLern. Copyright 000, ll rights reserved.
D ntroduction to Distriution Systems 43 +0-0 ground MAN 40 0 lck white green lck lck white Figure D.9 Residentil Distriution Pnel Wter Pipe Applince without sfety ground Applince with sfety ground Ground Figure D.0 Sfety Ground Principles All mterils re under copyright of PowerLern. Copyright 000, ll rights reserved.
D ntroduction to Distriution Systems 44 A specil circuit reker, clled Ground Fult nterrupter (GF), cn e used to provide more protection to the user. As shown in Figure D., the GF senses the mount of current flowing to the pplince(s). f the mount of current flowing on the hot (lck) wire is not equl to the current flowing in the white wire, or if there is ny current flow through the green wire, then the circuit reker is opened. Note tht mny residences hve GF instlled in the wll ox insted of in the distriution pnel. The circuit reker in the distriution ox is then norml circuit reker. The Ntionl Electricl Code (NEC) requires GFs in ny room where the user my come in contct with wter pipe or the wter tles (kitchen, th, outdoors, etc.). Consider tht lmost ll wter pipes re metl nd re connected to ground through smll resistnce compred to most uilding mterils. The concern is tht the pplince my ecome defective nd the current would return to the ground through the user. Figure D. shows the protection provided y GF when user uses metl instrument (knife) to remove red from toster. Most users would prefer GF to strt the dy in etter wy. Wter Pipe Applince without sfety ground Applince with sfety ground nd GF Ground Figure D. Appliction of Ground Fult nterrupter The student should consider lterntive grounding prolems such s the stry voltge prolem. The stry voltge prolem hs occurred in mny frm fcilities. This prolem occurs when the ground is of high resistnce compred to lterntive current pths. An lterntive current pth my e uilding, nturl gs pipes, rilrod trcks, etc. Metl structures within the uilding re the primry prolem. Consider wht would hppen if n niml or humn touched hndril or metl feeder which is t higher voltge thn the ground underneth the feet. The electrons lwys find the lest resistnce pth independent of the discomfort to the iologicl entity. This is often serious prolem, since nimls cnnot sy tht they feel the power. The result is often cow tht will not et or llow nyone to hook up the milking mchine. f frm equipment is improperly instlled the currents cn e sufficiently high to kill. D. Two Port Circuits The tretment elow is lrgely dpted from []. n performing circuit nlysis of single phse distriution systems, it is often convenient to mke use of two-port theory. A two-port network is one with two pirs of terminls emerging from it such tht, for ech pir of terminls, the current entering one terminl is the sme s the current leving the other terminl. Figure D. illustrtes generlized two-port network. We require tht there e no independent sources in the network (dependent sources re llowed), nd there cn e no energy stored in the network. All mterils re under copyright of PowerLern. Copyright 000, ll rights reserved.
D ntroduction to Distriution Systems 45 + - Two Port Network + - Fig. D.: Generlized Two-Port Network Pssive two-port networks re commonly specified in terms of their network prmeters, ccording to the following expressions: z z Z (impednce) prmeters Z (D.7) z z Y (dmittnce) prmeters y y y y Y (D.8) H (hyrid) prmeters h h h h H (D.9) G (inverse hyrid) prmeters g g g g G (D.0) (trnsmission) prmeters A (D.) (inverse trnsmission) prmeters B (D.) n using two ports, it is ssumed tht only voltges nd currents t the terminls re of interest, i.e., there is no interest in computing currents nd voltges within the circuit. Such emphsis on terminl ehvior is common when deling with opertionl mplifiers. t is lso common when deling with trnsformers nd trnsmission lines. n fct, the -prmeters re used in distriution system nlysis nd re commonly referred to s the ABCD prmeters. Typiclly, the ABCD prmeters re defined with the current of Fig. D. defined in the opposite direction, s shown in Fig. D.3, where A B T C D (D.3) The -trnsmission prmeters my e relted to the ABCD-trnsmission prmeters ccording to A=, B=-, C=, nd D=-. All mterils re under copyright of PowerLern. Copyright 000, ll rights reserved.
D ntroduction to Distriution Systems 46 + - Two Port Network + - Fig D.3: Generlized two port network with direction reversed Given two-port network, ny prticulr set of prmeters my e computed from mesurements if we re clever out the conditions under which we perform the mesurements. For exmple, the z-prmeters my e computed from mesurements of so-clled short-circuit conditions. Specificlly, z z z z (D.4) 0 0 0 0 Ech expression in (D.4) my e inferred from (D.7) y solving for the desired prmeter nd then setting to zero the current or voltge necessry to eliminte the remining prmeter. For exmple, from (D.7), we see tht: z z z z Here, we see if =0, then z which is the first expression of (D.4). The expressions for the other z- prmeters, nd indeed for ll of the prmeters in the other two-port models, my e found in similr fshion. The expressions for the dmittnce, hyrid, inverse hyrid, trnsmission, nd inverse trnsmission re s follows: y y y y (D.5) h g 0 0 0 0 0 h g 0 0 0 0 0 h g 0 0 0 0 0 0 h (D.6) 0 g (D.7) 0 (D.8) 0 (D.9) 0 The ABCD prmeters cn e derived using eqs. (D.8) for given topology. There re three topologies tht re frequently encountered in power system models, s shown in Figs. D.4-D.6, nd it is useful to e le to derive their ABCD prmeters. The student should mke these derivtions for the three topologies shown. All mterils re under copyright of PowerLern. Copyright 000, ll rights reserved.
D ntroduction to Distriution Systems 47 0 Y Y Z Z Fig. D.4: ABCD prmeters for circuit YZ Z Z YZZ Y Y YZ Fig. D.5: ABCD prmeters for T circuit Z YZ Z Y Y Y Y ZY Y Y Z Fig. D.6: ABCD prmeters for circuit One importnt reson for studying two ports is tht lrger system contining multiple two-ports my e efficiently nlyzed ccording to how the vrious two-ports re interconnected. Connection possiilities include series, prllel, nd cscde. A hyrid connection is lso possile []. llustrtion of the series, prllel, nd cscde connections re given elow together with the reltions etween the pproprite prmeters []. So the pproch to deling with interconnected two-ports is:. dentify the connection type.. Otin the two-port prmeters pproprite to the connection type (Z for series, Y for prllel, nd T for cscde) 3. Otin the composite prmeters y performing the opertion pproprite to the connection type (ddition of Z for series, ddition of Y for prllel, nd multipliction for cscde). All mterils re under copyright of PowerLern. Copyright 000, ll rights reserved.
D ntroduction to Distriution Systems 48 Z Z Z Z z z Z z z Z z z z z Fig. D.7: Series connection of two-port networks Fig. D.5: Prllel connection of two-port networks Y Y Y Y Y Y y y y y y y y y Fig. D.6: Cscde connection of two-port networks T T T T A T C T A C B D B D n the nlysis of opertionl mplifier circuits, it is frequently of interest to compute input impednce, current gin, nd/or voltge gin. Such computtions re fcilitted y comining the two port-reltions with the constrint reltions ssocited with the input nd output quntities of the two-port. Such nlysis my e found in mny circuit nlysis texts such s []. Power system pplictions of two ports include clcultion of voltge regultion, e.g, exmple D. my e solved using two-port networks. References [] P. Anderson, Anlysis of Fulted Power Systems, 973, ow Stte University Press, Ames, ow. [] J. Neilsson, Electric Circuits, second edition, Addison-Wesley Pulishing Compny, Reding, Mss., 986. All mterils re under copyright of PowerLern. Copyright 000, ll rights reserved.
D ntroduction to Distriution Systems 49 P R O B L E M S Prolem A single-phse distriution system is modeled using the circuit elow. ) Compute the voltge required t the source to hold voltge of 4k t the lod us. ) Determine the dditionl cpcitive rective power required t the lod us to correct the power fctor t the lod to 0.98. Prolem A single-phse distriution feeder supplies lod of 80 kw, 0 kars. The impednce of the feeder is Z = (5+j) ohms. t is required to hold voltge of k t the lod. Compute the minimum sending-end voltge tht will stisfy the lod-end voltge requirement t this lod level. Also compute reg. Prolem 3 A lnced three-phse industril lod hs two components: 00 ka t 0.5 power fctor lgging 00 kw resistive lod The two lods re supplied y 3. k feeder. ) Find the mgnitude of the feeder current ) Determine the cpcitive ARs required to correct the power fctor of the entire lod to 0.80 lgging. Prolem 4 A lrge industril customer is connected to the system vi three-phse distriution circuit. The customer consumes 30 MW t 0.95 power fctor lgging during pek conditions. The voltge t the customer is 4 k. n order to otin lower electric energy rte form the supplier, the customer must correct the power fctor to 0.98 lgging. So the customer hs decided to instll shunt cpcitnce. Compute the necessry correction in terms of: ) 3-phse rective power ) susceptnce B c c) cpcitnce C Prolem 5 A lrge industril fcility is consuming 0MW t 0.90 power fctor lgging nd 5MW t 0.85 power fctor lgging. Compute the cpcitive ARs necessry to correct the overll plnt power fctor to 0.95 lgging. All mterils re under copyright of PowerLern. Copyright 000, ll rights reserved.
D ntroduction to Distriution Systems 50 Prolem 6 A lrge industril lod tht consumes 8 MW nd 6 MAR must correct its power fctor to 0.95. How much dditionl rective power is necessry to do this? Prolem 7 A single-phse distriution feeder supplies lod of 00+j400. The impednce of the feeder is Z = (5+j) ohms. t is required to hold voltge of k t the lod. Compute the voltge regultion of this feeder. Prolem 8 Three lods re connected in prllel cross three-phse supply hving line-to-line voltge of.47 k. These lods re specified s Lod : nductive Lod, 60 kw nd 660 kar Lod : Cpcitive Lod, 40 kw t 0.8 power fctor Lod 3: Resistive Lod of 60 kw () Find the totl complex power consumed y ll three lods, nd the current nd power fctor s seen from the supply. Be sure to indicte whether the power fctor is leding or lgging. () A Y-connected cpcitor nk is connected in prllel with the lods. Find the totl kar required from the cpcitor to improve the power fctor to 0.8 lgging. Prolem 9 A customer t the end of distriution feeder is experiencing voltge regultion prolems. Specificlly, under high demnd, when the customer s power fctor is 0.90 lgging, the voltge mgnitude t the customer s meter is too low. The customer comes to you, the engineer, suggesting tht it my e fesile to correct this voltge mgnitude prolem y instlling shunt cpcitor t the customer s site. Descrie () How shunt cpcitor cn help the voltge mgnitude prolem under the indicted conditions? () Wht clcultions you would mke to determine whether this is in fct fesile solution? Prolem 0 An industril fcility is consuming 00 kw nd 48.4 kar t voltge of 480 volts line-to-line. () Compute the power fctor of the lod. ndicte whether it is leding or lgging. () Compute the dditionl rective power necessry from cpcitor nk to correct the power fctor to 0.95 lgging. (c) Compute the per-phse rectnce of the cpcitors needed to perform this correction. Assume the cpcitors will e Y-connected t 480 volts. Prolem A lnced three phse industril fcility consists of two prllel lods, s follows: o 00 ka t 0.5 power fctor lgging o 00 kw (entirely resistive lod) The two lods re supplied y three phse, distriution feeder circuit hving impednce of 4+j ohms per phse. The lod voltge is 3. k line-to-line.. Find the mgnitude nd ngle of the feeder current. Find the mgnitude of the line-to-line voltge t the sending end of the distriution feeder circuit. c. Determine the cpcitive ARS required to correct the power fctor of the entire lod to 0.80 lgging. d. Determine the susceptnce of the cpcitor necessry to supply these vrs t the stted lod voltge. All mterils re under copyright of PowerLern. Copyright 000, ll rights reserved.
D ntroduction to Distriution Systems 5 Prolem Show tht ABCD prmeters for the following circuits re given s indicted.. Shunt or circuit s given in Figure D.4.. T-circuit s given in Figure D.5. c. Pi-circuit s given in Figure D.6. Note the circuit configurtion nd mtrix when Y =Y =0. Prolem 3 Consider the following circuit prtitioned ccording to P,,P4, where 3-phse rdil feeder is supplying 5 customer lods. The line-to-line voltge of the lst us (#5) is k. Line impednces nd lod vlues re given in the figure. This prolem is solved in Exmple D. using KCL nd KL. n the exercise elow, the gol is to use ABCD prmeters to solve it, i.e., otin src given 5=,000/sqrt(3). P P P3 P4. P: Compute 5. Get ABCD prmeters. Compute [ 3 4] T.. P: Compute s3 nd then 3. Get ABCD prmeters. Compute [ 3] T. (Note tht you will lredy hve 3 so relly this clcultion just gives you ). c. P3: Compute s nd then. Get ABCD prmeters. Compute [ ] T. (Note tht you will lredy hve so relly this clcultion just gives you ). d. P4: Compute s nd then. Get ABCD prmeters. Compute [ src ] T. (Note tht you will lredy hve so relly this clcultion just gives you src). All mterils re under copyright of PowerLern. Copyright 000, ll rights reserved.