m_pbs - r111 EE740 Name pledge"N0 aid given received...

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Unformatted text preview: r111 EE740 Name _____________ __ pledge "N0 aid given, received, or observed" Department of Electrical Engineering The Ohio State University I 4W”- Problem 1. A {3% phase distribution feeder has two loads : Load#1 is rated 133.4 kva with power factor of 0.0 leading. Load #2 is rated 100 kW with power factor of 0.6 lagging. Both loads are connected to the same load bus. 1f the bad voltage is to be maintained at 4.4 kV( line value), compute the power factor at the load bus and the direction of active and reactive power )flow I . {15$ “fliLa—faz'iaow mew-win P=9(o;e. u-alaat’w " 3km”! Kw“ ._ f’ _ 9.- . I S _ “Emit J“: A“ “31:33”!st l?" zo-ouudlma 9‘ 3H4,“ L51” km —-—--- nfi-.....___fi...._ . _ . =lovrw+$133qq 61: lag-sq |--% :0 -3 I334”. giaiszlhbuw-Va ‘33.!!! _‘|?3-?:‘ = ‘00 VW tit-3‘ ") Problem '2. A balanced wye connected generator rated 460 volts (line Value) is supplying a balanced wye connected resistive load of 10 ohms through a balanced transmission line with 25 = 8 +j10, zab=zbc=zca=zm= )4 (ohms). If the generator is grounded and load is not grounded , draw the positive, negative and zero sequence F I 6 I . networks. i; L 1 o “EL—fl ._.___%:;i~.-fi._ ~ Ha (I) i r WW __ .- q 1. 1‘50 Ell W‘fi ' y+u ‘ i 57%“ . / - / .. _ .. _ _‘ " mg ' " ’7'" 9"3'3’ i'” 2-": 25+?2m =9'rsm’ f 21:25.. 2...: east. \7) Problem 3 Same as problem 2, but both the generator and loads are grounded. Compute the Symmetrical components of currents. £1; 5 "265-811.? .._£é_5:.§if:‘i = I ~ “lo I 3 1.- ‘gflb -— "L154, HAIL J‘W 5"vv'uu'l' M I I . .‘ Problem‘l. A three phase transformer rated 63 kV (wye connected and grounded)/ 20 kV(de1ta connected) .100 WA and 10% short circuit reactance supplied from a grounded power system with short circuit capacity of 1000 MVA. Assume a balanced three phase fault on 20 kV side while the 63 kV side is supplied from the power system at the rated voltage. Compute the short circuit capacity of 20 kV bus. \ooomfi‘r 63/20,” loom“: io‘l §CC_ :- |oou_ a flit—3‘ [Lu loo . '0 xth .— |U I 9-: l g I 7.". .1: 0” 3'11) f. 3'2 f 6 C~C. P.“ '~ 9f.” g‘c‘c Mr» = 9X|oo '1: 9m, x" /. / Problem 1. Consider the power system given below. Assume the following cases: Case 1): two phases are short circuited at bus 2 and then connected to ground. Case 2): one phase connected to ground at the mid secn'on of 20 ohm transmission line between bus 6 and bus 7. Case 3): three phases are short circuited at bus 5 and then connected to ground. Determine whether the ground current will now by indicating "yes" or "no" in the Table given below. Location Case] Case 2 Case 3 Ground reactance of trans. T3 0 LV side Y¢$ Y“ N0 ________ -—----------—---—-—---—In“.C-In-u---uuni-CHI--.Chin—OQUCCCIQ-utttue Ground reactance of of generator 61 Ground reactance of motor M -------- -—---_....---------_—_-_-—_-----*-_-"w~n¢-—--oco—a—oun—n“up—nu. Ground reactance of generator 62 -------------------------- .— _._... - .. — - -- - --- ----- -——- - -_------ - - -- - -.------ _ Ground reactance of trans.T-4 on HV side Problem (-3.- A three phase transformer rated 63 kV ( grounded wye) f20 kV(de1ta ) ,100 WA and 10% short circuit reactance supplied from a grounded power system with short circuit capacity of 1000 MVA. Assume a balanced three phase fault on 20 kV side while the 63 kV side is supplied from the power system at the rated voltage. Compute the short circuit capacity of 20 kV bus. WON-Ian- 51-91- % 90098? 595th» gin... “M: I) CamvpszCus 17loch 2.) Com ‘Sa cc:be vmot‘hrl}: 3) W 4'??— M/ 4?; - U.“ m l‘km'l‘fanmg C0 an f. Hmmasszmw (435,“. 5) W 4-!“ Sulfa; mmbr Pomr supflu/ i=0 flu mfwwk- m -’-! —l0 “(gas * [-uf 9 -9 Ieus—‘Vaus‘yeug —10 —5 19 VI ---)L" n a a 6 __ VL'S': i1 :0 V; —”-' 0 avg}; 1:3 “U Vg-‘LD fi?;_: 9 ’ __ a ‘5 j u —S 15 AF}; :9 P, -:. I-o f u EE740 Name ............. __ pledge "No aid given, received, or observed" Department of Electrical Engineering The Ohio State University Fall 2000 Problem}. A three phase transformer rated 138 kV ( grounded wye} £118 kV(grounded wye),100 MVA and 10% short circuit reactance supplied from a grounded power system with short circuit capacity of 1000 MVA. The power system bus voltage is 138 IN. The 13.8 kV side of transformer is fully loaded with a power factor of 0.0 lagging at rated voltage. Compute the fault current for a single line to ground fault at 13.8 kV bus for the following conditions: a) if the load is connected as a grounded wye . b) if the load is connected as an ungrounded wye . Assume that the impedance of positive, negative and zero sequence networks are equal. C-AUGFE bajg “cliff/185 qj .n___— LLH S5 : (GOA/TVA - ' Us.“ : £3 KV Vb; :2Urc1/ %_ \..._ DD tit? SLW‘E C(‘rcuL‘ (“a/Vac“? of 19A:- Sryrwtem :5 Wflen a CAUIK (36(qu if ZULU b'b/il-aj-{J «If 6aj7éawrgr~ £61 3-— £3+YT :0.f+0./ 3 F C" 2 ~50 “fl/U? gfior‘f Chm/vi?! Capacmflg ,‘5 Chum lime :Ja(u83 as 3—4.? 5L 9; : {00nd I/fl I 35 : Ub,i::r;8kl/. Ub.L:f3.8KV 'V . I r 33540}? umeedmlae is -__— W “J: 0 I f9 U H —j 9 x : ~j 36.2}Xr03r‘ =-324.1§KA I6: Tr D b) [396th r‘5 (orFfI€({£-'c/ 613 an unjlouflr/flf w if“ VLF—4w? Grid “Pam‘hyé 0”“? jzp Same n: I {[2] )1 fl '3 g H . 'rmxrrfiufi ?ta 2 25 43K: : 30. 2 > a 2;“ X7 25} J 4- 0 ‘ . L I =I a“: '7 'J'IWP” Compute the fault current for a single Ime to ground fault st 13.8 kV following conditions: a) if the load is - J E. {+1 {0) W MB s; z 275.}. : j0-2//Jl : new? Pu gang: “if 5* a“ M?)& w‘lll‘ PM “I “igflwfi 7wfév+jarm~jl k 3 I” I' T "l " ' " ' 3; J! i .T ‘l 3.7f' 1. H II Ila rg‘l Q - f '1? 5—12-50 I It QT. If” - I U U E‘— — {fluYr-g ‘ I“ 33 7“ - 327"? K“ T - :5 ,(LL) Co: Heir-le +£1£ {4-} 01d (_) ((Sufna; if.” «HIE (“3L9 a: H g) 3._ A is: 3 0 .. n. n+0 : DO 4 YT :3 do 4 s. D Um at S” I : I ‘ I :5 4' flfij‘ffif‘" : 0 __.____L_____J.__ 3H. (H. ‘ (+3.. 1' 01 L L, : 3' Problem'i A transformer bank is composed of three single- phase transformers supplying a three-phase load consistin of three identical 100 Ohm reactors. Ea single transfcrrner . rated at 100 MVA .138 W“ 3.8 W has a leakage reactance of 12% .The loads are connected in delta through a transmission line with .1+j.01 Ohm impedance. Assume transformers are connected in wyeldelta. Perform the following: Agmmpute the single - phase equivalent circuit . B Compute per unit uivalent circuit for a base of 100MVA and a voltage base of 13.8 V. A} for transformer annealed in Y/Q V” : Ta Xlééi : zgq rev VL : KU a _ (138de __ from LV§:JE,XT-— 0'2 X I'D!)ng - 0,12? 1 1.111 Yefm‘uaien‘i PERM-cian of U1; [and £5 XL : ig—D : 35.3.51 _ . x . c. So 'fLLO ffiuiucifiii CIVCM‘E "far i '1 Sid! ‘S 3.3!“! float? Hm; 33.3-; FM —' it.ti Yr. 570m iii/“ale 4‘1 WWW: (Mm km; 2375! jello 39pm i338-»_ Y~ all?“ Ewfiéiih r —’m‘“‘"” “’14 '6 21 I x7 VELMQ XL I élrtzi HJIIiYIW _ 50+ 33/1 *1 - :3 i *- =9933£l .. I 3 “3.8m i): P3} Sip—rug MUA, "JbzliRVU 27$: : J: 6 r )L '0' 'o yT: Pk” 42_ -i+).ul ‘ a-Lme : “TQM” I ‘Ogfl j-UUE PI“ Y 2 "'—3—5--—5-~ “'* l - '- lam ~- '7 S P” 1.0 w LI$NWfi VHV 3364+ Problem "1. A three phase transformer rated 63 kV ( grounded wye) £20 kV(delta ) .100 MVA and 10% short circuit reactance supplied from a grounded power system with short circuit capacity of 1000 MVA. Assume a balanced three phase fault on 20 kV side while the 63 kV side is supplied from the power system at the rated voltage. Compute the short circuit capacity of 20 kV bus. Have ‘0 f 5'C‘Cgpdk - Wv i+h=‘,!; 35‘”) Y1. ‘ U... ‘3‘?» smer a” I95 0‘) 3.. x, Haw/Mm EE74O Name ............. -_ pledge "N0 aid given, received, or observed" Department of Electrical Engineering The Ohio State University EE 740 —— Midterm Exam Problem 1. Consider a power system given below. The internal impedance of generators is: Zgl = j 0.10 Zg2 - j 0.20 All values are in p.u. 1.) Compute YBUS model for power flow studies. 2.) Compute ZEUS model for short circuit studies. Problem 2. For Problun Lassume the following load and generation schedules: Loads: s1 -'- 0.5 + 5.0.5 p.u.132 - 0.2 — j0.5 p.u. Generators: Generator one is the Swing generator and generator 2 scheduled power is P=0.5 p.u. Use Gauss-Seidel and compute Load Voltage at BUS 2. Assume V1 - 1.10 p.u. (Give your results for one iteration only.) Problem 3. For Probleml; assume a three phase fault at BUS 2. Compute the fault current seen by circuit breaker A. lgawvr.+ht Loads. Mill EE740 Algorithm for calculations of balanced and unbalanced fault currents. Step 1: Build the positive sequence Zbua matrix with reference to ground. Step 2: Build the negative sequence Z bus matrix. This is normally the same as the positive sequence. Step 3: Build the corresponding zero sequence Z buflo) matrix. Step 4: For specified fault types, connect the appropriate network sequences and compute the voltage and current flow at the faulted bus. 4 Zbusu) 1: positive 2: negative zbus(2) 0: zero zbus(0) l calculate the fault current based on a given fault type print the results A. Keyhani A. Keyhani EE740 Y fl Fault- Phase A to ground D (bus) E (bus) Generator Generator (-) (0) JIILDlII a; u 1 :5 it I W. Consider a as shown below: 1100(9) KVA = 133.34 load #3 Pf: 0,0 (leading) VR KVA = 100 100 KW KVA = 100 load #7 _ . 108d #4 pf = 0-0 “ad #1 p.f = .6 (leading) 1” ' '6 (1mm) { (lagging) Compute the following: 1) The sauce voltage V3, if V}; is to be maintained at 4.4 V (VR=4.4 kV line value). 2) The source current and power factor at the source. 3) The total complex power supplied by the source 4) How much reactive power should be connected to the source bus for obtaining unity power factor at the source bus? ‘ v | a.- a“. _ _ I _ ___ .‘. i d it": __L-J-‘LC opt:- a; Sksfir ‘- y 3 1 _h t 5”“? /- r, 1- H'- t: _ -‘I ‘ 4'_*"_‘ "' ‘ -—~———- 0a -' é " .. H _ I. .1 ""——-+ r: d. F :- rl J ‘ a ‘ ‘ ‘7 f! .- (.- r'r" ~ "‘“"'-~———-— -._- 'u.‘- “1—. ._ - '._'..r I’ _fltr} 1-. l“~.:_r_b “H .v C - I s: g ' ‘2 60K»; ‘6 3°ka‘( ‘ «:99 '5 fig) 9 car—F I. . " 31""57-‘w 93°-r:133-3L Li he .. I“. " L We "‘ ‘r 1m 33"" ‘33-'39 - . -—-—-"*—"""“" " ~ _ ’3 "3-7. ___:Eruw-Q' ._..._ I) c 3'32“ ‘e v '33" -; “ V1 is m be maintained a .5 Q— 4“! "' __ F ‘_ ‘ ’r a- I """"'_"' l ( 5"”?‘3 ‘3‘???“- “5 POh‘E‘r factor- at the 5".“E:?'C:?. it'LCI‘. Ci'.'n£'lflx Dwar- tfie gour.:eo warm-'7‘; HF»: mar: rE'Lct'Ix'»: power shoflc‘ be; committed to the ... ;: ' ., ;+.,. .-. -.. .. . "a. an ., pawer facux at +5: so;r':~} ins-:3 a n 5,. =5...) IH-i- 2943 Le- -. on a. In (lo-1'1".) = . I sfiufi $.11“ '11., v.3. ML; l3 q 33.: G . V? In... V; 6‘31 15:16-91:151” ‘54 VI was 94,: Lostl£- 3 4-51-13- 15"“! I5 F" on". ~¢M-. ---0 Problem 4. For the power system given below: 1.} Assume V1 = 1 p.u. Commute V2 and V3 (Use Gauss- Seidel. First iteration only.) 2.) If after ten iterations of Gauss-Seidel V2 - 1.07 p.u. and V3 :- 0.91 p.u, Compute: 2.1) Power mismatch at busl, bus2, and bus3 2.2) Power loss of transmission systems. ...
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m_pbs - r111 EE740 Name pledge"N0 aid given received...

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