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643_mid2 - ece643-exam#l — au/09 AIL T P/w(1 WI...

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Unformatted text preview: ece643-exam #l — au/09 . AIL T: P/w (1 WI imttfi‘yéiflgl’m (i) (a) (12 points) A typical torque-speed characteristic is shown below for a 3gb induction motor when operated from a fixed frequency - fixed voltage source. Consider a variable voltage — 1. (20 points total) variable frequency drive used to control that 34) induction motor. On each graph, sketch the new tor ue-s eed curve or the ollowin conditions: (i) voltage reduced by 50%, (ii) both voltage and fre uenc reduced b 50% . >- (i) voltage reduced by 50% (ii) both voltage and frequency reduced by 50 % twisty: W («1. It _ (b) (8 points) With the use of a variable voltage — variable frequency supply for an induction 2 / machine, wh is it essential to control the volts er hert ? Eggnog). Tibet” 15 how \Vv 9N mth THE Towbuél lg’th: Quit) \b The» Zfigfil/ /,_,_4 \‘i i .0 ’ e , ,4 i _ , ii , Wait/i alarm!) Viol/ill”) fififi/z‘alm/ fiM/gg 59319,, Mm, MM alumni (mm a, \HNM ,_._ ”\W Page 2 of 7 ece643-exam #1 — au/09 .3 We. aim/13‘ ‘ h v“; ,¥,r (aoF’ 2. (20 points total) The current waveforms to the right represent the stator currents and the field current of a 3 s nchrous rator folloWing a 3-phase short circuit on the stator terminals. Prior to the short circuit the mm d” p current um onent machine had been operating unloaded at rated ' voltage. M fiBiE§ES it possible for the 2g Mm onent in ene 0 the hases o the stator ”um urre «be/era? Explain why or why not, k‘98 M ZQ‘WHB Shmoinsm$ Mpg-I ‘ _ 0 Field in n! 06 o D OTV—K (/5 W6 Phxes ’ 5' 8 dc component I Time £1?! i‘;:\a L‘lfiw :;%cgr?>‘ndii?r¥:izhi::éfield currents followingasudgen short circuiton an initiallyu 7 ~ (1‘ {I 5th {439$ 2’ L4? ii “'bg Q, ” *1!) ~34: (”0’3 t0]: ‘4" ®() (“9411‘ Q‘fi‘lb . t» Y ‘ l _. 3,,” R «rm;i:;'\,4“, I +V (6 mints)“ e ac com onent o the teld current results rom (i) the changing peak-peak value of the ac component of stator current [which in turn is caused by transitioning from Xd” to Xd’ to Xd ] 0R he yoflponents of the stator current. g m (QR/cunts) Which one 0 the ollowin windin S rovides the dominate e ect on them / current waveforms durin the subtransient eriod: ./ I ‘ Hot! Laa = 7 d.- 2*cos(20) mHenries Lab = I; 3 + 2*cos(26 - 21t/3) mHenries K$$ Lay—“— 8*cos(6) mHenries 'Lff = 12 mHenries / 2,, Zia: :9 3 /‘ RR 0 Z L37L éb \\ ece643-exam #1 _ au/O9 MW!) filfi fllwkei) of}: / 3. (20 points) The following inductances mare given for a salient pole _. synchrous machine; determine the values of direct-axis reactance (X4) and transient direct-axis reactance. (3/2. '3: (a? "l” l ”E; C13 [ll/é) ’2! 3%4m LMb (0 La.\_ “" 3] L3: Lam (Llano 44/31.) ece643-exam #l — au/09 Tend to increase 0R 0 Have no tendency to change 0R 0 Tend to decrease. ’J\Cb9\(6 points) This question compares the angle between the magnetic axis of the rotor 4/ produced flux and the magnetic axisof the stator produce 0R stays the same 0R decreases MS points) Consider the equal-area criteria / analysis associated with a synchrous machine for' the condition of an abrupt change in loading. Illustrate the “equal areas” for a generator operating at (P0, 50) when the electrical load is abruptly disconnected and then a short time later, reconnected [same value of load reconnected]. ece643-exam #1 - au/09 GM, . . . . . . / \5,\ (20 pomts) Cons1der a 4 pole, 60 Hz Inductlon Woperatmg at Sllp of -0.05. Determine the followin : “ u g HAZE?) 1; $660 %\ Frequency of the rotor current (in Hert Ugo) £075) : a; ( Q Sgeed of the rotor roduced flux relative to the rotor (in rev. per min.) 70 a Sgeed of the stator roduced flux relative to stator (in rev. per min.) ? W Eerie/4 1 LC A "\ A? \ Mechanical sgeed of the rotor (in rev. per min.) 1! (if ‘1’, / 3 / l O o] W A” $2: 0+ ,0§ “SUD F m gme/fl/df LU if \ p Page 6 of 7 / (i . , ’. 54% ’Za?‘ Ifibfkr /7 J 7%? [2’21” , ECE 643 Exam #2 answers. au109 1.(a)(i) Torque 0L voltage, therefore ifV @ 1/2 then T @ %; same nsync, (ii) nSync reduced to 1/z of original value, T reduced also. (b) from Faraday’s law, v(t) = d)\(t) / dt & A = turns*d> = turns*area*B -) thus volts per hertz represents the level of magnetic flux density, so if volts/hertz toooooo high then, will saturate. 2.(a) at any given instant of time, ia(t) + ib(t) + ic(t) =0 , and all phase currents are the sum of a decaying exponential (i.e., dc offset) plus a ”symmetrical” component. Refer to the homework problem and the difference in the two ”symmetrical” waveforms as 00 was varied. Since the generator was initially unloaded, i.e., stator current = 0, then the sum of the dc offset and the ”symmetrical” component at t = 0" must also be zero. Any one phase may have a zero dc offset which means that the dc offset of the other two phases must have equal magnitudes with opposite signs. (b) The ac component of field current is a result of the DC components of the stator currents; (c) the amortisseur windings (damper windings) on the rotor dominate the subtransient time regime. 3. Laao = 6, La. = 1, ng = 2, Lane = 8, Lfi = 12 all values in millihenries xd = 27:de = 27tf*[1 + (3/2)*(6 + 2)]*10'3 = 2m‘*[13]*10‘3 = 4.9 ohms xd '= 27thd’ = 2nf*[13 - (3/2)*(82/12)]*1O‘3 = 2m‘*[5]*10'3 = 1.885 ohms 4. (a) when fault (short-circuit) occurs, all electrical power output goes to zero, whereas the mechanical power in stays the same, thus that mechanical power now goes to accelerating the machine. (b) The orientation of the magnetic axis of the stator and rotor of a DC machine are fixed in space by construction, thus does not change with load. However, for both synchrous and induction, angle between these two magnetic axis increase with load. (c) Initially machine operating at Pelee (P0,50), thus mechanical power input = P0, neglecting losses. Then when load removed, operating point ”jumps” to (0,230) and machine starts to accelerate, thus 8 increases. Then when load reconnected, operating point ”jumps” to sin(6) curve and 5 continues to increase for a bit. To be stable, energy used for acceleration, i.e., area A1 must be less than or equal to energy used to decelerate, i.e., area A2. 5. (a) frequency of rotor current = lslipl *frequency of stator current: 3 Hz; (b) speed of rotor produced flux relative to rotor = slip*nsync = -90 rpm, where nsync = 120*frequency/poles; (c) nsync = 1800 rpm; (d) nsync = 1800 rpm; (d) speed of rotor = (1 —— slip)*nsync = 1890 rpm. ammmm mu v ‘y'wi K. a / ? ece643-exam #2 —- au/08 l. (20 points total) Misc. questions [each question of equal value]. (a) A 39 synchronous generator is operating fully loaded and is a part of a arge interconnected system when suddenly a 3¢ short circuit occurs on its terminals. Obviously ere can be no i instantaneous change in speed of the machine, however in the first few electrical cycles following the short circuit what will be the trend related to the rotor speed: slow down or stay the same or ”If ’t mm (b) A 3Q synchronous motor is operating fully loaded and is connected to an infinite bus when suddenly a 3th short circuit occurs on its terminals. Obviously there can be no instantaneous change in speed of the machine, however in the first few electrical cycles following the short circuit what will be the trend related to the rotor w d: slow down -' or s eed u (c) The following is Kirchhoff’s Voltage Law in the direct axis for a synchronous machine in the qu reference frame. What .. of that e ouation re resents What is referred t% = a .fiéfl (d) Refer to the direct axis voltage given in (0) above. Unders steady state loaded conditions, which one, if any, of the ter on the ri ht hand side 0 the e uation would be era? Page 2 of 6 i ece643-exam #2 — au/os P g: (20 points) You are given two 31) ac motors; one is an induction machine and the other a synchronous machine, though you do not know which is which; each machine is a 6 pole, 60 Hz machine with comparable ratings and connected to a 60 Hz source. For various operating % conditions you have a‘limited amount of information available, either the speed or power factor information. For each case, identifg with a check mark all gossible machines for which this operating condition could satisfl. Note: there will be at least one answer for each case and for some there can be multiple answers — for full credit MUST indicate all. ' WWW Speed (in rpm) WE leading , I 637 ,J‘awrwa'hnw m~1~v¥h¢r4wwwwvmvu=i -_. t ~. s M} \ I _ 1M 3 1w ‘4 11;; 35:5 \\ SW7“ ”1" tr“ flgflw xi th‘b C16 6‘ ' Page3of6 ece643-exam #2 - tau/08 [1:3120 points total) Consider a salient pgle 3Q , 60~Hz, 2-pgle, sygchronous machine with the . ‘ k, , fo’Ilbwing inductances: L, = 9.15 milli-henries Lq = 6.0 milli-henries Laf= 6.6 milli—henries Lff= 8.0 milli-henries XA 3 3% C L51 What is the direct-axis transient reactance of the machine? . ,L. Q‘ 451, L‘“ 3‘7???) WW (m) M" ID Page 4 of6 V 7 ece643 -exam #2 ~ au/08 4_I_._ (20 points) A non-salient gale 3 Q synchronous generator has the following parameters: Xi = 1.2 per unit Xd’ = 0.6 per unit X4”= 0.25 per unit. Assume the machine is operating unloaded at rated terminal voltage, when a 3¢ short circuit occurs on the terminal of the machine. The resulting “symmetrical” short circuit for phase a IS shown below; also, it can be assumed that the dc offset for phase a is zero for this situation. Subtransient period Transient period Steady-state period Short-circu it current Extrapolation of envelope steady value Extrapolation of» .. transient envelope l (b) The circuit breaker that is installed for the generator and ust interrupt this current has a 5 / cycle interrupting time. What IS the “s metrical” rms curr nt at the time of mterru tion? bl. f/ . ‘ i C. } [pf HM T Q ”/4“. m; j g (c) What is the worst case eak current following the short circuit? This will require you / neglect the effect of the subtransient time constatnt. f i} X. (7; 7g 2. s; f e m » 0.1 V Page 5 of 6 ece643-exam #2 -— au/08 6 §_._ (20 points total) Misc. questions. [Each question of equal value] (a) Consider an induction motor that [8 driving a load that can be represented by a torque s eed curve of T= K *nz. Will ar reduction in the terminal voltage of the machine d__e_________crease ofihé the sli at which the machine' is o eratin ? (b) Consider an induction motor that IS driving a load that can be represented by a torque speed curve of T= K *nz. Assume this motor is supplied by a variable fieguencx, variable voltage supply. If the fr uenc is decreased with the volta e sta in the same, will thet teorgu delivered by the motor increase or decrease? (c) Consider the case for a 3 s chronous enerator o eratin at rated conditions when a 3g short circuit occurs on its terminals. As a result, there will be an ac current for a short period of time in the field circuit. Why? - WU; 155"" IA {15” {[514 iii-”f7 4-}! ‘ - '4;de t1~t€4 :19“ ii, J“4x‘~-i. in) dWirE, “lie“; {pg Hm; (LL (/unuf‘ tug fit, 5.“ "it it“ (ifwl'f‘ 30C GWSai“ pmitsfe Mimi t?! ”AWL l5 set» is» Spmma‘b win F’Wévm‘fj; VW across W’W ' (d) Under steady-state conditions, thes eed of the rotor roduced flux of a 34’ ac machine is rotatin_ in the o - uosite direction of the r Induction or or Generator? Page 6 of 6 ece643—exam #2 —— au/08 l. (20 points total) Misc. questions [each question of equal value]. (a) A 3Q synchronous generator is operating fully loaded and is a part of a large interconnected system when suddenly a 34) short circuit occurs on its terminals. Obviously there can be no instantaneous change in speed of the machine, however in the first few electrical cycles following the short circuit what Will be the trend related to the rotor speed: slow down or stay the same or sgeed up (b) A 39; synchronous motor is operating fully loaded and is connected to an infinite bus when suddenly a 3(1) short circuit occurs on its terminals. Obviously there can be no instantaneous change in speed of the machine, however in the first few electrical cycles following the short circuit what will be the trend related to the rotor speed: 510W down or stay the same 01‘ sgeed HE (C) The following is Kirchhoff” 5 Voltage Law in the direct axis for a synchronous machine in the qu reference frame. What part of that equation represents what is referred to as the speed voltage term and what part represents the transformer voltage term? d/i dk/dt -) transformer, a)?» 9 speed voltage v —- R *z’ +————d — * h d "' a d dt me q (d) Refer to the direct axis voltage given in (c) above. Under steady state loaded conditions, which one, if any, of the terms on the ri ht hand side 0 the e nation would be era? dA/dt 9 would go to zero in 3.3. because flux in both d & q axis are constant. Induction machine Synchronous machine Speed (in rpm) Power factor Motor Generator Motor unknown -- --— _—- -— -- -— X X X Page 1 of 2 ece643—exam #2 — au/08 3; (20 points total) Consider a salient pole 3g; , 60-Hz, 2-pole, synchronous machine with the following inductances: L1 = 9.15 milli—henries Lq z 6.0 milIi—henries Laf= 6.6 milli-henries Lff= 8.0 milli-henries What is the direct-axis transient reactance of the machine? Ld’ = L; —~ (3/2)*(Laf2/Lff) =0.9825 10'3 henries -) Xd’ = ZaiLd’ = 0.370 £2 £ (20 points) A non-salient ole 3 s nchronous enerator has the following parameters: X; = 1.2 per unit Xd’ = 0.6 per unit Xd” = 0.25 per unit. (a) What is the dc otZset [or phase b? When ia @ 0 going positive, ib @ —\/3/2 of peak & ic @ +\/3/2 of peak. Therefore the offset must be +\/3/2 of peak for phase b so that item] 9 zero; the peak is given by \/2* l/Xd” = 5.66 -) 11335561: 4.90 per unit. (b) The circuit breaker that is installed for the generator that must interrupt this current has a 5 cycle interrupting time. What is the “symmetrica ” rms current at the time of interruption? This is during the transient time 9 l/Xd’ = 1.67 per unit (0) What is the worst case peak current foTlowing the short circuit? This will require you neglect the effect of the subtransient time constant, i.e., Td” -) 00. 2 "E \/2* l/Xd” = 11.32 per unit. i (20 points total) Misc. questions. [Each question of equal value] , (a) Consider an induction motor that is driving a load that can be represented by a torque speed curve of T = K *nz. Will a reduction in the terminal voltage of the machine decrease or increase the slip at which the machine is operating? (b) Consider an induction motor that is driving a load that can be represented by a torque speed curve of T = K *nz. Assume this motor is supplied by a variable frequency, variable voltage supply. If the frequency is decreased with the voltage staying the same, will the torgue delivered by the motor increase or decrease? (c) Consider the case for a 3g; synchronous generator operating at rated conditions when a 33 short circuit occurs on its terminals. As a result, there will be an ac current for a short period of time in the field circuit. Why? From dc component of current in the stator produced flux, i.e., a stationary field on the stator, and the rotor is rotating, thus an ac voltage is induced in field. ((1) Under steady-state conditions, the speed of the rotor produced flux of a 3(1) ac machine is rotating in the opposite direction of the rotor. Is the machine Induction or synchronous AND is it a Motor or Generator? [or Ind. machine in “braking” region] Page 2 of 2 /, k ECE 643 Exam #2 answers au/07 1. (a) NO; since short circuit (so) occurs simultaneously across all phases, and generator initially balanced, then initial conditions for each phase will always be different and it is the initial conditons, i.e., the voltage at the instant of sc, that determines the DC offset. No DC offset occurs if 30 occurs at voltage angle = 90°; cannot occur simultaneously on all 3 phases. (b) All currents must add to zero at all times, even during transients, thus offset of phase b must be equal magnitude and opposite sign to that for phase c. (c) Xd" < Xd' < Xd Xline’ Xline 2- (a) Xgen := 1.0 Xline := 0.25 Xeq := Xgen + ———-— Xeq = 1.125 Xline + Xlinc Va: 1 P2: 1 pf ;= 1 therefore 121;: 1 Eat" :2 Va + Ia~j~Xeq IEafI = 1.505 arg(Eaf) = 48.366 deg or could have used: 5 z: asin fl\ 8 = 48.366deg N“ [EafI‘IVaU (b) Note that with one line out the equailvaent raeactance changes; however the mangitude of Eaf will remain the same since the field current does not change. in the steady state, the current will be different, though the real power will remain constant. _ , P-Xeq_new \ Xeq_new := Xgen + the 6new:= asm(-——-——— Snew = 56.145deg lEati - M J (0) immediately after the one line is opened, the power transfer equation given by Pelec = [|Eaf|\Va|/Xeq}sin(6)] gives that Pelec < Pmech, therefore the machine will speed up. 3. deg= 1.25 Xq:= 0.85 Am: 1.0 1%=1.0 m: 1.0 Need angle 6 for the reluctance torque calculation Eafp :2 Va + j~Xq $3.212: arg(Eafp) 6 = 40365ng 2 Preluctance := wsinfib) Z-Xd‘Xq Preluctancc = 0.186 4.. -2- / 17' BONUS: speed of rotor poduced fluxwngLM,(&~«ef‘SlVl‘)“WTthT”s‘tat‘6r IS synchroususpeed“ I [get ' mfléflfif 6%«4‘3; b. / ._ ._ ._ ._ ._ ‘ 1 - 4. ‘Laao ._ 1.0 VLal ._ 0.1 ng ._ 0.25 'Laf.e 1.2 m Lff.— 13 IL)” Lilli“ j’ M “m - , “NM“. 2 \”’ - i 3 L f ’ ’7'“ ’ Ld ;= La1+ gaaao + ng) Ld = 1.975 Ldp := Ld — 2' 1:1“ Ldp = 0.535 01 k ‘4' ' m 5. (a) N voltage magnitude will reduce quickly (less than a second) to zero because the magnetic energy stored in the machine when the stator voltage removed is finite. For the SM with a PM rotor, the magnetic energy supply is "infinite" and thus as long as there is rotation, there will be a voltage; the voltage will be directly related to speed. The frequency of the OC voltage for both machines is directly related to the speed of the machines; the amount of kinetic energy available at the time when the stator voltage is removed determines how long it will take for the speed to go to zero. (b) T = 2*Tmax / [ s / smaxT + smaxT / s] .,...,xmr:m”T-M~~—v’ """""""" / / / ...
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