Survey

* Your assessment is very important for improving the work of artificial intelligence, which forms the content of this project

Transcript

5.1 The full-load torque angle of a synchronous motor at rated voltage and frequency is 35 electrical degrees. Neglect the effects of armature resistance and leakage reactance. If the field current is held constant, how would the full-load torque angle be affected by the following changes in operating condition? a. Frequency reduced 10 percent, load torque and applied voltage constant. b. Frequency reduced 10 percent, load power and applied voltage constant. c. Both frequency and applied voltage reduced 10 percent, load torque constant. d. Both frequency and applied voltage reduced 10 percent, load power constant. ANS: 由公式(5.1)知 T poles 2 ( 2 ) 2 R F f sin RF 因此 T R F f sin RF 並且塲電流與 F f 同為常數 所以 R V sin RF Vt , 則T t f f 故 P f T Vt sin RF (a) 減少 31.1 (b) 不用改變 (c) 不用改變 (d) 增加 39.6 5.2 The armature phase windings of a two-phase synchronous machine are displaced by 90 electrical degrees in space. a. What is the mutual inductance between these two windings? b. Repeat the derivation leading to Eq.5.17 and show that the synchronous inductance is simply equal to the armature phase inductance; that is, Ls Laa0 La1 ,where Laa0 is the component of the armature phase inductance due to space-fundamental air-gap flux and La1 is the armature leakage inductance. ANS: (a) The windings are orthogonal and hence the mutual inductance is zero. (b) Since the two windings are orthogonal, the phases are uncoupled and hence the flux linkage under balanced two-phase operation is unchanged by currents in the other phase. Thus, the equivalent inductance is simply equal to the phase self-inductance. 5.3 Design calculations show the following parameters for a three-phase, cylindrical-rotor synchronous generator: Phase-α self-inductance Laa 4.83 mH Armature leakage inductance Lal 0.33 mH Calculate the phase-phase mutual inductance and the machine synchronous inductance. Sol : 1 Lab Lba L ac Lca Lbc Lcb Laa 0 2 Laa Laa 0 Lal Laa0 Laa Lal 1 1 Lab ( Laa Lal ) (4.83 0.33) 2.25 mH 2 2 3 ( Laa Lal ) Lal 2 3 1 Laa Lal 2 2 3 1 (4.83) (0.33) 2 2 7.245 0.165 7.08 mH Ls 5.4 The open-circuit terminal voltage of a three-phase, 60-Hz synchronous generator is found to be 15.4 KV rms line-to-line when the field current is 420 A. Calculate the stator-to-rotor mutual inductance Laf. b. Calculate the open-circuit terminal voltage if the field current is held constant while the generator speed is reduced so that the frequency of the generated voltage is 50 Hz. a. Sol : part (a): eaf af d af dt Laf I f cos( t 0 ) eaf d ( Laf I f cos( t 0 )) dt Laf I f sin( t 0 ) sin cos( 90 0 ) eaf Laf I f cos( t 0 90 0 ) E af ,l l ,rms Laf Laf I f 2 3 2 E af ,l l ,rms 3 I f 2 15.4 3 2 60 420 part (b): 50 Voltage 15.4 kV 12.8 kV 60 79.4 mH 5.5 A 460V，50-KW，60HZ，three-phase synchronous motor has a synchronous reactance of X s 4.15 and an armature-to-field mutual inductance， Laf 83mH .The motor is operating at rated terminal voltage and an input power of 40KW.Calculate the magnitude and phase angle of the line-to-neutral generated voltage E af and field currect I f if the motor is operating at （a）0.85 power factor lagging （b）unity power factor，and （c）0.85 power factor leaging Sol： 40 103 （a）a 相電流 E af Va jX s I a Ia 59.1A 0.85 3 460 功因 0.85 落後，所以相角 cos 1 0.85 31.8 I a I a e j 59.1e j 31.8 E af Va jX s I a If 2 E af Laf 460 3 j 4.15 59.1e j 31.8 136 56.8 V 11.3 A （b）同（a）pf=1 相角 cos 1 1 0 I a I a e j 59.1e j 0 59.1 E af Va jX s I a 266 38.1 V If 2 E af Laf 15.3 A （c）同（a）pf=0.85 超前 相角 cos 1 0.85 31.8 I a I a e j 59.1e j 31.8 E af Va jX s I a 395 27.8 V If 2 E af Laf 20.2 A 5.6 The motor of Problem 5.5 is supplied form a 460-V，three-phase source through a feeder whose impedance is Z f 0.084 j 0.82 . Assuming the system（as measured at the source）to be operating at an input power of 40KW，Calculate the magnitude and phase angle of the line-to-neutral generated voltage E af and field currect I f if the motor is operating at （a）0.85 power factor lagging （b） unity power factor，and （c）0.85 power factor leaging Sol： Z f jX s 0.084 j 0.82 j 4.15 0.084 j 4.97 解法同 5.5 I f 1 2.2 A I f 16.3 A I f 22.0 A （a） E af 106 66.6 V （b） E af 261 43.7 V （c） E af 416 31.2 V 5.8 The manufacturer’s data sheet for a 26-kv, 750MVA, 60Hz, three-phase synchronous generator indicates that it has a synchronous reactance Xs=2.04 and a leakage reactance Xal=0.18,both in per unit on the generator base. Calculate (a) the synchronous inductance in mH, (b) the armature leaking inductance in mH,and (c)the armature phase inductance Laa in mH and per unit sol : (a) Z base Ls 2 Vbase (26 10 3 ) 2 0.901 Pbase 750 10 6 X s , pu Z base 2.04 0.901 4.88 mH 2 60 (b) La1 X a1, pu Z base 0.18 0.901 0.43 mH 2 60 (c ) 先求三相同步電感 L s 定義為 Ls 3 2 L aa0 L a1 L aa0 L aa0 2 ( L s L a1 ) (1) 3 代表氣隙磁通的基波分量所產生的自感成分 L a1 代表電樞漏磁通所產生的電感分量 將(1)式代入電樞相電感公式得 L aa L aa0 l a1 2 3 (4.48 0.43) 0.43 3.40mH 5.9 The following reading are taken from the results of an open-and a short-circuit test on an 800-MVA,three-phase,Y-connected,26-kV,two-pole,60-Hz turbine generator driven at synchronous speed: Field current, A Armature current, short-circuit test, kA Line voltage, open-circuit characteristic, kV Line voltage,air-gap line,kV 1540 9.26 26.0 29.6 2960 17.8 (31.8) (56.9) The number in parentheses are extrapolations based upon the measured data. Find (a) the short-current ratio,(b)the unsaturated value of the synchronous reactance in ohms per phase and per unit ,and (c) the saturated synchronous reactance in per unit and in ohms per phase sol : (a) 短路比SCR AFNL 1540 0.52 AFSC 2960 (b) Z base 2 Vbase (26 10 3 ) 2 0.845 Pbase 800 10 6 X s ,u 2960 29.6 2.19 pu 1.85 1540 26 (c) 由課本5 - 29式可發現, SCR是飽合同步電抗標么值得倒數 1 1 Xs 1.92 pu 1.62 SCR 0.52 5.10 The following readings are taken from the results of an open and a short circuit test on a 5000-kW, 4160-V, three-phase, four-pole, 1800-rpm synchronous motor driven at rated speed. The armature resistance is 11 mOhm/phase. The armature leakage reactance is estimated to be 0.12 p.u. on the motor rating as base. Find (a) the short-circuit ratio, (b) the unsaturated value of the synchronous reactance in ohms per phase and p.u., (c) the saturated synchronous reactance in per unit and in ohms per phase. Sol: part(a) SCR AFNL 1.14 AFSC part(b) Z base 4160 2 Xs (5000 10 3 ) 3.46 1 1.11 pu 3.86 SCR part(c) X s ,u AFSC 0.88 pu 3.05 AFNL, ag 5.12 Consider the motor of Problem 5.10. a. Compute the filed current required when the motor is operating at rated voltage, 4200kW input power factor leading. Account for saturation inder load by the method described in the paragraph relating to Eq. 5.29. b. In addition to the data given in Problem 5.10, additional points on the open-circuit characteristic are given below. If the circuit breaker supplying the motor of part (a) is tripped, leaving the motor suddenly open-circuited, estimate the value of the motor terminal voltage following the trip. Sol: part( a) The total power is S I a 67029.5 0 and P 4200kW 4828kVA pf 0.87 Z s 0.038 j 4.81 4349 I f A F N (L ) 3 0 6A 4 1 6 /0 3 part(b) If the machine speed remains constant and the field current is not reduced, the terminal voltage will increase to the value corresponding to 306 A of field current on the open-circuit saturation characteristic. Interpolating the given data shows that this corresponds to a value of around 4850 V line-to-line. 5-14 Loss data for the motor of Problem 5.10 are as follows： Open-circuit core loss at 4160V = 37kW friction and windage loss = 46kW field-winding resistance at 75℃=0.279Ω Compute the output power and efficiency when the motor is operating at rated input power, unity power factor, and rated voltage. Assume the field-winding to be operating At a temperature of125℃ Ans: At rated power, unity power factor, the armature current will be Ia =5000 kW/(√3 4160 V) = 694 A. The power dissipated in the armature winding will then equal Parm = 3× 6942 × 0.011 = 15.9 kW.The field current can be found from |Eaf | = |Va − ZsIa| = |4160√3 − ZsIa| = 2394 V, line-to-neutral If = AFNL_2394/4160/√3) = 238 A At 125◦C, the field-winding resistance will be Rf = 0.279(234.5 + 125/234.5+ 75)= 0.324 Ω and hence the field-winding power dissipation will be Pfield = I2f Rf = 18.35 kW.The total loss will then be Ptot = Pcore + Parm + Pfriction/windage + Pfield = 120 kW Hence the output power will equal 4882.75 kW and the efficiency will equal 4882.75/5000 = 0.976 = 97.6%. 5-16 What is the maximum per-unit reactive power that can be supplied by a synchronous machine operating at its rated terminal voltage whose synchronous reactance is 1.6 per unit and whose maximum field current is limited to 2.4 times that required to achieve rated terminal voltage under open circuit conditions? Ans: For Va = 1.0 per unit, Eaf,max = 2.4 per unit and Xs = 1.6 per unit Qmax = (Eaf,max − Va)/ Xs= 0.875 per unit 5.19 A synchronous machine with a reactance of 1.28 per unit is operating as a generator at a real power loading of 0.6 per unit connected to a system with a series reactance of 0.07 per unit. An increase in its filed current is observed to cause a decrease in armature current. a. Before the increase, was the generator supplying or absorbing reactive power from the power system? b. As a result of this increase in excitation, did the generator terminal voltage increase or decrease? c. Repeat parts (a) and (b) if the synchronous machine is operating as a motor. SOL: part (a) : It was underexcited, absorbing reactive power. part (b) : It increased. part (c) : The answers are the same. 5.22 A four-pole, 60-Hz, 24-kV, 650-MVA synchronous generator with a synchronous reactance of 1.82 per unit is operating on a power system which can be represented by a 24-kV infinite bus in series with a reactive impedance of j0.21Ω. The generator is equipped with a voltage regulator that adjusts the field excitation such that the generator terminal voltage remains at 24kV independent of the generator loading. a. The generator output power is adjusted to 375MW. (i) Draw a phasor diagram for this operating condition. (ii) Find the magnitude (in kA) and phase angle (with respect to the generator terminal voltage) of the terminal current. (iii) Determine the generator terminal power factor. (iv) Find the magnitude (in per unit and kV) of the generator excitation voltage Eaf. b. Repeat part (a) if the generator output power is increased to 600 MW. part (a): (i) (ii) Vt = V∞ =1.0 per unit. P = 375/650 = 0..577 per unit. Thus t sin 1 ( and Ve Iˆ a t PX ) 12.6 VtV j t V jX 0.5783.93 per unit I base Pbase /( 3Vbase ) 15.64kA and Thus I a 9.04 kA (iii) The generator terminal current lags the terminal voltage by t / 2 and thus the power factor is pf cos 1 t / 2 0.998 lagging (iv) Eˆ af V j ( X X s ) Iˆa 1.50 per unit 36.0 kV , line to line part (b): (i) (ii) Same phasor diagram Iˆa =0.928∠6.32° per-unit. Ia=14.5 kA. (iii) pf=0.994 lagging (iv) Eaf=2.06 per unit =49.4 kV, line-to-line 5.25 Repeat Example 5.9 assuming the generator is operating at one-half of its rated KVA at a lagging power factor of 0.8 and rated terminal voltage. ( X d 1.0 p.u, X q 0.6 p.u ) Solution: 0 LetVˆa Va e j 0.0 Va power factor = 0.8 lagging cos 1 0.8 36.9 Iˆ I e j 0.8 j 0.6 1.0e j 36.9 a a Eˆ Vˆa jX q I q 1 j 0.6(1.0e j 36.9 ) 1.44e j19.4 ' 19.4 19.4 (36.9) 56.3 I d Iˆa sin( ) 1.0 sin 56.3 0.832 I q Iˆa cos( ) 1.0 cos 56.3 0.555 Iˆd 0.832e j ( 9019.4) 0.832e j 70.6 Iˆq 0.555e j19.4 Eˆ af Vˆa jX d I d jX q I q 1 j 0.5(0.832e70.4 ) j 0.6(0.555e j19.4 ) 1.281 j 0.452 1.358e j19.4 5.30 What maximum percentage of its rated output power will a salient-pole motor deliver without loss of synchronism when operating at its rated terminal voltage with zero field excitation ( Eaf 0 ) if X d 0.90 per unit and X q 0.65 per unit ? Compute the per unit armature current and reactive power for this operating condition. Solution: For E af 0 LetVt 1.0 p.u Pmax Vt 2 2 1 1 0.21 21% X X q d Maximum power for 45 V cos so I d t 0.786 p.u Xd Iq Vt sin 1.088 p.u Xq I a I d2 I q2 1.34 p.u S Vt I a 1.34 p.u Q S 2 P 2 1.32 p.u 5.31 if the synchronous motor of Problem 5.30 is now operated as a synchronous Generator connected to an infinite bus of rated voltage, find the minimum Per-unit field excitation (where 1.0 per unit is the field current Required to achieve rated open-circuit voltage) for which the generator will remain synchronized at (a) half load and (b) full load Sol: p V Eaf Xd sin V2 1 1 ( ) sin 2 2 Xq Xd 發電機要保持同步只要 Pmax＞P 就可以，但我們可以用 MATLAB 就可以輕易的 解題 而且可以滿足任何特定的條件 所以 Part (a) For P = 0.5, must have Eaf ≥ 0.327 per unit. Part (b) For P = 1.0, must have Eaf ≥ 0.827 per unit. 5.32 A salient-pole synchronous generator with saturated Synchronous reactances Xd=1.57 per unit and Xq=1.34 per unit is connected to an infinite bus of rated voltage Through an external impedance Xbus =0.11 per unit. The generation is Supplying its rated MAV at 0.95 power factor Lagging, as measured at the generator terminals. a. Draw a phasor diagram indicating the infinite-bus voltage the armature current the generator terminal voltage the excitation voltage and the rotor angle b. Calculate the per-unit terminal and excitation voltage and the rotor angle in degrees Sol: Part(a) 向量圖如下 在 P=0.95 之下必須先滿足下列兩公式 p VVt sin t X bus Ia Vt V jX t 同時帶入 Vt Vt t 在指定功因角為提供一定功率，而且要考慮其發電機端電壓。 使用 MATLAB 來快速求解算出 Vt = 1.02 per-unit; Eaf = 2.05 per-unit; δ = 46.6 5-34. A two-phase permanent-magnet ac motor has a rated speed of 3000 r/min and a six-pole rotor.Calculate the frequency (in Hz) of the armature voltage required to operate at this speed. ANS: f n poles 3000 6 150 Hz 120 120 5-35. A 5_kW，three-phase，permanent-magnet synchronous generator produces an open-circuit voltage of 208 V line-to-line，60-Hz，when driven at a speed of 1800 r/min。when operating at rated speed and supplying a resistive load，its terminal voltage is observed to be 192 V line-to-line for a power output of 4.5 kW。 A. B. Calculate the generator phase current under this operating condition Assuming the generator armature resistance to be negligible，calculate the generator 60-Hz synchronous reactance。 C. Calculate the generator terminal voltage which will result if the motor generator load is increased to 5kW(again purely resistive) while the speed is maintained at 1800 r/min ANS: (a) I a P 4500 13.5 A 3 *Va 3 *192 (b) Eaf 208 120 V 3 Xs (c ) Ia Xs Eaf2 Va2 Ia P 3 * Va Eaf Va ( X s I a ) 2 2 3.41 5000 15 3 * 192 1202 110.82 3.07 15 The easiest way to solve this is to use MATLAB to iterate to find the required load resistance.If this is done,the solution is Va =108 V(line-to-neutral)=187 V (line-to-line).