Electric Machinery Fundamentals Power & Energy_14 doc

281 (a) What is the line voltage of the two loads? (b) What is the voltage drop on the transmission lines? (c) Find the real and reactive powers supplied to each load. (d) Find the real and reactive power losses in the transmission line. (e) Find the real power, reactive power, and power factor supplied by the generator. S OLUTION To solve this problem, first convert the two deltas to equivalent wyes, and get the per-phase equivalent circuit. + - 277 ∠ 0° V Line 0.090 Ω j 0.16 Ω 1 φ Z 2 φ Z Ω°∠= 87.365.2 1 φ Z Ω°−∠= 2067.1 2 φ Z load, φ V + - (a) The phase voltage of the equivalent Y-loads can be found by nodal analysis. ,load ,load ,load 277 0 V 0 0.09 0.16 2.5 36.87 1.67 20 j φφφ −∠° ++= +Ω ∠°Ω∠−°Ω VVV () ( ) ()() ,load ,load ,load 5.443 60.6 277 0 V 0.4 36.87 0.6 20 0 φφφ ∠−° −∠°+∠− ° +∠° =VVV () ,load 5.955 53.34 1508 60.6 φ ∠− ° = ∠− °V ,load 253.2 7.3 V φ =∠−°V 282 Therefore, the line voltage at the loads is 3 439 V L VV φ = . (b) The voltage drop in the transmission lines is line ,gen ,load 277 0 V 253.2 -7.3 41.3 52 V φφ ∆=−=∠°−∠°=∠°VV V (c) The real and reactive power of each load is () 2 2 1 253.2 V 3 cos 3 cos 36.87 61.6 kW 2.5 V P Z φ θ == °= Ω () 2 2 1 253.2 V 3 sin 3 sin 36.87 46.2 kvar 2.5 V Q Z φ θ == °= Ω () () 2 2 2 253.2 V 3 cos 3 cos -20 108.4 kW 1.67 V P Z φ θ == °= Ω () () 2 2 2 253.2 V 3 sin 3 sin -20 39.5 kvar 1.67 V Q Z φ θ == °=− Ω (d) The line current is line line line 41.3 52 V 225 8.6 A 0.09 0.16 Zj ∆∠° == =∠−° +Ω V I Therefore, the loses in the transmission line are ()( ) 2 2 line line line 3 3 225 A 0.09 13.7 kWPIR== Ω= ()( ) 2 2 line line line 3 3 225 A 0.16 24.3 kvarQIX== Ω= (e) The real and reactive power supplied by the generator is gen line 1 2 13.7 kW 61.6 kW 108.4 kW 183.7 kWPPPP=++= + + = gen line 1 2 24.3 kvar 46.2 kvar 39.5 kvar 31 kvarQQQQ=++= + − = The power factor of the generator is gen -1 1 gen 31 kvar PF cos tan cos tan 0.986 lagging 183.7 kW Q P − == = A-3. Figure PA-2 shows a one-line diagram of a simple power system containing a single 480 V generator and three loads. Assume that the transmission lines in this power system are lossless, and answer the following questions. (a) Assume that Load 1 is Y-connected. What are the phase voltage and currents in that load? (b) Assume that Load 2 is ∆ -connected. What are the phase voltage and currents in that load? (c) What real, reactive, and apparent power does the generator supply when the switch is open? (d) What is the total line current L I when the switch is open? (e) What real, reactive, and apparent power does the generator supply when the switch is closed? (f) What is the total line current L I when the switch is closed? (g) How does the total line current L I compare to the sum of the three individual currents 123 III++? If they are not equal, why not? 283 S OLUTION Since the transmission lines are lossless in this power system, the full voltage generated by 1 G will be present at each of the loads. (a) Since this load is Y-connected, the phase voltage is 1 480 V 277 V 3 V φ == The phase current can be derived from the equation 3cosPVI φφ θ = as follows: ()() 1 100 kW 133.7 A 3 cos 3 277 V 0.9 P I V φ φ θ == = (b) Since this load is ∆ -connected, the phase voltage is 2 480 VV φ = The phase current can be derived from the equation 3SVI φ φ = as follows: () 2 80 kVA 55.56 A 3 3 480 V S I V φ φ == = (c) The real and reactive power supplied by the generator when the switch is open is just the sum of the real and reactive powers of Loads 1 and 2. 1 100 kWP = ( ) ( ) ( ) 1 1 tan tan cos PF 100 kW tan 25.84 48.4 kvarQP P θ − == = °= ( ) ( ) 2 cos 80 kVA 0.8 64 kWPS θ == = ( ) ( ) 2 sin 80 kVA 0.6 48 kvarQS θ == = 12 100 kW 64 kW 164 kW G PPP=+= + = 12 48.4 kvar 48 kvar 96.4 kvar G QQQ=+= + = (d) The line current when the switch is open is given by 3 cos L L P I V θ = , where 1 tan G G Q P θ − = . 11 96.4 kvar tan tan 30.45 164 kW G G Q P θ −− == =° ()() 164 kW 228.8 A 3 cos 3 480 V cos 30.45 L L P I V θ == = ° 284 (e) The real and reactive power supplied by the generator when the switch is closed is just the sum of the real and reactive powers of Loads 1, 2, and 3. The powers of Loads 1 and 2 have already been calculated. The real and reactive power of Load 3 are: 3 80 kWP = ( ) ( ) ( ) 1 3 tan tan cos PF 80 kW tan 31.79 49.6 kvarQP P θ − == = 123 100 kW 64 kW 80 kW 244 kW G PPPP=++= + + = 123 48.4 kvar 48 kvar 49.6 kvar 46.8 kvar G QQQQ=++= + − = (f) The line current when the switch is closed is given by 3 cos L L P I V θ = , where 1 tan G G Q P θ − = . 11 46.8 kvar tan tan 10.86 244 kW G G Q P θ −− == =° ()() 244 kW 298.8 A 3 cos 3 480 V cos 10.86 L L P I V θ == = ° (g) The total line current from the generator is 298.8 A. The line currents to each individual load are: ()() 1 1 1 100 kW 133.6 A 3 cos 3 480 V 0.9 L L P I V θ == = () 2 2 80 kVA 96.2 A 3 3 480 V L L S I V == = ()() 3 3 3 80 kW 113.2 A 3 cos 3 480 V 0.85 L L P I V θ == = The sum of the three individual line currents is 343 A, while the current supplied by the generator is 298.8 A. These values are not the same, because the three loads have different impedance angles. Essentially, Load 3 is supplying some of the reactive power being consumed by Loads 1 and 2, so that it does not have to come from the generator. A-4. Prove that the line voltage of a Y-connected generator with an acb phase sequence lags the corresponding phase voltage by 30°. Draw a phasor diagram showing the phase and line voltages for this generator. S OLUTION If the generator has an acb phase sequence, then the three phase voltages will be 0 an V φ =∠°V 240 bn V φ =∠− °V 120 cn V φ =∠− °V The relationship between line voltage and phase voltage is derived below. By Kirchhoff’s voltage law, the line-to-line voltage ab V is given by ab a b =−VVV 0 240 ab VV φφ =∠°− ∠− °V 1333 2222 ab VVjVVjV φ φφφφ =−− + = − V 31 3 22 ab Vj φ =− V 330 ab V φ =∠−°V 285 Thus the line voltage lags the corresponding phase voltage by 30 ° . The phasor diagram for this connection is shown below. V an V bn V cn V ab V bc A-5. Find the magnitudes and angles of each line and phase voltage and current on the load shown in Figure P2- 3. S OLUTION Note that because this load is ∆ -connected, the line and phase voltages are identical. 120 0 V 120 120 V 208 30 V ab an bn =−=∠° ∠ °=∠°VVV 120 120 V 120 240 V 208 90 V bc bn cn -=−=∠−° ∠ °=∠°VVV 120 240 V 120 0 V 208 150 V ca cn an -=−=∠−° ∠°=∠°VVV 286 208 30 V 20.8 10 A 10 20 ab ab Z φ ∠° == =∠° ∠°Ω V I 208 90 V 20.8 110 A 10 20 bc bc Z φ ∠− ° == =∠−° ∠°Ω V I 208 150 V 20.8 130 A 10 20 ca ca Z φ ∠° == =∠° ∠°Ω V I 20.8 10 A 20.8 130 A 36 20 A aabca =−= ∠° ∠°=∠°II I 20.8 110 A 20.8 10 A 36 140 A bbcab =−= ∠−° ∠°=∠ °II I 20.8 130 A 20.8 -110 A 36 100 A ccabc -=−= ∠° ∠ °=∠°II I A-6. Figure PA-4 shows a small 480-V distribution system. Assume that the lines in the system have zero impedance. (a) If the switch shown is open, find the real, reactive, and apparent powers in the system. Find the total current supplied to the distribution system by the utility. (b) Repeat part (a) with the switch closed. What happened to the total current supplied? Why? S OLUTION (a) With the switch open, the power supplied to each load is () kW 86.9530 cos 10 V 804 3cos3 2 2 1 =° Ω == θ φ Z V P ( ) 2 2 1 480 V 3 sin 3 sin 30 34.56 kvar 10 V Q Z φ θ == °= Ω () kW 04.4636.87 cos 4 V 277 3cos3 2 2 2 =° Ω == θ φ Z V P () 2 2 2 277 V 3 sin 3 sin 36.87 34.53 kvar 4 V Q Z φ θ == °= Ω kW 105.9 kW 46.04 kW 86.59 21TOT =+=+= PPP TOT 1 2 34.56 kvar 34.53 kvar 69.09 kvarQQQ=+= + = The apparent power supplied by the utility is 22 TOT TOT TOT 126.4 kVASPQ=+= The power factor supplied by the utility is 287 -1 1 TOT TOT 69.09 kvar PF cos tan cos tan 0.838 lagging 105.9 kW Q P − == = The current supplied by the utility is ()() TOT 105.9 kW 152 A 3 PF 3 480 V 0.838 L T P I V == = (b) With the switch closed, 3 P is added to the circuit. The real and reactive power of 3 P is () () kW 090 cos 5 V 277 3cos3 2 2 3 =° Ω == - Z V P θ φ ( ) () 2 2 3 277 V 3 sin 3 sin 90 46.06 kvar 5 V P- Z φ θ == °=− Ω TOT 1 2 3 59.86 kW 46.04 kW 0 kW 105.9 kWPPPP=++= + + = TOT 1 2 3 34.56 kvar 34.53 kvar 46.06 kvar 23.03 kvarQQQQ=++= + − = The apparent power supplied by the utility is 22 TOT TOT TOT 108.4 kVASPQ=+= The power factor supplied by the utility is -1 1 TOT TOT 23.03 kVAR PF cos tan cos tan 0.977 lagging 105.9 kW Q P − == = The current supplied by the utility is ()() TOT 105.9 kW 130.4 A 3 PF 3 480 V 0.977 L T P I V == = (c) The total current supplied by the power system drops when the switch is closed because the capacitor bank is supplying some of the reactive power being consumed by loads 1 and 2. 288 Appendix B: Coil Pitch and Distributed Windings B-1. A 2-slot three-phase stator armature is wound for two-pole operation. If fractional-pitch windings are to be used, what is the best possible choice for winding pitch if it is desired to eliminate the fifth-harmonic component of voltage? S OLUTION The pitch factor of a winding is given by Equation (B-19): 2 sin υ ρ = p k To eliminate the fifth harmonic, we want to select ρ so that 0 2 5 sin = ρ . This implies that () n 180 2 5 °= ρ , where n = 0, 1, 2, 3, … or () ,144 ,72 5 1802 °°= ° = n ρ These are acceptable pitches to eliminate the fifth harmonic. Expressed as fractions of full pitch, these pitches are 2/5, 4/5, 6/5, etc. Since the desire is to have the maximum possible fundamental voltage, the best choice for coil pitch would be 4/5 or 6/5. The closest that we can approach to a 4/5 pitch in a 24-slot winding is 10/12 pitch, so that is the pitch that we would use. At 10/12 pitch, 966.0 2 150 sin = ° = p k for the fundamental frequency ()( ) 259.0 2 150 5 sin = ° = p k for the fifth harmonic 289 B-2. Derive the relationship for the winding distribution factor k d in Equation B-22. S OLUTION The above illustration shows the case of 5 slots per phase, but the results are general. If there are 5 slots per phase, each with voltage Ai E , where the phase angle of each voltage increases by γ° from slot to slot, then the total voltage in the phase will be AnAAAAAA EEEEEEE ++++++= 54321 The resulting voltage A E can be found from geometrical considerations. These “n” phases, when drawn end-to-end, form equally-spaced chords on a circle of radius R. If a line is drawn from the center of a chord to the origin of the circle, it forma a right triangle with the radius at the end of the chord (see voltage 5A E above). The hypotenuse of this right triangle is R, its opposite side is 2/E , and its smaller angle is 2/ γ . Therefore, R E 2/ 2 sin = γ ⇒ 2 sin 2 1 γ E R = (1) The total voltage A E also forms a chord on the circle, and dropping a line from the center of that chord to the origin forms a right triangle. For this triangle, the hypotenuse is R, the opposite side is 2/ A E , and the angle is 2/ γ n . Therefore, R En A 2/ 2 sin = γ ⇒ 2 sin 2 1 γ n E R A = (2) Combining (1) and (2) yields 290 2 sin 2 1 2 sin 2 1 γγ n EE A = 2 sin 2 sin γ γ n E E A = Finally, 2 sin 2 sin γ γ n n nE E k A d == since d k is defined as the ratio of the total voltage produced to the sum of the magnitudes of each component voltage. B-3. A three-phase four-pole synchronous machine has 96 stator slots. The slots contain a double-layer winding (two coils per slot) with four turns per coil. The coil pitch is 19/24. (a) Find the slot and coil pitch in electrical degrees. (b) Find the pitch, distribution, and winding factors for this machine. (c) How well will this winding suppress third, fifth, seventh, ninth, and eleventh harmonics? Be sure to consider the effects of both coil pitch and winding distribution in your answer. S OLUTION (a) The coil pitch is 19/24 or 142.5°. Note that these are electrical degrees. Since this is a 4-pole machine, the coil pitch would be 71.25 mechanical degrees. There are 96 slots on this stator, so the slot pitch is 360 ° /96 = 3.75 mechanical degrees or 7.5 electrical degrees. (b) The pitch factor of this winding is 947.0 2 5.142 sin 2 sin = ° == ρ p k The distribution factor is 2 sin 2 sin γ γ n n k d = The electrical angle γ between slots is 7.5 ° , and each phase group occupies 8 adjacent slots. Therefore, the distribution factor is [...]... rated conditions? (b) What is the voltage regulation of this generator at the rated conditions? (c) Sketch the power- versus-torque-angle curve for this generator At what angle δ is the power of the generator maximum? (d) How does the maximum power out of this generator compare to the maximum power available if it were of cylindrical rotor construction? SOLUTION (a) At rated conditions, the line and phase... (c) The power supplied by this machine is given by the equation P= 3Vφ E A Xd 3Vφ2  X d − X q    sin 2δ sin δ + 2  Xd Xq    297 3(7621)(9890) 3(7621)  0.62 − 0.40  P= sin δ +   (0.62 )(0.40 )  sin 2δ  0.62 2   P = 364.7 sin δ + 77.3 sin 2δ MW 2 A plot of power supplied as a function of torque angle is shown below: The peak power occurs at an angle of 70.6°, and the maximum power that... − X q    sin 2δ 2  Xd Xq    2 3(277)  0.25 − 0.18  P=   sin 2δ = 179 sin 2δ kW 2  (0.25)(0.18)    P= At δ = 45° , 179 kW can be drawn from the motor 300 Appendix D: Errata for Electric Machinery Fundamentals 4/e (Current at 10 January 2004) Please note that some or all of the following errata may be corrected in future reprints of the book, so they may not appear in your copy of the... (c) The power supplied by this machine is given by the equation P= 3Vφ E A Xd sin δ + 3Vφ2  X d − X q    sin 2δ 2  Xd Xq    3(277 )(322) 3(277)  0.25 − 0.18  sin 5.61° +   sin 11.22° 0.25 2  (0.25)(0.18)    P = 104.6 kW + 34.8 kW = 139.4 kW 2 P= The cylindrical rotor term is 104.6 kW, and the reluctance term is 34.8 kW, so the reluctance torque accounts for about 25% of the power in... is the internal generated voltage of this machine when it is operating at rated conditions? How does this value of E A compare to that of Problem 5-2b? (c) What fraction of this generator’s full-load power is due to the reluctance torque of the rotor? SOLUTION (a) If the no-load terminal voltage is 480 V, the required field current can be read directly from the opencircuit characteristic It is 4.55... (a) The pitch factor of this winding is k p = sin ρ 2 = sin 150° = 0.966 2 (b) The coils in each phase group of this machine cover 4 slots, and the slot pitch is 360/48 = 7.5 mechanical degrees or 15 electrical degrees Therefore, the distribution factor is nγ (4)(15°) sin 2 = 2 = 0.958 kd = 15° γ 4 sin n sin 2 2 sin (c) The frequency of the voltage produces by this winding is fe = nm P (1800 r/min... produced by this generator at these conditions? SOLUTION There are 6 slots per pole × 6 poles = 36 slots on the stator of this machine Therefore, there are 36 coils on the machine, or 12 coils per phase The electrical frequency produced by this winding is fe = n m P (1200 r/min )(6 poles) = = 60 Hz 120 120 The phase voltage is Vφ = 2πN P k p k d φf e = 2π (96 turns )(0.981)(0.956)(0.02 Wb )(60 Hz ) = 480... SOLUTION (a) The pitch of this winding is 8/9 = 160°, so the pitch factor is k p = sin 160° = 0.985 2 The phase groups in this machine cover three slots each, and the slot pitch is 20 mechanical or 20 electrical degrees Thus the distribution factor is (3)(20°) nγ sin 2 2 = = 0.960 kd = γ 20° n sin 3 sin 2 2 sin The phase voltage of this machine will be Vφ = 2πN P k p k d φf e = 2π (6 coils )(60 turns/coil... power that the generator can supply is 392.4 MW (d) If this generator were non-salient, PMAX would occur when δ = 90°, and PMAX would be 364.7 MW Therefore, the salient-pole generator has a higher maximum power than an equivalent non-salint pole generator C-3 Suppose that a salient-pole machine is to be used as a motor (a) Sketch the phasor diagram of a salient-pole synchronous machine used as a motor (b)... stator pitch is 12/15 = 4/5, so ρ = 144° , and k p = sin 144° = 0.951 2 293 Each phase belt consists of (180 slots)/(12 poles)(6) = 2.5 slots per phase group The slot pitch is 2 mechanical degrees or 24 electrical degrees The corresponding distribution factor is nγ (2.5)(24°) sin 2 = 2 = 0.962 kd = 24° γ 2.5 sin n sin 2 2 sin Since there are 60 coils in each phase and 8 turns per coil, all connected in . the real and reactive powers supplied to each load. (d) Find the real and reactive power losses in the transmission line. (e) Find the real power, reactive power, and power factor supplied. the power- versus-torque-angle curve for this generator. At what angle δ is the power of the generator maximum? (d) How does the maximum power out of this generator compare to the maximum power. degrees or 7.5 electrical degrees. (b) The pitch factor of this winding is 947.0 2 5 .142 sin 2 sin = ° == ρ p k The distribution factor is 2 sin 2 sin γ γ n n k d = The electrical