(McGraw-Hill) (Instructors Manual) Electric Machinery Fundamentals 4th Edition Episode 2 Part 5 ppsx

275 The resulting torque-speed characteristic is shown below: 10-5. A 220-V, 1.5-hp 50-Hz, two-pole, capacitor-start induction motor has the following main-winding impedances: 1 R = 1.40 Ω 1 X = 2.01 Ω M X = 105 Ω 2 R = 1.50 Ω 2 X = 2.01 Ω At a slip of 0.05, the motor’s rotational losses are 291 W. The rotational losses may be assumed constant over the normal operating range of the motor. Find the following quantities for this motor at 5 percent slip: (a) Stator current (b) Stator power factor (c) Input power (d) AG P (e) P conv (f) out P (g) ind τ (h) load τ (i) Efficiency S OLUTION The equivalent circuit of the motor is shown below 276 1.4 Ω j1.9 Ω + - V = 220∠0° V I 1 R 1 jX 1 s R 2 5.0 j0.5X 2 j0.5X M j1.90 Ω j30 Ω s R −2 5.0 2 jX 2 j1.90 Ω j0.5X M j100 Ω { { { { Forward Reverse 0.5Z B 0.5Z F ()() 22 22 / / M F M Rs jX jX Z R s jX jX + = ++ ()() 30 1.90 100 26.59 9.69 30 1.90 100 F jj Zj jj + ==+Ω ++ () () () 22 22 /2 /2 M B M R s jX jX Z R s jX jX = −+ + ()() 0.769 1.90 100 0.741 1.870 0.769 1.90 100 B jj Zj jj + ==+Ω ++ (a) The input stator current is 1 11 I 0.5 0.5 FB RjX Z Z = ++ + V ()( )( ) 1 220 0 V I 13.0 27.0 A 1.40 1.90 0.5 26.59 9.69 0.5 0.741 1.870jjj ∠° ==∠−° ++ ++ + (b) The stator power factor is PF cos 27 0.891 lagging =°= (c) The input power is ()() IN cos 220 V 13.0 A cos 27 2548 WPVI θ == °= (d) The air-gap power is () ()( ) 2 2 AG, 1 0.5 13.0 A 13.29 2246 W FF PIR== Ω= () ()( ) 2 2 AG, 1 0.5 13.0 A 0.370 62.5 W BB PIR== Ω= AG AG, AG, 2246 W 62.5 W 2184 W FB PP P=−= − = 277 (e) The power converted from electrical to mechanical form is () ( )( ) conv, AG, 1 1 0.05 2246 W 2134 W FF PsP=− =− = () ( )( ) conv, AG, 1 1 0.05 62.5 W 59 W BB PsP=− =− = conv conv, conv, 2134 W 59 W 2075 W FB PP P=−= −= (f) The output power is OUT conv rot 2134 W 291 W 1843 WPPP=−= − = (g) The induced torque is () AG ind sync 2184 W 6.95 N m 2 rad 1 min 3000 r/min 1 r 60 s P τ π ω == = ⋅ (h) The load torque is ()( ) OUT load 1843 W 6.18 N m 2 rad 1 min 0.95 3000 r/min 1 r 60 s m P τ π ω == = ⋅ (i) The overall efficiency is OUT IN 1843 W 100% 100% 72.3% 2548 W P P η =× = × = 10-6. Find the induced torque in the motor in Problem 10-5 if it is operating at 5 percent slip and its terminal voltage is (a) 190 V, (b) 208 V, (c) 230 V. ()() 22 22 / / M F M Rs jX jX Z R s jX jX + = ++ ()() 30 1.90 100 26.59 9.69 30 1.90 100 F jj Zj jj + ==+Ω ++ () () () 22 22 /2 /2 M B M R s jX jX Z R s jX jX = −+ + ()() 0.769 1.90 100 0.741 1.870 0.769 1.90 100 B jj Zj jj + ==+Ω ++ (a) If T V = 190∠0° V, 1 11 I 0.5 0.5 FB RjX Z Z = ++ + V ()( )( ) 1 190 0 V I 11.2 27.0 A 1.40 1.90 0.5 26.59 9.69 0.5 0.741 1.870jjj ∠° ==∠−° ++ ++ + () ()( ) 2 2 AG, 1 0.5 11.2 A 13.29 1667 W FF PIR== Ω= () ()( ) 2 2 AG, 1 0.5 11.2 A 0.370 46.4 W BB PIR== Ω= AG AG, AG, 1667 W 46.4 W 1621 W FB PP P=−= − = 278 () AG ind sync 1621 W 5.16 N m 2 rad 1 min 3000 r/min 1 r 60 s P τ π ω == = ⋅ (b) If T V = 208∠0° V, 1 11 I 0.5 0.5 FB RjX Z Z = ++ + V ()( )( ) 1 208 0 V I12.327.0 A 1.40 1.90 0.5 26.59 9.69 0.5 0.741 1.870jjj ∠° ==∠−° ++ ++ + () ()( ) 2 2 AG, 1 0.5 12.3 A 13.29 2010 W FF PIR== Ω= () ()( ) 2 2 AG, 1 0.5 12.3 A 0.370 56 W BB PIR== Ω= AG AG, AG, 2010 W 56 W 1954 W FB PP P=−= − = () AG ind sync 1954 W 6.22 N m 2 rad 1 min 3000 r/min 1 r 60 s P τ π ω == = ⋅ (c) If T V = 230 ∠ 0 ° V, 1 11 I 0.5 0.5 FB RjX Z Z = ++ + V ()( )( ) 1 230 0 V I 13.6 27.0 A 1.40 1.90 0.5 26.59 9.69 0.5 0.741 1.870jjj ∠° ==∠−° ++ ++ + () ()( ) 2 2 AG, 1 0.5 13.6 A 13.29 2458 W FF PIR== Ω= () ()( ) 2 2 AG, 1 0.5 13.6 A 0.370 68 W BB PIR== Ω= AG AG, AG, 2458 W 68 W 2390 W FB PP P=−= − = () AG ind sync 2390 W 7.61 N m 2 rad 1 min 3000 r/min 1 r 60 s P τ π ω == = ⋅ Note that the induced torque is proportional to the square of the terminal voltage. 10-7. What type of motor would you select to perform each of the following jobs? Why? (a) Vacuum cleaner (b) Refrigerator (c) Air conditioner compressor (d) Air conditioner fan (e) Variable-speed sewing machine (f) Clock (g) Electric drill S OLUTION (a) Universal motor—for its high torque (b) Capacitor start or Capacitor start and run—For its high starting torque and relatively constant speed at a wide variety of loads (c) Same as (b) above 279 (d) Split-phase—Fans are low-starting-torque applications, and a split-phase motor is appropriate (e) Universal Motor—Direction and speed are easy to control with solid-state drives (f) Hysteresis motor—for its easy starting and operation at sync n . A reluctance motor would also do nicely. (g) Universal Motor—for easy speed control with solid-state drives, plus high torque under loaded conditions. 10-8. For a particular application, a three-phase stepper motor must be capable of stepping in 10° increments. How many poles must it have? S OLUTION From Equation (10-18), the relationship between mechanical angle and electrical angle in a three-phase stepper motor is 2 me P θθ = so 60 2 2 12 poles 10 e m P θ θ ° == = ° 10-9. How many pulses per second must be supplied to the control unit of the motor in Problem 10-7 to achieve a rotational speed of 600 r/min? S OLUTION From Equation (10-20), pulses 1 3 m nn P = so ()( ) pulses 3 3 12 poles 600 r/min 21,600 pulses/min 360 pulses/s m nPn== = = 10-10. Construct a table showing step size versus number of poles for three-phase and four-phase stepper motors. S OLUTION For 3-phase stepper motors, °= 60 e θ , and for 4-phase stepper motors, °= 45 e θ . Therefore, Number of poles Mechanical Step Size 3-phase ( 60 e θ =°) 4-phase ( 45 e θ =°) 2 60 ° 45° 4 30 ° 22.5 ° 6 20 ° 15 ° 8 15 ° 11.25° 10 12 ° 9 ° 12 10 ° 7.5° 280 Appendix A: Review of Three-Phase Circuits A-1. Three impedances of 4 + j3 Ω are ∆-connected and tied to a three-phase 208-V power line. Find I φ , I L , P, Q, S, and the power factor of this load. S OLUTION Z φ Z φ Z φ + - 240 V I L I φ Ω+= 43 jZ φ Here, 208 V L VV φ == , and 4 3 5 36.87 Zj φ =+ Ω=∠ °Ω, so 208 V 41.6 A 5 V I Z φ φ φ == = Ω () 3 3 41.6 A 72.05 A L II φ == = () 2 2 208 V 3 cos 3 cos 36.87 20.77 kW 5 V P Z φ θ == °= Ω () 2 2 208 V 3 sin 3 sin 36.87 15.58 kvar 5 V Q Z φ θ == °= Ω 22 25.96 kVASPQ=+= PF cos 0.8 lagging θ == A-2. Figure PA-1 shows a three-phase power system with two loads. The ∆-connected generator is producing a line voltage of 480 V, and the line impedance is 0.09 + j0.16 Ω . Load 1 is Y-connected, with a phase impedance of 2.5 ∠36.87° Ω and load 2 is ∆-connected, with a phase impedance of 5∠-20° Ω. 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 [...]... winding factor k w ? SOLUTION (a) The 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 = 0.9 62 kd = 24 ° γ 2. 5 sin n sin 2 2 sin Since there are 60 coils in each phase and... Vcn A -5 Find the magnitudes and angles of each line and phase voltage and current on the load shown in Figure P23 SOLUTION Note that because this load is ∆-connected, the line and phase voltages are identical Vab = Van − Vbn = 120 ∠0° V - 120 ∠ - 120 ° V = 20 8∠30° V Vbc = Vbn − Vcn = 120 ∠ − 120 ° V - 120 ∠ - 24 0° V = 20 8∠ - 90° V Vca = Vcn − Van = 120 ∠ − 24 0° V - 120 ∠0° V = 20 8∠ 150 ° V 28 5 I ab = Vab 20 8∠30°... 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, 150 ° = 0.966 2 (5) ( 150 °) = 0 . 25 9 k p = sin 2 k p = sin for the fundamental frequency for the fifth harmonic 28 8 B -2 Derive the relationship for the winding distribution factor kd in Equation B -22 SOLUTION The above illustration shows the case of 5 slots... = 3 Vφ 2 cos θ = 3 (27 7 V )2 cos (-90°) = 0 kW Z 5 2 Vφ ( 27 7 V )2 sin -90° = −46.06 kvar sin θ = 3 P3 = 3 ( ) Z 5 PTOT = P + P2 + P3 = 59 .86 kW + 46.04 kW + 0 kW = 1 05. 9 kW 1 QTOT = Q1 + Q2 + Q3 = 34 .56 kvar + 34 .53 kvar − 46.06 kvar = 23 .03 kvar The apparent power supplied by the utility is STOT = PTOT + QTOT = 108.4 kVA The power factor supplied by the utility is 2 PF = cos tan -1 2 QTOT 23 .03... factor is k p = sin 160° = 0.9 85 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 = 2 N P k p k d φf e = 2 (6 coils )(60 turns/coil )(0.9 85) (0.960) φ (50 Hz ) Vφ = 75 621 φ The desired phase voltage... − 24 0° V - 120 ∠0° V = 20 8∠ 150 ° V 28 5 I ab = Vab 20 8∠30° V = = 20 .8∠10° A 10 20 ° Ω Zφ I bc = Vbc 20 8∠ − 90° V = = 20 .8∠ − 110° A Zφ 10 20 ° Ω I ca = Vca 20 8∠ 150 ° V = = 20 .8∠130° A Zφ 10 20 ° Ω Ia = Iab − I ca = 20 .8∠10° A - 20 .8∠130° A = 36∠ - 20 ° A Ib = I bc − I ab = 20 .8∠ − 110° A - 20 .8∠10° A = 36∠ - 140° A Ic = Ica − Ibc = 20 .8∠130° A - 20 .8∠-110° A = 36∠100° A A-6 Figure PA-4 shows a small 480-V distribution... (b) Repeat part (a) with the switch closed What happened to the total current supplied? Why? SOLUTION (a) With the switch open, the power supplied to each load is 2 V (480 V )2 cos 30° = 59 .86 kW P1 = 3 φ cos θ = 3 Z 10 Ω Vφ 2 (480 V )2 sin 30° = 34 .56 kvar sin θ = 3 Q1 = 3 10 Ω Z 2 V (27 7 V )2 cos 36.87° = 46.04 kW P2 = 3 φ cos θ = 3 Z 4Ω 2 2 Vφ 27 7 V ) ( sin θ = 3 sin 36.87° = 34 .53 kvar Q2 = 3 4Ω... (b) 3 = 3464 V, so 3464 V = 0.046 Wb 75 621 The fifth harmonic: k p = sin (5) (160°) = 0.643 The seventh harmonic: k p = sin 2 (7)(160°) = −0.3 42 2 29 2 Since the fundamental voltage is reduced by 0.9 85, the fifth and seventh harmonics are suppressed relative to the fundamental by the fractions: 5th: 0.643 = 0. 653 0.9 85 7th: 0.3 42 = 0.347 0.9 85 In other words, the 5th harmonic is suppressed by 34.7% relative... triangle For this triangle, the hypotenuse is R, the opposite side is E A / 2 , and the angle is nγ / 2 Therefore, nγ E A / 2 sin = R 2 ⇒ 1 EA 2 R= nγ sin 2 Combining (1) and (2) yields 28 9 (2) 1 1 E EA 2 = 2 γ nγ sin sin 2 2 nγ sin EA 2 = γ E sin 2 Finally, nγ sin EA 2 kd = = nE n sin γ 2 since k d is defined as the ratio of the total voltage produced to the sum of the magnitudes of each component voltage... 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 k p = sin ρ = sin 2 1 42. 5 = 0.947 2 The distribution factor is kd = sin nγ 2 n sin γ 2 The electrical angle γ between slots is 7 .5 , and each phase group occupies 8 adjacent slots Therefore, the distribution factor is 29 0 . is () 2 2 1 25 3 .2 V 3 cos 3 cos 36.87 61.6 kW 2. 5 V P Z φ θ == °= Ω () 2 2 1 25 3 .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. below 27 6 1.4 Ω j1.9 Ω + - V = 22 0∠0° V I 1 R 1 jX 1 s R 2 5. 0 j0.5X 2 j0.5X M j1.90 Ω j30 Ω s R 2 5. 0 2 jX 2 j1.90 Ω j0.5X M j100 Ω { { { { Forward Reverse 0.5Z B 0.5Z F ()() 22 22 / / M F M Rs. T V = 20 8∠0° V, 1 11 I 0 .5 0 .5 FB RjX Z Z = ++ + V ()( )( ) 1 20 8 0 V I 12. 327 .0 A 1.40 1.90 0 .5 26 .59 9.69 0 .5 0.741 1.870jjj ∠° ==∠−° ++ ++ + () ()( ) 2 2 AG, 1 0 .5 12. 3 A 13 .29 20 10