ELG3311: Assignment 3

Transcription

1 LG33: ssignent 3 roble 6-: The Y-connected synchronous otor whose naeplate is shown in Figure 6- has a perunit synchronous reactance of 0.9 and a per-unit resistance of 0.0. (a What is the rated input power of this otor? (b What is the agnitude of at rated conditions? (c f the input power of this otor is 0 W, what is the axiu reactive power the otor can siultaneously supply? s it the arature current or the field current that liits the reactive power output? (d How uch power does the field circuit consue at the rated conditions? (e What is the efficiency of this otor at full load? (f What is the output torque of the otor at the rated conditions? xpress the answer both in newton-eters and in pound-feet. Solution: The base quantities for this otor are: V V, base S T, base, base base 6600V VT V L, base rated T L F 3(6600(404( W

2 (a The rated input power of this otor is in 3 V T L F 3(6600(404( W (b t rated conditions, V, pu.0 0 pu and, pu.0 0 pu, so is given in perunit quantities as, pu V R j, pu, pu, pu S, pu pu nd the is ( 0 (0.0(.0 0 j(0.90( 0 V ( ( V, pu, base (c Fro the capability diagra, we know that there are two possible constraints on the axiu reactive power the axiu stator current and the axiu rotor current. We will have to check each one separately, and liit the reactive power to the lesser of the two liits. The stator apparent power liit defines a axiu safe stator current. This liit is the sae as the rated input power for this otor, since the otor is rated at unity power factor. Therefore, the stator apparent power liit is 6.05 V. f the input power is 0W, then the axiu power that still protects the stator current is Q S (6.05 (0. 6VR Now we ust deterine the rotor current liit. The per-unit power supplied to the otor is 0 W / 6.05 W The axiu is 5067 V or.33 pu, so with set to axiu and the otor consuing 0 W, the torque angle (ignoring arature resistance is δ sin S sin (0.90(0.63 (.0( t rated voltage and 0 W of power supplied, the arature current will be V R + j j0.90 S pu n actual aps, this current is

3 , base, pu (404( The reactive power supplied at the conditions of axiu and 0 W power is Q 3 V sinθ 3(380(93(sin VR Therefore, the field current liit occurs before the stator current liit for these conditions, and the axiu reactive power that the otor can supply is 3.68 VR under these conditions. (d t the rated conditions, the field circuit consues field VF F ( 5(5. 650W (e The efficiency of this otor at full load is η out in (000hp(746W / hp 00 % 00% 97.6% 6.05W (f The output torque in S and nglish units is load load out ω 55 n (000hp(746W / hp 4700N (00r / in(in/ 60s(πrad /r out (55(000hp 990lb ft 00r / in roble 6-5: 00-hp, 440-V, 0.8-F-leading, -connected synchronous otor has an arature resistance of 0. Ω and a synchronous reactance of 3.0 Ω. ts efficiency at full load is 89 percent. (a What is the input power to the otor at rated conditions? (b What is the line current of the otor at rated conditions? What is the phase current of the otor at rated conditions? (c What is the reactive power consued by or supplied by the otor at rated conditions? (d What is the internal generated voltage of this otor at rated conditions? (e What are the stator copper losses in the otor at rated conditions? (f What is conv at rated conditions? (g f is decreased by 0 percent, how uch reactive power will be consued by or supplied by the otor? 3

4 Solution: (a The input power to the otor at rated conditions is out η (00hp(746W / hp in 8 kw (b The line current to the otor at rated conditions is 83.8k L 37, θ cos F cos F 3(440(0.8 T The phase current to the otor at rated conditions is L (c The reactive power supplied by this otor to the power syste at rated conditions is Qrated 3 V sinθ 3(440(79.4(sin kVR (d The internal generated voltage at rated conditions is V R j V S (0.( j(3.0( (e The stator copper losses at the rated conditions are cu 3 R 3(79.4 ( kW (f conv at rated conditions is conv in cu 83.8k 4.6k 79. 6kW (g f is decreased by 0%, the new value of is 90% (0.9( Then the quantity sinδ is constant as changes, so 4

5 δ sin sinδ sin (603sin( Therefore V j j3.0 S The reactive power supplied by the otor to the power syste will be Q 3 V sinθ 3(440(70.5(sin kVR roble 7-7: 08-V, two-pole, 60-Hz, Y-connected wound-rotor induction otor is rated at 5-hp. ts equivalent circuit coponents are R 0.00Ω R 0.0Ω 5.0Ω 0.40Ω 0.40Ω ech 50W isc 0 core 80W For a slip of 0.05, find (a The line current L (b The stator copper losses SCL (c The air-gas power G (d The power converted fro electrical to echanical for conv (e The induced torque ind (f The load torque load (g The overall achine efficiency (h The otor speed in revolutions per inute and radians per second Solution: The equivalent circuit of this induction otor is shown below: 5

6 (a The easiest way to find the line current (or arature current is to get the equivalent ipedance Z F of the rotor circuit in parallel with j, and then calculate the current as the phase voltage divided by the su of the series ipedances, as shown below: The equivalent ipedance of the rotor circuit in parallel with j is Z F.0 + j Ω + + j Z j j0.4 The phase voltage is V VT / 3 08/ 3 0V So the line current L is V R + j + R + j j j0.745 L F F (b The stator copper losses are SCL 3 R 3(44.8 (0.0 05W R (c The air gas power is G 3 3 RF s (Note that R 3 RF is equal to 3, since the only resistance in the original rotor s circuit was R /s, and the resistance in the Thevenin equivalent circuit is R F. The power consued by the Thevenin equivalent circuit ust be the sae as the power consued by the original circuit. R G 3 3 RF 3(44.8 ( kW s 6

7 (d The power converted fro electrical to echanical for is conv ( s G ( 0.05(3.4k. 73kW (e The induced torque in the otor is ind G 3.4k 35. 5N ω 0(60 sync ( (in/ 60s(πrad /r (f The output power of this otor is out conv ech core isc.73k kW The output speed is n 0(60 ( s nsync ( 0.05( 340r / in Therefore the load torque is load out.3k 34. 3N ω (340r / in(in/ 60s(πrad /r (g The overall efficiency is η out in out.3k 00 % 00% 00% 84.5% cosθ 3(0(44.8(cos 5.5 (h The otor speed in revolutions per inute is 340 r/in. The rotor speed in radians per second is ω ( 340r / in(πrad /r (in/ 60s 358rad / s roble 7-4: 440-V, 50-Hz, two-pole, Y-connected induction otor is rated at 75-kW. The equivalent circuit paraeters are 7

8 R 0.075Ω 0.7Ω R 0.065Ω 0.7Ω F & W. 0kW isc 50W core.kw For a slip of 0.04, find (a The line current L (b The stator power factor (c The rotor power factor (d The stator copper losses SCL (e The air-gas power G (f The power converted fro electrical to echanical for conv (g The induced torque ind (h The load torque load (i The overall achine efficiency η (j The otor speed in revolutions per inute and radians per second Solution: 7.Ω The equivalent circuit of this induction otor is shown below: (a The easiest way to find the line current (or arature current is to get the equivalent ipedance Z F of the rotor circuit in parallel with j, and then calculate the current as the phase voltage divided by the su of the series ipedances, as shown below: 8

9 The equivalent ipedance of the rotor circuit in parallel with j is Z F j Ω + + j Z j j0.7 The phase voltage is V VT / 3 440/ 3 54V So the line current L is V R + j + R + j j j0.364 L F F (b The stator power factor is F S cos( lagging (c To find the rotor power factor, we ust find the ipedance angle of the rotor θ R tan R / s tan Therefore the rotor power factor is F R cos( lagging (d The stator copper losses are SCL 3 R 3(49.4 ( W R (e The air gas power is G 3 3 RF s (Note that R R 3 F is equal to 3, since the only resistance in the original rotor s circuit was R /s, and the resistance in the Thevenin equivalent circuit is R F. The power consued by the Thevenin equivalent circuit ust be the sae as the power consued by the original circuit. 9

10 R G 3 3 RF 3(49.4 ( kW s (f The power converted fro electrical to echanical for is ( s ( 0.04(03k 98. conv G 9 (g The induced torque in the otor is kw ind G 03k 37. 9N ω 0(50 sync ( (in/ 60s(πrad /r (h The output power of this otor is 98.9k.0k.k out conv ech core isc 6 The output speed is n 0(50 ( s nsync ( 0.05( 880r / in Therefore the load torque is kw load out 98.9k 37. 6N ω (880r / in(in/ 60s(πrad /r (i The overall efficiency is η out in out 96.6k 00 % 00% 00% 89.4% cosθ 3(54(49.4(cos8.3 (j The otor speed in revolutions per inute is 880 r/in. The rotor speed in radians per second is ω ( 880r / in(πrad /r(in/ 60s 30.6rad / s 0

11 roble 7-9: 460-V, four-pole, 50-hp, 60-Hz, Y-connected, three-phase induction otor develops its full-load induced torque at 3.8 percent slip when operating at 60 Hz and 460 V. The perphase circuit odel ipedance of the otor are R 0.33Ω 30Ω 0.4Ω 0.4Ω echanical, core, and stray losses ay be neglected in this proble. (a Find the value of the rotor resistance R. (b Find ax, s ax and the rotor speed at axiu torque for this otor. (c Find the starting torque of this otor. (d What code letter factor should be assigned to this otor? Solution: The equivalent circuit for this otor is The Thevenin equivalent of the input circuit is Z V j R + R ( R + j j( + j + j( + ( j30( j j Ω j( j30 V ( V j( (a f losses are neglected, the induced torque in a otor is equal to its load torque. t full load, the output power of this otor is 50 hp and its slip is 3.8%, so the induced torque is 0(60 n ( s nsync ( 0.038( 73r / in 4 (50hp(746W / hp ind load 05.7N (73r / in(in/ 60s(πrad /r

12 The induced torque is given by the equation ind ω sync [( R R / s + R / s + ( + ] Substituting known values and solving for R /s yields 3(6 R / s 05.7 (88.5[(0.3+ R / s + ( ] R / s 0.56,4.53 R Ω,0.7Ω These two solutions represent two situations in which the torque-speed curve would go through this specific torque-speed point. The two curves are plotted below. s you can see, only the 0.7Ω solution is realistic, since the Ω solution passes through this torque-speed point at an unstable location on the back side of the torque-speed curve. (b The slip at pullout torque can be found by calculating the Thevenin equivalent of the input circuit fro the rotor back to the power supply, and then using that with the rotor circuit odel. The Thevenin equivalent of the input circuit was calculated in part (a. The slip at pullout torque is

13 s R 0.7 ax R + ( + (0.3 + ( The rotor speed at axiu torque is n 0(60 ( sax nsync ( 0.9( 454r / in 4 pullout nd the pullout torque of the otor is 0.9 ax ax ω sync [( R 448N + R + ( + ] (88.5[(0.3+ 3(6 (0.3 + ( ] (c The starting torque of this otor is the torque at slip s. t is ind ind ω sync [( R 99N R / s + R / s + ( + 3(6 (0.7 ] (88.5[( ( ] (d To deterine the starting code letter, we ust find the locker-rotor kv per horsepower, which is equivalent to finding the starting kv per horsepower. The easiest way to find the line current (or arature current at starting is to get the equivalent ipedance Z F of the rotor circuit in parallel with j at starting conditions, and then calculate the starting current as the phase voltage divided by the su of the series ipedances, as shown below. The equivalent ipedance of the rotor circuit in parallel with j at starting conditions (s is 3

14 Z Ω + + j Z j j0.4 F, start j The phase voltage is V VT / 3 460/ 3 66V So the line current L is V R + j + R + j j j0.45 L, start F F Therefore, the locker-rotor kv of this otor is S 3 VT L, start 3(460(74 8kV nd the kv per horsepower is 8 kv / hp 4.36kV/ hp 50 This otor would have starting code letter D, since letter D covers the range roble 7-3: would-rotor induction otor is operating at rated voltage and frequency with its slip rings shorted and with a load of about 5 percent of the rated value for the achine. f the rotor resistance of this achine is doubled by inserting external resistors into the rotor circuit, explain what happens to the following: (a Slip s (b otor speed n (c The induced voltage in the rotor (d The rotor current (e ind (f out (g RCL (h Overall efficiency η Solution: 4

15 (a The slip s will increase. (b The otor speed n will decrease. (c The induced voltage in the otor will increase. (d The rotor current will increase. (e The induced torque will adjust to supply the load s torque requireents at the new speed. This will depend on the shape of the load s torque-speed characteristic. For ost loads, the induced torque will decrease. (f The output power will generally decrease: out ω. ind (g The rotor copper losses (including the external resistor will increase. 5