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EE 255 - Chapter 9g

# EE 255 - Chapter 9g - The resulting torque-speed...

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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

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276 1.4 Ω j 1.9 Ω + - V = 220 0° V I 1 R 1 jX 1 s R 2 5 . 0 j 0.5 X 2 j 0.5 X M j 1.90 Ω j 30 Ω s R - 2 5 . 0 2 jX 2 j 1.90 Ω j 0.5X M j 100 Ω { { { { Forward Reverse 0.5 Z B 0.5 Z F ( )( ) 2 2 2 2 / / M F M R s jX jX Z R s jX jX + = + + ( )( ) 30 1.90 100 26.59 9.69 30 1.90 100 F j j Z j j j + = = + Ω + + ( ) ( ) ( ) 2 2 2 2 / 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 j j Z j j j + = = + Ω + + (a) The input stator current is 1 1 1 I 0.5 0.5 F B R jX 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 1.870 j j j ∠ ° = = ∠ - ° + + + + + (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 W P VI θ = = ° = (d) The air-gap power is ( ) ( ) ( ) 2 2 AG, 1 0.5 13.0 A 13.29 2246 W F F P I R = = Ω = ( ) ( ) ( ) 2 2 AG, 1 0.5 13.0 A 0.370 62.5 W B B P I R = = Ω = AG AG, AG, 2246 W 62.5 W 2184 W F B P P P = - = - =
277 (e) The power converted from electrical to mechanical form is ( ) ( )( ) conv, AG, 1 1 0.05 2246 W 2134 W F F P s P = - = - = ( ) ( )( ) conv, AG, 1 1 0.05 62.5 W 59 W B B P s P = - = - = conv conv, conv, 2134 W 59 W 2075 W F B P P P = - = - = (f) The output power is OUT conv rot 2134 W 291 W 1843 W P P P = - = - = (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. ( )( ) 2 2 2 2 / / M F M R s jX jX Z R s jX jX + = + + ( )( ) 30 1.90 100 26.59 9.69 30 1.90 100 F j j Z j j j + = = + Ω + + ( ) ( ) ( ) 2 2 2 2 / 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 j j Z j j j + = = + Ω + + (a) If T V = 190 0 ° V, 1 1 1 I 0.5 0.5 F B R jX 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 1.870 j j j ∠ ° = = ∠ - ° + + + + + ( ) ( ) ( ) 2 2 AG, 1 0.5 11.2 A 13.29 1667 W F F P I R = = Ω = ( ) ( ) ( ) 2 2 AG, 1 0.5 11.2 A 0.370 46.4 W B B P I R = = Ω = AG AG, AG, 1667 W 46.4 W 1621 W F B P P P = - = - =

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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 1 1 I 0.5 0.5 F B R jX Z Z = + + + V ( ) ( ) ( ) 1 208 0 V I 12.3 27.0 A 1.40 1.90 0.5 26.59 9.69 0.5 0.741 1.870 j j j ∠ ° = = ∠ - ° + + + + + ( ) ( ) ( ) 2 2 AG, 1 0.5 12.3 A 13.29 2010 W F F P I R = = Ω = ( ) ( ) ( ) 2 2 AG, 1 0.5 12.3 A 0.370 56 W B B P I R = = Ω = AG AG, AG, 2010 W 56 W 1954 W F B P P 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 1 1 I 0.5 0.5 F B R jX 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 1.870 j j j ∠ ° = = ∠ - ° + + + + + ( ) ( ) ( ) 2 2 AG, 1 0.5 13.6 A 13.29 2458 W F F P I R = = Ω = ( ) ( ) ( ) 2 2 AG, 1 0.5 13.6 A 0.370 68 W B B P I R = = Ω = AG AG, AG, 2458 W 68 W 2390 W F B
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