Three Phase Induction Motors

# 54 also if the rotational losses are neglected the

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Unformatted text preview: d the stator current I1 will remain the same, and the machine will develop the same torque (Equation (4.54)). Also, if the rotational losses are neglected, the developed torque is the same as the load torque. Therefore, for unity turns ratio, Example 4.6 The following test results are obtained from three phase 100hp,460 V, eight pole star connected induction machine No-load test : 460 V, 60 Hz, 40 A, 4.2 kW. Blocked rotor test is 100V, 60Hz, 140A 8kW. Average DC resistor between two stator terminals is 0.152 Ω (a) Determine the parameters of the equivalent circuit. (b) The motor is connected to 3ϕ , 460 V, 60 Hz supply and runs at 873 rpm. Determine the input current, input power, air gap power, rotor cupper loss, mechanical power developed, output power and efficiency of the motor. 268 Chapter Four (c) Determine the speed of the rotor field relative to stator structure and stator rotating field Solution: From no load test: (a) Z NL = RNL = 460 / 3 = 6.64 Ω 40 PNL 4200 = = 0.875 Ω\ 2 3 * I1 3 * 40 2 X NL = 6.64 2 − 0.8752 = 6.58 Ω Then, X 1 + X m = 6.58 Ω From blocked rotor test: RBL = R1 = 8000 = 0.136 Ω 3 *140 2 0.152 = 0.076 Ω (from resistance between two stator 2 terminals). Z BL = 100 / 3 = 0.412 Ω 140 X BL = 0.412 2 − 0.136 2 = 0.389 Ω ′ Then, X 1 + X 2 = 0.389 Ω 269 Three-Phase Induction Machine ′ Then, X 1 = X 2 = 0.389 = 0.1945 Ω 2 X m = 6.58 − 0.1945 = 6.3855 Ω R = RBL − R1 = 0.136 − 0.076 = 0.06 Ω 2 0.1945 + 6.3855 ′ Then, R2 = * 0.06 = 0.0637 Ω 6.3855 0.076 Ω j0.195 Ω j0.195 Ω j6.386 Ω (b) ns = s= 0.0637 s 120 f 120 * 60 = = 900rpm 8 P ns − n 900 − 873 = = 0.03 900 ns ′ R2 0.0637 = = 2.123 Ω 0.03 s Input impedance Z1 Z1 = 0.076 + j 0.195 + I1 = ( j 6.386)(2.123 + j 0.195) = 2.121∠27.16o 2.123 + j (6.386 + 0.195) V1 460 / 3 = = 125.22∠ − 27.16o Z1 2.12∠27.16 270 Chapter Four Input power: Pin = 3 * ( ) 460 *125.22 cos 27.16o = 88.767 kW 3 Stator CU losses: Pst = 3 *125.22 2 * 0.076 = 3.575 kW Air gap power Pag = 88.767 − 3.575 = 85.192 kW Rotor CU losses P2 = sPag = 0.03 * 85.192 = 2.556 kW Mechanical power developed: Pmech = (1 − s ) Pag = (1 − 0.03) * 85.192 = 82.636 kW Pout = Pmech − Prot From no load test: Prot = PNL − 3I12 * R1 = 4200 − 3 * 40 2 * 0.076 = 3835.2W Pout = 82.636 *103 − 3835.2 = 78.8 kW Then the efficiency of the motor is: η= Pout 78.8 *100 = *100 = 88.77 % 88.767 Pin Three-Phase Induction Machine 271 Example 4.7 A three phase, 460 V 1450 rpm, 50 Hz, four pole wound rotor induction motor has the following parameters per ′ ′ phase ( R1 =0.2Ω, R2 =0.18 Ω, X 1 = X 2 =0.2Ω, X m =40Ω). The rotational losses are 1500 W. Find, (a) Starting current when started direct on full load voltage. Also find starting torque. (b) (b) Slip, current, power factor, load torque and efficiency at full load conditions. (c) Maximum torque and slip at which maximum torque will be developed. (d) How much external resistance per phase should be connected in the rotor circuit so that maximum torque occurs at start? Solution: (a) V1 = 460 = 265.6 V / phase 3 Z1 = 0.2 + j 0.2 + Then, I st = (b) s = j 40 * (0.18 + 0.2) = 0.55∠46.59o Ω 0.18 + j 40.2 V1 265.6 = == 482.91 ∠ − 46.3o Ω o I1 0.55∠46.59 1500 − 1450 = 0.0333 1500 272 Chapter Four ′ R2 0.18 = = 5.4 Ω s 0.0333 Z1 = 0.2 + j 0.2 + Then I1 FL = j 40 * (5.4 + j 0.2) = 4.959 ∠10.83o Ω 5.4 + j 45.4 265.6 = 53.56∠ − 10.83o A 4.959∠10.83o o Then the power factor is: cos 10.83 = 0.9822 lag. ω sys = Vth = 1500 * 2π = 157.08 rad / sec . 60 265.6 * ( j 40 ) = 264.275 ∠0.285o V (0.2 + j 40.2) Then, Z th = j 40 * (0.2 + j 0.2 ) = 0.281432 ∠45.285o = 0.198 + j 0.2 Ω 0.2 + j 40.2 Then, 3 * (264.275) * 5.4 T= = 228.68 Nm 2 2 157.08 * (0.198 + 5.4 ) + (0.2 + 0.2 ) 2 Then, Pag = T * ω sys = 228.68 *157.08 = 35921.1W Then, P2 = sPag = 0.0333 * 35921.1 = 1197 W And, Pm = (1 − s )Pag = 34723.7W Then, Pout = Pm − Prot = 34723.7 − 1500 = 33223.7W 273 Three-Phase Induction Machine Pin = 3 * 265.6 * 53.56 * 0.9822 = 41917 W Then, η = Pout 33223.7 = = 79.26 % 41914 Pin Then, Tm = sTmax [ 3 * (264.275)2 ( 2 *188.5 0.198 + 0.198 + (0.2 + 0.2 ) 0.18 = = 0.4033 2 1/ 2 2 0.198 + (0.2 + 0.2 ) [ (d) sTmax = 1 = )] 2 1/ 2 2 = 862.56 Nm [0.198 ′ ′ R2 + Rext 2 2 1/ 2 + (0.2 + 0.2 ) ′ ′ Then, R2 + Rext = 0.446323 ′ Then, Rext = 0.446323 − 0.18 = 0.26632 Ω Example 4.8 The rotor current at start of a three-phase, 460 volt, 1710 rpm, 60 Hz, four pole, squirrel-cage induction motor is six times the rotor current at full load. (a) Determine the starting torque as percent of full load torque. (b) Determine the slip and speed at which the motor develops maximum torque. (c) Determine the maximum torque developed by the motor as percent of full load torque. 274 Chapter Four Solution: Note that the equivalent circuit parameters are not given. Therefore equivalent circuit parameters cannot be used directly for computation. (a) The synchronous speed is From Equation (4.57) 2 2 I 2 R2 I 2 R2 α , Thus, T= sω syn s...
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