Determine i full load speed ii speed at maximum

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Unformatted text preview: From Equation(4.72) Three-Phase Induction Machine 275 From Equation (4.72) Example 4.9 A 4 pole 50 Hz 20 hp motor has, at rated voltage and frequency a starting torque of 150% and a maximum torque of 200 % of full load torque. Determine (i) full load speed (ii) speed at maximum torque. 276 Chapter Four Solution: T 1.5 Tst T = 1.5 and max = 2 then, st = = 0.75 TFL TFL Tmax 2 From above and equation (4.72) 2 sTmax Tst = = 0.75 2 Tmax 1 + sTmax 2 Then, 0.75 sTmax − 2 sTmax + 0.75 = 0 Then sTmax = 2.21525 (unacceptable) Or sTmax = 0.451416 Also from Equation 4.72 2 2 Tmax sTmax + s FL = =2 TFL 2sTmax * s FL But sTmax = 0.451416 2 Tmax 0.4514162 + s FL Then = =2 TFL 2 * 0.451416 * s FL 2 s FL − 4 * 0.451416 s FL + 0.451416 2 = 0 2 s FL − 1.80566 s FL + 0.203777 = 0 s FL = 1.6847 (unacceptable) or s FL = 0.120957 ns = 120 * 50 = 1500 rpm 4 Three-Phase Induction Machine 277 then (a) nFL = (1 − s FL ) * ns nFL = (1 − 0.120957 ) *1500 = 1319 rpm (b) ( ) nTmax = 1 − sTmax * ns = (1 − 0.451416) *1500 = 823 rpm Example 4.10 A 3φ, 280 V, 60 Hz, 20 hp, four-pole induction motor has the following equivalent circuit parameters. ′ ′ R1 = 0.12 Ω, R2 = 0.1 Ω, X 1 = X 2 = 0.25 Ω, and X m = 10 Ω The rotational loss is 400 W. For 5% slip, determine (a) The motor speed in rpm and radians per sec. (b) The motor current. (c) The stator cu-loss. (d) The air gap power. (e) The rotor cu-loss. (f) The shaft power. (g) The developed torque and the shaft torque. (h) The efficiency. Solution: 120 * 60 = 1800 rpm 4 1800 ωs = * 2π = 188.5 rad / sec 60 ns = 278 Chapter Four 0.12 Ω j0.25 Ω j0.25 Ω 0.1 =2 0.05 j10 Ω Z1 = 0.12 + j 0.25 + Re + X e Z1 = 0.12 + j 0.25 + V1 = I1 = 208 j10 * (2 + j 0.25) = 2.1314∠23.55o Ω 2 + j10.25 = 120.1 V 3 120.1 = 2.1314∠ − 23.55o A o 2.1314∠23.55 2 (c) P = 3 * 56.3479 * 0.12 = 1143.031W 1 (d) ( ) Ps = 3 *120.1 * 56.3479 * cos − 23.55o = 18610.9794 W Pag = Ps − P = 17467.9485 W 1 (e) P2 = sPag = 0.05 *17467.9785 = 873.3974 W (f) Pm = (1 − s ) Pag = 16594.5511W (g) T = Pag 188.5 = 17467.9485 = 92.6682 N .m 188.5 Three-Phase Induction Machine Tshaft = (h) η = Pshaft 188.5 Pshaft Ps = 279 16194.5511 = 85.9127 Nm 188.5 *100 = 87.02% For operation at low slip, the motor torque can be considered proportional to slip. Example 4.11 A 30, 100 WA, 460 V, 60 Hz, eight-pole induction machine has the following equivalent circuit parameters: ′ ′ R1 = 0.07 Ω, R2 = 0.05 Ω, X 1 = X 2 = 0.2 Ω, and X m = 6.5 Ω (a) Derive the Thevenin equivalent circuit for the induction machine. (b) If the machine is connected to a 30, 460 V, 60 Hz supply, determine the starting torque, the maximum torque the machine can develop, and the speed at which the maximum torque is developed. (c) If the maximum torque is to occur at start, determine the external resistance required in each rotor phase. Assume a turns ratio (stator to rotor) of 1.2. 280 Chapter Four Solution: Xm 6.5 * V1 = * 265.6 = 257.7 V X1 + X m 0.2 + 6.5 Vth = Rth + jX th = ( j 6.5) * ( j 0.2 + 0.07 ) = 0.06589 + 0.07 + j 0.2 + j 6.5 Ω 0.06589 j0.1947 Ω j 0.1947 Ω j0.2 Ω 0.05 s 257.7V (b) Tst = 3 * 257.7 2 * 0.05 [ 94.25 (0.06589 + 0.05) + (0.1947 + 0.2 ) Tmax = 2 2 = 624.7 Nm 3 * 257.7 2 [ 2 * 94.25 0.06589 + 0.06589 2 + (0.1947 + 0.2 )2 = 2267.8 Nm sTmax = Speed ( 0.05 0.06589 + (0.1947 + 0.2 ) 2 2 in ) rpm for which = 0.1249 max torque = 1 − sTmax * ns = (1 − 0.1249 ) * 900 = 787.5 rpm (c) sTmax = ′ R2 R12 ′ + ( X1 + X 2 ) 2 ′ α R2 occurs 281 Three-Phase Induction Machine ′ or R2 start = sstart = 1 sTmax ′ * R2 = 1 0.1249 * 0.05 = 0.4 Ω Then Rext = (0.4 − 0.05) / 1.2 2 = 0.243 Ω 1-speed characteristic of an induction machine We have seen that a 3-phase squirrel-cage induction motor can also function as a generator or as a brake. These three modes of operation-motor, generator, and brake-merge into each other, as can be seen from the torque-speed curve of Fig.4.18. This curve, together with the adjoining power flow diagrams, illustrates the overall properties of a 3-phase squirrel-cage induction machine. We see, for example, that when the shaft turns in the same direction as the revolving field, the induction machine operates in 282 Chapter Four either the motor or the generator mode. But to operate in the generator mode, the shaft must turn faster than synchronous speed. Similarly, to operate as a motor, the shaft must turn at less than synchronous speed. Finally, in order to operate as a brake, the shaft must turn in the opposite direction to the revolving Fig.4.18 Complete torque speed curve of a 3-phase induction machine...
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This document was uploaded on 03/12/2014 for the course ENGINEERIN electrical at University of Manchester.

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