Question

5.8. A given load is driven by a 460 V four-pole 60 Hz 100 hp three-phase induction motor, data for which are as follows: N-m TAL-0.07 . ω, X, = large, For a speed of 1775 rpm, perform a comprehensive analysis (fnd the slip, currents, powers, torques, effciency, and pf). 3. Problem 5.8 (pg. 276)-ALSO DRAW CIRCUIT MODEL USEDⅢ Answers: PRL- Pout = Pin- TRL = TL =

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Answer 1

Given speed = 1775 r p m f = 60 Hz p = 4 N_s = 120 f/p = 120 times 60/4 = 18000 r pm slip = N_s - N/N_s = 1800 - 1775/1800 = 0.0138 I_a = 460/squareroot 3/0.04 + 0.04 + 2.84 + j 0.4 (as x_m is large we can neglect that branch) = 460/squareroot 3/2.92 + j 0.4 = 90.11 angle - 7.8001 A Therefore power factor = cos (- 7.8001 degree) = 0.99 lag Therefore power input = squareroot 3 V_L I_L cos Phi = squareroot 3 times 460 times 90.11 times 0.99 = 71076.599 w rotor input power = power input stator loss = 71076.599 - 3 times (90.11)^2 times 0.04 = 70102.22155 w \r\n rotor output = rotor input - rotor copper loss = 70102.2215 - 3 times (90.11)^2 0.04 = 69127.8441 watts Torque developed = 69127.8441/2 pi times 1775/60 = 371.899 n m torque cost sue to t = rotational loss is T_RL = 0.07 w = 0.07 times 2 pi times 1775/60 = 13.0114 N