HW8sol

# HW8sol - Homework #8 ME 576 1. 1 kW=1.34 HP 2 ∴ 2 HP= =...

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Unformatted text preview: Homework #8 ME 576 1. 1 kW=1.34 HP 2 ∴ 2 HP= = 1.492 kW 1.3 (a). Back induced e.m.f. during idling ( ⎛ 1 min ⎞ rad Vi = kb⋅n = 1 (V ⋅ s )(600 rpm) ⎜ ⎟ 2π sec ⎝ 60 sec ⎠ = 20π ⋅ (V ) ) Power consumption during idling P = VΑ ⋅ J Α = ( RA ⋅ J A + Vi ) I A = 1, 492 RA I A2 + (20π ) I A -1, 492 = 0 ∴ IA = (10π ) −10π ± 2 + 10 × 1492 10 = 9.47(Amp). For speed control, IA is maintained at a constant level and so is Tmax. ( Tmax = KT ⋅ I A = 1 kg ⋅ m A ) ⋅ 9.47( A) = 9.47kg ⋅ m = 92.8 N − m. Maximum acceleration is attained when the motor starts running 0 TM = J M ωmax = 0 d ω (t ) + Bω (t ) + TL dt Tmax Jm ) ( m2 9.47( kg ⋅ m ) 9.47 × 9.8 kgm ⋅ sec 2 = = 0.1( kg ⋅ m 2 ) 0.1 ( kgm ⋅ m 2 ) ( = 928 rad sec 2 ) = 147.7 ( rev sec ) 2 (b). During idling, the power generated by the motor is consumed to overcome this dissipative energy due to damping P = Tm ⋅ ω 0 0 ⎛ dw (t ) ⎞ = ⎜ Jm + Bω ( t ) + TL ⎟ ω (t ) dt ⎝ ⎠ 1, 492(W ) P . B= = = 0.38 N ⋅ m 2 2 rad / sec ω (t ) ( 20π ) (c). VA = RA I A + Vi = RA I A + K bω (t ) ∴ ω (t ) = VA - RA I A Kb 250 − 10 × 9.47 = 155.3 rad = 1, 483 rpm sec 1 Pmax = VA max ⋅ I A max ωmax = = 250 × 9.47 = 2,368W = 3.17 HP 2. (a) VA = RA ⋅ I A + Vi When motor starts rotation, Vi = 0 ∴ RA = 200 = 10(ohms ) 20 at ω (t) = 600 rpm 50(V ) min ×1 600( rpm) × 2π rad rev 60 sec = 0.796 V ⋅ sec. ∴ kb = ( ) ( ) For constant torque up to 1000 rpm, IA must be maintained at [email protected] V A = RA ⋅ I A + k b ⋅ ω ( t ) = 10 × 20 + 0.796 ⋅ 1000 x 2π 60 = 283V (b) w(t)α 1000 : 1 φ 1 φ1000 1000 = = 1500 : 1 φ1500 φ1500 φ1000 φ1500 1000 = = 0.667 φ1000 1500 1500 ∴ 33.3% (c) 0 d ω (t ) + Bw (t ) + TL dt Tmax = K T ⋅ I A max Tm = J m = ⋅1 × 20 = 20 N ⋅ m. ∴ ωmax = = Tm - TL Jm 20 − 5( N ⋅ m ) 10 ( kg ⋅ m 2 ) = 1.5 rad sec 2 3. R r = 10 Ω L r =0.06 H Vf =220V n=1, p=3 ( ωp =60 × 2π rad sec = 376 rad sec ωf = ) ωp = 60 × 2π rad sec n 2 p ⋅ nV sR r Tm = ωp ( R r 2 + s2ωp 2 L r 2 ) = 3 × 1 × (220)2 ⋅ (10) ⋅ s (60 × 2π ) ⎡102 + s2 ⋅ (60 × 2π ) 2 × (0.06) 2 ⎤ ⎣ ⎦ = 38.5s 1, 452,000 ⋅ s = 2 37, 699 + 192, 884 s 1 + 511 s2 s=1- ωm ω ⋅ n = 1− m ωp ωp or ωm =(1-s)ωp ...
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## This document was uploaded on 12/28/2011.

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