A7 solution - Study reference PDF

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University of Waterloo ME 269 - Winter 2016

Assignment 7

Induction Motors 1- At what speed will a 12-pole 60Hz induction motor operate if the slip is 6%? Solution: ns = 120f/p = 120*60/12 = 600 rpm nm= (1-s)ns = (1-0.06)*600 = rpm 2- A 60-Hz induction motor runs at 860 r/min at full load. Determine a. the synchronous speed. b. the frequency of the rotor currents. and c. the rotor speed relative to the revolving field. Solution: a. 900 rpm b. s = (ns-nm)/ns= (900-860)/900 = 0.044 fr = sfs = 0.044*60 = 2.6667 Hz c. nr= ns – nm = 900 – 860 = 40 rpm 3- if the EMF in the stator of a six-pole induction motor has a frequency of 60 Hz, and that in the rotor of 2 Hz, at what speed is the motor running and what is the slip? Solution: s = fr/fs = 2/60 = 0.033 ns = 120f/p = 120*60/6 = 1200 rpm nm = (1-s) ns = 1160 rpm 4- A three-phase 20-hp 220-V 60-Hz six-pole Y-connected squirrel-cage induction motor has the following parameters per phase in stator terms: R1 = 0.126 Ω, R2=0.094 Ω, Xe=0.46 Ω (Xe=X1+X2), Xm=9.8 Ω. The rotational losses are 560 W. for a slip of 3%, find: a. The line current and power factor. b. the horsepower output and shaft torque, and c. the efficiency Solution: a) Ze = Z1 + Z2||jXm = (R1+jX1)+jXm((R2/s)+jX2)/((R2/s)+j(X2+Xm)) = 2.7+j1.7 I1 = V1/Ze = (220/√3)/(2.7+17) = 39.8 /-32.2 A b) PAG = Pin - Pscl = √3 VLI1cos(θ) - 3IL2 R1 = √3 *220*39.8*cos(32.2) - 3*39.82*0.126 = = 12833 - 598 = 12235 W Pconv = (1-s)PAG = (1-0.03)*12235 = 11867 W

Pout = Pconv – PF+W = 11867 – 560 = 11307 W = 15.16 hp ns = 120f/p = 120*60/6 = 1200 rpm nm = (1-s)ns = 1164 rpm Tout = Pout/ωm = 11307/(2π/60)*1164 = 92.76 N.m c) η = (Pout/Pin) *100% = (11307/12861)*100 = 87.9% 5- A three-phase induction motor is supplying 100 hp to a coupled mechanical load. At this point of operation, the rotational losses are 2046 W, the total core losses are 3320 W, the total stator coil cupper losses are 2690 W, and the slip is 3%. Determine a. the efficiency at this point of operation and b. the total rotor coil cupper losses. Solution: a) Pout = 100*746 = 74600 W Pconv = Pout + PF+W = 74600+2046 = 72554 W PAG = Pconv/(1-s) = 74798 W Pin = PAG + Pscl + Pcore = 74798 +3320+2690 =80808 W η = Pout/Pin *100% = (149200/161933)*100 = 92.3% b) Prcl = PAG – Pconv = 74798-72554 = 2244 W 6- A three-phase, 50 hp, 460 V, 60 Hz, 865 r/min induction motor is operating at rated conditions and has Pin= 43.75 kW and I1=61 A. It is known that R1=0.15 Ω and rotational losses at rated speed are 1050 W. Determine a. full-load power factor, b. full-load efficiency, c. total rotor coil ohmic losses, and d. total core losses. Solution: a) pf = Pin/ √3 V LI1 = 43750 / (√3 *460*61) = 0.9 b) η = (Pout/Pin)*100% = (50*746/43750)*100 = 85.2% c) ns = 900 rpm s = (900-865)/900 = 0.039 Pconv = Pout + PF+W = 50*746 + 1050 = 38350 W PAG = Pconv/(1-s) =39906 W Prcl = sPAG = 0.039*37721 = 1556 W d) Pscl = 3I12R1 = 3*612*0.15 = 1674.5 W Pcore = Pin – Pscl – PAG = 2169.5 W...