Electric Motor Control
Electric Motor Control
10th Edition
ISBN: 9781305177611
Author: Herman
Publisher: Cengage
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Chapter 44, Problem 4SQ
To determine

Find the series resistance required to limit the current to 30 A across a 230V circuit at 50 percent of its rated speed.

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Complete the wiring diagrams as per Worksheets 5-A and 5-B then solve for "A" through "G" for the DC motor on Worksheets 5-A and 5-B when it is running at capacity and its specifications are as follows:Rated input = 125 volts, 16 ampsArmature Resistance = 0.3 ohms(Ω)Shunt Field Resistance = 50 ohms (Ω)Series Field Resistance = 0.6 ohm (Ω)Rheostat Resistance = 12.5 ohm (Ω)A) Shunt field current (I-shunt)a) 2.5 A b) 13.5A c) 16 A d) 18.5 AB) Armature current (I-arm)a) 2.5 A b) 13.5A c) 16 A d) 18.5 AC) Voltage drop of shunt field (I-shunt)a) 4.05 V b) 120.95 V c) 125 V d) 129.5 VD) Voltage drop of armature (IR-arm)
A 150 HP, 3-phase, 6 pole, 460 V, 60 Hz induction motor draws 195 A line current at full load speed of 1170 rpm and 0.85 pf. Solve for the percent slip at full load.                (a). 2.5%                            (b). 3.0%                (c). 2.8%                             (d). 2.0%
Complete the wiring diagrams as per Worksheets 7-A and 7-B then solve for "A" through "K" for the DC motor on Worksheets 7-A and 7-B when it is running at capacity and its specifications are as follows:Rated input = 125 volts, 16 ampsArmature Resistance = 0.3 ohms(Ω)Shunt Field Resistance = 50 ohms (Ω)Series Field Resistance 0.6 ohms (Ω)Rheostat Resistance = 12.5 ohms (Ω)A)power consumed by series field(P-series)a)56.58W b)153.6W c)265.42W d)1523.WB)power consumed by shunt field(P-shunt)a)56.58W b)153.6W c)265.42W d)1523.56WC)power consumed by armature resistance(P-arm)a)56.58W b)153.6W c)265.42W d)1523.56WD)percent efficiency of the machine( % eff) if rotational loses are 850 Wa)30% b)64% c)86% d)99%
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