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BASIC ELECTRICAL TECHNOLOGY DET 211/3 Chapter 7 Direct Current (DC) Motor & Direct Current (DC) Generator DC Motor There are four major types of DC motor in general use: • Separately excited DC motor • Shunt DC Motor • Series DC Motor • Compounded DC Motor Speed Regulations • DC motors are often compared by their speed regulations. • Speed Regulations (SR) of the a motor is defined by nl fl SR x 100% fl n nl n fl SR x 100% n fl The Equivalent Circuit of a DC Motor RA Armature circuit represented by voltage source, EA and a resistor RA. The brush voltage drop represented by battery, Vbrush opposing the direction of current flow in the machine. The field coils, which produce the magnetic flux are represented by inductor LF and resistor RF. The separate resistor, Radj represents an external variable resistor used to control the amount of current in the field circuit. The Equivalent Circuit of a DC Motor The brush voltage drop is often only a very tiny fraction of the generated voltage in a machine so the voltage drop may be left out or approximately included in the value of RA. The internal resistor in the field coils is sometimes lump together with the variable resistor, and the total is called RF. The Equivalent Circuit of a DC Motor The internal generated voltage in DC motor is E A K The induce torque developed by DC motor is ind KI A These two equations, the KVL equation of the armature circuit and the machine’s magnetization curve are all the tools necessary to analyze the behavior and performance of the dc motor. Separately Excited and Shunt DC motors VF IF RF VT EA IA R A IL IA Separately excited DC motor VT IF RF VT EA IA R A IL IA IF Shunt DC motor Example A 50hp, 250V, 1200rpm DC shunt motor with compensating windings has an armature resistance of 0.06Ω. Its field circuit has a total resistance of 50Ω, which produces a no load speed of 1200rpm. There are 1200 turns per pole on the shunt field winding. Find: 1. 2. 3. 4. the speed of this motor when its input current is 100A the speed of this motor when its input current is 200A the speed of this motor when its input current is 300A. the induced torque of this motor for above conditions. Solution VT 250V n 1200rpm 2n 2 x 1200 125.67rads 1 60 60 E A K The relationship between the speeds and internal generated voltages of the motor at two different load conditions is E A1 K1 (1) EA2 K2 (2) The flux is constant and no armature reaction Solution E A1 K1 E A 2 K2 At no load, n 1200rpm IA = 0A EA1 VT 250V 1) VT 250 IA IL IF IL 100 95A RF 50 Solution EA at this load will be EA VT IA R A 250 95(0.06) 244.3V The resulting speed of this motor is E A1 K1 E A 2 K2 E A 2 1 244.3x125.67 2 122.8rads 1 E A1 250 60 60 x122.8 n2 1173rpm 2 2 Answer 2) IA = 195A, EA = 238.3V, n2 = 1144rpm 3) IA = 295A, EA = 232.3V, n2 = 1115rpm Solution 4) Pconv E A I A ind E A IA ind At IL = 100A ind 244.3x95 189 Nm 122.8 At IL = 200A ind 388Nm At IL = 300A ind 587Nm Series DC Motor VT E A I A (R A R s ) I L I A IS Compounded DC Motor A compounded DC motor is a motor with both a shunt and a series field. Current flowing into magnetomotive force. a dot produces a positive If current flows into the dots on both field coils, the resulting magnetomotive forces add to produces a larger total magnetomotive force. It is called cumulative compounding. Equivalent circuit of compounded DC motor VT E A I A (R A R s ) IA IL IF VF IF RF Long shunt connection The net magnetomotive force Fnet F F FSE - FAR Effective shunt field current I*F I F N SE F I A AR NF NF Short shunt connection Applications of DC motor types Separately-excited dc motor applications: i) Golf cars (buggy) ii) Forklift iii) Aerial lift equipment DC Generator There are five major types of DC generators: 1. Separately excited generator. In a separately excited generator, the field flux is derived from a separately power source independent of the generator itself. 2. Shunt generator. In a shunt generator, the field flux is derived by connecting the field circuit directly across the terminals of the generator. 3. Series generator. In a series generator, the field flux is produced by connecting the field circuit in series with the armature of the generator. 4. Cumulatively compounded generator. In a cumulatively compounded generator, both a shunt and a series field are present, and their effects are additive. 5. Differentially compounded generator. In a differentially compounded generator, both a shunt and a series field are present, but their effects are subtractive. DC Generator DC generators are compared by their voltages, power ratings, efficiencies, and voltage regulations. Voltage regulation, VR is defined by Vnl Vfl VR x 100% Vfl Separately Excited Generator IL IA IF VF RF VT EA IA R A Example If no load voltage of a separately-excited dc generator is 135V at 850 r/min, what will be the voltage if the speed is increased to 1000 r/min? Assume constant field excitation Solution V1 135V n1 850rpm n2 1000rpm V2 ? V Constant field excitation means; if1 = if2 or constant flux; 1 = 2 E A1 Kn1 n1 E A 2 Kn 2 n 2 n2 EA 2 EA1 n1 1000 ( )135 158.8V 850 Shunt DC Generator IA IF IL VT EA IA R A Series DC Generator I A IS I L VT E A I A (R A R S ) The Cumulatively Compounded DC Generator Total Magnetomotive force Fnet F F FSE - FAR N F I*F N F I F N SE I A FAR I*F I F Cumulatively compounded dc generator with a long shunt connection N SE I A FAR NF NF IA IF IL VT E A I A (R A R S ) VT IF RF Cumulatively compounded dc generator with a short shunt connection Differentially Compounded DC Generator With a long shunt connection Fnet F F FSE - FAR N I N F I F N SE I A FAR * F F Equivalent shunt field current, I eq N SE I A FAR NF NF N SE I A FAR I IF NF NF * F Example A short-shunt compound generator delivers 50A at 500V to a resistive load. The armature, series field and shunt field resistance are 0.16, 0.08 and 200, respectively. Calculate the armature current if the rotational losses are 520W, determine the efficiency of the generator Solution Pu 520W Pout 500Vx 50A 25000W 500 If 2. 5A 200 Ia If IL 2.5A 50A 52.5A Armature Copper Loss: Pca (Ia )2 Ra (52.5)2 (0.16) 441W Series Field Copper Loss: Pcf 2 Shunt Field Copper Loss: (Ia )2 Rf 2 (52.5)2 (0.08) 220.5W Pcf 1 (If )2 Rf 1 (2.5)2 (200) 1250W Friction + Stray + windage + etc: So,Total Losses = Pu 520W ( 441 220.5 1250 520) 2431.5W Continued… Efficiency, η = Pout Pout Pin Pout Totallosse s 25000 0.9113 @ 91.13% 25000 2431.5 Assignment 6 QUESTION 1 The equivalent circuit of the separately-excited dc motor Figure above shows fixed field voltage VF of 240V and armature voltage VA that can be varied from 120 V to 240 V. What is the no-load speed of this separately-excited dc motor when Radj = 175 and a) VA = 120V, b) VA = 180V, c) VA = 240V ? Assignment 6 QUESTION 2 The equivalent circuit of the shunt dc motor a) If the resistor Radj is adjusted to 175, what is the rotational speed of the motor at no-load conditions? b) Assuming no armature reaction, what is the speed of the motor at full load? What is the speed regulation of the motor? The magnetization curve for the dc motor of Question 1 and Question 2. This curve was made at a constant speed of 1200 r/min.