Lec 16 DC Motors II(pdf) - KFUPM Open Courseware

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Figure 2 DC motor starter.Example 2A 10 HP 220-V 1000 rpm DC shunt motor has a full load current of 40 A. The motor isstarted by a four-point starter. The armature resistance is 0.3406 Ω and the resistances of thesteps in the starting resistor are 1.375, 0.694, and 0.3406 Ω in the order in which they are cutout . When the armature current has dropped to its rated value, the starter is switched to itsnext point. This means eliminating a step at time of the resistor. Find:(i) the initial and final value of the armature current at each step.(ii) the corresponding speed at each step.SolutionStep # 1 Resistance in armature circuit, R t1 = 1.375+0.694+0.39+0.3406 =2.75 ΩStep # 2 Total resistance in armature circuit, R t2 = 0.694+0.39+0.39 =1.375ΩStep # 3 Total resistance in armature circuit, R t3 = 0.3406+0.3406 =0.6812 ΩStep # 4 Total resistance in armature circuit, R t4 =0.3406 Ω1. At starting, the generated back emf, E a , is zero. All the resistance is in circuit. The startingarmature current I stVa−Ea Va−0 220Ist= = = = 80.0ARt1 Rt12.75When the armature drops to its rated value of 40 A, the corresponding emf isEa1 = Va− IaRa= 220 − 40x2.75 = 110VWhen the motor delivers rated load at rated speed of 1000 rpm, the emfEa = Va− IaRa= 220 − 40x0.3406 = 206.376VThus the speed corresponding to en emf (E a1 ), 110 isE110a1N1= xNfull= x1000 = 533 rpmEa206.3762. The 1.375 Ohm is cut of the circuit. The remaining resistance is R t2 = 1.375 ΩThe speed at this stage is 533 rpm and the emf is still 110IV− E220 −110 80 Aa a1st= = =Rt21.375When the armature drops to its rated value of 40 A, the corresponding emf is
Ea2 = Va− IaRa= 220 − 40x1.375 = 165.5VThus the speed corresponding to en emf (E a2 ), 165.5 isE165.5a 2N2= xNfull= x1000 = 801.9rpmEa206.3763. The remaining resistance is R t3 = 0.6812 ΩThe speed at this stage is 801.9 rpm and the emf is still 165.5 V.IV− E220 −165.5 80 Aa a2st= = =Rt20.6812When the armature drops to its rated value of 40 A, the corresponding emf isEa3 = Va− IaRa= 220 − 40x0.6812 = 192.75VThus the speed corresponding to en emf (E a3 ), 192.75 isE192.75a3N3= xNfull= x1000 = 933.97 rpmEa206.3764. The remaining resistance is R t4 = 0.3406 ΩThe speed at this stage is 923.6 rpm and the emf is still 192.75 V.IV− E220 −192.75 80 Aa a3st= = =Rt20.3406When the armature drops to its rated value of 40 A, the corresponding emf isEa4 = Va− IaRa= 220 − 40x0.3406 = 206.376VThus the speed corresponding to en emf (E a3 ), 204.4 isE206.376 1000 1000a 4N4= xNfull= x = rpmEa206.376Table 1 provides a summary of the resultsstep InitialCurrent (A)FinalCurrent(A)Speed(rpm)1 80 40 5332 80 40 801.93 80 40 933.974 80 40 10003. Efficiency of DC machinesThe efficiency of the DC machine, whether motor or generator, is expressed asoutput Input − losses lossesEfficieny = = 1− (5)input input inputThe losses in a DC machine are made of the following components:a. Electrical or copper losses
Page 1 and 2: EE 360 ELECTRIC ENERGYLECTURE 16DC
Page 3: P = E I = 230x40 = 9200Wdev a aPdev
Page 7: Pout100000Load current I L =400 AV

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