高等電機機械期末報告 研究生: 陳禹達 學號: ma320104 指導教授:秦純. 2 例題 7.3 For each of the following changes in operating condition for a dc shunt motor, describe how the armature.

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高等電機機械期末報告 研究生: 陳禹達 學號: ma 指導教授:秦純

2 例題 7.3 For each of the following changes in operating condition for a dc shunt motor, describe how the armature current and speed will vary: a. Halving the armature terminal voltage while the field flux and load torque remain constant. b. Halving the armature terminal voltage while the field current and load torque remain constant. c. Doubling the field flux while the armature terminal voltage and load torque remain constant. d. Halving both the field flux and armature terminal voltage while the load power remains constant. e. Halving the armature terminal voltage while the field flux remains constant and the load torque varies as the square of the speed. Only brief quantitative statements describing the general nature of the effect are required, for example. ”speed approximately doubled.”

3 例題 7.3 解 : 驗證這個 (a) : 減半, : 常數。 (b) : 減半, : 加倍。 (c) : 減半, : 減半。 (d) : 常數, : 加倍。 (e) : 減半, : 減少四分之一。

4 例題 7.4 The constant-speed magnetization curve for a 25-kW, 250-V dc machine at a speed of 1200 r/min is shown in Fig7.27. This machine is separately excited and has an armature resistance of This machine is to be operated as a dc generator while driven from a synchronous motor at constant speed. a. What is the rated armature current of this machine? b. What the generator speed held at 1200 r/min and if the armature current is limited to its rated value, calculate the maximum power output of the generator and the corresponding armature voltage for constant field currents of (i) 1.0A,(ii) 2.0A and (iii)2.5A. c. Repeat part (b) if the speed of the synchronous generator is reduced to 900r/min.

5 例題 7.4 解: Part (a): Rated armature current = 25 kW/250-V = 100 A Part (b): 由圖可知,當轉速 1200 時, If=1.0 時, Ea=150V If=2.0 時, Ea=240V If=2.5 時, Ea=270V 又 Va=Ea+IaRa , Ia=100 故 If=1.0 時, Va=164V If=2.0 時, Va=254V If=2.5 時, Va=284V 其功率 If=1.0 時, P=16.4KW If=2.0 時, P=25.4KW If=2.5 時, P=28.4KW Part(c): 轉速降至 900 時,其 Ea 也依比例下降, If=1.0 時, Ea=112V If=2.0 時, Ea=180V If=2.5 時, Ea=202V 又 Va=Ea+IaRa , Ia=100 故 If=1.0 時, Va=126V If=2.0 時, Va=194V If=2.5 時, Va=216V 其功率 If=1.0 時, P=12.6KW If=2.0 時, P=19.4KW If=2.5 時, P=21.6KW

6 例題 7.9 When operated from a 230-V dc supply, a dc series motor operates at 975 r/min with a line current of 90A. Its armature-circuit resistance is 0.11 series-field resistance is Due to saturation effects, the flux produced by an armature current of 30A is 48 percent of that an armature current of 90A. Find the motor speed when the armature voltage is 230V and the armature current is 30A.

7 例題 7.9

8 討 論