高等電機機械 學生:林易德 學號:MA320110 指導教授:秦純

7-16 A 26-kW,230-V shunt motor has an armature resistance of 0.11Ω and a field resistance of 117Ω. At no load and rated voltage ,the speed is 2150r/min and the armature current is 6.35A.At full load and rated voltage the armature current is 115A and, because of armature reaction ,the flux is 6 percent less than its no-load value. What is the full-load speed? 一分激式馬達，功率26KW，額定電壓230V，電樞電阻0.11Ω和場電阻117Ω 無載，速度為2150r/ min，且電樞電流是6.35A. 滿載，電樞電流是115A和電樞反應的原因，磁通是比無載小6％ 滿載的轉速為？ Vt = Ea + IaRa 無載, Ea_n = Vt – IaRa 230 − 6.35 × 0.11 = 229.3 V. 滿載, Ea_f = Vt – IaRa 230 − 115 × 0.11 = 217.4 V. Ea ∝ nΦ 屬正比例關係 滿載轉速 = 無載轉速(Ea_f/Ea_n)(Φ_n/Φ_f) = 2150(217.4/229.3/)(1/0.94) = 2168 r/min

7-21 A 230-V dc shunt motor has an armature-circuit resistance of 0.23 ohms. When operating from a 230-V supply and driving a constant-torque load, the motor is observed to be drawing an armature current of 60 A. An external resistance of 0.1 ohms is now inserted in series with the armature while the shunt field current is unchanged. Neglecting the effects of rotational losses and armature reaction, calculate a. the resultant armature current and b. the fractional speed change of the motor. 一分激式電機230V，有0.23歐姆的電樞電阻。 230V供電電壓和定轉矩 60 A的0.1歐姆的外部電阻現在串聯插入與電樞電流而分流激磁電流是不變 忽略的旋轉損耗和電樞反應的影響，計算a.電樞電流 b.速度比  

7-23 A 25-kW,230-V shunt motor has an armature resistance of 0.064Ωand a field-circuit resistance of 95Ω.The motor delivers rated output power at rated voltage when its armature current is 122A. When the motor is operating at rated voltage , the speed is observed to be 1150 r/min when the machine is loaded such that the armature current is 69.5A. a.Calculate the rated-load speed of this motor. In order to protect both the motor and the dc supply under starting conditions, an external resistance will be connected in series with the armature winding (with the field winding remaining directly across the 230-V supply). The resistance will then be automatically adjusted in steps so that the armature current does not exceed 200 percent of rated current .the step size will be determined such that, until all the external resistance is switched out, the armature current will not permitted to drop below rated value. In other words, the machine is to start with 200 percent of rated armature current and as soon as the current falls to rated value, sufficient series resistance is to be cut out to restore the current to 200 percent. This process will be repeated until all of the series resistances has been eliminated. b.Find the maximum value of the series resistance. c.How much resistance should be cut out at each step in the starting operation and at what speed should each step change occur

一個25KW，230V的分激電機具有0.064Ω的電樞電阻和場電阻95Ω，提供額定輸出功率在額定電壓下 當它的電樞電流是122A。當電機工作在額定電壓時，速度1150rpm，當電機被加載，使得電樞電流是69.5A。 計算該電機的額定負載速度。