Principles of Alternating-Current Machinery

This historic book may have numerous typos and missing text. Purchasers can download a free scanned copy of the original book (without typos) from the publisher. Not indexed. Not illustrated. 1920 Excerpt: ...generator action. Maximum Possible Motor Excitation with Fixed Impressed Voltage and Fixed Resistance and Reactance.--In order that the machine shall run as a motor Pm must be negative. The limiting value of Ea' will be that value which makes Pm zero. E nr VE' aximum rm = = 0 z, z, Em'r, = Ea'V z.2 z, The maximum ... Read More

This historic book may have numerous typos and missing text. Purchasers can download a free scanned copy of the original book (without typos) from the publisher. Not indexed. Not illustrated. 1920 Excerpt: ...generator action. Maximum Possible Motor Excitation with Fixed Impressed Voltage and Fixed Resistance and Reactance.--In order that the machine shall run as a motor Pm must be negative. The limiting value of Ea' will be that value which makes Pm zero. E nr VE' aximum rm = = 0 z, z, Em'r, = Ea'V z.2 z, The maximum possible motor voltage is equal to the impressed voltage multiplied by the ratio of the impedance to the resistance. Maximum Motor Activity with Fixed Impressed Voltage and Fixed Reactance and Resistance.--dPm n d tEa'he Ejv dEa' U _ dEa' z, z. J 2Ea're v The maximum motor power, therefore, occurs when the motor voltage has one-half of its maximum possible value. This corresponds to Jacobi's law for a direct-current motor operating with a constant field. Substituting the value of Ea' just found in equation (98) gives V P =-----This is the greatest possible motor power. CHAPTER XXVI Hunting; Damping; Stability; Methods Of Starting Synchronous Motors Hunting.--All synchronous machines in which a change in load is accompanied by a change in phase are subject to hunting. Consider the case of a synchronous motor operating under constant excitation and load. Under this condition there will be a perfectly definite phase angle between the impressed and excitation voltages. Suppose the load it carries is increased The motor will now be developing less power than is demanded by the load and, as a result, it will immediately start to slow down. It will continue to slow down, thereby changing its phase, until the phase displacement between its impressed and excitation voltages corresponds to that required for the load. This slowing down may last several cycles, but unless the load exceeds the maximum load the motor can carry, the change in speed will not last ... Read Less