. A text-book of electrical engineering;. sed almost exclusivelyin America, whereas, on the Continent they are quite the exception. InEngland, both systems have been largely adopted, and their relative meritsare continually under discussion. T. E CHAPTER XIII 96. The diagram for an generator.—97. The ampere-turns diagram for generator.—98. Calculation of armature reaction.—99. Experimental deter-mination of armature reaction and armature leakage.—100. Predeteriiiination ofexciting current and pressure regulation.—loi. Effect of polar leakage. 96. The Diagram for an A.
. A text-book of electrical engineering;. sed almost exclusivelyin America, whereas, on the Continent they are quite the exception. InEngland, both systems have been largely adopted, and their relative meritsare continually under discussion. T. E CHAPTER XIII 96. The diagram for an generator.—97. The ampere-turns diagram for generator.—98. Calculation of armature reaction.—99. Experimental deter-mination of armature reaction and armature leakage.—100. Predeteriiiination ofexciting current and pressure regulation.—loi. Effect of polar leakage. 96. The Diagram for an Generator. If the assumption be made that the coefficient of self-induction of thearmature of a machine has a constant value for all conditions of load andexcitation, the vector diagram is extremely simple. The electromotiveforce £1, induced in the armature by the rotating magnetic flux, forms thehypotenuse of a right-angled triangle, one side of which is equal to the total B *^ X ^__ a \ 5__ Vp \^ A EgBxt. Esint 0 Fig. 307. Fig. 308 inductive pressure drop, both internal and external, while the other sideis equal to the sum of the internal and external ohmic pressure drops. Thedirection of the current vector in Fig. 307 is vertically upwards. The self-induction induces an lagging 90° behind the current, which must becounterbalanced by a component OA of the OA consists of two parts,one Esint lost in the armature, and the other E^ext overcoming the back the external circuit. The component AB consists similarly of a part ADlost in the machine, and a part DB which overcomes the external ohmicresistance. To find the terminal pressure we must complete the rectangleFDA and join OF and FB. The latter represents the terminal pressure V,for it is the hypotenuse of a right-angled triangle with sides equal to theexternal inductive and ohmic pressure components. The line OF representsthe total drop of pressure in the generator. We have already seen in
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