The Magnetic Circuit

Front Cover
McGraw-Hill book Company, 1911 - Electric machinery - 285 pages
0 Reviews
 

What people are saying - Write a review

We haven't found any reviews in the usual places.

Other editions - View all

Common terms and phrases

Popular passages

Page 30 - ... principle used in some practical cases when it is desired to modify the relation between the flux and the mmf by providing a highly saturated magnetic path in parallel with a feebly saturated one...
Page 116 - Karapetoff 6 from which I quote in part; "In order to calculate the permittance (capacity) of a given dielectric, or to find the flux densities and stresses in different parts of it, proceed as follows: The field is mapped out into small cells by lines of force and equipotential surfaces, drawing them to the best of ones judgment, the total permittance is calculated by properly combining the permittances of the cells in series and in parallel. Then the assumed directions are somewhat modified, the...
Page v - The advanced student must be taught to treat every electrical machine as a particular combination of electric and magnetic circuits, and to base its performance upon the fundamental theoretical relations, rather than upon a separate 'theory' established for each machine, as is so often done.
Page 42 - The eddy-current loss is the loss due to circulating currents in the core iron, caused by the magnetic flux in the iron cutting the iron, which is a conductor. The eddy-current loss is proportional to the square of the frequency and the square of the flux density. In order to reduce the eddy-current lo>the core is built up of thin laminations (usually 0.014 in.
Page 30 - Gap length 0.25 cm ^^- shunted through the cast-iron part, but as the flux density increases the cast iron becomes saturated, and a larger portion of the flux is deflected into the air gap. What percentages of the total flux in the yoke are shunted through the cast iron when the flux density in the air gap is 1,000 and 7,000 lines per sq cm, respectively?
Page 59 - E=lHv (42) that is, the electromotive force induced in a wire / centimeters long, moving sidewise at a velocity of v centimeters per second across a uniform magnetic field of intensity H is equal to the product IHv.
Page 141 - ... this gives the induced voltage in the machine. Knowing from the no-load saturation curve the required net excitation at this voltage, and correcting it for the effect of the armature reaction, the necessary field ampere-turns are obtained. The result of such calculations for different values of the armature current and for various power factors are represented by the load-characteristic curves.
Page 135 - If such a conductor is connected to the terminals of a seat of emf of negligible internal resistance, the circuit equation becomes The form of this equation is the same as that of Eq.
Page 29 - FIG. 73. total flux is shunted through the cast-iron path, but as the flux density increases, the cast iron becomes saturated, and a larger portion of the flux is diverted...
Page 247 - ... He found, in passing a relatively large alternating current through a non-electrolytic, liquid conductor contained in a trough, that the liquid contracted in cross-section and flowed up hill lengthwise of the trough, climbing up upon the electrodes. With a further increase of current, he found that this contraction of cross-section became so great at one point that a deep depression was formed in the liquid with steeply-inclined sides like the letter V. This depression extended in the case of...

Bibliographic information