. Design of a direct current generator and switchboard for a farm power plant. of turns of each have been determined. Yoke:- In the yoke the flux has two paths as in the 1,165,000 armature, therefore the flux per path is 2 = 699,000. The yoke should be wide enough to extend out over the arm-ature windings to protect them. The armature is 5 in. longand as shown under Armature Resistance the windings extend outbeyond the end of the armature 3/4+ 1/2 p = 3/4+ each end. Therefore the width of the yoke should be atleast 5+2 x = in. Make it inches wide. A suit-able depth
. Design of a direct current generator and switchboard for a farm power plant. of turns of each have been determined. Yoke:- In the yoke the flux has two paths as in the 1,165,000 armature, therefore the flux per path is 2 = 699,000. The yoke should be wide enough to extend out over the arm-ature windings to protect them. The armature is 5 in. longand as shown under Armature Resistance the windings extend outbeyond the end of the armature 3/4+ 1/2 p = 3/4+ each end. Therefore the width of the yoke should be atleast 5+2 x = in. Make it inches wide. A suit-able depth of yoke would be 2 in. which gives the area to be _ 2 x = 29 sq. in. Therefore the density is Z$24,100 lines per sq. in. Using cast iron for the yoke and re-ferring to the iron curves it is found to take perinch of length. The distance from the center of the armature to thecenter of the yoke is composed of lengths as follows:-Armature and teeth 5 inches, air gap 3/16 in., pole piece 3/4inpole core 5 in., yoke 1 in. making in all 12 inches. The lengt. 12 x 2 x TT of the flux path through the yoke per pole is then — = d X Tt inches. The total ampere turns for the yoke are x = 166 per pole. Having determined the ampere turns per pole for eachpart of the magnetic circuit they will now be summed up. Part Gap 2910Teeth 8Armature body 22Pole Core 114Yoke 166Total at no load 3220Shunt Wire:- Knowing the total number of for theshunt field, we can now find the size of wire needed and checkon the length of pole core used. The four shunt coils are tobe inseries, therefore the volts per coil are —j- = * e be the voltage impressed per coil, p= the specificresistance in ohms per mil foot equal to 12, 1=length of thewire in feet, T s the number of turns, A = the number of amperesflowing, a = the area of the wire in circular mils and R = the total resistance of one coil. Then, e e r * A t 1 a Q or —t— = and s
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Keywords: ., bookcentury1900, bookdecade1910, booksubjecttheses, bookyear1912
