. A text-book of electrical engineering;. K, and with a radius M^G. This circle meets the line GM-^ producedat the point K vertically below F; it also cuts the line AG in the point horizontal is drawn through P to meet the ordinate AH s± the point is evident from the geometry of the figure that the angles a^, a^ and a^are all equal, and that the triangles AHF, PGA are similar. From this itfoUows that jiQ HP ap AP=FA^^FA = ^--Multiplying these equations together, we have AC = = HFAana^ («). In the triangle OAF, we have Ij^==FA^ + ]o^ + , and since FA^^GF. HF = ^.HF, T I
. A text-book of electrical engineering;. K, and with a radius M^G. This circle meets the line GM-^ producedat the point K vertically below F; it also cuts the line AG in the point horizontal is drawn through P to meet the ordinate AH s± the point is evident from the geometry of the figure that the angles a^, a^ and a^are all equal, and that the triangles AHF, PGA are similar. From this itfoUows that jiQ HP ap AP=FA^^FA = ^--Multiplying these equations together, we have AC = = HFAana^ («). In the triangle OAF, we have Ij^==FA^ + ]o^ + , and since FA^^GF. HF = ^.HF, T I^^=-> + Io + T 414 Electrical Engineering Solving this equation for HF, we get J-i -o HF = - Substituting this in the above equation for AC, and using equation (193), we have ^C^i^^.tan«3=(-^^^^^^ •in .(—:) We shall now neglect the term Ig^.R, which is always very small, moreespecially as we can allow for it later when correcting for the no-load the primary copper loss, we have therefore ^ = Fig. 411 By simple subtraction we find the power transmitted from the statorto the rotor to be P, = 3 . V^-CH = 3 . Fj-Pg. To determine the effect of the rotor copper loss we draw the line GL inFig. 412, so that tanyi _tan«,= (i±^f)!,^#^^=C ,^Fi .(194). This is the same expression as we used in Section 125 (equation (186))for determining the shp. The vertical through M cuts this Une in the pointM^. With the centre M^ and the radius M^G we draw the circle before, the angles yn Y2 ^^^ Ya ^.re equal. A horizontal through R meetsAH in S. Then j^^ ^5 jjp ^ = tany3and^=^. 129. The most Convenient Form of the Circle-diagram 415 Multiplying these equations together, we have 4S = flFtany3 = HFtanyi (6). We saw above that AC = HF tan a^, whence CS = AS — AC = HF tan yj — HF tan a^ = HF (tan y^ — tan a^). FAIf we substitute for HF its equivalent ^^y-?, and for tan yj — tan a^ its value from equation (194), we get r^_FA^ (I + Ti)^■
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