HW4s - ME?“ 634‘ Pstw snwfions 2‘ ‘ , “.V’J...

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Unformatted text preview: ME?“ 634‘ Pstw snwfions 2‘ ‘ , “.V’J 3?: H . ._~'I.'._ P“ PROBLEM é.1 Part a 1' '59 8'1 V's-1W Ki With stator current acting alone the situation is as depicted at the right. Recognizing by symmetry that Hrswfi-fl- -Hrs(¢) we use the contour shown and Ampere's law to get 4 W+n N i zancw): - [w (2—337; sin w'umng _ Naiscoaw (Lgtour ‘oF1_ [away-dink from which. Nsiscosw Rm ('1’) ‘ 2g v- .4 I A; and n c uoNsiscosw ‘ _r 28 snow - Part b w -—> (xv-e) Part c The flux density varies around the periphery and the windin ‘thus a double integration is required to find inductances, from stored energy or from flux linkages. The total radial flux density is gs are distrihuted whether they ere fofind We will use flux linkages. Ll . I I. -—2 ' I Br Bra + Brr 28 [Nsiscosw + Nrircos(v—9)] PROBLEM 4.1 (Continued) Taking first the elemental coil on the stator having sides of angular span ow at positions W and ¢+n as illustrated. This coil links an amount of flux NS lp+1r ' - ——-———— _ ' I 9' number of turns in flux linking one turn elemental coil of elemental coil - uoN8(R+g)L w+u - _ V I_ I dks —7;iEr————-sinwdw 1w [Nsiscosw + Nrircos(w 63d¢ uoNa(R+s)£ axe - zgv sinY[Nsissinw + Nfirsin(w-6)]dw To find the total flux linkage with the stator coil we add up A11 of the contributions uoNa(R+g)£ u 18- 18 J0 sinWINsissinW + Nrirsin(w-6)]dw u N (R+g)£ ‘ w n - L{— .— As _ 2.8 2 Nsi8 + 2 Nrircose] This can be written as A -Li +Mi cose s s s r where 1m NZRJE L _ o s s 4lg fluoNaNrR£ M - 4's and we have written R+g % R because 3 << R. When a similar process is carried out for the rotor winding, it yields A - L i + Hi c036 r r r s where 2 - nuoNrkz r 4.3 and M is the same as calculated before. L PROBLEM 4.7 _ Part a Cocncrgy is .'_l2_1_2 wm(is'1r’e) 2 L915 + 2 Lrir + Lsr(eziair e aw;(1s,1r,e) V dLsr(6) 'r-—————-—--11——————- s r d6 38 e T - ~isirffilsin6 + 3H sin36] 3 Part b With the giveh constraints e . T -I5Irsinmst ainwrt[Hlsin(wmc+Y)+ 3M3sin3(wmt+y)] Repeated applicaeien of trigedgmctric identiiies leads to: e MlIsIr . T - — ——7r~—- sin[(wm+ws~wr)t+y]+ sin[(wm—ws+wr)t+yl -sin[(wm+ms+wr)t+y]- sin[(wm-ws-wr)t+7{; r - ———7r-—- g;in[(3mm+ms—wr)t+3y]+ sin[(3u$ru%+wr)t+371 -sin[(3wm+ma+wr)t+3Y]4 sin[(3wh-ms-mr)t+3yig lvitz .To have a time-average torque, ode of the coefficients of time gust equal zero. This leads to the eight possible mechanical speeds w + w 8"]? For For For For ...
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HW4s - ME?“ 634‘ Pstw snwfions 2‘ ‘ , “.V’J...

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