# CH06 - Problem 6.15 part(a For R1 = 0 Elm = O and thus from...

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Unformatted text preview: Problem 6.15 part (a): For R1 = 0, Elm = O and thus from E—q. 6.34 R- , r 2 = (Aleq'l—A2) SmaxT From Eq. 6.36. T = 0.5-nphvﬁeq U4'5 [Al—133:1 + 2(2) and from Eq.6.33 with 5 = 1 'n'Phl'fEeqRQ Tstart = gr—rz -W~S[R2 +(A1_eq + A2] ] Noting that TIM,C _ 2.20 = 1.63 Tstart — 1 we can take the ratio of the above. equations . , , . R A Tm 2163 2 25+ [Af_eq+)x2]2 2 (X1 Ema) +1 Tstart RQIIArl‘eq + 2(2) Xi’gi—XQ which can be. solved to give smMT = 0.343 = 34.300. part [13): From Eq. 6.33 with Rechl = 0 and with s = smted. r2 ’ nphl' 13:1 [ Eb;If 5rated] Trﬁ. e = i t d |'-"-“s[(-R2,’lgrated); + (Alec; + Aﬂlzl and thus Tmax = = 0-5llR2fl3ratedlg 'l' ( req.1 ‘l‘ Ar2l2l Tratecl (RﬂfsrateclMAreqi + 2(2] _ + (Slnafollsratecﬂgl — Smaij'llsrated This can be. solved to give 5111th = 0-240511mxT = = part. [0}: .nteut R2 + le‘X'Oqll + X1?) [A1311 + AF2::'[5111:-\xT + j} Rg'fsl-awul + jigsreq'l + X2} [X13514 + X2]{StuarcTEISN-tted + firmed = T hunk liQ-Startl = ISIIIBXTfilfsratBCII+ = = = |I2.1'3t0d| lsmaxT + + ...
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CH06 - Problem 6.15 part(a For R1 = 0 Elm = O and thus from...

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