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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 = 0.5nphvﬁeq 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 = = 05llR2fl3ratedlg 'l' ( req.1 ‘l‘ Ar2l2l
Tratecl (RﬂfsrateclMAreqi + 2(2]
_ + (Slnafollsratecﬂgl
— Smaij'llsrated This can be. solved to give 5111th = 0240511mxT = = part. [0}: .nteut R2 + le‘X'Oqll + X1?) [A1311 + AF2::'[5111:\xT + j} Rg'fslawul + jigsreq'l + X2} [X13514 + X2]{StuarcTEISNtted + firmed = T hunk liQStartl = ISIIIBXTfilfsratBCII+ = = = I2.1'3t0d lsmaxT + + ...
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 Spring '07
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