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C6.Träng lîng thiÕt bÞ vµ ht tµu (P02 vµ P03)?
C¸c thµnh phÇn TL thiÕt bÞ P02vµ TL ht tµu P03 thùc chÊt ®îc t¸ch ra tõ TL th©n tµu P01 vµ ®îc x® nh sau:
P02 = Ptb = p02.D = q02LBH
P03 = Pht = p03.D = q03LBH
+ NÕu tµu thiÕt kÕ ®ßi hái ph¶i trang bÞ c¸c ht vµ tbÞ míi th× cÇn ph¶i ®iÒu chØnh 2 tphÇn TL nµycña tµu mÉu tríc lóc cã ý ®Þnh sö dông chóng ®Ó tÝnh to¸n tkÕ
+ NÕu tµu tkÕ > h¬n tµu mÉu 1 c¸ch ®¸ng kÓ th× cÇn lÊy TL ®¬n vÞ mµ ë ®ã TLTP t¨ng chËm h¬n. Lóc nµy sö dông c«ng thøc sau:
P02 = Ptb = p02’.D2/3 = q02’.(LBH)2/3
P03 = Pht = p03’.D2/3 = q03’.(LBH)2/3
Tµu hµng kh«: p02’ = 0,49 ± 0,06
q02’ = 0,21 ± 0,04
Tµu dÇu: p03’ = 0,3 ± 0,03
q03’ = 0,35 ± 0,05
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