1 / 5

Example 1.3 . Uranium fuel rod in nuclear reactor 核子反應爐內的鈾燃料棒

Example 1.3 . Uranium fuel rod in nuclear reactor 核子反應爐內的鈾燃料棒 Before insertion 插入前 , rod length 棒長 = 3.241 m After insertion 插入後 , rod length 棒長 = 3.249 m Q 問 : What is the increase in length 長度增加多少 ? A: 3.249 m  3.241 m = 0.008 m = 8 mm

vanida
Télécharger la présentation

Example 1.3 . Uranium fuel rod in nuclear reactor 核子反應爐內的鈾燃料棒

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. Example 1.3. Uranium fuel rod in nuclear reactor 核子反應爐內的鈾燃料棒 Before insertion 插入前 , rod length 棒長 = 3.241 m After insertion 插入後 , rod length 棒長 = 3.249 m Q 問 : What is the increase in length 長度增加多少 ? A: 3.249 m  3.241 m = 0.008 m = 8 mm Accuracy 精確度 = 1 mm Error 誤差 =  0.001 m =  1 mm  Increase in length is 長度增加 8 mm ( 1 sig. dig. 1 個有效數字) Any intermediate results must have at least 1 extra sig. dig. to avoid rounding errors. 每個中介結果必需保留多一個有效數字才能避免四捨五入的誤差 Caclulator: apply round-off & truncation only at the end. 計算機 : 全部算完才四捨五入和捨棄非有效數字。

  2. Significant Figures • Significant figures (digits) • of an integer: all digits between the leftmost & rightmost non-zero digits. • Trailing zeros are ambiguous. • of a real number: all digits except leading zeros. Examples: Numbers with 5 sig. dig. : 001000500000, 123.45, 0.0012345, 0.010000 Note: 001000500000 may be taken as having 10 sig. dig. Caution: An integer sometimes denotes infinite accuracy (  sig. dig. ). e.g., 2 in the formulae C = 2  R & A =  R2.

  3. Accuracy & Significant Figures means 2.94 is between 1.6 & 1.8 i.e. or • Accuracy worsens after each calculation. • Result has accuracy of the least accurate member. • /  : Number of significant digits = that of the least accurate member. • + /  : result is rounded off to the rightmost common digit.

  4. Bridge= 1.248 km ( accuracy = 0.001 km ) Ramp = 65.4 m = 0.0654 km ( acc = 0.0001 km ) Overall length = 1.248 km + 0.0654 km = 1.3134 km Overall acc = 0.001 km, error =  0.001 km  Overall length = 1.313 km  = 3.14159 ( # sig. dig. = 6 ) RE = 6.37 106 m ( # sig. dig. = 3 ) 2  RE = 40.0238566106 m Overall # sig. digits = 3  2  RE = 40.0106 m

  5. Example 1.3. Uranium fuel rod in nuclear reactor Before insertion, rod length = 3.241 m After insertion, rod length = 3.249 m Q: What is the increase in length? A: 3.249 m  3.241 m = 0.008 m = 8 mm Accuracy = 1 mm Error =  0.001 m =  1 mm  Increase in length is 8 mm ( 1 sig. dig. ) Any intermediate results must have at least 1 extra sig. dig. to avoid rounding errors. Caclulator: apply round-off & truncation only at the end.

More Related