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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

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

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  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.

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