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[考題] NBOD之計算
Aug 10th 2013, 06:23, by sandy18932

作者sandy18932 (周昀忻)

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標題[考題] NBOD之計算

時間Sat Aug 10 06:23:04 2013

【出處】鼎茂出版社《環境化學與環境微生物分類題庫》二版,P 220 【94年成大環工所考題】原為原文,試著翻成中文,翻得不好,請包涵。 請說明某溶液(含有Glucose 180 mg/L 與 NH4+ - N 28 mg/L), 在被好氧菌降解的十天期間, 其BODc與NBOD之生物需氧量各為多少? 【問題】 1.我自己解的答案與書上答案不同,有疑惑中。 2.書上的解答是這樣: C6H12O6 + 6O2 ==> 6 CO2 + 6H2O 180 --------- * 6 * 32 = 192 mg/L = BODc 180 4NH4+ + 9O2 ==> 4 HNO3 + 6 H2O 28*(18/14) 9 --------------- * ------ * 32 = 144 mg/L= NBOD 18 4 ^^^^^^^^^^^^ 依照這個算式算出的答案確實是144。但,為何這個分子項是28*(18/14)? 而不是28*(14/18),也就是說為什麼不是 28*(14/18) -------------- 18 若您願意的話,還請諸位先進不吝指教與指導,謝謝您 ^^~ ※ 編輯: sandy18932 來自: 219.84.116.48 (08/10 06:25)

chinlun:推推 急氣分享喜氣~ 08/10 19:15

chinlun:推錯文 SORRY~ 08/10 19:16

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