流力問題 - 土木
By Yuri
at 2013-04-15T10:07
at 2013-04-15T10:07
Table of Contents
僅就第一題進行回覆:
此類牽涉到時間的問題用積分方法解最為適當,一般單純的課本例題通常不考慮
細管的摩擦係數,然此題有給也就必須考量主要損失(hL)。利用伯努力定理:
h = V^2/2g + hL,又hL = f*(L/D)*V^2/2g。計算可得h = 3*(V^2/2g),若要將
變數與時間扯上關係,則須導入流量的想法:
Q = [(pi*D^2)/4]*(-dh/dt) = [(pi*d^2)/4]*V ,其中(-dh/dt)代表水面下降
速度,負值表示其方向性。經過移項,將邊界條件取(h:3 → 1.5 ;t: 0 → T)
小弟所算出的時間 T = 158.53 sec,煩請各位前輩不吝提出指教。
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