102年高考流力討論 - 高考
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By Isla
at 2013-07-10T17:28
at 2013-07-10T17:28
Table of Contents
※ 引述《prevoc (衝吧)》之銘言:
: 小弟今年沒參加高考,但基於練習的心態把今年高考流力練習...
: 期盼高手一起討論...
: 小弟不才,有錯請指出
: https://docs.google.com/file/d/0Bwyy1b2D1231al9HSjFINll4MkU/edit?usp=sharing
第一題
(2)的管口正對噴流 量到的應該是總水頭
P2 V^2 P3
也就是 —— = —— = H = —— + y = h(1-SG) 所以 h = 0
γw 2g γw
第三題
前半段動量方程式求反力 作法和prevoc大一樣
求得水流對載具的反力 R(V) = -2Aρ(V^2)-ρgh
然後由加速度公式 (Vo)^2 + V^2 = 2as
(Vo)^2 + V^2 R(V)
ds = [——————]dV , a = [——]dV
2a M
100 [(Vo)^2 + V^2]
所以 S = ∫ds = ∫ -M { ———————— } dV
900 [4Aρ(V^2)-ρgh]
--
: 小弟今年沒參加高考,但基於練習的心態把今年高考流力練習...
: 期盼高手一起討論...
: 小弟不才,有錯請指出
: https://docs.google.com/file/d/0Bwyy1b2D1231al9HSjFINll4MkU/edit?usp=sharing
第一題
(2)的管口正對噴流 量到的應該是總水頭
P2 V^2 P3
也就是 —— = —— = H = —— + y = h(1-SG) 所以 h = 0
γw 2g γw
第三題
前半段動量方程式求反力 作法和prevoc大一樣
求得水流對載具的反力 R(V) = -2Aρ(V^2)-ρgh
然後由加速度公式 (Vo)^2 + V^2 = 2as
(Vo)^2 + V^2 R(V)
ds = [——————]dV , a = [——]dV
2a M
100 [(Vo)^2 + V^2]
所以 S = ∫ds = ∫ -M { ———————— } dV
900 [4Aρ(V^2)-ρgh]
--
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高考
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By Skylar Davis
at 2013-07-15T11:58
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