圖片參考：http://imgcld.yimg.com/8/n/HA00270495/o/701103010022713873385720.jpg
雖然之前有c 兄解過,但小弟仍有地方唔太明:
P1=static pressure the pt. below static tube along the stream arrow
P2=static pressure the pt. at mouth of total tube
Suppose that ρ is the density of the liquid contained in the tube, then using the Bernoulli's equation:
P1 + ρv12/2 + ρgh1 = P2 + ρv22/2 + ρgh2
Since the ends of the tube are levelled, h1 = h2 and hence
P1 + ρv12/2 = P2 + ρv22/2
Assume that the air stops immediately as soon as it reaches the front mouth of the tube (open horizontally), i.e. v1 = 0, we have:
P1 = P2 + ρv2/2
v = √[2(P1 - P2)/ρ]
Also since P1 - P2 = ρgh, we have:
v = √(2gh)
Q: P1 - P2 = ρgh 係係邊度黎?係唔係用Bernoulli's equation?
P1 - P2 同兩支tube既高度又有咩關係?點解可以咁砌出來?
請師兄解答
更新1:
果然又係天同師兄,多謝你既解答,但小弟想了解多d點解pressure diff= ρgh ,條式點整出黎?
