Yahoo Answers: Answers and Comments for Fluid mechanics [其他  科學]
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Fri, 20 Jun 2014 23:52:34 +0000
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Yahoo Answers: Answers and Comments for Fluid mechanics [其他  科學]
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https://hk.answers.yahoo.com/question/index?qid=20140620000051KK00141
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From 天同: Consider point A and the nozzle, apply Bernoul...
https://hk.answers.yahoo.com/question/index?qid=20140620000051KK00141
https://hk.answers.yahoo.com/question/index?qid=20140620000051KK00141
Sat, 21 Jun 2014 19:42:15 +0000
Consider point A and the nozzle, apply Bernoulli's Equation,
Pa + (1/2)p(Va)^2 = Pn + (1/2)p(Vn)^2 + pg(1.1)
where Pa and Pn are the pressures at A and at the nozzle respectively,
p os the density of water (= 1000 kg/m^3)
Va and Vn are the speeds of water at A and at the nozzle respectively
g is the acceleration due to gravity
Because the water is at free flow at the nozzle, Pn = 0
thus, (55 x 10^3) + (1000/2)(Va)^2 = (1000/2)(Vn)^2 + 1000 x 9.81 x 1.1
simplifying, 88.42 + (Va)^2 = (Vn)^2  (1)
By equation of continuity,
(Aa).(Va) = (An).(Vn)
where Aa and An are the crosssectional areas of the pipe at point A and at the nozzle respectively.
hence, [pi(0.2/2)^2].(Va) = [pi.(0.1/2)^2].(Vn)
0.04(Va) = 0.01(Vn)
i.e. Va = Vn/4  (2)
Substitute Va from (2) into (1),
88.42 + (Vn/4)^2 = (Vn)^2
solve for (Vn)^2 gives (Vn)^2 = 94.31 (m/s)^2
By conservation of mechanical energy,
gain in potential energy of water at top of water jet = loss of kinetic energy of water at nozzle,
thus, pgh = (1/2)p(Vn)^2
h = (Vn)^2/(2g) = 94.31/(2x9.81) m = 4.81 m