3. The figure at the right shows a tank that is open to the atmosphere at the top and has a pipe at the bottom, through which liquid exits.(a) Show that the expression for the speed of the liquid leaving the pipe is?1 = √2?ℎ(b) Calculate the speed v1 of the liquid leaving the pipe if the difference in height is h = y2 – y1 = 1.5 m.Hint: Assume that the liquid behaves as an ideal fluid (non-viscous). Therefore, we can apply Bernoulli’s equation, and in preparation of doing so, we locate two points in the liquid. Point 1 is just outside the efflux pipe, and point 2 is at the top surface of the liquid. The pressure at each of these points is equal to the atmospheric pressure (P1 = P2 = atmospheric pressure), a fact that will be used to simplify Bernoulli’s equation.Solution:(a) Show that the expression for the speed of the liquid leaving the pipe is?1 = √2?ℎFrom the Bernoulli’s equation?1 +12??12 + ???1 = ?2 +12??22 + ???2?1 − ?2 +12??12 + ???1 =12??22 + ???2Since the pressures at points 1 and 2 are the same, P1 = P2, then P1 – P2 = 0.0 +12??12 + ???1 =12??22 + ???212??12 + ???1 =12??22 + ???2 -------- continue until you arrive at ?1 = √2?ℎAdditional hint: h = y2 – y1 and if the tank is very large, the liquid level changes only slowly, and the speed at point 2 can be set equal to zero (v2 = 0). 3. The figure at the right shows a tank that is open to the atmosphere at thetop and has a pipe at the bottom, through which liquid exits.(a) Show that the expression for the speed of the liquid leaving the pipe isV₁ = √2gh(b) Calculate the speed v₁ of the liquid leaving the pipe if the difference inheight is h = y₂ - Y₁ = 1.5 m.Hint: Assume that the liquid behaves as an ideal fluid (non-viscous).Therefore, we can apply Bernoulli's equation, and in preparation ofdoing so, we locate two points in the liquid. Point 1 is just outside theefflux pipe, and point 2 is at the top surface of the liquid. The pressure at each of these points is equal to theatmospheric pressure (P1 = P2 = atmospheric pressure), a fact that will be used to simplify Bernoulli'sequation.Solution:(a) Show that the expression for the speed of the liquid leaving the pipe isV₁ = √√2ghFrom the Bernoulli's equation1P₁ + pv² + pgy₁ = P₂ + pv² + pgy₂11P₁-P₂ + pv² + pgy₁ = pv² + pgyz{Since the pressures at points 1 and 2 are the same, P1 = P2, then P1 P2 = 0.10++ 1/ 200² ·+ pgy₁ ==pv² + pgyz11pv² - +pgy₁ = pv² +pgy₂continue until you arrive atV₁ = √√2ghAdditional hint: h= y2-y₁ and if the tank is very large, the liquid level changes only slowly, and the speedat point 2 can be set equal to zero (v₂ = 0).

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Answered: 2 The figure at the right shows a tank…

College Physics

11th Edition

ISBN:9781305952300

Author:Raymond A. Serway, Chris Vuille

Publisher:Raymond A. Serway, Chris Vuille

Chapter1: Units, Trigonometry. And Vectors

Section: Chapter Questions

Problem 1CQ: Estimate the order of magnitude of the length, in meters, of each of the following; (a) a mouse, (b)...

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3. The figure at the right shows a tank that is open to the atmosphere at the top and has a pipe at the bottom, through which liquid exits. (a) Show that the expression for the speed of the liquid leaving the pipe is ?1 = √2?ℎ (b) Calculate the speed v1 of the liquid leaving the pipe if the difference in height is h = y2 – y1 = 1.5 m. Hint: Assume that the liquid behaves as an ideal fluid (non-viscous). Therefore, we can apply Bernoulli’s equation, and in preparation of doing so, we locate two points in the liquid. Point 1 is just outside the efflux pipe, and point 2 is at the top surface of the liquid. The pressure at each of these points is equal to the atmospheric pressure (P1 = P2 = atmospheric pressure), a fact that will be used to simplify Bernoulli’s equation. Solution: (a) Show that the expression for the speed of the liquid leaving the pipe is ?1 = √2?ℎ From the Bernoulli’s equation ?1 + 1 2 ??1 2 + ???1 = ?2 + 1 2 ??2 2 + ???2 ?1 − ?2 + 1 2 ??1 2 + ???1 = 1 2 ??2 2 + ???2 Since the pressures at points 1 and 2 are the same, P1 = P2, then P1 – P2 = 0. 0 + 1 2 ??1 2 + ???1 = 1 2 ??2 2 + ???2 1 2 ??1 2 + ???1 = 1 2 ??2 2 + ???2 -------- continue until you arrive at ?1 = √2?ℎ Additional hint: h = y2 – y1 and if the tank is very large, the liquid level changes only slowly, and the speed at point 2 can be set equal to zero (v2 = 0).

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