2 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 v₁ = √2gh (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.
2 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 v₁ = √2gh (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.
Principles of Physics: A Calculus-Based Text
5th Edition
ISBN:9781133104261
Author:Raymond A. Serway, John W. Jewett
Publisher:Raymond A. Serway, John W. Jewett
Chapter15: Fluid Mechanics
Section: Chapter Questions
Problem 46P
Related questions
Question
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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