Explanation Given: The tank in the given figure is a cylinder of radius 4 m and height 15 m . Assume that the tank is half-filled with water and that water leaks through a hole in the bottom of area B = 0.001 m 2 . Calculation: Length of the upper surface of water is L = 15 m . So, width of the upper surface of water is: W = 2 R 2 − ( R − y ) 2 = 2 R 2 − ( R 2 + y 2 − 2 R y ) = 2 R 2 − R 2 − y 2 + 2 R y = 2 2 R y − y 2 So, the area of cross-section of the upper surface of water at a distance y from the bottom is: A ( y ) = L × W = 30 2 R y − y 2 According to Toricelli’s Law, the velocity of the water flowing through a hole is: v = − 2 g y , where g = 9.8 m s 2 . So, the differential equation for the height of the water in the tank is: d y d t = B v ( y ) A ( y ) , where A ( y ) is the cross-sectional area of the tank at height y. Now, d y d t = B v ( y ) A ( y ) = B ( − 2 g y ) 30 2 R y − y 2 = B ( − 2 g ) 30 2 R − y = − 0.001 2 × 9.8 30 8 − y = − 0.0001 19.6 3 8 − y So, the separation of variables and then integrating: 8 − y d y = − 0

3rd Edition

ISBN: 9781464125263

Author: Jon Rogawski, Colin Adams

Publisher: W.H. Freeman & Co

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Chapter 10.1, Problem 51E

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The water level y(t) and the time te when the tank is empty.

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Chapter 10.1, Problem 50E

Chapter 10.1, Problem 52E

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The water level y ( t ) and the time t e when the tank is empty.

Solution Summary: The author calculates the area of cross-section of the upper surface of water at a distance y from the bottom.

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The water level y(t) and the time te when the tank is empty.

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