Problem 4.49

# Problem 4.49 - t 2 ft ⋅ 45.6 s = t 1 ft ⋅ 105 s = Solving for t t 2 y ⋅ g D d ⎛ ⎝ ⎞ ⎠ 2 ⋅ 1 y y − ⎛ ⎜ ⎝ ⎞ ⎠ ⋅ = Hence

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Problem 4.49 [Difficulty: 3] Given: Data on draining of a tank Find: Times to a depth of 1 foot; Plot of drain timeversus opening size Solution: 0 = + CS CV A d V V d t r r ρ Basic equation Assumptions: 1) Uniform flow 2) Incompressible flow 3) Neglect air density Treating the tank as the CV the basic equation becomes t 0 y y ρ A tank d ρ V A opening + 0 = or ρ π 4 D 2 dy dt ρ π 4 d 2 V + 0 = Using V2 g y = and simplifying dy dt d D 2 2g y 1 2 = Separating variables dy y 1 2 d D 2 2g dt = and integrating 2y 1 2 y 0 1 2 d D 2 2g t = Using the given data t2f t
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Unformatted text preview: t 2 ft ⋅ ( ) 45.6 s = t 1 ft ⋅ ( ) 105 s = Solving for t t 2 y ⋅ g D d ⎛ ⎝ ⎞ ⎠ 2 ⋅ 1 y y − ⎛ ⎜ ⎝ ⎞ ⎠ ⋅ = Hence for the first drop of 1 foot ∆ t t 2 ft ⋅ ( ) = ∆ t 45.6 s = For the second drop of 1 foot ∆ t t 1 ft ⋅ ( ) t 2 ft ⋅ ( ) − = ∆ t 59.5 s = This is because as the level drops the exit speed, hence drain rate, decreases. 0.1 0.2 0.3 0.4 0.5 5 10 15 d (in) Drain Time (min)...
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## This note was uploaded on 10/19/2011 for the course EGN 3353C taught by Professor Lear during the Fall '07 term at University of Florida.

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