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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.
 Fall '07
 Lear
 Fluid Mechanics

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