15) Given a 2D, non-‐divergent flow on an f-‐plane with a velocity fiel

<strong>15</strong>) <strong>Given</strong> a <strong>2D</strong>, <strong>non</strong>-­‐divergent <strong>flow</strong> on an f-­‐plane with a velocity field in which u = −y + y 3
and v = x − x 3 :
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a. Verify that the <strong>flow</strong> is <strong>non</strong>-­‐divergent.
€ ∂u
To show that the <strong>flow</strong> is <strong>non</strong>-­‐divergent, we must show that
∂x + ∂v
∂y = 0.
∂
∂x (−y + y 3 ) + ∂ ∂y (x − x 3 ) = 0 + 0 = 0
b. Determine the vorticity at t = t 0
.
€ We know ζ = ∂v
∂x − ∂u . Solving for ζ, we get:
∂y
ζ = ∂v €
∂x − ∂u
∂y = ∂ ∂x (x − x 3 ) − ∂ ∂y (−y + y 3 ) =1− 3x 2 +1− 3y 2 = 2 − 3(x 2 + y 2 )
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c. Determine the local time derivative of vorticity at t = t 0
.
The <strong>2D</strong> Lagrangian vorticity equation can be written as:
€ Dη
Dt = ∂ζ
∂t + ∂f
∂t + u∂ζ ∂x + u∂f ∂x + v ∂ζ
∂y + v df . However, with the assumption of an
dy
€
Dζ
f-­‐plane, f = f 0
the equation reduces to
Dt = ∂ζ
∂t + u∂ζ ∂x + v ∂ζ and given that
∂y
Dζ
∂ζ ∂ζ
€ the <strong>flow</strong> is <strong>non</strong>-­‐divergent = 0, which leaves = −u
Dt ∂t ∂x − v ∂ζ . Now we
∂y
€ ∂ζ
solve for
€
∂t .
∂ζ
€
€
∂t = (y − y 3 ) ∂ ∂x [2 − 3(x 2 + y 2 )] − (x − x 3 ) ∂ ∂y [2 − 3(x 2 + y 2 )] = 6xy(y 2 − x 2 )
d. In what
€
compass directions from the origin are the points where the vorticity is
increasing with time?
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€
€
∂ζ
For
∂t = 6xy(y 2 − x 2 ), lines along which x=y,
∂ζ
∂ζ
= 0. Regions in which is positive
∂t ∂t
(negative) are shown on the plot to the left
and indicated by + (-­‐).
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19) <strong>Given</strong> the ten instantaneous measurements below, solve for τ x
= −ρ s
u ʹ′ w ʹ′ , where
ρ s
=1.2kg m -­‐3 .
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u 8 6 11 14 13 9 8 9 <strong>15</strong> 12
€
w 0 1 0 -1 0 -1 -2 1 0 -2
First calculate separately the average zonal and vertical winds speeds to get and u =10.5
ms -­‐1 and w = −0.4 ms -­‐1 . Next calculate u ʹ′ , w ʹ′ , and u ʹ′ w ʹ′ .
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u' -2.5 -4.5 0.5 3.5 2.5 -1.5 -2.5 -1.5 4.5 1.5
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w' 0.4 1.4 0.4 -0.6 0.4 -0.6 -1.6 1.4 0.4 -1.6
u'w' -1 -6.3 0.2 -2.1
€ €
1
€
0.9 4 -2.1 1.8 -2.4
Now we can solve for u ʹ′ w ʹ′ , which u ʹ′ w ʹ′ = −0.6 .
Finally, we have τ x
= −ρ s
u ʹ′ w ʹ′ = −1.2 kg m -3 (−0.6 m 2 s -2 ) = 0.72 (kg m s -2 )m -2 N m -­‐2 .
Note that we have solved for a stress – a tangential force per unit area.
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