ATOC/ASEN 5235
Homework #3
Due Thursday, 23 September
Please staple all pages together
!!!
Each problem is worth 25 points
1. Petty 2.3.
The atmospheric boundary layer is that region near the surface that is “wellmixed” by
mechanical and/or convective turbulence originating at the surface. Its thickness may range from
a few meters to several kilometers. Heat added by conduction from the surface is typically
distributed quickly throughout the boundary layer.
At a certain location in the tropics, the sun rises at 06 Local Solar Time (LST), is directly
overhead at noon, and sets at 18 LST (=6 PM). Assume that, during the twelve hours that the sun
is up, the net flux of solar energy absorbed by dry vegetation and immediately transferred to the
overlying air is F(t) = F
0
cos[
π
(t12)/12], where t is the time of day (in hours), and F
0
=500 W m
2
.
(a) Ignoring other heating and cooling terms, compute the total solar energy (in J m
2
) added to
the boundary layer over one 24hour period.
(b) If the boundary layer depth
Δ
z = 1 km, its average air density is
ρ
a
= 1 kg m
3
, and the heat
capacity at constant pressure is c
p
= 1004 J kg
1
K
1
, compute the temperature increase
Δ
T
implied by your answer to (a).
(c) If, instead, the boundary layer depth started out at sunrise only 10 m deep and remained at
that depth throughout the day, what would be the corresponding change of temperature? Why is
it much more likely that the boundary layer would deepen quickly after sunrise?
Answer:
(a) F(t) = F
o
cos[
π
(t12)/12], t=local time (hrs), F
o
= 500 W m
2
Let
Q
= Energy per unit area (J m
2
):
(
)
[
]
∫
∫
−
=
=
18
6
18
6
dt
12
/
12
cos
dt
)
(
t
F
t
F
Q
o
π
Solve by substitution: u =
π
(t12)/12
du/dt =
π
/12, so dt = 12 du /
π
:
7
.
3819
24
2
sin
2
sin
12

sin
12
du
cos
12
18
6
18
6
=
=
⎟
⎠
⎞
⎜
⎝
⎛
−
−
=
=
=
=
=
=
=
∫
π
π
π
π
π
π
o
o
t
t
o
t
t
o
F
F
u
F
u
F
Q
The units on
Q
as expressed above: W m
2
hr = J s
1
m
2
hr
To convert to J m
2
:
Q
= 3819.7 J s
1
m
2
hr
×
3600 s hr
1
= 1.38
×
10
7
J m
2
(b)
Δ
z = 1 km,
ρ
a
= 1 kg m
3
,
c
p
= 1004 J kg
1
K
1
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