Physical Meteorology, MTR3440
Spring 2008
Homework #2
Due Wed., Feb. 20
1. Consider an idealized cloud consisting of spherical droplets with a uniform radius of 20
μm and number concentration of 1.2 cm-3.
a. How long a path through such a cloud would be required to deplete a beam of
visible radiation by a factor of e (neglecting multiple scattering)?
This is the definition of unit optical depth. That is, an optical depth of 1 is
defined as the distance into a medium in which the radiation is depleted by a
factor of e. Another way of saying this is that an optical depth of 1 occurs when
the radiation is equal to e-1 (=1/e=0.367, or 36.7%) times the incident radiation.
In this problem, the appropriate equation to use for optical depth is
τ = Qextπ r 2 Nz
where Qext is the extinction efficiency, r is the radius of the particles, N is the
number concentration, and z is the distance. At visible wavelengths, Qext ~ 2
because the Mie Size parameter is > 50 (i.e., 2π*20 μm / 0.5 μm ~ 240 ) and we
are in the geometric optics limit. Setting τ=1 and solving for z, we get
1
1
m
z=
=
= 332 m
*
2
−4
2
−3
Qextπ r N 2π (20 x10 cm) (1.2 cm ) 100 cm
b. If the cloud layer were 1 km thick, what would be the optical depth at (visible
wavelengths) of the layer (again, neglecting multiple scattering)?
Here we are solving for optical depth using the equation above:
105cm
τ = Qextπ r 2 Nz = 2π (20 x10−4 cm)2 (1.2 cm -3 )
*1km = 3.0
km
c. Assume that the cloud now has particles of 10 μm radius, but that the liquid water
content (LWC, LWC=mass of cloud particles per unit volume) remains
unchanged. Calculate the optical depth as in b.
LWC in part a of this problem is
4
4
LWC = N ρV = N ρ π r 3 = (1.2 cm −3 )(1 g cm -3 ) π (20 x10−4 cm)3 = 4x10-8 g cm -3
3
3
Now we need to find what N is when we change r but keep LWC the same
LWC 3 LWC 3
4x10-8g cm -3
N=
=
=
= 9.6 cm -3
3
-3
-4
3
4 ρπ r
4 (1g cm )π(10x10 cm)
ρV
Finally, we calculate the optical depth
105cm
τ = Qextπ r 2 Nz = 2π (10 x10−4 cm) 2 (9.6 cm-3 )
*1km = 6.0
km
d. How do your answers in b and c compare? That is, for a cloud with a fixed LWC,
which cloud attenuates more visible radiation? Why?
The optical depth for the case with small 10 μm particles is larger than the case
with 20 μm particles. Thus, the cloud with the smaller particles attenuates more
visible radiation for a given LWC. That is because, for a given mass of particles
(or LWC), there is more surface area to “block” the radiation if all of the
particles are small, compared to if they are large.
2. Calculate the radiative heating rate (in °C hr-1) within the 650-500mb atmospheric layer,
given the following flux densities at the top and bottom of the layer.
F↓ = 780 Wm-2
F↓ = 700 Wm-2
F↑ = 420 Wm-2
F↑ = 680 Wm-2
500 mb
650 mb
Here we solve for the radiative heating rate of the layer, using the equation
dT g dF ( z ) g ΔF ( z )
=
=
dt c p dp
c p Δp
=
9.8 m s -1
780 W m -2 - 420 W m -2 + 680 W m -2 - 700 W m -2
= 2.2 x 10-4 K s -1
-1
-1
1004 J K kg
65000Pa-50000Pa
⎛ 3600 s ⎞
−1
-1
= 2.2 x 10-4 K s -1 ⎜
⎟ = 0.79 K hour = 0.79 °C hour
⎝ hour ⎠
3. The Solar Radiation and Climate Experiment (SORCE, http://lasp.colorado.edu/sorce/)
satellite has been making measurements of the solar input into the Earth system from low
earth orbit since 2003. One of the instruments aboard SORCE is the Spectral Irradiance
Monitor (SIM). SIM measures the solar spectral irradiance (i.e., the monochromatic
irradiance impinging on a surface at the mean Earth-Sun distance and oriented
perpendicular to the Earth-Sun direction) from 300 to 2400 nm.
a. Using SIM data, compute the solar irradiance (normal to the surface of the Earth)
in the near IR band (0.7 – 1.5 μm) at the top of the atmosphere over Denver at
solar noon on Dec 21, 2007. On this day, the Earth is at 0.984 Astronomical
Units (AU, 1 AU = the mean Earth-Sun distance). SIM data can be downloaded
from http://lasp.colorado.edu/cgi-bin/ion-p?page=input_data_for_spectra.ion.
Note that these data follow the definition of solar spectral irradiance given above.
Use a program such as MATLAB, IDL, or Excel to compute your answer, and
attach any relevant code or data to show your work.
The SIM solar spectral irradiance integrated over the wavelength range 0.7 – 1.5
μm is 545 W m-2. On this part, all you had to do was multiply the SIM value (W
m-2 nm-1) by the spacing between wavelength points (nm), and sum them to get
the total value (in units W m-2).
SIM values are defined at 1 AU. Because the Earth is at 0.984 AU on this day,
the solar flux density in the near IR is given by
2
⎛ 1 AU ⎞
-2
545 W m -2 ⎜
⎟ =563 W m
⎝ 0.984 AU ⎠
This value (563 W m-2) is the solar flux density at a surface oriented
perpendicular to the Earth-Sun vector at the Earth’s distance on this day.
However, we need to calculate the solar irradiance at Denver on this day. To do
this, we must calculate the SZA. On Dec. 21 (the Winter Solsctice), δ = -23.5º.
We could calculate SZA using our equation, or just recognize that at solar noon
on the Winter solstice, the SZA is the latitude plus the (absolute value of) the
declination, or 40 + 23.5 = 63.5. So, our final answer for solar irradiance is (563
W m-2)*(cos 63.5) = 251 W m-2
b. Water vapor accounts for most of the absorption in the near IR due to the many
rotational-vibrational lines in this region. Assume that for the near IR band, a
representative value for the absorption coefficient for water vapor is 0.5 cm2 g-1.
Determine the optical depths for water vapor in the 0-500mb and 500-840mb
layers. Assume that the mean value of the mass mixing ratio is 0.1 g kg-1 for the
upper layer and 2.5 g kg-1 for the lower layer.
Optical depth for a layer is defined as
z2
τ = ∫ kr ρ dz
z1
Writing this in terms of the hydrostatic equation, dp = − ρ gdz ,
τ=
p1
kr
dp
g
p2
∫
Assuming k and r are constant with pressure, we can solve for the optical depth
of layer 1:
p1
kr
kr
(0.5cm2 g −1 )(1m 2 /1002 cm 2 )(0.1gkg −1 )
τ = ∫ dp = ( p1 − p2 ) =
(50, 000 Pa − 0 Pa ) = 0.026
g
g
9.8ms −2
p2
Solving for layer, 2:
kr
(0.5cm 2 g −1 )(1m 2 /1002 cm 2 )(2.5 gkg −1 )
(84, 000 Pa − 50, 000 Pa ) = 0.43
τ = ( p1 − p2 ) =
g
9.8ms −2
c. Determine the near IR band solar irradiance reaching 500 mb and 840 mb,
ignoring all other mechanisms for attenuating the beam.
The near IR solar irradiance reaching 500 mb is related to the irradiance
calculated in part a, and the optical depth from part b:
−τ 0−500 mb
μ
F500 mb = F0 e
= (251 W m -2 )e-0.026/cos 63.5 =237 W m -2
The solar irradiance reaching 840 mb is then calculated using the irradiance at
500 mb and the optical depth of the 500-840 mb layer:
F840 mb = F500 mb e
−τ 500−840 mb
μ
= (237 W m -2 )e-0.43/cos 63.5 =90 W m -2
d. Calculate the heating rate (K day-1) in each of the two layers due to absorption of
water vapor alone.
We will again use the heating rate equation in terms of pressure as we did in
problem 2. For the upper layer, we have:
dT g ΔF ( z )
9.8 m s -1
251 W m -2 - 237 W m -2
=
=
dt c p Δp
1004 J K -1 kg -1
50000 Pa - 0 Pa
= 2.73 x 10-6 K s-1
⎛ 86400 s ⎞
-1
= 2.73 x 10-6 K s -1 ⎜
⎟ = 0.24 K day
day
⎝
⎠
For the lower layer, we similarly have
dT g ΔF ( z )
9.8 m s -1
237 W m -2 - 90 W m -2
=
=
dt c p Δp
1004 J K -1 kg -1 84000 Pa - 50000 Pa
= 4.22 x 10-5 K s -1
⎛ 86400 s ⎞
-1
= 4.22 x 10-5 K s -1 ⎜
⎟ = 3.65 K day
day
⎝
⎠