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(mid2024:sol)=
# A301 midterm solutions

Name (Last, First):

Student Number:


Instructions: Answer all questions, clearly labeling each answer, showing all your work on the exam booklet.

For partial credit: If you're stuck on one part of a multipart question and need to move on, you can invent a reasonable set of numbers so that you can answer subsequent questions for full credit. Just be very clear in explaining what you're doing.

Closed book, equation sheet provided, calculators ok

+++

## Q1) (12) 

A satellite orbiting at an altitude of 36000 km observes the
surface in the $CO_2$ absorption band with a wavelength
range of 14 μm  $< λ < 16$ μm.


###  Q1a) (4 points) 

The atmosphere is 20 km thick, and has a density scale height of
$H_\rho$ = 9 km and a surface air density of
$\rho_{air}$ = 1.1 $kg\,m^{-3}$. The $CO_2$ mass
absorption coefficient is
$k_λ$ = 0.15 $m^2\,kg^{-1}$ at
λ = 15 μm and its mixing ratio is $4 \times 10^{−4}\,kg\,kg^{-1}$.
Find the:

- the atmosphere’s vertical optical thickness
    $τ_λ$ in the $CO_2$ band

- the atmosphere’s transmittance $t_\lambda$ in the $CO_2$
    band

for a pixel directly beneath the satellite.

+++

#### Q1a answer

$$
 \rho(z) &= \rho_0 r_{mix} \exp( -z/H_\rho) \\
 \tau_T &= \int_0^{20} \rho_0 r_{mix} \exp(-z/H) k_\lambda dz = - H \rho_0 r_{mix} k_\lambda (\exp(-20/9)-1)\\
 \tau_T &= 9000 \times 4 \times 10^{-4} *0.15 * (1 - 0.108) = 0.53\\
 t_T &= \exp(-\tau_T) = 0.589
$$

+++

### 

Q1b (4 points)
 
If the surface is a blackbody with a temperature of 290 K, and
        the atmosphere has an constant temperature of 270 K, find the

-   monochromatic radiance observed by the satellite at 15
        μm in ($W\,m^{-2}\,\mu m^{-1}\,sr^{-1}$)

-   the brightness temperature of the pixel in Kelvins for that
        radiance

+++

#### Q1b answer

  From the Planck diagram, at $\lambda=15\ \mu m$ I get:
  
  - $L_0 \approx 6\ W\,m^{-2}\,\mu m^{-1}\,sr^{-1}$
  - $B_\lambda \approx 4.6\ W\,m^{-2}\,\mu m^{-1}\,sr^{-1}$

Use equation (28) from the equation sheet, assume looking straight down
so $\mu = 1$:

$$
 L_\lambda(\tau/\mu)= B_\lambda(T_{skin}) \hat{t}_{tot} 
             +  (1 - \exp(-\tau/\mu) )B_\lambda(T)
$$
  
  $$L_\uparrow(\tau_T) = 6 \times 0.589   + (1 - 0.589) \times 4.6 = 5.41\ W\,m^{-2}\,\mu m^{-1}\,sr^{-1}$$
  
  From the Planck diagram 5.41 $W\,m^{-2}\,\mu m^{-1}\,sr^{-1}$ at 15 $\mu m$ is a brightness temperature of about 282 K.

+++

### Q1c (4 points)

- Given a pixel size $10\ km^2$, find:

    -   the imager field of view in steradians

    -   the flux, in $W\,m^{-2}$ at the satellite for the wavelength range between
        14-16 $\mu m$

+++

#### Q1C answer

 $\Delta \omega$ = $\frac{area}{R^2}$ = $\frac{10}{36000^2}$ = $7.7 \times 10^{-9}\ sr$
    
- $F = L_\uparrow(\tau_T) \Delta \lambda \,\Delta \omega = 5.41 \times 2 \times 7.7\times 10^{-9} = 8.4\times 10^{-8}\ W\,m^{-2}$

+++

## Q2 (8 points)

+++

- For the same atmosphere as Question 1, find the heating/cooling rate, in degrees K/day,
  showing all your work.

+++

### Q2 Answer

+++

Start with equation (30) from the equation sheet

$$
  F_\lambda &= \pi B_\lambda(T_{skin}) \exp(-1.66 \tau_{tot}) \nonumber \\
  & +   \pi B_\lambda(T)(1 - \exp(-1.66 \tau))
$$

For the heating rate, we're also going to need to integrate over frequency, which we can do using Stefan-Boltzman

$$
  F = \sigma T_{skin}^4 \exp(-1.66 \tau_{tot}) 
   +   \sigma T^4_{layer}(1 - \exp(-1.66 \tau_{tot}))
$$
where we've assumed that $\tau$ is constant with wavelength.

Some numbers:  top of layer is level 2, bottom of layer is level 1
downward flux is positive.  We want to find

$$
\frac{dF_n}{dz} \approx \frac{\Delta F_n}{\Delta z}
$$

At the base level 1:

$F_\downarrow = \epsilon \sigma T_{layer}^4$ = 176 $W\,m^{-2}$

$F_\uparrow = F_{skin} = -\sigma T^4_{skin}$ = -401 $W\,m^{-2}$

$F_{n1} = F\uparrow + F_\downarrow = -224.7\ w\,m^{-2}$

At the top level 2

$F_{n2} = F_{layer} + (1 - \epsilon)F_{skin}$ = -388.9 $W\,m^{-2}$

$$
\Delta F = F_{n2} - F_{n1} = -164.2\ W\,m^{-2}
$$

Total mass M is the integral of the density from quesition 1

$$
M = 1.1 \times 9000 \times  (1 - 0.108) = 8831\ kg\,m^{-2}
$$

$$
\frac{\Delta T}{\Delta t} = \frac{1}{M c_p}\frac{\Delta F_n}{\Delta z} = 10^{-4}\ K/day
$$

```{code-cell} ipython3
import numpy as np
sigma = 5.67e-8
Tlayer = 270
Tskin=290


tautot=0.53
transtot = np.exp(-1.66*tautot)
epsilon = 1 - transtot
Flayer = epsilon*sigma*Tlayer**4
Fskin = sigma*Tskin**4
Fn1 = Flayer - Fskin  #level 1 net
Fn2 = -Flayer - tautot*Fskin
deltaF = Fn2 - Fn1
deltaZ = 20000
cp = 1004.
M = 9000*1.1*(1 - 0.108)
seconds_day = 24*3600.
dTdt = (1/(M*cp))*(deltaF/deltaZ)*seconds_day
print(f"{Flayer=:.1f}")
print(f"{Fskin=:.1f}")
print(f"{transtot=:.1f}")
print(f"{epsilon=:.1f}")
print(f"{Fn1=:.1f}")
print(f"{Fn2=:.1f}")
print(f"{deltaF=:.1f}")
print(f"{M=:.1f}")
print(f"{dTdt=:.5f}")
```

## Q3 (6 points)

Starting with the Schwartzchild equation:

$$
  dL_\lambda = -L_\lambda\, d\tau/\mu  + B_{\lambda}(z) d\tau/\mu
$$

Derive the integrated solution for an isothermal layer:

$$
  L_\lambda(\tau/\mu)= B_\lambda(T_{skin}) \hat{t}_{tot}
             +  (1 - \exp(-\tau/\mu) )B_\lambda(T)
$$

stating all your assumptions.

+++

### Q3 answer

Follow  {ref}`schwartz:constant`:

$$
\frac{dL_\lambda}{L_\lambda - B_\lambda (T_{layer})} = -\frac{d\tau_\lambda}{\mu}
$$ (schwart2_sol)

Use the change of variables as in {ref}`schwartz:change`


$$
U^\prime &= L^\prime_\lambda - B_\lambda \\
dU^\prime &= dL^\prime_\lambda\\
\frac{dL^\prime_\lambda}{L^\prime_\lambda -
     B_\lambda} &= \frac{dU^\prime}{U^\prime} = d\ln U^\prime
$$

given $dB_\lambda = 0$ since the temperature is constant.

+++

Solve this by integrating a perfect differential from $U^\prime=U_0,\ \tau^\prime=0$ to $U^\prime = U,\ \tau^\prime = \tau_T$ :

$$
\int_{U_0}^U d\ln U^\prime = -\int_0^{\tau_T} \frac{d\tau^\prime}{\mu} = \ln \left (\frac{U}{U_0} \right ) =  \ln \left (\frac{L_\lambda - B_\lambda}{L_{\lambda 0} - B_\lambda}
      \right ) = - \frac{\tau_{T}}{\mu}
$$ (constTc_sol)

+++

Taking the $\exp$ of both sides:

$$
L_\lambda - B_\lambda = (L_{\lambda 0} - B_\lambda) \exp (-\tau_{T}/\mu)
$$ (constTd_sol)

or rearranging and recognizing that the transmittance is $\hat{t_{tot}} = \exp(-\tau_{T}/\mu )$:

$$
L_\lambda = L_{\lambda 0} \exp( -\tau_{ T}/\mu  ) + B_\lambda (T_{layer})(1- \exp( -\tau_{T}/\mu ))
$$ (rad_constant_sol)

+++

## Planck curves

```{figure} figures/a301_radiance_planck.png
---
width: 100%
---
Planck function
```