Study Question Set 03

Radiation

  1. The highest yearly total \(I_{ex}\) (extraterrestrial irradiance) is found at the Equator (\(\theta = 0^{\circ}\)) with \(13.2\,\rm{GJ}\,\rm{m}^{-2}\,\rm{year}^{-1}\) and the lowest yearly total \(I_{ex}\) is at the Poles \(\theta = 90^{\circ}\), with \(5.5\,\rm{GJ}\,\rm{m}^{-2}\,\rm{year}^{-1}\). What are the implications of this?

This means that there is an energy gradient from the Equator to the Poles that creates a global circulation exchanging energy from low to high latitudes (ocean currents, general atmospheric circulation)

  1. The highest daily total \(I_{ex}\) is found at the poles with \(47\) \(\,\rm{MJ}\,\rm{m}^{-2}\,\) \(\rm{day}^{-1}\) during the summer solstice. Why?

This is because the solar irradiance reaches the poles during the full 24h cycle.

  1. For Vancouver, the highest instantaneous \(I_{ex} \approx 1190\,\rm{W}\,\rm{m}^{-2}\) is found on the day of the summer solstice (Jun 22) at 12 LAT. When is the lowest value?

The lowest daily maximum would be on the winter solstice. But the annual minimum is (\(0\,\rm{W}\,\rm{m}^{-2}\)), which occurs during every nighttime period.

  1. Using the radiation conservation equation we know that radiation of wavelength \(\lambda\) incident upon a substance must either be transmitted through it, be reflected from its surface, or be absorbed. If a body is opaque and has an absorptivity coefficient for of 0.75 for \(\lambda=0.55 \mu m\): A. what is \(\alpha_\lambda\) for \(\lambda=0.55 \mu m\), and B. what is \(\alpha_\lambda\) for \(\lambda=0.65 \mu m\)>

\[ \zeta_\lambda + \Psi_\lambda + \alpha_\lambda = 1 \tag{1}\]

A. If the body is opaque, then \(\Psi_{\lambda} = 0\) and: \[ \begin{eqnarray} \alpha_{\lambda} + \zeta_{\lambda} &=& 1 \nonumber \\ \alpha_{\lambda} &=& 1 - \zeta_{\lambda} \end{eqnarray} \]

so the reflectivity is \(\alpha_{\lambda} = 1 - 0.75 = \underline{0.25}\).

B. There isn’t enough information to answer the questions - the conservation equation applies per wavelength. It would be a reasonable assumption that the reflectivity at \(\lambda = 0.65 \mu m\) is similar to \(\lambda = 0.55 \mu m\) if you don’t have any further information, but there is no guarantee they are the same.

  1. If incident PAR (Photosynthetically Active Radiation) is \(800\,\mu \rm{mol}\,\rm{s}^{-1}\) and plant canopy has the following characteristics: \(\Psi_{PAR}\) = 0.08 and \(\alpha_{PAR}\) = 0.11. How many mols of PAR will 1 m-2 the plant canopy absorb in one hour?

We use Equation 1 and solve for absorptivity: \[\begin{eqnarray} \zeta_{\rm{PAR}} = 1 - (\Psi_{\rm{PAR}} + \alpha_{\rm{PAR}}) \label{E3} \end{eqnarray}\] This results in \(\zeta_{\rm{PAR}} = 1 - (0.08 + 0.11) = 0.81\). Hence, the fraction of the incident radiation of \(800\,\mu \rm{mol}\,\rm{s}^{-1}\,\rm{m}^{-2}\) that is absorbed is 81%: \[\begin{eqnarray*} 0.81 \times 800\,\mu \rm{mol}\,\rm{s}^{-1}\,\rm{m}^{-2} = \underline{648\,\mu \rm{mol}\,\rm{s}^{-1}\,\rm{m}^{-2}} \\ \end{eqnarray*}\]

There are 3600 seconds in an hour and 1,000,000 \(\mu mol\) per mol, so assuming PAR doesn’t change over this period:

\[ 648\,\mu \rm{mol}\,\rm{s}^{-1}\,\rm{m}^{-2} * 3600 \frac{s}{h} * \frac{1 mol}{1000000 \mu mol} = 2.3328\, mol\, \rm{m}^{-2} \]

  1. If \(I_{ex} = 562\,\rm{W}\,\rm{m}^{-2}\), \(SW_\downarrow = 298\,\rm{W}\,\rm{m}^{-2}\) and the sun angle \(\beta = 23.74^{\circ}\), determine the bulk transmissivity of atmosphere:

\[ SW_\downarrow = I_{ex} \Psi_a^{m} \tag{2}\]

where

\[ m = \small\frac{1}{\cos(Z)} = \small\frac{1}{\sin(\beta)} \]

we use \(\beta = 23.74^{\circ}\) and \(\sin \beta = 0.403\) from Study Question Set 4, Question 4 and get \(m\) = 2.481. We insert this in Equation 2:

\[\begin{eqnarray} \Psi_a &=& \left( \frac{298\,\rm{W}\,\rm{m}^{-2}}{562\,\rm{W}\,\rm{m}^{-2}} \right) ^{\frac{1}{2.481}} = \underline{0.77} \end{eqnarray}\]

  1. We use \(\Psi_a = 0.77\) from determine \(SW\downarrow\) at the surface if \(I_{ex} = 439 \,\rm{W}\,\rm{m}^{-2}\) and the zenith angle \(Z = 71.6^{\circ}\):

Following Equation 2:

\[\begin{eqnarray} K_{\downarrow} &=& 439 \,\rm{W}\,\rm{m}^{-2} \times 0.77^{3.17} = \underline{191 \,\rm{W}\,\rm{m}^{-2}} \end{eqnarray}\]