9.9 Appendix I

Calculation of the Right-Hand Limit of the Climate Space
In their original paper, Porter and Gates (1969) state with regard to Equation (9.3), “… An estimate was made of $\overline{Q_a}$ as generally related to air temperature for value of absorptivity from 0.2 to 1.0.” Gates (1977) shows how to calculate $S$ and $s$ as a function of latitude, time of year, and time of day. A representative value of 40° was chosen for latitude and then values of $S$, $s$ were calculated under clear sky conditions in the late morning and afternoon. For $r$, a representative value of 0.15 was probably used. Curves of $T_a$ and $T_g$ as a function of time of day such as shown in Figure 9.2 were then taken from weather bureau statistics. $R_a$ and $R_g$ could then be calculated using the Stefan-Boltzmann law. All the numbers necessary to estimate $\overline{Q_a}$ are then available. The final step is to choose pairs of $T_a$ and $Q_a$ that are to be used. In Figure 9.3, 9:00-10:00 are hours of the day when this is true. This procedure is repeated at several latitudes and times of the year, from which the right-hand boundary can be derived.

Campbell (1977 pp.89-92) presents simplified equations to calculate the left- and right-hand boundaries of the climate space. He includes a correction factor to average the longwave radiation from the ground and the sky for the left-hand boundary. The direct beam and diffuse shortwave radiation fluxes are simply given for the right-hand boundary. The reflected shortwave component seems to be included in the diffuse term which at 25% of the direct flux is higher than Gates (1978) gives. The absorbed shortwave radiation is then added to the left-hand boundary values. Therefore as $T_a$ increases the shortwave component is constant using Campbell’s equations but using Gates’ method the shortwave flux increases.

9.9.1 Appendix II

Derivation of the Constant $k_s$ for the Convection Coefficient
Mitchell (1976) reported that the best overall relationship between the Reynolds and Nusselt numbers is given by

$$\begin{equation} Nu = 0.34Re^{0.6} \tag{9.19} \end{equation}$$ Recalling that Reynolds number $Re$ is the ratio of interim forces to viscous forces in the fluid, we write

$$\begin{equation} Re = \frac{VL}{\nu} \tag{9.20} \end{equation}$$

where
$V$ = the fluid velocity (m s^-1)
$L$ = the characteristic length (m), and
$\nu$ = the kinematic viscosity (m² s^-1)

The Nusselt number is a way to scale the rate of heat transfer as a function of wind velocity, size of the organism and fluid thermal diffusivity.
It can also be expressed as

$$\begin{equation} Nu = \frac{h_cL}{k} \tag{9.21} \end{equation}$$

where
$h_c$ = heat transfer coefficient (W m^-2 °C^-1),
$L$ = characteristic length (m), and
$k$ = thermal conductivity W m^-1 °C^-1).

Mitchell (1976) defined the characteristic length as

$$\begin{equation} L = (\frac{M_b}{\rho})^{1/3} \tag{9.22} \end{equation}$$

where $M_b$ is the mass kg
$\rho$ is the mass density kg m^-3.

Using these expressions we can solve for the heat transfer coefficient as a function of weight and wind velociy. Rearranging Equation (9.21) and substituting the Equation (9.19) for $Nu$, we have $$h_c = \frac{k}{L}(0.34Re^{0.6})$$ We can use Equations (9.20) and (9.22) to incorporate wind velocity and weight, respectively.

$$\begin{align*} h_c &= 0.34\frac{k}{L}(\frac{VL}{\nu})^{0.6} \\ &= \frac{0.34k}{\nu^{0.6}}V^{0.6}L^{-0.4} \\ &= \frac{0.34 \times k}{\nu^{0.6}}V^{0.6}(\frac{M_b}{\rho})^{-0.133} \\ &= \frac{0.34 \times k \times \rho^{-0.133}}{\nu^{0.6}}V^{0.6}M_b^{-0.133} \\ \end{align*}$$

Letting $k_s = \frac{0.34 \times k \times \rho^{0.133}}{\nu^{0.6}}$ if $\rho = 1 \times 10^3$ kg m^-3 which is the density of water $$k = 2.57 \times 10^{-2} W m^{-1} K^{-1}$$ $$\nu = 1.51 \times 10^{-5} m^2s^{-1}$$ at 20 °C then $k_s = 17.24$.

To see the error of assumng $k$ and $\nu$ at 20 °C we can compare the ratio of $\frac{k}{\nu^{0.6}}$.

Air temperature (°C)	$\frac{k}{\nu^{0.6}}$
-10	20.79
20	20.07
50	19.51

The difference over the 20 °C value is $\frac{1.28}{20.79} = 6.4$ percent.