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

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

\(V\) = 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

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

\(h_c\) = 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.