Solar radiation incident on a large body of clean water $\left(k=0.61
\mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\right.\( and \)\left.\alpha=0.15
\times 10^{-6} \mathrm{~m}^{2} / \mathrm{s}\right)$ such as a lake, a river,
or a pond is mostly absorbed by water, and the amount of absorption varies
with depth. For solar radiation incident at a \(45^{\circ}\) angle on a
\(1-\mathrm{m}\)-deep large pond whose bottom surface is black (zero
reflectivity), for example, \(2.8\) percent of the solar energy is reflected
back to the atmosphere, \(37.9\) percent is absorbed by the bottom surface, and
the remaining \(59.3\) percent is absorbed by the water body. If the pond is
considered to be four layers of equal thickness \((0.25 \mathrm{~m}\) in this
case), it can be shown that \(47.3\) percent of the incident solar energy is
absorbed by the top layer, \(6.1\) percent by the upper mid layer, \(3.6\) percent
by the lower mid layer, and 2.4 percent by the bottom layer [for more
information, see Cengel and Ozi?ik, Solar Energy, 33, no. 6 (1984), pp.
581-591]. The radiation absorbed by the water can be treated conveniently as
heat generation in the heat transfer analysis of the pond.
Consider a large \(1-\mathrm{m}\)-deep pond that is initially at a uniform
temperature of \(15^{\circ} \mathrm{C}\) throughout. Solar energy is incident on
the pond surface at \(45^{\circ}\) at an average rate of $500 \mathrm{~W} /
\mathrm{m}^{2}$ for a period of 4 h. Assuming no convection currents in the
water and using the explicit finite difference method with a mesh size of
\(\Delta x=0.25 \mathrm{~m}\) and a time step of \(\Delta t=15 \mathrm{~min}\),
determine the temperature distribution in the pond under the most favorable
conditions (i.e., no heat losses from the top or bottom surfaces of the pond).
The solar energy absorbed by the bottom surface of the pond can be treated as
a heat flux to the water at that surface in this case.
