McKay, Donald. "Front matter" Multimedia Environmental Models ...

©2001 CRC Press LLC
C 1 = C O/(1 + Vk/(Uhw)) = C O/(1 + kL/U)
Note the consistency of the dimensions, kL/U being dimensionless. The group (1 +
kL/U) has a value of 1.0175, thus C 1 is 0.983C O. 1.7% of the chemical is lost in
each segment. The same equation applies to the second reach, thus C 2 is 0.983C 1
or 0.983 2 C O. In general, for the nth reach, C n is (0.983) n C O or C O/(1 + kL/U) n .
One mile is reached when n is 8, and 10 miles corresponds to n of 80, thus C 8
is 0.983 8 C O or 4.45, and C 8 is 1.29. The half distance will occur when 0.983 n is 0.5,
i.e., when n is log 0.5/log 0.983 or 40, corresponding to 8000 m or 5 miles.
The Eulerian answer is thus slightly different. It could be made closer to the
Lagrangian result by carrying more significant figures or by decreasing L and
increasing n. An advantage of the Eulerian system is that it is possible to have
segments with different properties such as depth, width, velocity, volume, and
temperature. There can be additional inputs. The general equation employing the
group (1 + kL/U) n will not then apply, each segment having a specific value of this
factor. The mathematical enthusiast will note that L/U is t/n, where t is the flow time
to a distance nL m. The Lagrangian factor is thus also (1 + kt/n) n , which approaches
exp(kt) when n is large. It is good practice to do the calculation in both systems
(even approximately) and check that the results are reasonable. Some water quality
models of rivers and estuaries can have several hundred segments, thus it is difficult
to grasp the entirety of the results, and mistakes can go undetected.
2.6 STEADY STATE AND EQUILIBRIUM
In the previous section, we introduced the concept of “steady state” as implying
unchanging with time, i.e., all time derivatives are zero. There is frequent confusion
between this concept and that of “equilibrium,” which can also be regarded as a
situation in which no change occurs with time. The difference is very important and,
regrettably, the terms are often used synonymously. This is entirely wrong and is
best illustrated by an example.
Consider the vessel in Figure 2.1A, which contains 100 m 3 of water and 100 m 3
of air. It also contains a small amount of benzene, say 1000 g. If this is allowed to
stand at constant conditions for a long time, the benzene present will equilibrate
between the water and the air and will reach unchanging but different concentrations,
possibly 8 g/m 3 in the water and 2 g/m 3 in the air, i.e., a factor of 4 difference in
favor of the water. There is thus 800 g in the water and 200 g in the air. In this
condition, the system is at equilibrium and at a steady state. If, somehow, the air
and its benzene were quickly removed and replaced by clean air, leaving a total of
800 g in the water, and the volumes remained constant, the concentrations would
adjust (some benzene transferring from water to air) to give a new equilibrium (and
steady state) of 6.4 g/m 3 in the water (total 640 g) and 1.6 g/m 3 in the air (total
160 g), again with a factor of 4 difference. This factor is a partition coefficient or
distribution coefficient or, as is discussed later, a form of Henry’s law constant.
During the adjustment period (for example, immediately after removal of the air
when the benzene concentration in air is near zero and the water is still near 8 g/m 3 ),