Page 205 - Soil and water contamination, 2nd edition

192 Soil and Water Contamination

Example 11.1 Load calculation
-1
The mean annual ortho-phosphate concentration in a river is 0.04 mg l . The mean
-1
3
annual discharge of the river is 8.2 m s . Calculate the annual load of dissolved
-1
inorganic phosphorus in tonnes y .
Solution
First, calculate the inorganic phosphorus concentration from the ortho-phosphate
3-
concentration (compare Equation 11.2). The molar mass of ortho-phosphate (PO )
4
is 31 (from the P; see Appendix I) + 4 × 16 (from O) = 95. The inorganic phosphorus
load is
-3
3 -1
Dissolved inorganic P load = 31/95 × 0.04 g m × 8.2 m s = 0.107 g s -1
= 0.107 × 31.536 = 3.37 tonnes y -1
Note that the concentration may fluctuate considerably during the year, often depending
on the discharge of the river. If this is the case, multiplying the average concentration by
the average discharge gives inaccurate results for the annual load . For more details on this,
see Section 18.3.

In Chapter 10 we saw that the conservation law of mass imposes that changes in the total
mass in a store , or control volume, should be accounted for in terms of inputs and outputs .
A control volume is a soil or water body with clearly defined boundaries and, as mentioned
before, can consist of a functional unit, a morphological unit, or an arbitrary box anywhere
on Earth. It can also be as small as an infinitesimally thin slice of water in a river. Water can
be viewed as a conservative substance with only inputs and outflows from an adjacent water
body. Thus, the same principles apply for other conservative substances (i.e non-reactive or
inert) transported by water, such as chloride . Changes in the total mass of non-conservative
substances may also be due to physical and biogeochemical transformation . A mass balance is
simply the accounting of mass inputs, outputs, reactions, and change in storage:

ΔStorage = Inputs – Outflows ± Transformations (11.3)

If we consider, for example, the simple reservoir lake system (see Section 10.1), the
transformation term is zero, because water is a conservative substance. If we assume that the
specific density of water remains the same over time, the mass balance for water in the lake
could be expressed by the following algebraic difference equation:

V (Q - Q ) t (11.4)
in out
3
where ΔV = change in storage volume [L ], Q = input from feeding streams, groundwater,
in
3 -1
and precipitation [L T ], Q = output by discharging stream, seepage to groundwater, and
out
3
-1
evaporation [L T ], Δt = time increment [T]. If we assume that the surface area of the lake
does not change with the water level, the change in lake water depth H [L] can be obtained
2
by dividing both sides of the equation by the surface area A [L ]:
V (Q - Q )
H in out t (11.5)
A A
This difference equation can be made into a differential equation if we divide both sides of
the equation by Δt and take the limit as Δt → 0:

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