Page 206 - Soil and water contamination, 2nd edition

Substance transport 193
dH (Q in - Q out )
(11.6)
dt A

The output flux through the stream was given by Equation (10.1), so the equation can be
rewritten as:

dH Q in kV Q in (11.7)
- - kH
dt A A A
where H = the mean water depth [L]. Note that this equation is a first-order differential
equation, in which H represents the state variable, Q the upstream boundary condition, and
in
k a parameter that is constant within the system. This problem has an analytical solution ,
provided that Q is either constant over time or described by a continuous function of time.
in
If Q is constant over time, the analytical solution is (see Box 11.I):
in
H ) t ( Q in H Q in e kt (11.8)
k A 0 k A
The system is in steady state if the inputs and outputs are in balance and the water depth
does not change in time, thus:
dH
0 (11.9)
dt
Hence, the analytical steady state solution of the lake water depth in the reservoir lake reads:
Q in
H (11.10)
eq
k A
Example 11.2 Lake water depth and discharge

3
-1
A lake with a surface area of 4 ha is fed by a stream with a discharge of 3.8 m s . The
lake water depth is 5 m.
-5
a. Given k = 2·10 s will the lake level rise or fall?
b. Calculate the lake water depth after the response time .
c. Calculate what the discharge of the feeder stream must be to maintain a lake water
depth of 5 m.
Solution
a.
Use Equation (11.7) to calculate the change in the lake water depth:

dH 3.8 5 - 6 -1
2 10 5 5 10 m s
dt 4 10 4

The change is negative, so the lake water level will drop.
Another approach is to calculate the equilibrium lake water depth (Equation 11.10):
3.8
H eq . 4 75 m
2 10 5 - 4 10 4

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