1. (soluble pesticide discharged to well-mixed pond)

(volatilization flux density)

(volume flux into and out of pond)

Mass Balance:

( )

Since we are assuming instantaneously well-mixed conditions, we can find the initial

concentration as follows:

( )

( )( )

The eigenvalue is given by

( )

( )( ) ( )

( )

Thus,

[ ] [ ]

The 95% response time, i.e., the time it takes for the concentration to be reduced by 95%, is

found as follows:

( )

( ) ( ) ( )

( ) ( )

( )

2. Well-mixed lake

( ) (steady-state concentration of total

phosphorous (TP) pre-1994)

(TP load starting in 1994)

( )

( )

( )

Pre-1994 Mass Balance:

A background load must be introduced because the steady-state concentration is nonzero. It

could be upstream loading (i.e., an advective influx), but its nature is not important. From

above,

( )

The hydraulic residence time , eigenvalue of the homogeneous problem , and background

loading are found as follows:

( )

( )

( )

( )

( )( )( )

1994-On Mass Balance:

Since , , , and are independent of time, we may rewrite this as

(

) (

)

(

)

where we have made use of the expression for found above, and the subscript indicates

the initial value.

Rearranging, we find

( )

( )

( )( ) ( )

[ ] ( [ ]) [ ]

3.

( )

( )

( )

( )

[ ] ( )( )

The pre-1997 balance here is the same as the pre-1994 balance found above, so we may carry

our conclusions over:

1997-On Mass Balance:

5 10 15 20t years since 1994

5

10

15

c g L

The general solution to this equation is the sum of the general solution to the homogeneous

equation and a particular solution to the inhomogeneous equation. We already know the

solution to the homogeneous equation:

( )

where is the initial value of the particular inhomogeneous solution.

To find the inhomogeneous solution, we assume it is of the form

( ) ( )

We see that

( )

Thus,

( )

( )

[ (

( ) )]

( )

The full solution is therefore

( )( )

( )

( )

( )( ) ( )

[ ] ( [ ] [ ])

4. ( )

(first-order decay rate)

peak discharge on Oct. 1

minimum discharge on April 1

mean discharge = ( )

discharge range = ( )

We model the pollutant discharge as a sinusoid:

5 10 15 20t years since 1997

50

100

150

200

c g L

2 4 6 8 10t months since January

1

2

3

4

L g yr

Mass Balance:

Again, the equation is solved by adding the general homogeneous and a particular

inhomogeneous solution. The general homogeneous solution is

( )

A particular inhomogeneous solution is found by assume a solution of the form

:

( )

( ) ( ) (

) ( )

We see that

From the last two equations we find

(

)

(

)

Thus,

(

)(

)

(

)

[ (

(

))] (

(

))

Thus,

( )

(

)[

( ) ]

( )( )

( )

( )

( )( )

(

)

( )

( ) [ ( )

] ( )

[ ] ( [ ]) [ ( [ ] [ ]) [ ]]

In the steady-state limit, the transients (terms ) have vanished, leaving

(

)(

)

2 4 6 8 10t years

50

100

150

200

250

c g yr

The average value is given by

⟨ ⟩

In the following plot we compare

(

)(

) to

(

) . In the second expression, we have retained only the dominant term in the

sinusoidal portion:

We see that the solution rapidly converges to the second expression above, and we may neglect

the sine term in approximating the phase shift and amplitude of the steady-state response.

5. d

2 4 6 8 10t years

50

100

150

200

250

c g yr