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202                                                  Soil and Water Contamination


                     Box 11.IV  (continued) Physics of surface water flow

                     the case in rivers. Using Q = u ⋅ A, Equations (11.IVa) and (11.IVb) can be solved for Q.
                     Because these Saint-Venant equations are not amenable to analytical solutions except in a
                     few simple cases, the equations are usually solved numerically.



                       Example 11.3  Advection

                                                             -1
                       In a river with a mean flow velocity of 0.80 m s , the chloride  concentration increases
                       due to increasing wastewater discharge from an upstream salt mine at location x = 0 km.
                                                                                -1
                                                                                           -1
                       At x = 4 km, the chloride concentration increases linearly from 20 mg l  to 140 mg l
                       between 4 a.m. (04.00 h) until 4 p.m. (16.00 h). Calculate the chloride concentration at
                       x = 40 km and 8 p.m. on the same day.
                       Solution
                                                      -1
                                                             -1
                       First, convert the flow velocity from m s  to km h :
                                 -1
                                           -1
                                                    -1
                       ū  = 0.80 m s  = 0.80 m s  × 3600 s h  = 2.880 km h -1
                        x
                       Next, assume that t  is the time that the chloride  concentration at location x = 4 km is
                                      0
                       equal to the concentration at x = 40 km and 8 p.m. Thus, the initial condition at x = 4
                       km and t = t  is:
                                0
                       C  , 4 ( t  )    C  ) 4 (
                           0       0
                       Use equation (11.15) to define the concentration at x = 40 km at t = t (8 p.m.):
                       C ( 40 ,t )    C 0  ( 40  . 2  880  (t  t  0 ))
                       Because we have assumed that C(4,t ) = C(40,t), we may write:
                                                  0
                       C 0  ) 4 (     C 0 ( 40  . 2  880  (t  t 0 ))
                       4     40  . 2  880  (t  t  0 )

                        . 2  880  (t  t  )    36
                                0
                       t  t     12  5 . h
                          0
                       Thus, it takes 12.5 hours for the water to travel from location x = 4 km to location x = 40
                       km. This can also be calculated directly from Δt = Δx/ū . From the calculation it appears
                                                                  x
                       that t  is 12.5 hours before 8 p.m. = 7.30 a.m., which is 3.5 hours after the chloride
                           0
                       concentration started increasing. At location  x  = 4 km, the chloride concentration
                                                                            -1
                                                    -1
                                                                         -1
                                         -1
                       increases from 20 mg l  to 140 mg l  in 12 hours, so 10 mg l  h . This implies that
                                                                       -1
                                                                                       -1
                                                                     -1
                       after 3.5 hours, the concentration had increased by 10 mg l  h  × 3.5 h = 35 mg l . The
                       chloride concentration at location x = 0 and t = t  is
                                                            0
                                             -1
                       C  , 4 ( t  )    20    35    55 mg l .
                           0
                       This is also the concentration at x = 40 km and 8 p.m.






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        Soil and Water.indd   214                                                           10/1/2013   6:44:51 PM
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