Respuesta :
Answer
given,
Q u  = 8.7 m³/s
Q d= 0.9 m³/s
BOD concentration = 50 mg/L
a) BOD concentration at the down stream
[tex]C_{down}=\dfrac{0.9\times 50}{8.7+0.9}[/tex]
[tex]C_{down}=\dfrac{0.9\times 50}{9.6}[/tex]
        = 4.69 mg/L
b) discharge = 9.6 m³/s
cross sectional area = 10 m²
velocity steam = [tex]\dfrac{8.7+0.9}{10}[/tex]
            = 0.96 m/s
time taken to move 50 km down stream =[tex]\dfrac{50 \times 1000}{0.96}[/tex]
                                 = 52083.3 s
                                 = [tex]\dfrac{52083.3}{3600\times 24}[/tex]
                                  = 0.6 days
now,
[tex]C_t=C_0e^{-kt}[/tex]
[tex]C_t=4.6875\ e^{-0.2\times 0.6}[/tex]
[tex]C_t = 4.16 mg/L[/tex]
