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(a)  Yes (21.33 mg/L ≈ 21 mg/L)

(b) 15.5 metric ton/d ≈ 16 metric ton/d

Explanation:

q1= Volumetric flow rate of stream

q2= Volumetric flow of road runoff pipe

p1= Chloride concentration of stream

p2 =Chloride concentration of road runoff pipe

p3 = Chloride concentration of mixture downstream ( mixture of stream and road runoff pipe)

p4 = Standard Chloride concentration

From the question:

q1= 10 m3/s or 10,000 L/s      p1 = 12 mg/L          p3 = ?

q2= 5 m3/s or 5,000 L/s        p2=  40 mg/L        p4 = 20 mg/L

(a) Using the Law of Conservation of Mass (Chloride):

(p1 x q1) + (p2 x q2) = (q1 + q2) p3 (1)

( 12 x 10,000) + (40 x 5,000) = ( 10,000 + 5,000) x p3

120,000 + 200,000 = 15,000 x p3

320,000 = 15,000 x p3

p3  = 320,000/15,000

     =  21.33

    ≈ 21 mg/L

This concentration is above the standard quality concentration for Chloride.

(b) Rearranging  equation (1), and making  the Road Runoff pipe (q1 x p1)  the subject of the formula, we have  :

Runoff pipe (max) =  p4 x (q1 +q2) - (q1 x p1)

                             =  20 x (10,000 + 5,000) -  (10,000 x 12)

                             =  20 x ( 15,000) - 120,000

                            =   300,000 -  120,000

                            =  180,000 mg/s

                            =  180 g/s

                            = 0.18kg/s

                           =0.00018 metric ton/ s

To convert  to metric ton/d  we multiply by 86,400

                          = 0.00018 x 86,400 metric ton/d

                         = 15.552 metric ton/d

                         ≈ 16 metric ton/d