Answer:
(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
