Question

) In the winter, a stream flows at 10 m3/s and receives discharge from a pipe that contains road runoff. The pipe has a flow of 5 m3/s. The stream's chloride concentration just upstream of the pipe's discharge is 12 mg/L, and the runoff pipe's discharge has a chloride concentration of 40 mg/L. Chloride is a conservative substance. (a) Does wintertime salt usage on the road elevate the downstream chloride concentration above 20 mg/L? (b) What is the maximum daily mass of chloride (metric tons/day) that can be discharged through the road runoff pipe without exceeding the water quality standard? Answer: yes (21 mg/L), 16 metric ton/d

Answer 1

(a) Yes (21.33 mg/L ≈ 21 mg/L)

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

Step-by-step 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