Question

A 200 g/l solution of common salt was discharged into a stream at a constant rateof 0.025 cms. Thebackground concentration of the salt in the stream water was found to be 10 ppm. At a downstreamsection where the solution was believed to have completely mixed, the salt concentration was found toreach an equilibrium value of 45 ppm.

Required:
Estimate the discharge in stream

Answer 1

142.825 m³/sec

Step-by-step explanation:

Let the discharge of this stream = Q1

200g/L = 200x10³mg/L

Q2 = 200x10³ ppm

C1 = 10

C2 = 0.025

Qmix * Cmix = c1Q1 + c2Q2

(Q1+0.025)45 = 10xQ1 + 0.025*200x10³

45Q1 + 1.125 = 10Q1 + 5000

45Q1 -10Q1 = 5000 - 1.125

35q1 = 4998.875

Q1 = 4998.875/35

Q2 = 142.825 m³/sec

142.825 m³/sec is therefore the estimated discharge from the stream.