Question

A system of seven river segments is shown. Given: $Q_1=5$ m$^3$/s, $Q_2=15$ m$^3$/s, $Q_4=3$ m$^3$/s (outflow), $Q_6=8$ m$^3$/s; $C_1=8$ kg/m$^3$, $C_2=12$ kg/m$^3$, $C_6=10$ kg/m$^3$. Assume complete mixing at junctions, no gains/losses otherwise, steady state. Find the pollutant concentration in segment $R_7$ (in kg/m$^3$, rounded to two decimals). \includegraphics[width=0.5\linewidth]{image555.png}

Hint:

For conservative pollutants at steady state, use mass conservation at each junction:
$Q_{\text{out}}C_{\text{mix}}=\sum Q_{\text{in}}C_{\text{in}}$.
Withdrawals simply reduce flow but keep the same concentration as the upstream segment.

Answer 1

Step 1: Mix $R_1$ and $R_2$ to form $R_3$.
$Q_3=Q_1+Q_2=5+15=20$ m$^3$/s.
$C_3=\dfrac{Q_1C_1+Q_2C_2}{Q_1+Q_2} =\dfrac{5\cdot 8+15\cdot 12}{20} =\dfrac{40+180}{20}=11$ kg/m$^3$.

Step 2: Account for withdrawal $Q_4$ to form $R_5$.
Outflow $Q_4=3$ m$^3$/s leaves with concentration $C_3$ (same water).
$Q_5=Q_3-Q_4=20-3=17$ m$^3$/s, $C_5=C_3=11$ kg/m$^3$.

Step 3: Mix with inflow $R_6$ to form $R_7$.
$Q_7=Q_5+Q_6=17+8=25$ m$^3$/s.
$C_7=\dfrac{Q_5C_5+Q_6C_6}{Q_7} =\dfrac{17\cdot 11+8\cdot 10}{25} =\dfrac{187+80}{25} =\dfrac{267}{25}=10.68$ kg/m$^3$.
\[ \boxed{C_{R_7}=10.68\ \text{kg/m}^3} \]