Answer:
the initial deficit of oxygen of the mixture just downstream of the discharge point = 2.44 mg/L
Step-by-step explanation:
From the diagram ; we determine the dissolved oxygen just downstream by using the equation
![DO = (Q_wL_w+Q_rL_r)/(Q_w+Q_r)](https://img.qammunity.org/2021/formulas/engineering/college/gfuwqvsl1jymymeu0puil1d63n7702wzgg.png)
where;
= volumetric flow rate of waste water = 1 m³/s
= volumetric flow rate of river just upstream of discharge point = 10 m³/s
= ultimate BOD of waste water = 4 mg/L
= ultimate BOD of mixture of the stream water and waste water = 8 mg/L
∴ DO =
![((1 \ m^3/s)(4 \ mg/L ) + ( 10 \ m^3/s) ( 8 \ mg/L))/((1 \ m^3/s)+(10 \ m^3 / s))](https://img.qammunity.org/2021/formulas/engineering/college/hfjaag7mcifzbw5enbits9dht1ykomrkzd.png)
DO = 7.64 mg/L
Now; to calculate the initial deficit of oxygen; we use the formuale
![D_0 = DO _(sat) - DO](https://img.qammunity.org/2021/formulas/engineering/college/ip3n80iw0qrim18hq7t5ugirksoi1edp6t.png)
where;
= the saturated value of the dissolved oxygen
since no presence of chloride in the stream; chloride concentration = 0 mg/L
However, we obtain the saturated dissolved oxygen from Table - 5.10 " Solubility of oxygen in water at 1 atm corresponding to the temperature at 15° C and 0 mg/L concentration of chloride = 10.08 mg/L
SO,
![D_0 = DO _(sat) - DO](https://img.qammunity.org/2021/formulas/engineering/college/ip3n80iw0qrim18hq7t5ugirksoi1edp6t.png)
![D_0 = 10.08 \ mg/L - 7.64 \ mg/L](https://img.qammunity.org/2021/formulas/engineering/college/ozr51fg5cjci50z4mrdj3u6jebhy6wc2y7.png)
![D_0 = 2.44 \ mg/L](https://img.qammunity.org/2021/formulas/engineering/college/u9zmn935nh2xy607azrvf0m7h3qt5xme3p.png)
Thus; the initial deficit of oxygen of the mixture just downstream of the discharge point = 2.44 mg/L