Answer:
the initial deficit of oxygen of the mixture just downstream of the discharge point = 2.44 mg/L
Explanation:
From the diagram ; we determine the dissolved oxygen just downstream by using the equation
[tex]DO = \frac{Q_wL_w+Q_rL_r}{Q_w+Q_r}[/tex]
where;
[tex]Q_w[/tex] = volumetric flow rate of waste water = 1 m³/s
[tex]Q _r[/tex] = volumetric flow rate of river just upstream of discharge point = 10 m³/s
[tex]L_w[/tex] = ultimate BOD of waste water = 4 mg/L
[tex]L_r[/tex] = ultimate BOD of mixture of the stream water and waste water = 8 mg/L
∴ DO = [tex]\frac{(1 \ m^3/s)(4 \ mg/L ) + ( 10 \ m^3/s) ( 8 \ mg/L)}{(1 \ m^3/s)+(10 \ m^3 / s)}[/tex]
DO = 7.64 mg/L
Now; to calculate the initial deficit of oxygen; we use the formuale
[tex]D_0 = DO _{sat} - DO[/tex]
where;
[tex]DO_{sat}[/tex] = 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,
[tex]D_0 = DO _{sat} - DO[/tex]
[tex]D_0 = 10.08 \ mg/L - 7.64 \ mg/L[/tex]
[tex]D_0 = 2.44 \ mg/L[/tex]
Thus; the initial deficit of oxygen of the mixture just downstream of the discharge point = 2.44 mg/L
