1. [Oxygen depletion, Streeter-Phelps equation]. A company intends to discharge a treated wastewater at a volumetric flow rate of 25 % of the average river flow at a specific discharge point. The effluent is estimated to contain 25 mg/L BOD5 and 2 mg/L of dissolved oxygen. Assume there are no other discharges into the river. At a location just upstream the discharge point, the BOD5 in the river is 3 mg/L and the dissolved oxygen is 9 mg/L. The average velocity of the river water is 1.0 m/min and the average flow rate is 15 m³/min. The river has a mean depth of 1.5 m and the average river temperature is 17°C. The BOD degradation rate constant (k₁) is 0.230 d-¹ at 20°C. a) For the data given above, calculate the minimum dissolved oxygen concentration and the distance this occurs downstream from the discharge point; [Ans: Domin = 3.79 mg/L; distance @ DO min = 7.88km]

Introduction to Chemical Engineering Thermodynamics
8th Edition
ISBN:9781259696527
Author:J.M. Smith Termodinamica en ingenieria quimica, Hendrick C Van Ness, Michael Abbott, Mark Swihart
Publisher:J.M. Smith Termodinamica en ingenieria quimica, Hendrick C Van Ness, Michael Abbott, Mark Swihart
Chapter1: Introduction
Section: Chapter Questions
Problem 1.1P
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Required formulae
• Re-aeration:
f (mm/h) = 78200 · [U (m/s)] 0.67 · [H(mm)]−0.85
fr = f20°C 0.98(20-T (°C))
K₂,7(h¯¹) =
• Substrate degradation:
K₁,T₂ = k₁,T₁ 0(T2-T₁), with T in °C, and 0=1.047 for T>20 °C and 0=1.135 for T<20 °C;
Calculation of DO at saturation:
DO* (3)=
▪
D(t)
=
Dmax
tmax
• Streeter-Phelps equation and derived equations
k₁ BODu,c
k₂-k₁
fr (mm/h)
H(mm)
-
▪
0.0686+0.00188432 T+0.00000612.T²
[e-k₁t - e
1
k₁. BODu,c
k₂
· e-k₂t] + Dc •
e-k₁tmax
·· e-k₂t
1
k₂
(k₂-k₁) Dc
-K₂ - k, lm / 1 (1 - K₁ - BOD auc
In
:)]
-
"
where T is in °C;
Transcribed Image Text:Required formulae • Re-aeration: f (mm/h) = 78200 · [U (m/s)] 0.67 · [H(mm)]−0.85 fr = f20°C 0.98(20-T (°C)) K₂,7(h¯¹) = • Substrate degradation: K₁,T₂ = k₁,T₁ 0(T2-T₁), with T in °C, and 0=1.047 for T>20 °C and 0=1.135 for T<20 °C; Calculation of DO at saturation: DO* (3)= ▪ D(t) = Dmax tmax • Streeter-Phelps equation and derived equations k₁ BODu,c k₂-k₁ fr (mm/h) H(mm) - ▪ 0.0686+0.00188432 T+0.00000612.T² [e-k₁t - e 1 k₁. BODu,c k₂ · e-k₂t] + Dc • e-k₁tmax ·· e-k₂t 1 k₂ (k₂-k₁) Dc -K₂ - k, lm / 1 (1 - K₁ - BOD auc In :)] - " where T is in °C;
1. [Oxygen depletion, Streeter-Phelps equation]. A company intends to discharge a treated
wastewater at a volumetric flow rate of 25 % of the average river flow at a specific discharge
point. The effluent is estimated to contain 25 mg/L BOD5 and 2 mg/L of dissolved oxygen.
Assume there are no other discharges into the river. At a location just upstream the discharge
point, the BOD5 in the river is 3 mg/L and the dissolved oxygen is 9 mg/L. The average velocity
of the river water is 1.0 m/min and the average flow rate is 15 m³/min. The river has a mean
depth of 1.5 m and the average river temperature is 17°C.
The BOD degradation rate constant (k₁) is 0.230 d-¹ at 20°C.
a) For the data given above, calculate the minimum dissolved oxygen concentration and the
distance this occurs downstream from the discharge point;
[Ans: DO min = 3.79 mg/L; distance @ DO min = 7.88km]
Transcribed Image Text:1. [Oxygen depletion, Streeter-Phelps equation]. A company intends to discharge a treated wastewater at a volumetric flow rate of 25 % of the average river flow at a specific discharge point. The effluent is estimated to contain 25 mg/L BOD5 and 2 mg/L of dissolved oxygen. Assume there are no other discharges into the river. At a location just upstream the discharge point, the BOD5 in the river is 3 mg/L and the dissolved oxygen is 9 mg/L. The average velocity of the river water is 1.0 m/min and the average flow rate is 15 m³/min. The river has a mean depth of 1.5 m and the average river temperature is 17°C. The BOD degradation rate constant (k₁) is 0.230 d-¹ at 20°C. a) For the data given above, calculate the minimum dissolved oxygen concentration and the distance this occurs downstream from the discharge point; [Ans: DO min = 3.79 mg/L; distance @ DO min = 7.88km]
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