Water Quality Management

Topics: Eutrophication, Water pollution, Algal bloom Pages: 10 (1826 words) Published: September 17, 2013
RIVER WATER QUALITY MANAGEMENT

River Ganga in upper Stretch

River Ganga in Middle Stretch

River Yamuna in upper Stretch

River Yamuna in Middle Stretch

Addition of several drains into the river Yamuna

Water Quality Management in Rivers

Dissolved Oxygen Depletion

Dissolved Oxygen Sag Curve

Mass Balance Approach
• Originally developed by H.W. Streeter and E.B. Phelps in 1925 • Oxygen is depleted by BOD exertion

• Oxygen is gained through reaeration

Selecting System Boundaries

Initial Mixing
Qw = waste flow (m3/s) DOw = DO in waste (mg/L) Lw = BOD in waste (mg/L) Qr = river flow (m3/s) DOr = DO in river (mg/L) Lr = BOD in river (mg/L) Qmix = combined flow (m3/s) DO = mixed DO (mg/L) La = mixed BOD (mg/L)

Remember: We have to Use Ultimate BOD for DO predictions

1. Determine Initial Conditions
a. Initial dissolved oxygen concentration Qw DOw  Qr DOr DO  Qw  Qr b. Initial dissolved oxygen deficit

D  DOs  DO

where D = DO deficit (mg/L) DOs = saturation DO conc. (mg/L)

Qw DOw  Qr DOr Da  DOs  Qmix

1. Determine Initial Conditions
DOsat is a function of temperature. Values can be found in Table (Gilbert Masters) c. Initial ultimate BOD concentration

Qw Lw  Qr Lr La  Qw  Qr

2. Determine Reaeration Rate
a. O’Connor-Dobbins correlation 1/ 2 3.9u kr  3 / 2 h where kr = reaeration coefficient @ 20ºC (day-1) u = average stream velocity (m/s) h = average stream depth (m) b. Correct rate coefficient for stream temperature

kr  kr , 20

T  20

where Θ = 1.024

Determine the De-oxygenation Rate
a. rate of deoxygenation = kdLt where kd = deoxygenation rate coefficient (day-1) Lt = ultimate BOD remaining at time (of travel downstream) t b. If kd (stream) = k (BOD test)

Lt  L0e
and

 kd t

rate of deoxygentation  kd L0e kd t

3. Determine the De-oxygenation Rate
c. Correct for temperature

kr  kr , 20

T  20

where Θ = 1.135 (4-20ºC) or 1.056 (20-30ºC)

4. DO as function of time
• Mass balance on moving element dD  kd Lt  kr D dt • Solution is kd La kd t kr t Dt  e e  Da e kr t kr  kd





 

5. Calculate Critical time and DO
 kr  k r  k d  1  tc  ln  1  Da kr  kd  kd  k d La   

kd La kd tc  k r tc  k r tc Dc  e e  Da e kr  ka





Example
• A city of 200,000 people discharges 1.32 m3/s of treated sewage having an ultimate BOD of 28.0 mg/L and 1.8 mg/L DO into a river with a flow of 9.0 m3/s and velocity of 0.36 m/sec. Upstream of the discharge point, the river has a BOD of 3.6 mg/L and a DO of 7.6 mg/L. The saturation DO is 8.5 mg/L, kd = 0.61 day-1, and kr = 0.76 day-1. Determine a) the critical DO and critical distance, and b) the DO at 17 km downstream.

1.Determine Initial Conditions
a. Initial dissolved oxygen concentration of mixed condn

Qw DOw  Qr DOr DO  Qw  Qr

1.81.32  7.69.0  6.85 mg DO  1.32  9 L
b. Initial dissolved oxygen deficit (just after mixing)

D  DOs  DO
mg Da  8.5  6.85  1.6 L

1. Determine Initial Conditions
c. Initial ultimate BOD concentration (Mix condition)

Qw Lw  Qr Lr La  Qw  Qr
La

281.32  3.69  6.75 mg 
10.32 L

It will decay by following first order kinetics

Step 1. Variations
• Saturation DO & BOD ult given

2. Determine Re-aeration Rate
• kr = 0.76 day-1 given • no need to calculate from stream geometry • assume given value is at the stream temperature (since not otherwise specified), so no need to correct

3. Determine the Deoxygenation Rate
• kd = 0.61 day-1 given • no need to calculate corrections from stream geometry • assume given value is at the stream temperature (since not otherwise specified), so no need to correct

5. Calculate Critical Time and DO
 kr  k r  k d  1  tc  ln  1  Da   kr  kd  kd  k d La   0.76  1 0.76  0.61  1  1.6  tc  ln   0.616.75  0.76  0.61  0.61  

tc  1.07 day

5....
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