Part 6 Surface Water Quality
Part 6 Surface Water Quality
Part 6 Surface Water Quality
- O2 removal (Deoxygenation)
- O2 supply from atmosphere (Reaeration)
Deoxygenation:
Rate..of . . Deoxygenation=K d Lt
Kd: is deoxygenation rate constant (day-1)
Lt: BOD remaining for t days after the waste enter the river (mg/L)
− Kd t
BUT
Lt =L0 e
−K d t
Rate . .of . . Deoxygenation=K d L 0 e
L0: BOD of a mixture of stream water and wastewater (mg/L)
Qw
Qr Lw
Lr
Modeled as
Plug flow
Q w Lw + Q r Lr
L0 =
Q w +Qr
Qw = 1.1 m3/s
Lw = 50 mg/L
BOD??
Qr =8.7 m3/s
Lr = 6.0 mg/L
Reaeration
Q w DO w +Q r DOr
D 0 =DO S −
Qw +Q r
D0: initial O2 deficit of the mixture of river and wastewater
DOS: saturated value of DO in water at river temperature
DOw: DO of wastewater
DOr: DO of river before discharge point
Example: if the wastewater DO is 2 mg/L and a discharge rate 1.1 m3/s. the
river DO is 8.3 mg/L and flow rate is 8.7 m3/s at to= 20oc. Assuming
complete mixing estimate the initial DO deficit of the mixture
Solution:
∂D −K t
=K d Lt −K r D=K d L0 e d −K r D
∂t
The final solution is
K d L0
D= (e− K t −e−K t ) +D0 e−K t
d r r
K r −K d
BUT
D = DOS-DO
DO = DOs – [D]
DO=DOs−
K r −K d
( e
(
K d L0 − K d t −K r t
−e ) + D0 e
−K r t
)
This equation is known as Oxygen
Sag equation
A plot of this equation gives an oxygen Sag Curve
At the critical point downstream, dissolved oxygen reaches its minimum value and
river condition are at the worst situation. Beyond the critical point, the remaining
organic matter in the river has diminished to the point where oxygen is being added to
the river by reaeration faster than it is being withdrawn by decomposition, and the
river begins to recover
NOTE: When Kd =Kr
−K d t
D=( K d L0 t+ D 0 )e
NOTE: t can be written in the form (x/u)
x: distance downstream
u: stream velocity
t: elapsed time between discharge and point and distance x downstream
t c=
1
K r −K d
Ln
Kr
Kd
1−
[ (
D 0 ( K r −K d )
Kd L0 )]
tc: is critical time
Temperature effect on the oxygen sag curve
Solution:
Water quality in Lakes:
All lakes gradually accumulate silt and organic matter (OM) as they undergo
a natural aging process known as Eutrophication
As the accumulated sediments causes the lake to get shallower and warmer,
more plants take root along the shallow edges and the lake slowly
transformed into a marsh or bog
This is a natural process that takes thousands of years. BUT the rate can
be accelerated through human activities “Cultural Eutrophication”.
Q (m3/s)
Cin (g/m3)
Q (m3/s)
A (m2) C (g/m3)
C (g/m3)
Q
A (m2)
Q
v
Settling rate ( s)
Using MB approach
Rate of addition of P = Rate of removing of P
Q Cin + S = Q C +vs A C
S: adding from point source
C: concentration in lake
Vs: settling rate velocity (m/s)
NOTE:
QC in +S
C=
Q+ v s A
To keep the Conc. of P in the lake at an acceptable level of 0.01 mg/L, we can find the
rate of P removal required at the wastewater treatment plant S
S=C [ Q+ v s A ]−QC in
Example: phosphorous loading in lake
A phosphorous –limited lake with surface area equal to 80x106 m2 is fed by a 15 m3/s
stream that has a phosphorous concentration of 0.01 mg/L. in addition, effluent from a
point source adds 1.0 gr/s of phosphorous. The phosphorus settling rate is estimated at 10
m/yr.
a) Estimate the average total phosphorus concentration
b) What rate of P removal at the wastewater treatment plant would be required to
keep the concentration of P in the lake at an acceptable level of 0.01 mg/L?
Solution: