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Oxygen Profile in Streams Assoc. Prof. Kozet YAPSAKLI
Oxygen depletion in streams
Two mechanisms are known to contribute oxygen to surface waters:
dissolution of oxygen from the atmosphere (reaearation)
production of oxygen by algal photosynthesis
Oxygen level in surface waters
Oxygen deficit is represented mathematically by
𝐷 = 𝑐𝑠 − 𝑐
For constant equilibrium conditions, i.e. Cs does not change, the rate of change in the deficit:
𝑑𝐷
𝑑𝑡= −
𝑑𝑐
𝑑𝑡
The deficit increases at the same rate that oxygen is used
The dissolved oxygen deficit is the main driving force for reaeration
Oxygen Deficit
DO sag definitions
Streeter-Phelps Model*
Mass Balance for the Model
Not a Steady-state situation
rate O2 accum. = rate O2 in – rate O2 out + produced – consumed
rate O2 accum. = rate O2 in – 0 + 0 – rate O2 consumed
Kinetics
Both reoxygenation and deoxygenation are 1st order
* Streeter, H.W. and Phelps, E.B. Bulletin #146, USPHS (1925)
The oxygen deficit in a stream is a function of
Oxygen utilization
Reaeration
𝑑𝐷
𝑑𝑡= 𝑟𝐷 + 𝑟𝑅
rD=Rate of BOD exertion
rR=Rate of reaeration
Sag Curve
Kinetics for Streeter-Phelps Model
• Deoxygenation
L = BOD remaining at any time
dL/dt = Rate of deoxygenation equivalent to rate of BOD removal
dL/dt = -k1L for a first order reaction
k1 = deoxygenation constant, f’n of waste type and temp.
kLdt
Ld
][
C
C
t
dtkL
dL
0 0
ktkt eLLeL
Lorkt
L
L 0
00
ln
Developing the Streeter-Phelps
Rate of reoxygenation = k2D
D = deficit in D.O.
k2 = reoxygenation constant*
2
3
21
)20(21
2
025.19.3
H
vk
T
Where
– T = temperature of water, ºC
– H = average depth of flow, m
– ν = mean stream velocity, m/s
D.O. deficit
= saturation D.O. – D.O. in the water
There are many correlations for this. The simplest one, used here, is from O’Connor and Dobbins, 1958
or
Combining the kinetics
OR
Net rate of change of oxygen deficiency, dD/dt
dD/dt = k1L - k2D
where L = L0e-k1t
dD/dt = k1L0e-k1t - k2D
Integration and substitution
tk
o
tktko eDeekk
LkD 221 )(
12
1
tk
oc
tk
o
eLk
kD
DkeLkdt
dD
1
1
2
1
21 0
o
oc
Lk
kkD
k
k
kkt
1
12
1
2
12
)(1ln
1
The last differential equation can be integrated to:
It can be observed that the minimum value, Dc is achieved when dD/dt = 0:
, since D is then Dc
Substituting this last equation in the first, when D = Dc and solving for t = tc: