16 Strea
Stream Water Quality Analysi
Dissolved Oxygen (DO)
Stream DO - DO Sag Curve
1
Water Quality Analysis_F11
16 Strea
Water Quality Analysis_F11
DO Deficit, D
D = Cs - C Do = Cs - Co Dc = Cs - Cc
C
where D = dissolved oxygen de icit, mg/L Cs = saturation concentra tion of dissolved oxygen, mg/L. See Table A-2 . Saturation values of dissolved oxygen in fresh water (DC, p. 868). C = actual concentration of dissolved oxygen, mg/L Do = initial DO deficit, g/L= DO deficit at t = 0 Dc = Critical DO deficit, mg/L Cc= Critical DO, mg/L Initial mixing of waste stream nd river 3
Qr = volumetric flow rate of the river, m /s 3 Qw = volumetric flow rate of wa stewater, m /s 3 Qm = volumetric flow rate of th river after mixing, m /s Cr = dissolved oxygen concentr tion in the river, mg/L Cw = Dissolved oxygen concent ration in the wastewater, mg/L Cm = dissolved oxygen concent ation in the river after mixing, mg/L Lr = ultimate BOD of the river, g/L Lw = ultimate BOD of the wast water, mg/L Lm = ultimate BOD of the river after mixing, mg/L Tr = temperature of the river, ºC Tw = temperature of the waste ater, ºC Tm = temperature of the river af er mixing, ºC
2
Qw Cw Lw Tw
Qr Cr Lr Tr
Qo Co Lo To
16 Strea
Water Quality Analysis_F11
Qw Cw Lw Tw
Mass Balance:
Mass in = Mass out For C, Qr Cr Lr Tr
Qw Cw + Qr Cr = Co (Qw + Qr) Qw Cw + Qr Cr Co = --------------------Qw + Qr
Qo Co Lo To
For L, Qw Lw + Qr Lr = Lo (Qw + Qr) Qw Lw + Qr Lr Lo = -----------------(Qw + Qr) For T, Qw Tw + Qr Tr To = -------------------Qw + Qr Heat balance H = m Cp
∆T
where H = change in enthalpy, J m = mass of substance, g Cp = specific heat at constant pressure, J/g K = 4.19 J/g K for natural water ∆T = change in temperature, K
Streeter-Phelps Model - The DO sag equation
D =
kd Lo ka
−
(e k
k t
− d
−
e
)
k t
− a
+
D (e
k t
− a
)
d
where D = dissolved oxygen de icit in river water after exertion of BOD at time t, mg/L. Do = initial deficit after ri ver and wastewater have mixed, mg/L. Lo = initial ultimate BO after river and wastewater have mixed, mg/L. -1 k d = deoxigenation rate c onstant, d . -1 k a = reaeration rate const nt, d . t = time of travel of wast water discharge downstream, d
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16 Stream Analysis_F11
Deoxygenation rate constant
k d = k + u η / H ........ Empirical equation where
-1
k d = deoxygenation rate constant at 20ºC, d -1 k = BOD rate constant determined in laboratory at 20ºC, d u = average stream velocity, m/s H = average depth of stream, m η = bed-activity coefficient (0.1 - 0.6) = 0.1 for stagnant or deep water = 0.6 for rapidly flowing streams Temperature Correction: k d, T = k d,20
θ T-20 where θ= 1.135
Reaeration Rate Constant, ka
k a = 3.9 u
0.5
/ H
1.5
where k a = (k r) = reaeration rate constant at 20ºC, day u = average stream velocity, m/s H = average depth, m Temperature Correction: k a,T = k a, 20 θ
T-20
where
θ = 1.024
Travel time, t
t=x/u where x = travel distance Time to the critical distance, t c
tc
=
1 k a −kd
k a ka − k d 1 D − o kd Lo kd
ln
4
-1
16 Stream Analysis_F11
Critical Deficit, Dc
Dc
=
kd Lo k a − k d
(
e
−
kd t c
−
e
−
ka tc
)
+
(
Do e
−
)
ka tc
Critical Dissolved Oxygen concentration, Cc
Cc = Cs - Dc
Critical Distance, xc
xc = tc u
Example 4-8 (DC, 305); Example 4-9 (DC, 307); Example 4-10 (DC, 311); Example 4-11 (DC, 312) . 3
The town of State College discharges 17,360 m /d of treated wastewater into the Bald Eagle Creek. -1 The treated wastewater has a BOD 5 of 12 mg/L and a k for BOD kinetics of 0.12 d in laboratory 3 study at 20°C. Bald Eagle Creek has a flow rate of 0.43m /s and an ultimate BOD of 5.0 mg/L. The DO of the river is 6.5 mg/L and the DO of the wastewater is 1.0 mg/L. Example 4-8 Calculate the DO and initial ultimate BOD after mixing. 2) (Example 4-9) Calculate the initial deficit of the Bald Eagle Creek after mixing with the wastewater from the town of State College. The stream temperature is 10ºC and the wastewater temperature is 10ºC. 3)
(Example 4-10) Determine the deoxigenation rate constant for the reach of Bald Eagle Creek below the wastewater outfall, discharge pipe. The average speed of the stream flow in the creek is 0.03 m/s. The depth is 5.0 m and the bed-activity coefficient is 0.35.
4) Determine the DO concentration at a point 5 km downstream from the State College discharge into the Balad Eagle Creek. Also determine the critical DO and the distance downstream at which it occurs (Example 4-11).
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16 Stream Water Quality Analysis_F11
(Solution) Given: 3
3
Qw = 17360 m /d (= 0.2 m /s) DOw = 1.0 mg/L Lw = (BOD5)w =12 mg/L Tw = 10ºC -1 k = 0.12 d @ 20 ºC)
3
Qr = 0.43 m /s Lr= 5 mg/L DOr =6.5 mg/L Tr = 10 ºC
Q = Qw + Qr = ? Lo = ? Co = ? To = ? 3
3
Qw = (17,360 m /d) (1 d/86,400 s/d) = 0.20 m /s
1) DO after mixing (Qw) (DOw) + (Qr)(DOr) Co = -----------------------------(Qw + Qr) 3
3
(0.2 m /s) (1.0 mg/L) + ( 0.43 m /s)(6.5 mg/L) = ---------------------------------------------------------- = 4.75 mg/L 3 3 ( 0.20 m /s + 0.43 m /s) -kt
BODt = BODL (1 - e ) where BODL= ultimate BOD L Convert BOD5 to BODL BODt 12 mg/L BODL = ------------- = -------------------- = 26.6 mg/L = L w - (0.12)(5) (1 - e-kt) 1-e
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16 Stream Water Quality Analysis_F11
Qw Lw + Qr Lr Lo = ------------------Qw + Qr 3
3
(0.20 m /s) (26.6 mg/L) + (0.43 m /s)(5.0 mg/L) Lo = -------------------------------------------------------------- = 11.86 mg/L 3 3 (0.20 m /s + 0.43 m /s) 2)
Given: Tr = 10ºC Tw = 10ºC To = 10ºC (Cs)r = 11.33 mg/L at To = 10ºC
(Table A-2; DC, p = 868)
(Co) = 4.75 mg/L from (1) Calculate the initial DO Deficit, Do: Do = Cs – Co = 11.33 mg/L - 4.75 mg/L = 6.58 mg/L
3) Deoxygenation rate constant, k d k d = k + (u η / H ) -1
where k d = deoxygenation rate constant at 20ºC, d -1 k = BOD rate constant determined in laboratory at 20ºC, d u = average velocity of stream flow, m/s H = average depth of stream, m η = bed-activity coefficient (0.1 - 0.6) = 0.1 for stagnant or deep water = 0.6 for rapidly flowing streams Given:
-1
k = 0.12 d u = 0.03 m/s H = 5.0 m η = 0.35
-1
k d = 0.12 d + (0.03 m/s)(0.35) / (5.0 m) = 0.1221 d
-1
Temperature Correction by: k d.T = k d,20 θ
T-20
where θ = 1.135 -1
k d,10 = (0.1221 d )(1.135)
10 - 20
(DC 293-294) -1
= (0.1221)(0.2819) = 0.03442 d
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16 Stream Water Quality Analysis_F11
Reaeration Rate Constant, ka
k a = 3.9 u
0.5
/ H
1.5
where
-1
k a = reaeration rate constant at 20 ºC,
day u = average stream velocity, m/s H = average depth, m Given: u = 0.03 m/s H = 5.0 m 0.5
1.5
k a,20 = 3.9 (0.03 m/s) / (5.0 m)
-1
= 0.0604 d
Temperature correction by k a,T = k r,20 θ
T-20
where
θ = 1.024
-1
10 - 20
k a,10 = (0.0604 d ) (1.024)
= 0.0477 d-1
Travel time, t
t=x/u Given: x = 5 km = 5000 m u = 0.03 m/s t = 5000 m/ [(0.03 m/s)(86400 s/d)] = 1.929 d 4) Calculate oxygen deficit (D) in river water after exertion of BOD for time, t, mg/L. D =
kd Lo k a−
(e k
k t
− d
−
e
−
ka t
)
+
Do ( e
−
ka t
)
d
where Lo = 11.86 mg/L k d = 0.03442 d
-1
-1
k a = 0.04766 d t = 1.929 d Do = 6.58 mg/L
D = (30.83) (0.9358 - 0.9122) + 6.58 (0.9122) = 6.7299 = 6.73 mg/L C = Cs - D = 11.33 mg/L - 6.73 mg/L = 4.60 mg/L (at 5 km downstream)
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16 Stream Water Quality Analysis_F11
Critical time, tc - time to the critical distance tc
=
1 ka
−
kd
k a ka − k d 1 − Do kd Lo kd
ln
Given: Lo = 11.86 mg/L -1 k d = 0.03442 d -1 k a = 0.04766 d Do = 6.58 mg/L tc = 6.45 day
Critical deficit, Dc
Dc
=
k d Lo k a − k d
(e
−
k d tc
−
e
−
k atc
)
+
Do ( e
−
)
k at c
Given: Lo = 11.86 mg/L -1 k d = 0.03442 d -1 k a = 0.04766 d Do = 6.58 mg/L tc = 6.45 day Dc = 6.85 mg/L Critical DO (Cc)
Cc = Cs - Dc = 11.33 mg/L - 6.85 mg/L = 4.48 mg/L
Critical Distance, xc
xc = tc u where tc = 6.54 days = (6.45 day)(86400 s/day) =557280 s u = 0.03 m/s = (0.03 m/s)( 1 km/ 1000 m) = 3 x 10 -5 km/s -5
xc = tc u = (557280 s)( 3 x 10 km/s) = 16.7 km The critical DO occurs downstream at a distance of 16.7 km from the wastewater discharge point.
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16 Strea
14.0
Effect of T T = 20 C
12.0
T = 10 C
) 10.0 L / g m 8.0 ( D
T = 30 C
6.0 4.0 2.0 0.0 0
50
100
150
200
Rive distance (km)
12
Effect f T
10 8 ) L / g m ( C
6
T = 20 C
4
T = 10 C T = 30 C
2 0 -2
0
50
100
150
200
-4 -6 -8
River di tance (km)
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Water Quality Analysis_F11
16 Strea
Analysis of Streeter-Phelps M del - The DO sag equation
D =
kd Lo ka
−
(e k
k t
− d
−
e
)
k t
− a
+
D (e
k t
− a
)
d
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Water Quality Analysis_F11
16 Strea
12
Water Quality Analysis_F11
