17-stream water quality analysis_f11

Stream Water Quality Analysis

Dissolved Oxygen (DO)

Stream DO - DO Sag Curve

16 Stream

1

Stream Water Quality Analysis

16 Stream Water Quality Analysis_F11

DO Deficit, D

where D = dissolved oxygen deficit, mg/L

Cs = saturation concentration of dissolved oxygen, mg/L. See Table A

of dissolved oxygen in fresh water (DC, p. 868).

C = actual concentration of dissolved oxygen, mg/L

Do = initial DO deficit, mg/L= DO deficit at t = 0

Dc = Critical DO deficit, mg/L

Cc= Critical DO, mg/L

Initial mixing of waste stream and river

Qr = volumetric flow rate of the river, m

Qw = volumetric flow rate of wastewater, m

Qm = volumetric flow rate of the river after mixing, m

Cr = dissolved oxygen concentration in the river, mg/L

Cw = Dissolved oxygen concentration in the wastewater,

mg/L

Cm = dissolved oxygen concentration in the river after

mixing, mg/L

Lr = ultimate BOD of the river, mg/L

Lw = ultimate BOD of the wastewater, mg

Lm = ultimate BOD of the river after mixing, mg/L

Tr = temperature of the river, ºC

Tw = temperature of the wastewater, ºC

Tm = temperature of the river after mixing, ºC

C

16 Stream

2

D = Cs - C

Do = Cs - Co

Dc = Cs - Cc

D = dissolved oxygen deficit, mg/L

= saturation concentration 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

= initial DO deficit, mg/L= DO deficit at t = 0

cal DO deficit, mg/L

Initial mixing of waste stream and river

Qr = volumetric flow rate of the river, m3/s

Qw = volumetric flow rate of wastewater, m3/s

Qm = volumetric flow rate of the river after mixing, m3/s

dissolved oxygen concentration in the river, mg/L

Cw = Dissolved oxygen concentration in the wastewater,

Cm = dissolved oxygen concentration in the river after

Lr = ultimate BOD of the river, mg/L

Lw = ultimate BOD of the wastewater, mg/L

Lm = ultimate BOD of the river after mixing, mg/L

Tw = temperature of the wastewater, ºC

Tm = temperature of the river after mixing, ºC

Qr

Cr

Lr

Tr


o

2. Saturation values

Qw

Cw

Lw

Tw

Qo

Co

Lo

To

Mass Balance:

Mass in = Mass out

For C,

Qw Cw + Qr Cr = Co (Qw + Qr)

Qw Cw + Qr Cr

Co = ---------------------

Qw + Qr

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 a ak t k t k td oo

a d

k LD e e D e

k k

− − −= − +

−

where

D = dissolved oxygen deficit in river water after exertion of

Do = initial deficit after river and wastewater have mixed, mg/L.

Lo = initial ultimate BOD after river and wastewater have mixed, mg/L.

kd = deoxigenation rate constant, d

ka = reaeration rate constant, d

t = time of travel of wastewater discharge downstream, d

16 Stream

3

= specific heat at constant pressure, J/g K = 4.19 J/g K for natural water

The DO sag equation

( )d a ak t k t k t

oD e e D e− − −

D = dissolved oxygen deficit in river water after exertion of BOD at time t, mg/L.

= initial deficit after river and wastewater have mixed, mg/L.

= initial ultimate BOD after river and wastewater have mixed, mg/L.

= deoxigenation rate constant, d-1

.

= reaeration rate constant, d-1

.

travel of wastewater discharge downstream, d

Qw

Cw

Lw

Tw

Qr Qo

Cr Co

Lr Lo

Tr To


BOD at time t, mg/L.

16 Stream Analysis_F11

4

Deoxygenation rate constant

kd = k + u η / H ........ Empirical equation

where

kd = deoxygenation rate constant at 20ºC, d-1

k = BOD rate constant determined in laboratory at 20ºC, d-1

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:

kd, T = kd,20 θ T-20

where θ= 1.135

Reaeration Rate Constant, ka

ka = 3.9 u 0.5

/ H 1.5

where

ka = (kr) = reaeration rate constant at 20ºC, day

-1


H = average depth, m

Temperature Correction:

ka,T = ka, 20 θ T-20

where θ = 1.024

Travel time, t

t = x / u

where x = travel distance

Time to the critical distance, tc

1ln 1a a d

c o

a d d d o

k k kt D

k k k k L

−= −

−

16 Stream Analysis_F11

5

Critical Deficit, Dc

( ) ( )cd o d a c a cc o

a d

k t k t k tk LD e e D e

k k

− − −= − +

−

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) .

The town of State College discharges 17,360 m3/d of treated wastewater into the Bald Eagle Creek.

The treated wastewater has a BOD5 of 12 mg/L and a k for BOD kinetics of 0.12 d-1

in laboratory

study at 20°C. Bald Eagle Creek has a flow rate of 0.43m3/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).


6

(Solution)

Given:

Qw = 17360 m3/d (= 0.2 m

3/s)

DOw = 1.0 mg/L

Lw = (BOD5)w =12 mg/L

Tw = 10ºC

k = 0.12 d-1

@ 20 ºC)

Qr = 0.43 m3/s Q = Qw + Qr = ?

Lr= 5 mg/L Lo = ?

DOr =6.5 mg/L Co = ?

Tr = 10 ºC To = ?

Qw = (17,360 m3/d) (1 d/86,400 s/d) = 0.20 m

3/s

1) DO after mixing

(Qw) (DOw) + (Qr)(DOr)

Co = ------------------------------

(Qw + Qr)

(0.2 m3/s) (1.0 mg/L) + ( 0.43 m

3/s)(6.5 mg/L)

= ---------------------------------------------------------- = 4.75 mg/L ( 0.20 m

3/s + 0.43 m

3/s)

BODt = BODL (1 - e

-kt)

where BODL= ultimate BODL

Convert BOD5 to BODL

BODt 12 mg/L

BODL = ------------- = -------------------- = 26.6 mg/L = Lw

(1 - e-kt) 1 - e - (0.12)(5)


7

Qw Lw + Qr Lr

Lo = -------------------

Qw + Qr

(0.20 m3/s) (26.6 mg/L) + (0.43 m

3/s)(5.0 mg/L)

Lo = -------------------------------------------------------------- = 11.86 mg/L

(0.20 m3/s + 0.43 m

3/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, kd

kd = k + (u η / H )

where kd = deoxygenation rate constant at 20ºC, d

-1

k = BOD rate constant determined in laboratory at 20ºC, d-1

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:

k = 0.12 d-1

u = 0.03 m/s

H = 5.0 m

η = 0.35

kd = 0.12 d-1

+ (0.03 m/s)(0.35) / (5.0 m) = 0.1221 d-1

Temperature Correction by:

kd.T = kd,20 θ T-20

where θ = 1.135 (DC 293-294)

kd,10 = (0.1221 d-1

)(1.135) 10 - 20

= (0.1221)(0.2819) = 0.03442 d-1


8

Reaeration Rate Constant, ka

ka = 3.9 u 0.5

/ H 1.5

where

ka = reaeration rate constant at 20ºC, day-1


H = average depth, m

Given: u = 0.03 m/s

H = 5.0 m

k a,20 = 3.9 (0.03 m/s)0.5

/ (5.0 m)1.5

= 0.0604 d-1

Temperature correction by

k a,T = k r,20 θ T-20

where θ = 1.024

k a,10 = (0.0604 d-1

) (1.024) 10 - 20

= 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 a ak t k t k td oo

a d

k LD e e D e

k k

− − −= − +

−

where

Lo = 11.86 mg/L

kd = 0.03442 d-1

ka = 0.04766 d-1

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)


9

Critical time, tc - time to the critical distance

1ln 1a a d

c o

a d d d o

k k kt D

k k k k L

−= −

−

Given:

Lo = 11.86 mg/L

kd = 0.03442 d-1

ka = 0.04766 d-1

Do = 6.58 mg/L

tc = 6.45 day

Critical deficit, Dc

( ) ( )d c a c a ck t k t k td oc o

a d

k LD e e D e

k k

− − −= − +

−

Given:

Lo = 11.86 mg/L

kd = 0.03442 d-1

ka = 0.04766 d-1

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

xc = tc u = (557280 s)(3 x 10-5

km/s) = 16.7 km

The critical DO occurs downstream at a distance of 16.7 km from the wastewater discharge point.

0.0

2.0

4.0

6.0

8.0

10.0

12.0

14.0

0 50 100

D(m

g/L

)

River distance (km)

Effect of T

-8

-6

-4

-2

0

2

4

6

8

10

12

0 50 100

C (m

g/L

)

River distance (km)

Effect of T

16 Stream

10

100 150 200

River distance (km)

Effect of T

T = 20 C

T = 10 C

T = 30 C

100 150 200

River distance (km)

Effect of T

T = 20 C

T = 10 C

T = 30 C


Analysis of Streeter-Phelps Model

( )d a ak t k t k td oo

a d

k LD e e D e

k k

− − −= − +

−

16 Stream

11

Phelps Model - The DO sag equation

( )d a ak t k t k t

oD e e D e− − −


16 Stream

12


17-stream water quality analysis_f11

Documents