CEE 577

8 April 2002

MID-TERM EXAM

 

Closed book, 1 sheet of notes allowed.

Answer both questions. Please state any additional assumptions you made, and show all work.

 

 

I.               (50%) Bovine Brook emerges from pristine headwaters and runs through an agricultural region before it reaches the city of Poultryville. The BOD of the headwaters is 4 mg/L, and the dissolved oxygen is 8.8 mg/L (saturation is 10.1 mg/L for 15oC). Starting at mile point zero, there is a significant non-point agricultural runoff of BOD amounting to 40 kg/mile/day. At mile point 4.0, the stream is met by the Poultryville WWTP outfall. Here a WW flow of 1 cfs is discharged with a CBOD of 150 mg/L, an ammonia-N concentration of 24 mg-N/L and a DO of 6 mg/L. Immediately past this outfall is 4 more miles of agricultural land. This is better managed than the upstream land, so that the runoff BOD averages 15 kg/mile/day. Calculate the dissolved oxygen concentration 4 miles downstream of the Poultryville WWTP outfall (T=15oC). Assume the flow is constant at 40 cfs from the headwaters to the end of the non-point agricultural runoff. You may also assume an SOD downstream of the WWTP outfall of 2.0 g/m2/d.

 

Additional Information:

 

U = 0.150 ft/sec = 2.45 miles/day T = 15oC

DOsat or Cs = 10.1 mg/L (at 15oC) H = 4 ft = 1.22 m

BOD deoxygenation rate (kN = kd) = 0.8 day-1 (at 15oC) for kN and kd, =1.047

CBOD settling rate (ks) = 0.080 day-1 (at 15oC) for reareation, =1.024

 


 


SOLUTION:

 

 

Calculate ka and adjust for temperature:

ka = 0.555 d-1

 

Divide stream into two reaches of 4 miles in length. The travel time for each is:

 

t = 1.63 d

 

The non-point CBOD load is:

 

 

 

 

Then determine the BOD and deficit at the end of the first reach:

 

 

Overall deficit = terms incorporating (CBOD point deficit + point BOD + distributed deficit + distributed BOD)

 

 

 

D = 0.526 + 1.639 + 0 + 0 + 0.978 0.467

D = 2.677

 

C = 10.1 2.677 = 7.423 mg/L

 

 

Calculate the BOD and deficit at the junction of the two reachs and the point source:

 

 

 

 

 

Finally, determine the BOD and deficit at the end of the second reach:

 

 

 

The non-point CBOD load is:

 

 

 

 

 

 

Overall deficit = terms incorporating (CBOD point deficit + point BOD + distributed deficit + distributed BOD)

 

 

 

D = 1.10 + 2.23 + 1.16 + 1.76 + 0.36 0.17

D = 6.44

 

C = 10.1 6.44 = 3.66 mg/L

 

 

 

 

 

 


Segment #1

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

General Parameters

 

 

 

 

 

 

 

 

 

 

 

U =

0.15

ft/s =

2.454545

mi/day =

0.045721

m/s

 

 

 

 

 

T =

15

deg-C

 

 

 

 

 

 

 

 

 

Cs =

10.1

mg/L

 

 

 

Options for ka

 

 

 

 

H =

4

ft =

1.219215

m

 

0.6825

/d @

20

deg-C

EPA SMWLA

 

kd =

0.8

/d @

15

deg-C

 

0.6245

/d @

20

deg-C

O'Connor-Dobbins

 

ks =

0.08

/d @

15

deg-C

 

 

 

 

 

 

 

kN =

0.8

/d @

20

deg-C

 

 

 

 

 

 

 

SOD =

0

g/m2/d @

20

deg-C

 

 

 

 

 

 

 

x =

4

miles =

21120

ft

 

 

 

 

 

 

 

slope =

3.5

ft/mi

 

 

 

 

 

 

 

 

Upstream Water

 

 

 

 

 

 

 

 

 

 

 

Qu =

40

cfs

Selected Reaeration Coefficient

 

 

 

 

 

Lou =

4

mg/L

ka =

0.6245

/d @

20.0000

deg-C

O'Connor-Dobbins

 

 

Lnou =

0

mg/L

 

 

 

 

 

 

 

 

 

Cu =

8.8

mg/L

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

Point Source

Carbonaceous

Nitrogeneous

 

 

 

 

 

 

 

 

Qw =

0

0

cfs

 

 

 

 

 

 

 

 

Low =

30

1.33

mg/L

 

 

 

 

 

 

 

 

Cw =

7.5

7.5

mg/L

 

 

 

 

 

 

 

Non-Point Source (Lrd)

 

 

 

 

 

 

 

 

 

 

runoff=

40

 

kg/mi/d

 

 

 

 

 

 

 

 

runoff=

1.003220979

 

mg/L/d

 

 

 

 

 

 

 

 

runoff=

1.003220979

 

mg/L/d

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

Parameter Calculations

 

 

 

 

 

 

 

 

 

 

Tdesign =

15

15

deg-C

 

 

 

 

 

 

 

 

deox rate =

0.8

0.635853

/d @

15

deg-C

 

 

 

 

 

 

tot. loss rate=

0.88

0.635853

/d @

15

deg-C

 

 

 

 

 

 

ka =

0.554683912

0.554684

/d @

15

deg-C

 

 

 

 

 

 

sb' =

0

 

mg/L/d @

15

deg-C

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

Initial Values (from mass balance)

 

 

 

 

 

 

 

 

 

Lo =

4

0

mg/L

0

 

 

 

 

 

 

 

Co =

8.8

 

mg/L

 

 

 

 

 

 

 

 

Do =

1.3

 

mg/L

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

x/U =

1.62962963

d

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

Dissolved Oxygen Deficit Terms

 

 

Net

CBOD Terms

Net

Desired x

time

---------

---------

---------

---------

---------

---------

Diss

---------

---------

CBOD

(miles)

(d)

Init-D

C-PS

N-PS

SOD

C-NPS-1

C- NPS-2

Oxygen

PS

NPS-1

 

---------

---------

---------

---------

---------

---------

---------

---------

---------

---------

---------

---------

0

0

1.30

0.00

0.00

0

0.00

0.00

8.80

4.00

0.00

4.00

4

1.62963

0.5265

1.6392

0.0000

0.0000

0.9783

-0.4672

7.4232

0.9533

0.8683

1.8217

 

 

 

 

Segment #2

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

General Parameters

 

 

 

 

 

 

 

 

 

 

 

U =

0.15

ft/s =

2.454545

mi/day

 

 

 

 

 

 

 

T =

15

deg-C

 

 

 

 

 

 

 

 

 

Cs =

10.1

mg/L

 

 

 

Options for ka

 

 

 

 

H =

4

ft =

1.219215

m

 

0.6825

/d @

20

deg-C

EPA SMWLA

 

kd =

0.8

/d @

15

deg-C

 

0.6245

/d @

20

deg-C

O'Connor-Dobbins

 

ks =

0.08

/d @

15

deg-C

 

 

 

 

 

 

 

kN =

0.8

/d @

20

deg-C

 

 

 

 

 

 

 

SOD =

2

g/m2/d @

15

deg-C

 

 

 

 

 

 

 

x =

2

miles =

10560

ft

 

 

 

 

 

 

 

slope =

3.5

ft/mi

 

 

 

 

 

 

 

 

Upstream Water

 

 

 

 

 

 

 

 

 

 

 

Qu =

40

cfs

Selected Reaeration Coefficient

 

 

 

 

 

Lou =

1.822

mg/L

ka =

0.6245

/d @

20

deg-C

O'Connor-Dobbins

 

 

Lnou =

0.000

mg/L

 

 

 

 

 

 

 

 

 

Cu =

7.4232

mg/L

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

Point Source

Carbonaceous

Nitrogeneous

 

 

 

 

 

 

 

 

Qw =

1

1

cfs

 

 

 

 

 

 

 

 

Low =

150

109.68

mg/L

24

mg-N/L

 

 

 

 

 

 

Cw =

6

6

mg/L

 

 

 

 

 

 

 

Non-Point Source (Lrd or Sd)

 

 

 

 

 

 

 

 

 

 

runoff=

15

 

kg/mi/d

 

 

 

 

 

 

 

 

runoff=

0.367032065

 

mg/L/d

 

 

 

 

 

 

 

 

runoff=

0.367032065

 

mg/L/d

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

Parameter Calculations

 

 

 

 

 

 

 

 

 

 

Tdesign =

15

15

deg-C

 

 

 

 

 

 

 

 

deox rate =

0.8

0.8

/d @

15

deg-C

 

 

 

 

 

 

tot. loss rate=

0.88

0.8

/d @

15

deg-C

 

 

 

 

 

 

ka =

0.554683912

0.554684

/d @

15

deg-C

 

 

 

 

 

 

sb' =

1.6404

 

mg/L/d @

15

deg-C

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

Lo =

5.435764835

2.675122

mg/L

 

 

 

 

 

 

 

 

Co =

7.388481053

 

mg/L

 

 

 

 

 

 

 

 

Do =

2.711518947

 

mg/L

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

x/U =

0.814814815

d

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

Dissolved Oxygen Deficit Terms

 

 

Net

CBOD Terms

Net

Desired x

 

---------

---------

---------

---------

---------

---------

Diss

---------

---------

CBOD

(miles)

 

Init-D

C-PS

N-PS

SOD

C-NPS-1

C- NPS-2

Oxygen

PS

NPS-1

 

---------

 

---------

---------

---------

---------

---------

---------

---------

---------

---------

---------

0

 

2.71

0.00

0.00

0.000

0.00

0.00

7.388

5.44

0.00

5.44

4

 

1.10

2.23

1.16

1.760

0.36

-0.17

3.663

1.30

0.32

1.61

 

 

 

 

 

 

 

 

 

 

 

 

 


 

 

II.                 (50%) Silver lake is an urban surface water that has received inputs of nickel for many years. Efforts to clean up the lake have resulted in the termination of all nickel loads. Lake levels have dropped due to decay (0.00005 d-1) and hydraulic flushing. Unfortunately, Senator Porkbarrel has succeeded in getting his brother's chrome fender business exempted from the nickel ban. The amount on nickel discharged is directly related to the amount of product produced. Due to changes in the automotive industry, production has dropped off since the business began (see table below). During this time nickel waste has been constant at 3.3 Kg Ni per ton of product. The output data below were collected for a few selected years. On January 1, 1985, the lake concentration of nickel was 0.13 mg/L. The lake has a volume of 12.5 million cubic meters, a surface area of 3 million square meters, and an outflow of 8000 cubic meters per day. Ignore any temperature effects.

 

Porkbarrel Plating Co. Data

Year

Output (tons/yr)

1965

580

1975

510

1985

440

1995

370

 

 

 

  1. Calculate the expected nickel concentration at the beginning of the year 2010.
  2. Determine the year when the maximum Ni concentration occurred or will occur.

 

 

 

Solution:

 

Loading from 1985

W(kg/yr) = (440-7t)*(3.3) = 1452 23.1t

 

Step load:

 

with linear decrease:

 

 


 

 

 

 

 

 

 

 

 

 

 

 


III. (50%) On a separate sheet of paper, answer any five (5) of the following questions.

 

A.                 Describe the difference between mechanistic and empirical modeling

B.                 Is a first order reaction always faster than a zero order reaction? Explain.

C.                 Describe the differences between point and non-point loading and give examples.

D.                 Explain the relationship between Secchi-disk depth and lake trophic state. Why is there such a relationship?

E.                  Describe 3 different methods for determining stream velocity.

F.                  Explain what the light and dark bottle method measures and how it works

G.                 Describe the factors that determine re-aeration in rivers, and contrast this with the factors that determine re-aeration in lakes. In your description, relate micro-scale processes (molecules) to macro-scale (bulk water or air)