CEE 577 8 April 2002

# MID-TERM EXAM

Closed book, 1 sheet of notes allowed.

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.

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

### 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)

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

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

#### 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:

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

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)