CEE 577

6 Nov 96


Open book, open notes.

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

Problem #1

I. (55%) Morgan Creek receives effluent from the Chapel Hill wastewater treatment plant (CH-WWTP) and the production facilities of HI-TECH, Inc. In addition, there is a significant non-point agricultural runoff of BOD beginning at the CH-WWTP and continuing downstream for 8 miles (to the HI-TECH outfall). Calculate the dissolved oxygen concentration 10 miles downstream of the CH-WWTP outfall.



Qw = 5 cfs

Qw = 5 cfs

(CBODU)w = 40 mg/L

(CBODU)w = 40 mg/L

Cw = 2 mg/L

Cw = 2 mg/L

Upstream Flow

Non-Point Source

Qu = 40 cfs

Lrd = 6 mg/L/day

(CBODU)u = 2.5 mg/L

Cu = 7.2 mg/L

Additional Information:

U = 0.40 ft/sec = 6.55 miles/day T = 15oC DOsat or Cs = 10.1 mg/L (at 15oC)

H = 6 ft = 1.83 m

In-stream CBOD deoxygenation rate (kd) = 0.264 day (at 15oC)

CBOD settling rate (ks) = 0.080 day-1 (at 15oC)


Divide Morgan Creek into 2 segments at the location of the second point discharge (mile 8). Then determine D and L at the end of the first segment, so that initial D and initial L for the second segment can be calculated. Finally, determine D for the end of the second segment (x=2).

This analysis gives a deficit (D) at the end of the first reach of 4.04 mg/L and a dissolved oxygen concentration of 6.06 mg/L. The final CBODU concentration is 10.37 mg/L.

Now repeat this for segment #2, recalculating the initial D and L.

This analysis gives a deficit (D) at the end of the second reach of 4.77 mg/L and a dissolved oxygen concentration of 5.33 mg/L.

Problem #2

II. (45%) Lake Perot receives discharge from the Texarcana WWTP as well as a substantial flow from the TexMex Chile factory. Texarcana has a current (as of November 6, 1996) population of about 125,000 and it is expected to grow exponentially such that it will reach 150,000 by November 6, 2000. The wastewater flow is 800 liters/capita /day. The average raw water sulfate concentration is 25 mg/L and the WWTP removes about 35% of it. The TexMex Chile factory currently has 4 identical production lines operating year round. They currently discharge to Lake Perot a total of 200 kg/day of sulfate. Due to a recent increase in production of chiles in Mexico, TexMex is forced to close one of these production lines down starting November 6, 1996. Once in Lake Perot, sulfate is lost to microbial uptake at a rate of 0.0002 day-1. If the current (i.e., November 6, 1996) concentration of sulfate in Lake Perot is 2.2 mg/L, what will it be on November 6, 2002?

Data on Lake Perot:


This requires a non-steady state lake model with a non-zero initial concentration, a step load and an exponential load. First the loading functions have to be determined. For the exponential loading this requires that the exponential coefficient be determined from the population growth.

P = Po*exp(Be*4)


Be = (ln(150,000/125,000))/4 = 0.4558 /yr

And the pre-exponential value comes from the loading at t=0.


= Po*(per capita flow)*(raw WW concentration)*(1-fraction removed)

= 125,000*(800 L/capita/day)*(25 mg/L)*(1-0.35)

= 1.625 x 109 mg/day

= 1625 Kg/day

So the final value for sulfate will be 1.7 mg/L.