A sewage treatment plant releases effluent into a stream that has a mean velocit

Question

A sewage treatment plant releases effluent into a stream that has a mean velocity of ~0.9 m s-1. The BOD of the effluent is 50 mg L-1. The water temperature (~15.5 °C) is such that the dissolved oxygen concentration at saturation cs is ~10 mg L-1. Assume naturally-introduced organic matter (e.g., decaying leaf litter, animal feces) imposes an additional (background?) BOD of ~ 0.4 mg L-1. Thus D0, the initial dissolved oxygen deficit, is 0.4 mg L-1, making B0 = 50.4 mg L-1. At ambient stream temperature the deoxygenation coefficient kB is 0.30 day-1 and the reaeration coefficient kA is 0.90 day-1.
a. What is the critical distance downstream (in km) at which the DO minimum occurs?
b. What is the DO minimum?
c. In general, many aquatic organisms require DO values greater than ~4 mg L-1. Specific requirements vary according to the particular organism, but as a rule colder water organisms (e.g., trout, salmon, and aquatic insects like stoneflies) require higher levels, often exceeding 6-7 mg L-1. Apparently even higher levels might be required during the spawning cycle. Warmer water organisms (e.g., bass, black fly, caddisfly, mayfly larvae) require lower concentrations, ~4 mg L-1. Some organisms (carp, catfish, mosquitoes) can tolerate DO levels less than ~2 mg L-1. Accordingly, as DO levels drop certain organisms will die off before others; thus a dominance of certain species may serve as an indicator of water quality. Typical lower limits of ~ 5 mg L-1 DO have been used as guideline in preserving water quality. How long (approximately, in hours) will conditions persist such that DO 5 mg L-1? (Remember distance and time are related through stream velocity.)
d. What is the maximum the effluent’s BOD can be in order to preserve DO levels of 5 mg L-1, that is cmin to be 5 mg L-1? You answer should be correct to the first decimal.

Hints: Since cs is 10 mg L-1 this requires Dcrit to be 5 mg L-1. I think the easiest way to approach this is by using Excel® (see above) and iterating manually the effluent BOD value until a Dcrit of 5 mg L-1 is obtained. (Note that you cannot calculate this value directly (e.g. Equation (8)) because you would need to know tcrit and a B0 value and the latter is what you are looking to find... hence an iterative approach is used.) Thus you can set up a pair of 3 linked cells one containing your guess at B0, one for calculating tcrit (depends on B0) and finally one for calculating Dcrit (depends on both B0 and tcrit) which needs to be 5 mg L-1. Alternatively you can repeat the exercise in Part c and vary B0 until the minimum allowable concentration cmin of 5 mg L-1 is found. If using the latter approach, you might need to increase the resolution in time (increments) near where the minimum concentration appears to be occurring in order to get a more precise value – what I am looking for. And once you’ve determined an acceptable value of B0 you not quite finished... Remember B0 is the total BOD, not that of the effluent.

Explanation / Answer

a. Critical distance xc=v*tcrit

where v=velocity of stream=0.9ms-1

tc=time taken in reaching the critical distance=?

First we need to calculate Tcrit

Tcrit=(1/k2-k1)*k2/k1(1-Do(k2-k1)/La*k1

where k1=deoxygenation rate=0.30day-1

k2=reaeration rate=0.90day-1

Do=initial dissolved oxygen deficit=0.4mgL-1

La=ultimate BOD=50mgL-1

Now Tcrit=(1/0.90-0.30) In ((0.90/0.30)(1-0.4(0.90-0.30)/50*0.30)

=(1/0.6) In((3)(1-(0.4*0.6/15)

=1.66 In(1-0.24/15)

=1.66 In(1-0.016)

=1.66 In0.984

=1.66*(-0.016)

=0.026days

Now critical distance=tcrit*v

tcrit=0.026days

v=0.9ms-1=77760mday-1

critical distance=0.026*77760=2021.76m

b.DO minimum=k1/k2 BODLek1(xc/v)

where k1=0.30day-1

k2=0.90day-1

xc=critical distance=2021.76m

v=velocity of river=0.9ms-1=77760mday-1

BODL=ultimate BOD=50mgL-1

Now DO minimum=0.30/0.90*50*e0.30(2021.76/77760)

=0.33*50e0.30*77760

=0.33*50e23328

=0.33*502.8

=165.9=166mgL-1

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.