CERM Enviro Ex. Problem Question

[roadwreck]

I'm wondering if you guys can help me out on this, I was reading the environmental section of my CERM last night (Chapter 28 - 10th edition) when I came to the example problem 28.3 on pages 28-11 & 28-12. It's a two part question where part (a) asks you to determine the distance downstream where the oxygen level is at a minimum (x[SIZE=8pt]c[/SIZE]) and part (b ) asks you to determine if the river can support fish life at that point. The solution for part (a) gives the distance downstream as 1.24 miles with a critical oxygen deficit of 2.0 mg/L, I would have assumed that these values would be used to determine the answer to part (b ), but they aren't. The solution in the CERM uses a distance downstream of 16.1 miles and a critical oxygen deficit of 2.8mg/L. WTF? What am I missing here? Where did 16.1 miles and 2.8mg/L come from?

Help a lowly transportation guy out, b/c I'm not understanding this crazy environmental stuff...

 

[owillis28]

This type of question is my reasoning for switching afternoon modules from Water Resource to Transportation. The environmental questions killed me on previous exams (third try is a charm, right?).

roadwreck, I would like to help you out on this problem but not sure if I can. I don't have my books with me at work but I will take a look at it tonight.

owillis

 

[Guest]

RoadWreck --

I don't have my CERM with me but I can provide some basic explanation and expand as necessary. The problem you described is what we typically call an oxygen sag curve problem. The reason: if you discharge a waste to a 'natural' stream, the waste will typically exert an oxygen demand (e.g. drop the concentration of dissolved oxygen) in the stream. If you think of that waste discharge point as an elemental body (volume), as it moves downstream it mixes with 'cleaner' water AND it becomes re-aerated by the atmosphere to where eventually the previous concentration of dissolved oxygen in the stream will be recovered.

If you think of a plot of Conc (O2) vs time (or distance) you will note that there is a sharp drop and then a slow (logarithmic) recovery. That curve is characteristic based on the Streeter-Phelps Equation. You can use that equation to determine what the Conc (O2) at various distances (or times) along the stretch of the river.

Keep in mind, there is a subtle difference in some of the terms you presented in your post:

Dissolved Oxygen = concentration of O2 in water

Deficit = Dissolved Oxygen - Oxygen Demand

Critical Deficit Oxygen = 2.0 mg/L should be the lowest value of dissolved oxygen in the stream that can maintain a HEALTHY flora/fauna conditions in the stream.

I will look at the problem tomorrow and provide you a step-by-step solution and discussion.

JR

 

[cantaloup]

This type of question is my reasoning for switching afternoon modules from Water Resource to Transportation. The environmental questions killed me on previous exams (third try is a charm, right?).
roadwreck, I would like to help you out on this problem but not sure if I can. I don't have my books with me at work but I will take a look at it tonight.

owillis

Me too, I don't really understand wastewater treatment as much as other subjects but other people have same problem too. I still stick with water resource. The only thing I can do is to go over and over the 3 chapters of wastewater treatment. For transportation, surveying is an important part, you'll do well if you're good at geometry & trigonometry. Try the problem below:

You're required to design a horizontal curve which goes through point A. Three points: PC, A, PT form a triangle. Given: the distance from PC to PT is 1154.3 ft, the distance from PC to point A is 461.0 ft and the distance from PT to point A is 700.0 ft. What is the nearest radius of curvature in feet ?

A/ 2652

B/ 2783

C/ 2886

D/ 3004

 

[IlPadrino]

Me too, I don't really understand wastewater treatment as much as other subjects but other people have same problem too. I still stick with water resource. The only thing I can do is to go over and over the 3 chapters of wastewater treatment.

I think there's very little wastewater in the Water Resources depth. Nothing much beyond a few plug and chugs and maybe a qualitative question or two. I spent *WAY* too much time preparing for wastewater and water supply (water chemistry, especially)

 

[Guest]

RoadWreck --

I am using CERM 8th Edition ld-025: that provides the same problem but apparently with different numbers. In my example,

Initial Oxygen Deficit D0 = 2.0 mg/L

Critical Time tc = 1.61 days

Critical Distance xc = 14.5 miles

Critical Oxygen Deficit Dc = 2.7 mg/L

Make sure for the problem you aren't confusing Initial Oxygen Deficit D0 with Critical Oxygen Deficit Dc. As far as the distance downstream - it should not change if it is the CRITICAL distance.

When you have time, if you (or some other kind soul) can scan/post those problem pages, I can take a look to see what is going on.

JR

 

[owillis28]

jr,

per your request, please see attachments

owillis

 

[Guest]

Owillis -- Thanks for the problem pages!

This problem has been worked out incorrectly. For starters, the Kr,20C was calculated based on the wrong equation (e.g. metric units instead of english).

Starting with Kr,20C = 3.3*v/d^1.33 = 0.287 d-1

Now, by adding the temperature correction, Kr,16.89C = 0.273 d-1

Calculations for BODu and D0 are correct.

Now that we know Kr,16.89C = 0.273 d-1, this change will cascade through the calculation of the CRITICAL VALUES.

tc = 1.61 days

xc = 14.5 miles <--- Correct Answer for Part (a)

Dc = 2.70 mg/L

Now consider Part (B) - Can Fish Life be Supported at that point (e.g. Critical Distance xc)

We know that the temperature in the unaffected, upstream direction is at ambient temperature (16 C), so if we look at Appendix 22.D for the degree of oxygen saturation in water, we find O2 = 9.95 mg/L. This represents the oxygen content in the water prior to the waste discharge. If the the critical deficit (e.g. the maximum point of O2 depletion in the stream) is calculated to be 2.70 mg/L then we have a matter of simple arithmetic -

Min Conc of Dissolved O2 = 9.95 mg/L - 2.70 mg/L = 7.25 mg/L

The basis of comparison is Dissolved O2 = 2.0 mg/L MINIMUM for propagation of aquatic life, so 7.25 > 2.0 therefore YES.

If you have any follow-up questions, please feel free to ask.

JR

 

[roadwreck]

Thanks JR.

I figured there was a mistake somewhere in their solution but I didn't have a clue where.

 

[thartley]

It seems that "the other board" has had problems with that particular problem for a while. If you check their Errata area, it was pointed out as having errors in both the first and second printings of both the Eighth and Ninth editions for what looks like the same errors. Now, from what is being posted here, it seems that they still haven't corrected it in the Tenth - it just hasn't been reported as being flawed yet.

It looks like others came up with a different temperature correction value but still came up with the same numbers for the critical values.

 

[roadwreck]

I had been to to the other board and looked for errata for the 10th edition, but they didn't have anything listed. I didn't think to look through the errata for previous versions though.