A circular clarifier handles a flow of 3400 m^3/d. The clarifier is
15(m) in diameter and 2.5 metres (m) deep. Find the weir overflow rate
in  m^3/d * m.

Answer 72.2 m^3 / d x m. 

Show me the work please and work me through as this is baffling to me.
I guaruntee 5 stars to mathtalk because I know he is brilliant and
will be able to help me, but hopefully I can get this answered
tonight. If not locked by mathtalk within an hour of posting or so, I
wouldn't mind if someone else gave it a shot.

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Request for Question Clarification by mathtalk-ga on 30 Jan 2004 00:06 PST

Hi, waldner-ga:

I didn't see your post until my brain cells were barely twitching
tonight, but I wrote up my best guess.  Perhaps this, combined with
your knowledge of the subject matter, will see your uncle through.

regards, mathtalk-ga

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Clarification of Question by waldner-ga on 30 Jan 2004 17:01 PST

Mathtalk, I am sure you have it right but I still do not understand what you did.

It has been a long time since I have taken math!

I have tried a number of methods using your formula m^3 / (d x m) but
have been unable to get the answer 72.2.

Can you please walk me through it in detail? 

The closest I get is 90.

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Clarification of Question by waldner-ga on 30 Jan 2004 17:47 PST

aht, 

it helps me understand it better, but I am still unsure how you got 72.15

Can you plug in the numbers and show me this step by step? 

sorry for being a pain

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Request for Question Clarification by aht-ga on 30 Jan 2004 18:26 PST

No problem, waldner-ga!

Try it this way, instead:

Circumference of weir:
----------------------

   Diameter, 'd': 15 m
   Pi:             3.141529 (approximately)

  so, Pi * d = 15 * 3.141529 = 47.1229 (and the unit is still metres)

Volume flowing over weir each day:
----------------------------------

   3400 m^3/d    (this is the volume that flows in, and therefore out, each day)

 divided by

   47.1229 m     (this is the total length of weir that the water flows over)

 equals

            m^3
   72.1517 ----- (the volume that flows over each metre of weir length per day)
           d * m


Does that make it clearer?

Regards,

aht-ga
Google Answers Researcher

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Clarification of Question by waldner-ga on 30 Jan 2004 18:32 PST

Awesome! That is exactly what I needed. Thanks so much. :)

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Request for Question Clarification by aht-ga on 30 Jan 2004 18:34 PST

Should I post it as the Answer, to help close the question?
Technically, mathtalk-ga did answer it first as you requested, so if
you prefer, we can leave it open for him to post the Answer instead.
It's all up to you, since you're the customer!

Regards,

aht-ga
Google Answers Researcher

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Request for Question Clarification by mathtalk-ga on 30 Jan 2004 18:36 PST

Yes, the m^3 over (d x m) is not a formula, but simply the units of
measurement for the answer.

The question asks, what is the ratio of (overflow rate) volume per
time, to the (weir circumference) length?

Units of overflow rate are given as cubic meters per day here, and (by
knowing the diameter of the circular weir) we compute the
circumference length.

Overflow rate = 3400 cubic meters per day (given)

Circumference = 15 meters * pi (formula for circle)

Ratio:

        3400/(15 pi) ~ 72.2 m^3 / (d x m)

since pi = 3.14159265...

Note that the "input" data is only known to two digits of accuracy
(3400 cubic meters per day is presumably a "round" figure), so it
would gilding the lily, so to speak to quote an answer with more than
three significant places.

The actual computation came out 72.1502..., so strictly speaking we
would round this to 72.2, which is precisely the figure you were
looking for.

best wishes, mathtalk-ga

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Clarification of Question by waldner-ga on 30 Jan 2004 19:59 PST

Thanks guys,

I initially thought that this would be a very simple answer and was
going to spend $2.50 for listing and $2.50 for tip. I guess I could
pay $10 and give you each a question? Please let me know if that is ok
and then I will list the second question.

---

Request for Question Clarification by mathtalk-ga on 30 Jan 2004 20:28 PST

Hi, waldner-ga:

I'd be very happy if aht-ga took the credit for getting you squared
away here.  If you have further math questions, I hope you'll consider
using Google Answers once again!

regards, mathtalk-ga

Hi waldner-ga:

As Mathtalk-ga is such a generous person, I will post the final Answer
to help close this question.

To summarize, the best way to think of the physical problem, is to
think of an inflatable child's wading pool. In this case, a very large
wading pool, with a diameter of 15 metres! Once the pool is filled to
the top with water, any more water that is put into the pool, will
force an equal amount of water to overflow over the edge of the pool.

In the problem statement, you are told that water flows into the
clarifier at a rate of 3400 m^3/day. That means that water must also
flow out at a rate of 3400 m^3/day, since the clarifier isn't able to
stretch to hold more water. Technically, you can say that the outflow
rate of the clarifier is therefore 3400 m^3/day. However, that isn't
what the question is asking for, unfortunately.

What the question does ask for, is the weir overflow rate, in
m^3/(d*m). When you read that out aloud, it's "cubic metres per day
per metre (of the weir)". Since the weir, in this case, is simply the
wall of the clarifier, the 'length' of the weir is equal to the
circumference of the clarifier. We are told that the diameter of the
clarifier is 15 metres. The formula for the circumference of a circle
is pi * diameter. So, the length of the weir is:

  Diameter, 'D':   15 metres
  Pi:               3.14159265...

  D * Pi = 15 * Pi = 47.123 metres

And now, the weir overflow rate is simply the total outflow rate of
the clarifier, divided by the length of the weir, to give the overflow
rate per metre of weir length:

  Total outflow:   3400 m^3/d

  divided by:        47.124 metres

                        m^3
  equals:        72.2  -----     (to 3 significant digits)
                       d * m


The 'extra' info provided in the problem, namely the depth of the
clarifier, is essentially a way to make you think twice about how to
solve the word problem. An attempt to distract you, basically, from
the real solution. In reality, the depth of the clarifier is
irrelevant once the clarifier is full, and the clarifier must be full
before any water can flow over the weir!


Glad that this was helpful to you, it was a lot of fun for me to help
you! And thanks to Mathtalk-ga, of course, for allowing me the
opportunity to complete this Answer for you! Speaking for myself, your
current list price is appropriate for this Question.

Regards,

aht-ga
Google Answers Researcher

Great research and effort to help me out! Highly appreciated!

I have no practical familiarit with the notion of "weir overflow
rate".  But from looking at the dimensions, my guess would be that it
is a measure of the flow (volume per day) over a width (length) of the
clarifier's circumference.

Since the clarifier has diameter 15 meters, its circumference would be
pi times that.  All that flows in must(?) flow out (in equilibrium),
so we divide the total flow by the circumference.  Let's see:

3400 (m^3/d) / (15 pi m) = 72.150... m^3 / (d x m)

Well, at least the numbers (roughly) and the units are correct.  Hope this helps!

regards, mathtalk-ga

mathtalk-ga:

You've modeled it correctly. For the definition of the weir 'length'
for a circular clarifier, you can refer to the 'weir diameter'
definition on the following page:

http://chppm-www.apgea.army.mil/dehe/pgm31/WaterDict.aspx?ltr=W

Regards,

aht-ga
Google Answers Researcher

waldner-ga:

What Mathtalk-ga has illustrated above is the following physical process:

Water enters the clarifier pool at a rate of 3400 cubic meters per
day. Water exits the clarifier pool by overflowing a weir that happens
to be the circumference of the pool. The total length of the weir, is
the circumference of the clarifier. Since the circumference of a
circle is (pi)*d  (the magic number Pi, times the diameter of the
circle), your formula is therefore:

3400 (m^3/d) divided by [15 m * Pi] = 72.15 m^3/(d*m)

sorry, hit the wrong key.

The reason why it is this formula, is because the circular clarifier
has a fixed volume. Once the clarifier is full, any water that goes
into it, must be balanced by an equal amount of water coming out of
it. The actual volume of the clarifier is irrelevant once the pool is
full. At the same time, no water will flow over the weir until the
pool is full. In the question, the depth of the clarifier is not
needed, but is provided to make you think a bit about how to approach
the problem.

Does this help?

aht-ga
Google Answers Researcher

The point aht-ga makes about the depth of the weir not being needed is
a good one.  The temptation is usually to think that every piece of
information provided must be important to the solution.  But in a
practical sense it is necessary to identify what data is needed to
obtain an answer, because one needs to ask the right questions to
solve real problems.  Throwing in an unused piece of information is an
"advanced" testing technique!

-- mathtalk

waldner-ga:

Thank you very much for the tip, and thank you for using Google Answers

Regards,

aht-ga
Google Answers Researcher