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Thread: Maths Mistake? (Excel 2003)

20040210, 20:30 #1
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Maths Mistake? (Excel 2003)
Hello all,
Please forgive me but this is only vaguely an Excel question as I suspect my problem is more to do with bad maths than bad Excel skills.
If you look at the attached sheet you will find a set of calculations relating to water droplets, their diameter, their volume & surface area. The purpose of the sheet is to calculate the total surface area of water that is created when 1 litre of water is broken up into droplets of a certain diameter.
The sheet shows that using droplets that have a diameter of 1mm makes (by my maths) 6m sqaured total surface area. I think my maths are right. I am using 4/3*PI*Radius Cubed for Volume and 4*PI*Radius Squared for Surface Area. The problem is that (as shown in the yellow cells) the textbooks and reference material I am working from tells me that the answer should not be 6 but infact 2.
The textbooks list further examples at 0.1mm and 0.01mm sizes and my answers are similarly wrong at these sizes.
Any thoughts?
Is it my use of brackets, PI or something in Excel?...
Is it my understanding of the maths and volume/area calculations?...
Is it that the rest of the world is wrong and I am right?...
If I have not given enough info please post for clarification.
Thanks
Jon Roberts

20040210, 20:59 #2
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Re: Maths Mistake? (Excel 2003)
The general formula for dividing a volume V into spheres of radius r:
The number of spheres is n = V/(4/3*Pi*r^3)
The total area is n * 4*Pi*r^2 = 3*V/r
This leads to the results you have. For example, V = 1 liter = 10^6 mm^3, and r = 0.5 mm, so area = 3*10^6/.5 = 6*10^6 mm^2 = 6 m^2.
In reality, water droplets are not perfect spheres; could that be the cause of the discrepancy? Doesn't seem likely to cause a factor 3...

20040210, 23:42 #3
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Re: Maths Mistake? (Excel 2003)
Based on the question as you pose it, I agree with Hans that your math is correct. I wonder if the textbook is asking a different question and requires a different solution and other assumptions. Either that or the text book mad a mistake.
Steve

20040211, 06:31 #4
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Re: Maths Mistake? (Excel 2003)
Guys,
Thanks for the backup on my maths. I was pretty sure I had that bit right but you know how it is with these problems  The error can be staring you in the face for days but it needs a fresh pair of eyes to see it.
I will have a search online and see if I can find a website with the "textbook" answers and will post a link asap.
Cheers

20040211, 07:51 #5
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Re: Maths Mistake? (Excel 2003)
Hi Hans,
Any other shape than a sphere would cause an increase in surface as you probably know. The most efficient (with respect to the ratio between surface and volume) container being a sphere.Jan Karel Pieterse
Microsoft Excel MVP, WMVP
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20040211, 07:58 #6
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Re: Maths Mistake? (Excel 2003)
<img src=/S/blush.gif border=0 alt=blush width=15 height=15> Of course. So it's down to Steve's questions.

20040211, 15:17 #7
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Re: Maths Mistake? (Excel 2003)
We are concerned about is the ratio of the Area to the volume:
In a sphere you get (4 pi r

20040211, 22:06 #8
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Re: Maths Mistake? (Excel 2003)
Steve,
Perhaps not such a "waste of time" if the question was actually trying to establish the surface area that might be "covered" by a litre of rain drops. In this case your ultrathin rectangular prism might be a representation of a "sheet" or thin film. And if this is the case, it's a rather illconceived question, since introducing spherical droplets into the mix is a waste of time and quite irrelevant to the crux of the problem.
<img src=/S/2cents.gif border=0 alt=2cents width=15 height=15>
Alan

20040212, 09:50 #9
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Re: Maths Mistake? (Excel 2003)
Edited by <!profile=unkamunka>unkamunka<!/profile> to reduce size of graphic and to prevent scrolling
Guys,
I have been unable to find a website that lists the answers as I have them, so I have scanned a page from the manual for the product that shows the "correct answers".
None of you have expressed a particular interest in why I want to answer this question but as you will see from the attachment I work in the Fire Protection industry and the question relates to using a very fine "mist" from a specially designed sprinkler head rather than a standard sprinkler head which produces much larger water drops. The increased surface area generated per litre of water has a massive additional cooling effect on a fire and works far better at putting the fire out than simply "getting it wet" with a normal sprinkler.
Thanks again for all your thoughts  even if the concept of an "ultrathin rectangular prism" has me totally lost!

20040212, 10:01 #10
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Re: Maths Mistake? (Excel 2003)
I think we may assume that mist droplets are roughly spherical. The results from the manual must be wrong, I fear.

20040212, 10:16 #11
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Re: Maths Mistake? (Excel 2003)
I agree with Hans. The manual is incorrect.
Steve

20040212, 13:53 #12
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Re: Maths Mistake? (Excel 2003)
If I read the table correctly, the mentioned surface has nothing to do with the water surface of the droplets themselves but something with the EFFECT of those droplets (it states Cooling Surface).
Maybe Steve's theory about an infinitely thin square applies after all <g>.Jan Karel Pieterse
Microsoft Excel MVP, WMVP
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20040212, 14:16 #13
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Re: Maths Mistake? (Excel 2003)
Not that it really adds much, but you might want to look at this website... http://www.fogtec.com/fogtec_relaunch/data.../watermist.html ... Fogtec are one of the manufacturers of the types of Watermist sprinklers that we are concerned with. If the link does not work just go to www.fogtec.com and click on the "Technology" link in the menu bar on the left and then click on "Principals of Water Mist".

20040212, 17:19 #14
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Re: Maths Mistake? (Excel 2003)
I still agree with what I said in my previous post. It is not the droplet surface area, but the EFFECT area. Not the same thing.
Jan Karel Pieterse
Microsoft Excel MVP, WMVP
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20040212, 17:45 #15
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Re: Maths Mistake? (Excel 2003)
Jan,
I have no idea if you are right or not but am confused as how the EFFECT area could be EXACTLY 1/3 of the Surface Area. After all we are dealing with water droplets and basic surface tension rules tell us that the droplet will be a sphere  or pretty close to one anyway.
Any thoughts?