My data is 0.9774 g/cm3 at 8.4C (temperature in Celsius)
The table of data I've taken from a methanol/water table is attached below. Can someone please drop a hint. Just stuck on how to get the interpolated wt%.
thanks in advance!
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Posted 04 December 2009 - 04:11 PM
Posted 04 December 2009 - 05:03 PM
Posted 05 December 2009 - 05:08 AM
Posted 05 December 2009 - 05:49 AM
Hi marcus
Let you try LINEST function , excel spreadsheet .
Regards
breizh
Posted 05 December 2009 - 11:09 AM
Posted 06 December 2009 - 03:04 PM
Posted 07 December 2009 - 09:14 AM
Posted 07 December 2009 - 09:32 AM
I think attached interp.xls is what you asked for, subject of course to criticism from other members.
It generally follows proposal of latexman, with some further simplifications. Right sg for 17% w/w content at 10 C seems to be 0.9739 as latexman suggested, not 0.9239.
Extrapolations and interpolations are needed, as explained on the spreadsheet.
Posted 07 December 2009 - 07:40 PM
Posted 08 December 2009 - 08:17 AM
Posted 08 December 2009 - 01:24 PM
Posted 09 December 2009 - 03:05 PM
Marcus,
If you have not reached the point in your course where you study interpolation and regression, then a useful technique is to draw a graph on paper and interpolate that way. Do students still draw graphs on paper? I hope so. Doing it this way gives you a good feeling for the numbers and will help avoid the gross errors like you nearly made using a method you did not yet understand.
See the attached (very rough) graph. Luckily no one is grading me on this effort, but it should serve to show you the technique.
Plot your data on the graph. See points A, B, C and D on the graph. Draw the lines through the families of data points and label them. Now measure the distance between the two lines at the target value, ie at density = 0.9774. To interpolate (linearly) between the 0 deg C and 10 deg C lines you simply measure the distance between the lines and then make a mark at 84% of the distance between them. 84% is (100* 8.4 / (10 - 0))
From my rough graph I get 14.35%. kkala got 14.30% by his Excel method. Using Excel I get 14.25%. Within the accuracy that we have here these are all affectively the same number, and much better than the 26.38% the other method gave you.
Posted 09 December 2009 - 05:38 PM
Posted 10 December 2009 - 10:48 AM
here is a great webpage for interpolation http://www.ajdesigne...on_equation.php
it even shows the math!
Posted 13 December 2009 - 09:44 AM
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