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<p class="MsoNormal"><span style="font-size:11.0pt;font-family:"Calibri",sans-serif;color:#1F497D">Dear Olivier,<o:p></o:p></span></p>
<p class="MsoNormal"><span style="font-size:11.0pt;font-family:"Calibri",sans-serif;color:#1F497D">Thanks for the question and the chance for me to correct another mistake.
<o:p></o:p></span></p>
<p class="MsoNormal"><span style="font-size:11.0pt;font-family:"Calibri",sans-serif;color:#1F497D">The physicsl diameter of the tubes were 1.27 cm = ½ inch.
<o:p></o:p></span></p>
<p class="MsoNormal"><span style="font-size:11.0pt;font-family:"Calibri",sans-serif;color:#1F497D"><o:p> </o:p></span></p>
<p class="MsoNormal"><span style="font-size:11.0pt;font-family:"Calibri",sans-serif;color:#1F497D">Why I used 2.54 cm to calculate the Reynolds Number, I do not know. It could be that I usually calculate the charcateristic length of cellular structures from
the radius and usually double the radius to get a characteristic length…and here I most likely mindlessly doubled the “diameter” which of course is wrong. It seems to me that the Reynolds numbers I calculated are twice as large as they should be: 12,700 should
actually be 6,350, which is greater than 2,300 and thus still indicative of turbulent flow.<o:p></o:p></span></p>
<p class="MsoNormal"><span style="font-size:11.0pt;font-family:"Calibri",sans-serif;color:#1F497D">Thanks,<o:p></o:p></span></p>
<p class="MsoNormal"><span style="font-size:11.0pt;font-family:"Calibri",sans-serif;color:#1F497D">Randy<o:p></o:p></span></p>
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<p class="MsoNormal"><b><span style="font-size:11.0pt;font-family:"Calibri",sans-serif">From:</span></b><span style="font-size:11.0pt;font-family:"Calibri",sans-serif"> Physics [mailto:physics-bounces@tuks.nl]
<b>On Behalf Of </b>O. Serret<br>
<b>Sent:</b> Thursday, November 10, 2016 2:31 PM<br>
<b>To:</b> physics@tuks.nl<br>
<b>Subject:</b> Re: [Physics] Arguments for or against the variable time<o:p></o:p></span></p>
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<p class="MsoNormal"><o:p> </o:p></p>
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<p class="MsoNormal"><span style="font-family:"Calibri",sans-serif;color:black">Dear Randy,<o:p></o:p></span></p>
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<p class="MsoNormal"><span style="font-family:"Calibri",sans-serif;color:black"> <o:p></o:p></span></p>
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<p class="MsoNormal"><span style="font-family:"Calibri",sans-serif;color:black">Thank you for your explanation. I will be very interested to know the result if you can do the experience with another fluid.<o:p></o:p></span></p>
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<p class="MsoNormal"><span style="font-family:"Calibri",sans-serif;color:black"> <o:p></o:p></span></p>
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<p class="MsoNormal"><span style="font-family:"Calibri",sans-serif;color:black">And I have a question on the experiment you did. You write :<o:p></o:p></span></p>
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<p class="MsoNormal"><span style="font-family:"Calibri",sans-serif;color:black">> page 298 : copper tubes ½ inch (= 1.27 cm)<o:p></o:p></span></p>
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<p class="MsoNormal"><span style="font-family:"Calibri",sans-serif;color:black">> page 298 : the tubes were five feet long with an internal diameter of 1.27 cm.<o:p></o:p></span></p>
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<p class="MsoNormal"><span style="font-family:"Calibri",sans-serif;color:black">> page 301 : the characteristic length or diameter of the pipe X = 2.54 × 10-2 m
<o:p></o:p></span></p>
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<p class="MsoNormal"><span style="font-family:"Calibri",sans-serif;color:black"> <o:p></o:p></span></p>
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<p class="MsoNormal"><span style="font-family:"Calibri",sans-serif;color:black">Which internal diameter did you use ? 2.54 or 1.27 cm ?<o:p></o:p></span></p>
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<p class="MsoNormal"><span style="font-family:"Calibri",sans-serif;color:black"> <o:p></o:p></span></p>
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<p class="MsoNormal"><span style="font-family:"Calibri",sans-serif;color:black">Thank you, and best regards<o:p></o:p></span></p>
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