Rules of Reynolds Averaging
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William Kett
le 2 Oct 2020
Réponse apportée : Alan Stevens
le 2 Oct 2020
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I tried to create an array, u:
u = [1,3,6,8,24];
uprime = diff(u)
which output a vector of differences (not sure if this is what I wanted).
Then I tried to take an average of uprime using the following code below:
mean(uprime)
According to the rule, I should have gotten 0, but I got an average of the numbers in the vector which was nonzero. Am I misinterpreting the rule or is the diff function not actually taking a derivative?
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Alan Stevens
le 2 Oct 2020
uprime doesn't mean du/dt here. It means u - umean.
u = [1,3,6,8,24];
umean = mean(u);
uprime = u-umean;
mean(uprime)
ans =
-3.5527e-16
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