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The instantaneous stream-wise velocity of a turbulent flow is given by \(u(x,y,z,t)=\bar{u}(x,y,z)+u'(x,y,z,t)\). The time-average of the fluctuating velocity \(u'(x,y,z,t)\) is
- u'/2
- −\bar{u}/2
- zero
- \bar{u}/2
Correct answer: zero
Solution
In Reynolds decomposition, the instantaneous velocity is split into a mean part and a fluctuation part. By definition, the time average of the fluctuation is zero.
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