Agreement between dissipation estimates

When two or more shear probes, in close proximity, are used to collect simultaneous data, the rate of dissipation derived from the simultaneous data will not agree exactly. Even for nearly flawless measurements, there will be disagreement for purely statistical reasons. Measuring a turbulent shear is sampling a statistical process. The sample variance will differ from the population variance and this difference will reduce with increasing length of data in the sample. The sampling uncertainty is distributed lognormally with a variance of

[math]\displaystyle{ \sigma^2_{\ln\varepsilon} = \frac{5.5}{1 + \left(\hat{L}_f/4\right)^{7/9}}\ \ ,\ \ \hat{L}_f \equiv \hat{L} V_f^{3/4} = \frac{L}{L_K} V_f^{3/4} }[/math]

where [math]\displaystyle{ L_K=\left(\nu^3/\varepsilon \right)^{1/4} }[/math] is the Kolmogorov length, and [math]\displaystyle{ V_f }[/math] is the fraction of the shear variance that is resolved by terminating the spectral integration at an upper wavenumber of [math]\displaystyle{ k_u }[/math] .

The 95% confidence interval on an individual dissipation estimate, [math]\displaystyle{ \varepsilon_1 }[/math] is thus

[math]\displaystyle{ \mathrm{CF_{95}}(\varepsilon_1) = \varepsilon_1\, \exp\left(\pm1.96\,\sigma_{\ln\varepsilon} \right) \ \ . }[/math]

The 95% confidence interval for the geometric mean of a pair of dissipation estimates is

[math]\displaystyle{ \mathrm{CF_{95}} = \sqrt{\varepsilon_1\,\varepsilon_2} \, \exp\left(\pm1.96\,\sigma_{\ln\varepsilon}/\sqrt{2} \right) \ \ . }[/math]