Analysis and treatment of errors due to high velocity gradients in particle image velocimetry

Abstract

This paper deals with errors occurring in two-dimensional cross-correlation particle image velocimetry (PIV) algorithms (with window shifting), when high velocity gradients are present. A first bias error is due to the difference between the Lagrangian displacement of a particle and the real velocity. This error is calculated theoretically as a function of the velocity gradients, and is shown to reach values up to 1 pixel if only one window is translated. However, it becomes negligible when both windows are shifted in a symmetric way. A second error source is linked to the image pattern deformation, which decreases the height of the correlation peaks. In order to reduce this effect, the windows are deformed according to the velocity gradients in an iterative process. The problem of finding a sufficiently reliable starting point for the iteration is solved by applying a Gaussian filter to the images for the first correlation. Tests of a PIV algorithm based on these techniques are performed, showing their efficiency, and allowing the determination of an optimum time separation between images for a given velocity field. An application of the new algorithm to experimental particle images containing concentrated vortices is shown.

Appendix

We seek an expression for the displacement Δr=r _f−r _i of a particle in the velocity field given by Eq. (3). At t=t _i, the particle is located at r _i, and at t=t _f at r _f. The origin of time (t=0) is given by the time at which one wishes to determine the velocity at the reference point 0 (of coordinates r=0). All derivatives are taken at this point and time. The particle trajectory r(t) is calculated by an iterative process as successive solutions of the differential equation

$$ \frac{{{\text{d}}{\mathbf{r}}}} {{{\text{d}}t}} = {\mathbf{\nu }}{\left[ {{\mathbf{r}}{\left( t \right)}} \right]} $$

(32)

at increasing orders of t and r.

At first order, the solution r ₁(t) of Eq. (32) [using Eq. (3) taken at order 0] is given by:

$$ {\mathbf{r}}_{1} {\left( t \right)} = {\mathbf{r}}_{{\text{i}}} + {\int_{t_{{\text{i}}} }^t {{\mathbf{\nu }}_{0} {\text{d}}{t}' = {\mathbf{r}}_{{\text{i}}} + {\mathbf{\nu }}_{0} (t - t_{{\text{i}}} )} } $$

(33)

Introducing this result into (3) leads, at first order, to:

$$ \frac{{{\text{d}}{\mathbf{r}}}} {{{\text{d}}t}} = {\mathbf{\nu }}{\left[ {{\mathbf{r}}_{1} {\left( t \right)}} \right]} = {\mathbf{\nu }}_{0} + {\mathbf{{\nu }'}}{\left( {{\mathbf{r}}_{{\text{i}}} + {\mathbf{v}}_{0} t} \right)} + t\partial _{{\text{t}}} {\mathbf{\nu }} $$

(34)

The solution of Eq. (34) is the approximation r ₂(t) of the trajectory to the second order:

$$ {\mathbf{r}}_{2} {\left( t \right)} = {\mathbf{r}}_{{\text{i}}} + {\mathbf{v}}_{0} {\left( {t - t_{{\text{i}}} } \right)} + \frac{{t^{2} - t^{2}_{{\text{i}}} }} {2}\partial _{{\text{t}}} {\mathbf{\nu }} + {\mathbf{{\nu }'r}}_{{\text{i}}} {\left( {t - t_{{\text{i}}} } \right)} + {\mathbf{{\nu }'\nu }}_{0} \frac{{{\left( {t - t_{{\text{i}}} } \right)}^{2} }} {2} $$

(35)

The third-order approximation r ₃(t) of the particle trajectory is found in the same way, the final result being

$$ \begin{array}{*{20}l} {{{\mathbf{r}}_{3} {\left( t \right)}} \hfill} & { = \hfill} & {{{\mathbf{r}}_{{\text{i}}} + {\int_{t_{{\text{i}}} }^t {\mathbf{v}} }{\left[ {{\mathbf{r}}_{{\text{2}}} {\left( {{t}'} \right)}} \right]}{\text{d}}{t}'} \hfill} \\ {{} \hfill} & { = \hfill} & {{{\mathbf{r}}_{{\text{i}}} + {\mathbf{v}}_{0} {\left( {t - t_{{\mathbf{i}}} } \right)} + {\mathbf{{v}'}}{\text{r}}_{2} {\left( {t - t_{{\text{i}}} } \right)}} \hfill} \\ {{} \hfill} & {{} \hfill} & {{ + {\mathbf{{v}'v}}_{0} \frac{{{\left( {t - t_{{\text{i}}} } \right)}^{2} }} {2} + {\mathbf{{v}'}}\partial _{{\text{t}}} {\mathbf{v}}{\left[ {\frac{{t^{3} - t^{3}_{{\text{i}}} }} {6} - \frac{{t^{2}_{{\text{i}}} }} {2}{\left( {t - t_{{\text{i}}} } \right)}} \right]} + {\mathbf{{v}'}}^{2} {\mathbf{r}}_{{\text{i}}} \frac{{{\left( {t - t_{{\text{i}}} } \right)}^{2} }} {2} + {\mathbf{{v}'}}^{2} {\mathbf{v}}_{0} \frac{{{\left( {t - t_{{\text{i}}} } \right)}^{3} }} {6}} \hfill} \\ {{} \hfill} & {{} \hfill} & {{ + \partial _{{\text{t}}} {\mathbf{v}}\frac{{t^{2} - t^{2}_{{\text{i}}} }} {2} + {\left( {\begin{array}{*{20}c} {{{\text{r}}^{\dag }_{{\text{i}}} {\mathbf{{v}'}}_{x} {\mathbf{r}}_{{\text{i}}} }} \\ {{{\text{r}}^{\dag }_{i} {\mathbf{{v}''}}_{y} {\mathbf{r}}_{{\text{i}}} }} \\ \end{array} } \right)}\frac{{t - t_{{\text{i}}} }} {2}} \hfill} \\ {{} \hfill} & {{} \hfill} & {{ + {\left( {\begin{array}{*{20}l} {{{\mathbf{v}}^{\dag }_{{\text{0}}} } \hfill} & {{{\mathbf{{v}''}}_{x} } \hfill} & {{{\mathbf{r}}_{{\text{i}}} + {\mathbf{r}}^{\dag }_{{\text{i}}} } \hfill} & {{{\mathbf{{v}''}}_{x} } \hfill} & {{{\mathbf{v}}_{0} } \hfill} \\ {{{\mathbf{v}}^{\dag }_{{\text{0}}} } \hfill} & {{{\mathbf{{v}''}}_{{\text{y}}} } \hfill} & {{{\mathbf{r}}_{{\text{i}}} + {\mathbf{r}}^{\dag }_{{\text{i}}} } \hfill} & {{{\mathbf{{v}''}}_{y} } \hfill} & {{{\mathbf{v}}_{0} } \hfill} \\ \end{array} } \right)}\frac{{{\left( {t - t_{{\text{i}}} } \right)}^{2} }} {4}} \hfill} \\ {{} \hfill} & {{} \hfill} & {{ + {\left( {\begin{array}{*{20}l} {{{\mathbf{v}}^{\dag }_{{\text{0}}} } \hfill} & {{{\mathbf{{v}''}}_{x} } \hfill} & {{{\text{v}}_{0} } \hfill} \\ {{{\mathbf{v}}^{\dag }_{{\text{0}}} } \hfill} & {{{\mathbf{{v}''}}_{{\text{y}}} } \hfill} & {{{\text{v}}_{0} } \hfill} \\ \end{array} } \right)}\frac{{{\left( {t - t_{{\text{i}}} } \right)}^{3} }} {6} + \partial _{{{\text{tt}}}} {\text{v}}\frac{{t^{3} - t^{3}_{{\text{i}}} }} {6}} \hfill} \\ {{} \hfill} & {{} \hfill} & {{ + \partial _{{\text{t}}} {\mathbf{{v}'r}}_{{\text{i}}} \frac{{t^{2} - t^{2}_{{\text{i}}} }} {2}} \hfill} \\ {{} \hfill} & {{} \hfill} & {{ + {\left( {\frac{{t^{3} - t^{3}_{{\text{i}}} }} {3} - \frac{{t^{2} - t^{2}_{{\text{i}}} }} {2}t_{{\text{i}}} } \right)}\partial _{{\text{t}}} {\mathbf{{v}'v}}_{{\text{0}}} } \hfill} \\ \end{array} $$

(36)

• Non-symmetric displacement: For the displacement corresponding to Fig. 1a, we have r _i=0, and the choice t _i=0 and t _f=Δt seems appropriate. Using Eq. (35), this leads to Eq. (4), showing that, in this case, the error between the measured velocity Δr/Δt, and the true velocity v ₀ at the measurement location and at the time of the first image is of second order in Δt. One could also choose the origin of time halfway between t _i and t _f [see Eq. (38) below], which means that the measured velocity field is associated with the instant between the exposures of the two images. In this case, the term proportional to ∂_t v in Eq. (4) would vanish, but the error, now given by Eq. (39), would still remain O(Δt ²).

• Symmetric displacement: For the displacement corresponding to the symmetric window shifting in Fig. 1b, the following relations hold:

$$ {\mathbf{r}}_{{\text{f}}} = - {\mathbf{r}}_{{\text{i}}} = \frac{{\Delta {\mathbf{r}}}} {2} $$

(37)

$$ t_{{\text{f}}} = - t_{{\text{i}}} = \frac{{\Delta t}} {2} $$

(38)

Introducing Eq. (38) into Eq. (35), we obtain for t=t _f:

$$ \Delta {\mathbf{r}} = {\mathbf{\nu }}_{0} \Delta t + {\mathbf{{\nu }'r}}_{{\text{i}}} \Delta t + {\mathbf{{\nu }'\nu }}_{0} \frac{{\Delta t^{2} }} {2} + O{\left( {\Delta t^{3} } \right)} $$

(39)

and, with Eq. (37) and I being the unit matrix

$$ {\left( {{\mathbf{I}} + \frac{{\Delta t}} {2}{\mathbf{{\nu }'}}} \right)}{\left( {\Delta {\mathbf{r}} - {\mathbf{\nu }}_{{\text{0}}} \Delta t} \right)} = O{\left( {\Delta t^{3} } \right)} $$

(40)

This results in

$$ \Delta {\mathbf{r}} = {\mathbf{\nu }}_{{\text{0}}} \Delta t + O{\left( {\Delta t^{3} } \right)} $$

(41)

showing that, for a symmetric displacement, the error is only of order Δt ³. The expression in Eq. (5) for this higher-order term is found by introducing Eqs. (37), (38) and (41) into Eq. (36).