V = 10 x 3 i ^ − 8 x 3 y j ^ V=10x^3 \hat{i}-8x^3y\hat{j} V = 10 x 3 i ^ − 8 x 3 y j ^ Gradient of velocity which is a second rank tensor is
L T = ∇ ⃗ V ⃗ = [ d V x d x d V x d y d V y d x d V y d y ] L^T=\vec\nabla \vec V=\begin{bmatrix}
\cfrac{dV_x}{dx} & \cfrac{dV_x}{dy} \\
\cfrac{dV_y}{dx} & \cfrac{dV_y}{dy}
\end{bmatrix} L T = ∇ V = ⎣ ⎡ d x d V x d x d V y d y d V x d y d V y ⎦ ⎤
L T = [ 30 x 2 0 − 24 x 2 y − 8 x 3 ] L^T=\begin{bmatrix}
30x^2 & 0 \\
-24x^2y & -8x^3
\end{bmatrix} L T = [ 30 x 2 − 24 x 2 y 0 − 8 x 3 ] Taking Its transpose
L = [ 30 x 2 − 24 x 2 y 0 − 8 x 3 ] L=\begin{bmatrix}
30x^2 & -24x^2y \\
0 & -8x^3
\end{bmatrix} L = [ 30 x 2 0 − 24 x 2 y − 8 x 3 ] Now strain rate tensor:
E = L + L T 2 E = 1 2 [ 60 x 2 − 24 x 2 y − 24 x 2 y − 16 x 3 ] E=\cfrac{L+L^T}{2}\\
E=\cfrac{1}{2}\begin{bmatrix}
60x^2 & -24x^2y \\
-24x^2y & -16x^3
\end{bmatrix} E = 2 L + L T E = 2 1 [ 60 x 2 − 24 x 2 y − 24 x 2 y − 16 x 3 ]
where, off-diagonal terms represent shear strain rate.
Now Spin Tensor:
W = L − L T 2 W = 1 2 [ 0 − 24 x 2 y 24 x 2 y 0 ] W=\cfrac{L-L^T}{2}\\
W=\cfrac{1}{2}\begin{bmatrix}
0 & -24x^2y \\
24x^2y & 0
\end{bmatrix} W = 2 L − L T W = 2 1 [ 0 24 x 2 y − 24 x 2 y 0 ] Since off-diagonal term in spin tensor are non zero hence flow is rotational.
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