Question #149100

A fluid flow is given by : V = 10x^3 i - 8x^3 yj. Find the shear strain rate and state whether the flow is rotational or irrotationsl.

Expert's answer

V=10x3i^−8x3yj^V=10x^3 \hat{i}-8x^3y\hat{j}

Gradient of velocity which is a second rank tensor is

LT=∇⃗V⃗=[dVxdxdVxdydVydxdVydy]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}

LT=[30x20−24x2y−8x3]L^T=\begin{bmatrix} 30x^2 & 0 \\ -24x^2y & -8x^3 \end{bmatrix}

Taking Its transpose

L=[30x2−24x2y0−8x3]L=\begin{bmatrix} 30x^2 & -24x^2y \\ 0 & -8x^3 \end{bmatrix}

Now strain rate tensor:

E=L+LT2E=12[60x2−24x2y−24x2y−16x3]E=\cfrac{L+L^T}{2}\\ E=\cfrac{1}{2}\begin{bmatrix} 60x^2 & -24x^2y \\ -24x^2y & -16x^3 \end{bmatrix}



where, off-diagonal terms represent shear strain rate.


Now Spin Tensor:

W=L−LT2W=12[0−24x2y24x2y0]W=\cfrac{L-L^T}{2}\\ W=\cfrac{1}{2}\begin{bmatrix} 0 & -24x^2y \\ 24x^2y & 0 \end{bmatrix}

Since off-diagonal term in spin tensor are non zero hence flow is rotational.

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