Determine the shearing stress for an incompressible Newtonian fluid with a velocity distribution of \(\mathbf{V}=\left(3 x y^{2}-4 x^{3}\right) \mathbf{i}+\) \(\left(12 x^{2} y-y^{3}\right) \mathbf{j}\).

Short Answer

Expert verified
The shear stress of the incompressible Newtonian fluid can be represented by two components in the i and j directions. These are \(τ_{i} = μ (3y^2 - 12x^2)\) and \(τ_{j} = μ (6xy)\), where μ represents the dynamic viscosity of the fluid.

Step by step solution

01

Extract the components of the velocity field

The first step is to extract the components of the velocity field which are \( u = 3xy^2 - 4x^3 \) and \( v = 12x^2y - y^3 \). These represent the i and j components of the vector, respectively.
02

Differentiate the components

Differentiate both u and v with respect to x and y, which will give us four results: \( du/dx, du/dy , dv/dx, dv/dy \). After differentiation, we get: \( du/dx = 3y^2 - 12x^2 \), \( du/dy = 6xy \), \( dv/dx = 24xy \), and \( dv/dy = 12x^2 - 3y^2 \).
03

Calculate the velocity gradient tensor

Next, we construct the velocity gradient tensor, which is a square matrix ordered as: \(\[ \frac{du}{dx} , \frac{du}{dy} ; \frac{dv}{dx} , \frac{dv}{dy} \] \). Substitution of the values obtained in step 2 into this matrix we get: \(\[3y^2 - 12x^2 , 6xy ; 24xy , 12x^2 - 3y^2 \] \). The off-diagonal elements of this tensor represent the shear rates.
04

Compute the Shearing Stress

We can now calculate the shearing stress for the fluid by multiplying the shear rates by the fluid's dynamic viscosity (μ). If we denote the dynamic viscosity as μ, the shear stress (τ) is given by: \( τ = μ * shear rate = μ * velocity gradient \). Thus, the components of shear stress in the i and j directions are \(τ_{i} = μ (3y^2 - 12x^2)\) in the i direction, and \(τ_{j} = μ (6xy)\) in the j direction. Remember, the dynamic viscosity (μ) should be given or assumed to calculate these values.

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