Constitutive relation: hypothesis 4

The fourth constitutive relation hypothesis states that the properties of a fluid are isotropic, i.e. independent of direction.

For example, an object moving in a fluid encounters the same resistance regardless of the direction of movement. From eq11, the properties of a fluid are described by β_ijkl, which must be isotropic.

As described in the articles on tensors, the general form of a fourth-order isotropic tensor is:

$\beta_{ijkl}=\lambda\delta_{ij}\delta_{kl}+\mu\delta_{ik}\delta_{jl}+\nu\delta_{il}\delta_{jk}\; \; \; \; \; \; \; (12)$

From eq11 of the previous article, τ_ijis symmetric. Therefore, τ_ij=τ_jiand

$\beta_{ijkl}=\beta_{jikl}$

Substitute eq12 in the above equation, we have

$\left ( \mu-\nu \right )\delta_{ik}\delta_{jl}=\left ( \mu-\nu \right )\delta_{il}\delta_{jk}$

$\mu=\nu\; \; \; \; \; \; \; (13)$

Substitute eq13 in eq12,

$\beta_{ijkl}=\lambda\delta_{ij}\delta_{kl}+\mu\left (\delta_{ik}\delta_{jl}+\delta_{il}\delta_{jk}\right )\; \; \; \; \; \; \; (14)$

Substitute eq14 in eq11,

$\tau_{ij}=\frac{1}{2}\lambda\delta_{ij}\delta_{kl}\left ( \frac{\partial u_k}{\partial x_l}+\frac{\partial u_l}{\partial x_k}\right ) +\frac{1}{2}\mu\left (\delta_{ik}\delta_{jl}+\delta_{il}\delta_{jk}\right )\left ( \frac{\partial u_k}{\partial x_l}+\frac{\partial u_l}{\partial x_k}\right )$

$\tau_{ij}=\frac{1}{2}\lambda\delta_{ij}\left ( \frac{\partial u_k}{\partial x_k}+\frac{\partial u_k}{\partial x_k}\right ) +\frac{1}{2}\mu\left (\delta_{ik}\frac{\partial u_k}{\partial x_j}+\delta_{ik}\frac{\partial u_j}{\partial x_k} +\delta_{il}\frac{\partial u_j}{\partial x_l}+\delta_{il}\frac{\partial u_l}{\partial x_j} \right )$

$\tau_{ij}=\lambda\delta_{ij} \frac{\partial u_k}{\partial x_k}+\mu\left (\frac{\partial u_i}{\partial x_j}+\frac{\partial u_j}{\partial x_i} \right )\; \; \; \; \; \; \; (15)$

Substitute eq15 in eq3,

$\sigma_{ij}=-p\delta_{ij}+\lambda\delta_{ij} \frac{\partial u_k}{\partial x_k}+\mu\left (\frac{\partial u_i}{\partial x_j}+\frac{\partial u_j}{\partial x_i} \right )\; \; \; \; \; \; \; (16)$

The values of μ and λ can only be determined through experiments. μ is known as the shear viscosity of the fluid while λ is the volume viscosity of the fluid, which is zero for an incompressible fluid. Eq16 becomes:

$\sigma_{ij}=-p\delta_{ij}+\mu\left (\frac{\partial u_i}{\partial x_j}+\frac{\partial u_j}{\partial x_i} \right )\; \; \; \; \; \; \; (17)$

Question

Show that μ in eq16 and eq7 are the same.

Answer

Consider the flow of an incompressible fluid where the flow velocity components are u = u(y), v = 0 and w = 0. Using eq16, the stress components are:

$\sigma_{11}=\sigma_{22}=\sigma_{33}=-p$

$\sigma_{12}=\sigma_{21}=\mu\frac{du}{dy}$

with the remaining components equal to zero.

Clearly, the components of stress in this case include the pressure p and the shear stress $\mu\frac{du}{dy}$ , which is the result of the derivation of Newton’s law of viscosity, eq7. Hence, μ in eq16 and eq7 are the same.

Constitutive relation: hypothesis 4

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