Question

The viscosity terms in x-momentum equation is \(\frac{\partial\tau_{xx}}{\partial x}+\frac{\partial\tau_{yx}}{\partial y}+\frac{\partial\tau_{zx}}{\partial z}\). In a more general form, this becomes div(μ gradu). Which of these relations is used for this transformation?

(a) Thermodynamic relations

(b) Stress-strain relations

(c) Fluid flow relations

(d) Geometric relations

The question was posed to me during an online interview.

The query is from Navier Stokes Equation topic in portion Governing Equations of Fluid Dynamics of Computational Fluid Dynamics

Answer 1

The correct answer is:

(b) Stress-strain relations

Explanation:

In fluid dynamics, the transformation of the viscosity terms in the x-momentum equation to the form div(μ∇u)div(μ∇u) comes from stress-strain relations in fluid mechanics, specifically the relation between the shear stress τijτij​ and the rate of strain (velocity gradient) in the fluid.

The term μ∇uμ∇u is derived from the stress tensor in the context of Newtonian fluids, where the shear stress is proportional to the velocity gradients (the strain rate). This relationship is fundamental in deriving the Navier-Stokes equations for incompressible flow.

The term ∂τxx/∂x+∂τyx/∂y+∂τzx/∂z∂τxx​/∂x+∂τyx​/∂y+∂τzx​/∂z is an expression for the divergence of the stress tensor in the x-direction, which describes how momentum is transferred due to viscous effects.

Thus, stress-strain relations are used to transform these viscosity terms in the momentum equations.