What would be the Stokes hypothesis in a 1D flow?

Question

In this paper, the author derives the Navier-Stokes equation for a Newtonian fluid starting from the Cauchy equation: $$\rho \frac{D\mathbf V}{Dt} = \rho \mathbf{f} + \nabla\cdot\mathbf{T}$$ where $\mathbf{T} = -p\mathbf I + \boldsymbol{\tau}_v$ is the stress tensor, $p$ being the pressure and $\boldsymbol\tau_v$ the viscous stress tensor. He then writes down the constitutive equation for the flow (equation (3) in the paper): $$\boldsymbol\tau_v = \lambda(\nabla\cdot\mathbf V)\mathbf I + 2\mu \mathbf E$$ and he divides $\mathbf E$ into its isotropic and deviatoric parts $\mathbf A$ and $\mathbf D$, to obtain $$\boldsymbol\tau_v = \kappa(\nabla\cdot\mathbf V)\mathbf I + 2\mu \mathbf D$$ where $\kappa = \lambda + \frac{2}{3}\mu$. Substituting the above into $\mathbf{T} = -p\mathbf I + \boldsymbol{\tau}_v$ yields the expression for $\mathbf T$ in terms of pressure and velocity gradients which can be used in equation (1) to obtain Navier-Stokes.

The author then states that the Stokes hypothesis assumes $\kappa = 0$, which is to say the viscous stress tensor has no isotropic part, or that:

isotropic dilatations of an elementary volume of fluid do not produce viscous stresses

Now, I want to write the 1D Navier-Stokes which respects the Stokes hypothesis using the information in the paper. There are two ways to do that:

In the first way I compute the stress tensor from equation (7) in the paper using $\kappa=0$, and substitute the result back in equation (1). It turns out that for a 1D flow all non diagonal terms of $\mathbf T$ are zero and $$T_{11} = -p + \frac{4}{3}\mu\frac{\partial V}{\partial x}, \quad T_{22} = T_{33} = -p - \frac{2}{3}\mu\frac{\partial V}{\partial x}.$$ Substituting this into (1) (and assuming $\mu$ is constant) yields

$$\boxed{\rho \frac{DV}{Dt} = \rho f - \frac{\partial p}{\partial x} + \frac{4}{3}\mu\frac{\partial^2 V}{\partial x^2}}.$$

In the second method, we start by reducing equation (1) itself into 1D space, so we write: $$\rho\frac{DV}{Dt} = \rho f + \frac{\partial T}{\partial x},$$ where $T$ here is just a scalar. Then we write $T$ as $$T = -p + \tau_v,$$ where $\tau_v$ is also a scalar, and its constitutive equation is $$\tau_v = \kappa \frac{\partial V}{\partial x},$$ which only contains a single Lamé parameter $\kappa$. Since $\tau_v$ is a scalar, it doesn't have any deviatoric part, and the Stokes hypothesis would simply be $\tau_v = 0$, so that the Navier-Stokes equation would be written as: $$\boxed{\rho \frac{DV}{Dt} = \rho f - \frac{\partial p}{\partial x}}$$ which is different than the equation obtained with the first method. So which is the correct one?

Edit: The kind of one-dimensional flow I'm considering is that in which all physical quantities depend on a single spacial coordinate ($x$-coordinate for example) and in which fluid motion is along this same coordinate (along the $x$-axis).

Answer 1

It turns out that for a 1D flow all non diagonal terms of T are zero

This is incorrect, a 1D flow is a flow where only one of the components of the velocity exists. That does not mean that the off diagonal terms of the velocity gradient like $\frac{\partial{v_x}}{\partial{y}}$ are zero. This is why your second derivation fails and you end up assuming that the fluid is inviscid without justification

I don't know if your first derivation is correct, but if you follow this: https://en.wikipedia.org/wiki/Navier–Stokes_equations

And use the last equation in the red rectangle right before "incompressible flow", and do every operation correctly, you should be good.

Edit: It doesn't really matter if you just restrict the variables to be only dependent on $x$, you cannot a priori remove terms from the stress tensor without the constitutive equation, because cauchy equations of motion deal with the divergent of the stress tensor, not the actual value of the stresses, which are determined by the constitutive equation. The best you can do with the equations of motion is:

$$ \nabla \cdot T = 0$$

Which means that the stress is overall uniform, and thats it, that doesn't mean there are no shears, or normal stresses, only that they do not vary in each point of the body. A priori there are infinite possible constitutive equations, they don't necessarily follow linear elasticity equations.

Keeping this always in mind, the mistake in your second derivation is assuming there are no deviatoric stresses. The splitting of the stress tensor into hydrostatic and deviatoric goes by:

$$\begin{bmatrix} T_{xx} & T_{xy} & T_{xz} \\ T_{yx} & T_{yy} & T_{yz}\\ T_{zx} & T_{xy} & T_{zz} \\ \end{bmatrix} = \begin{bmatrix} T_{hyd} & 0 & 0 \\ 0 & T_{hyd} & 0\\ 0 & 0 & T_{hyd} \\ \end{bmatrix} + \begin{bmatrix} T_{xx}-T_{hyd} & T_{xy} & T_{xz} \\ T_{yx} & T_{yy}-T_{hyd} & T_{yz}\\ T_{zx} & T_{xy} & T_{zz}-T_{hyd} \\ \end{bmatrix}$$

Given the constitutive equation, if all variables are dependent only on x, then the off diagonal terms are zero, then the result becomes: $$\begin{bmatrix} T_{xx} & 0 & 0 \\ 0 & T_{yy} & 0\\ 0 & 0 & T_{zz} \\ \end{bmatrix} = \begin{bmatrix} T_{hyd} & 0 & 0 \\ 0 & T_{hyd} & 0\\ 0 & 0 & T_{hyd} \\ \end{bmatrix} + \begin{bmatrix} T_{xx}-T_{hyd} & 0 & 0 \\ 0 & T_{yy}-T_{hyd} & 0\\ 0 & 0 & T_{zz}-T_{hyd} \\ \end{bmatrix}$$

So, just because no shear stress exist, that does not mean that no deviatoric stress exist.