Assumed width for 2D Navier-stokes and energy equation

Question

Let us consider the two-dimensional Navier-Stokes equation being solved for a channel of height $d$ and length $L$, wherein a fluid enters with an average velocity $u$. These dimensions are $L=40 \ mm, d=3 \ mm$. Now if I want to calculate the mass flow rate of the fluid $m$, entering the channel, it will be done using $m=\rho A_{cs} u$, where $A_{cs}$ is the channel cross-sectional area. My question is, how will this $A_{cs}$ be calculated ?

Will it be $A_{cs}=0.003*1 \ m^2$ ?

Similarly, for the two-dimensional energy equation, if a flux boundary condition of $q''=5000W/m^2$ is specified, what will the net heat input to the system be ?

Will it be $q=5000*0.04*1 W$ ?

In essence, what is the assumed width ?

Answer 1

Using 2D approximation of a problem, you're often implicitly assuming that the $3^{rd}$ direction you're neglecting is a homogeneous direction, i.e. the domain of the PDEs has a symmetry (translation symmetry, if you're using Cartesian coordinates, neglecting one of them, the 'out-of-plane' direction - let's call it $z$) and the solution doesn't depend on that coordinate as well.

Navier-Stokes in 2D. Namely, if you have a $z$ homogeneous direction, the velocity field is

$\mathbf{u}(\mathbf{r}) = u_x(x,y) \mathbf{\hat{x}} + u_y(x,y) \mathbf{\hat{y}} $.

You can evaluate a flux per unit-length, and a retrieve the flux exploiting the homogeneity of the $z$ coordinate.

As an example, if you're representing the flow between two large parallel flat plates, normal to the direction $y$, separated by a thin gap (so that the Reynolds number is low enough to avoid instabilities), the velocity field is approximately

$\mathbf{u}(\mathbf{r}) = 4 U \dfrac{y}{H}\left(1-\dfrac{y}{H} \right) \mathbf{\hat{x}}$,

being $U$ the maximum magnitude of the velocity field, $y=0$ and $y=H$ the coordinates of the flat plates, $x$ the direction of the flow, and $z$ the 'out-of-plane' homogeneous direction.

You can easily evaluate the mass flow per unit "depth" across a section of the channel with the integral

$\displaystyle \dot{m}_{2D} = \int_{y=0}^{H} \rho u_x(y) dy = \dfrac{2}{3} \rho U H = \rho U_{avg} H$, $\qquad $ with $\qquad[m_{2D}] = \frac{kg}{s \ m}$

being $U_{avg} = \frac{2}{3}U$ the average velocity of the profile.

If you need to evaluate the flux across a rectangular surface orthogonal to the velocity whose $z$-side has length $B$, you'd need to perform the integral

$\displaystyle \dot{m} = \int_{z=0}^B \int_{y=0}^{H} \rho u_x(y) dy$,

resulting in

$\displaystyle \dot{m} = \rho U_{avg} H B $, $\qquad $ with $\qquad[m] = \frac{kg}{s}$,

since the problem doesn't depend on $z$.

Heat flux in 2D. The same occurs if you consider a 2D heat conduction problem. The flux across a (one-dimensional) section of the boundary of the domain is a flux per unit length, and has the dimension of $\frac{W}{m}$.

Answer 2

Given an inlet mass flow rate in 3D as $\dot{m}_\text{3D}$, the 2D case would be given by, $$\dot{m}_\text{2D}=\dot{m}_\text{3D}\cdot\frac{H\cdot L_\text{ref}}{A_\text{3D}}$$ where $H$ is the height of the inlet, $L_\text{ref}$ a reference length (thickness) and $A_\text{3D}$ the inlet surface area of the 3D case. Some numerical packages/software (e.g., Fluent), define this reference length as 1 meter in non-axisymmetric simulations and $2\pi$ in axisymmetric cases; you probably would want to double check the reference docs for whatever numerical package you're using however.

For more complex geometries, you probably will have to compute the integral, $$m=-\int_{\partial\Omega}\rho\left(\mathbf{u}\cdot\mathbf{n}\right)\mathrm{d}S$$ for your system.