Drag and Lift of square cylinder

Question

I am trying to find the drag and lift of around a square cylinder. Shall I consider the boundaries of the whole domain for the integration or I can just take a portion of it? e.g. if i have considered $H=16D$, in order to calculate the lift shall I consider the integration on $y=-8D$ and $y=8D$ or I can find it over (say) $y=-2D$ and $y=2D$.

Answer 1

You want a domain that is sufficiently large that you can use easy boundary conditions, but also small and computationally cheap. Let's say that the center of the cylinder is in the origin, and that the domain is $x \in (-L,L)$ and $y\in (-H,H)$. If you will apply boundary conditions of uniform flow, you need to be sure that the exact solution is pretty close to the uniform flow in the boundary, i.e., at $x=\pm L$ and $y = \pm H$. AFAIK, there are no exact solutions for viscous flow around a square cylinder, so we have to be creative.

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I would consider the potential flow around a square cylinder and search for $L$ and $H$ such that the magnitude of the velocity is, for example, 99% of the far-field velocity. Naturally, that would be only an educated guess.

The solution for the square cylinder can be obtained from conformal mapping from the solution for the circular cylinder. However, we can get an idea of suitable dimensions for the domain by looking directly at the case of the circular cylinder. From the formulas in Wikipedia, we can obtain the magnitude of velocity in polar coordinates as $$ \left(\frac{\| \mathbf{V} \|}{U_\infty}\right)^2 = 1 - 2 \frac{R^2}{r^2} \cos 2 \theta + \frac{R^4}{r^4}, $$ in which $R$ is the radius of the cylinder. Along direction $x$ we have $r=x$ and $\theta = 0$. Defining $l=R^2/x^2$ for convenience, we have $$ l^2 - 2 l + 1 - \left(\frac{\| \mathbf{V} \|}{U_\infty}\right)^2 = 0, $$ $$ l = 1 - \frac{\| \mathbf{V} \|}{U_\infty}, $$ in which we chose this root so that $x>R$. See that for $\theta=0$, the velocity is smaller than the far-field velocity for all $x$. So, for $\| \mathbf{V} \|/U_\infty=0.99$, $l= 0.01$ and, therefore, $L/R = 1/\sqrt{l} = 10$. This means that, for the solution in potential flow, at a horizontal distance of 10 times the cylinder's radius, the magnitude of the velocity is 99% of the far-field velocity, and one could say that one can apply the boundary condition of uniform velocity there.

Along direction $y$, $r=y$ and $\theta = \pi/2$. Let $h = R^2/y^2$. Then: $$ h^2 + 2 h + 1 - \left(\frac{\| \mathbf{V} \|}{U_\infty}\right)^2 = 0, $$ $$ h = \frac{\| \mathbf{V} \|}{U_\infty} - 1. $$ Along $y$ direction the velocity is always larger than the far-field velocity. If we are looking for the position such that $\| \mathbf{V} \|/U_\infty = 1.01$, we have $h = 0.01$ and $H=10$.

Therefore, if you set the domain such that $x \in (-10 R, 10 R)$ and $y \in (-10 R, 10 R)$, the solution for potential flow in the circular cylinder will be between 99% and 101% of the actual value. Surely, this calculation was evaluated for potential flow around a circular cylinder, but it agrees with the rule of thumb of using the domain size one order of magnitude larger than the characteristic size of the object (for external flow).

Another "issue" with that approximation is that you probably will have a recirculation zone behind the cylinder. In that case, maybe you will need a larger domain in the region behind the cylinder than in the region in front of it.

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Finally, what actually is done for external flow (at least for more complicated geometries) is to compare the solution for progressively larger domains until the differences in the solutions are sufficiently small. Then you could run the calculation for a domain twice as large as the cylinder, then 5 times larger, and so on, until you stop noticing difference. This procedure is called domain independence test. Notice that the same is usually done with the mesh, i.e., the mesh is refined until no differences are noticed, in the procedure known as mesh independence test.