# Can the drag coefficient be calculated analytically in closed form?

I'm studying applied maths and physics and we did some experiments where we calculated the drag coefficients of different objects by measuring the drag force $$F_W$$ on the object and static pressure $$p_s$$ of the airflow. Then the drag coefficient $$c_W$$ could be calculated using: $$c_W=\frac{F_W}{Ap_s}$$ with $$A$$ being the projected Area of the object in the direction of the airflow.

We also used finite element analysis (COMSOL Multiphysics) to simulate the experiment and calculate $$c_W$$ that way.

Question: Are there setups where one could calculate $$c_W$$ analytically in closed form?

And related: Where do the $$c_W$$ values in the literature come from? From Experiments only?

I understand that once turbulent flow occurs, analytically calculating most things is out of the picture, but maybe it's possible for laminar or creeping flow?

Clarification: I'm looking for an equation in closed form for $$c_W$$, which does not use approximations.

• Sure it can be calculated analytically, if analytically includes numerical solution of the relevant (Navier-Stokes) equations. – Chet Miller Dec 13 '19 at 18:27
• @ChetMiller I edited the question to clarify what I meant by analytically. Thanks for making me realize the distinction. – NiveaNutella Dec 13 '19 at 18:33
• For small Reynolds numbers, the drag on a spherical object in viscous incompressible fluid can be calculated analytically, en.wikipedia.org/wiki/Stokes%27_law – Maxim Umansky Dec 13 '19 at 18:40
• @MaximUmansky Reading through the article, I came across this "In Stokes flow, at very low Reynolds number, the convective acceleration terms in the Navier–Stokes equations are neglected". So this is an approximation, is it not? If some part of the Navier-Stokes equation always have to be ignored to arrive at an analytic closed form solution, then I guess the answer to my question is "no". – NiveaNutella Dec 13 '19 at 19:12
• @NiveaNutella - formally, that solution is asymptotic, in the limit of infinitely small Reynolds number. But in practice it works quite well for finite (but small) Reynolds numbers. – Maxim Umansky Dec 13 '19 at 19:25

Furthermore as the drag force strongly depends on unsteady effects, most numerical methods will fail to predict it accurately. The current standard in the industry, RANS turbulence models (such as $$k-\epsilon$$ or $$k-\omega$$) apply time-averaging and are prone to horrendous errors in determining integral quantities such as drag and lift. More accurate numerical simulations such as large-eddy (LES) and hybrid RANS-LES models, that apply either of the two models depending on the distance to the nearest wall, are still computationally prohibitive in terms of number of mesh cells and time step size (they require huge computational clusters and still will be under-resolved) and thus experiments are still the cheapest and most accurate method to date for determining lift and drag coefficients for most applications. In cases where manufacturing a working prototype might be expensive and slow, such as in the automotive industry, first a virtual prototype and numerical methods will be used to iteratively improve the concept and then verified with actual measurements. If the precise magnitude of the corresponding quantities is secondary, e.g. when comparing different variants, even using cheap computational methods might give valuable information.
• @NiveaNutella The creeping flow is the limiting case of $Re := \frac{U L}{\nu} \to 0$, meaning the viscosity $\mu = \rho \nu$ dominates over inertial effects. The Stokes' solution only holds exactly for a vanishing inertia but we can apply it with good accuracy to flows $Re \lesssim 1$. – 2b-t Dec 13 '19 at 20:12
• @NiveaNutella It is exact in the limit $Re \to 0$. In fluid mechanics almost everything is an approximation: No substance will behave like a Newtonian fluid but still it is a reasonably accurate assumption for most fluids - materials will unlikely be even completely isotropic or have constant material values within the range of interest. Nothing is incompressible but for a small Mach number it might be a viable approximation. You won't find an ideal gas but under certain conditions it might be a good simplification or a necessary assumption to estimate parameters and obtain closed solutions. – 2b-t Dec 13 '19 at 20:24