I'm trying to calculate the drag force for a laminar flow around a cylinder.

We make the following assumptions :

  • the flow is horizontal
  • the flow is incompressible ($\vec{\nabla} \cdot \vec{v} = 0$)
  • the flow is stationnary
  • We are in 2D

Here's a picture of the situation :


I would like to start with the conservation of the momentum using Reynold Transport Theorem :

$$ \int_\Omega \rho v dS + \int_{\partial{\Omega}} \rho v^2 dl = -\vec{F}_d $$

However, in the lectures notes, there is no $ \int_\Omega \rho v dS $ term.

In fact, it says that $$ \int_{\partial{\Omega}} \rho v^2 dl = -\vec{F}_d \implies 2(\int_0^\xi -pU^2_\infty dl + \int_0^\xi pu^2 dl + \int_0^L \rho u v dl) = - \vec{F}d $$

I don't understand why ?


Your formula for the Reynold Transport theorem is false (you have forgotten a partial derivative in the first integral).

The correct version is shown below:

$$-\overrightarrow{F_d} = \frac{D}{Dt} \int \limits_{\Omega} \rho \overrightarrow{v}(\mathbf{x},t) dS = \int \limits_{\Omega} \rho \frac{\partial \overrightarrow{v}(\mathbf{x},t)}{\partial t} dS+ \int \limits_{\partial \Omega} \rho \overrightarrow{v}(\mathbf{x},t) (\overrightarrow{v(\mathbf{x},t)}\cdot \overrightarrow{n_{\partial \Omega}}) dl,$$

where $\overrightarrow{n_{\partial \Omega}}$ is the vector normal to the boundary $\partial \Omega$ (see for instance this Wikipedia page). If the flow is stationnary, the first term is zero, and you just have the second term remaining, that can be simplified if $\overrightarrow{v}$ is orthogonal to the surface $\partial \Omega$ into:

$$-\overrightarrow{F_d} = \int \limits_{\partial \Omega} \rho v(\mathbf{x},t) \overrightarrow{v}(\mathbf{x},t) dl$$

This can be seen as a macroscopic version of Newton's second law: for a stationary flow, the fluid exiting the surface $\Omega$ has a smaller momentum than the one entering it, so it must be that the cylinder is exerting an equivalent force $\overrightarrow{F_d}$ on the fluid such that $\overrightarrow{F_d} = \frac{d \overrightarrow{P_{\mathrm{fluid}}}}{dt}$ (and using the third law of motion, the fluid must exert a opposite force $-\overrightarrow{F_d}$ on the cylinder).


Is the drag not zero by d'Alembert's paradox? https://en.wikipedia.org/wiki/D%27Alembert%27s_paradox

  • $\begingroup$ I guess since the problem does not mention viscosity, the fluid can be viscous in general, so d'Alembert paradox does not hold here. Otherwise I agree that the force should be zero but the formula given in the question is still true (it simply shows that $\int \limits_{\partial \Omega} \rho \overrightarrow{v} v dl = \overrightarrow{0}$ for the stationary flow of a non-viscous fluid. $\endgroup$ Feb 26 '20 at 0:20
  • $\begingroup$ @QuantumApple: How does the formula in the Q show that $\int_{\partial \Omega} \rho {\bf v} ({\bf v}\cdot d{\bf S})=0$? I thought you said it was $-F_d$? And I agree with that. $\endgroup$
    – mike stone
    Feb 26 '20 at 12:25
  • $\begingroup$ Sorry for the confusion, I was answering to @mike stone. If the fluid is non-viscous (which it isn't in general), then d'Alembert's paradox says that $\overrightarrow{F_d} = \overrightarrow{0}$, so that the integral is also $0$. But the integral is indeed equal to $-\overrightarrow{F_d}$ in general. $\endgroup$ Feb 26 '20 at 12:52

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