# Laminar flow around a cylinder using conservation of momentum

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).

• 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. Feb 26, 2020 at 0:20
• @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. Feb 26, 2020 at 12:25
• 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. Feb 26, 2020 at 12:52