# Calculating momentum flux through a nozzle exit

I'd like to calculate the momentum flux in the x-direction through the exit of a tube (oriented normally to the x-direction).

I have a numerical solution of the flowfield in which I have flow quantities (density, velocity, etc) stored in a large number of cells making up this exit surface.

Momentum flux = $$\iint\rho ( \mathbf{V}\cdot\hat{n} )\mathbf{V}dA$$

To solve for momentum flux, I've been multiplying the mass flow rate by the bulk-averaged x-velocity at this surface.

Is this correct? I'd like to be able to show the steps logically from the integral equation to my numerical solution but I've been failing to get from Point A to B.

• Where is the cross-sectional area in your computation? – Ted Shifrin Sep 5 at 16:14
• Well, the momentum flux integral is a surface integral. I don't see that in your elementary computation. – Ted Shifrin Sep 5 at 16:28

Let $$S_e$$ denote the cross section at the exit of the tube. The exact momentum flux is

$$Q_{\text{momentum}} = \int_{S_e}\rho \mathbf{V} \mathbf{V} \cdot \mathbf{n} \, dA$$

In your solution for momentum flux as the product of the mass flow rate and the bulk-averaged x-velocity at the exit, you are implying that

$$\int_{S_e}\rho \mathbf{V} \mathbf{V} \cdot \mathbf{n} \, dA = \frac{ \int_{S_e}\rho \mathbf{V} \cdot \mathbf{n} \, dA \int_{S_e}\mathbf{V} \, dA}{ \int_{S_e}\, dA}$$

This need not be true as an equality but may serve as a reasonable approximation, for example, in turbulent flow where the velocity profile is relatively flat.

To examine this in the case of laminar flow, assume that cross section is a disk of radius $$R$$ normal to the x-direction and the velocity profile is that of steady axisymmetric Poiseuille flow,

$$\mathbf{V} = C\left[1- \left(\frac{r}{R}\right)^2\right]\mathbf{e_x}$$

In this case we have $$\mathbf{n} = \mathbf{e_x}$$ and

$$\int_{S_e}\rho \mathbf{V} \mathbf{V} \cdot \mathbf{n} \, dA = 2\pi \rho C^2\int_0^R\left[1- \left(\frac{r}{R}\right)^2\right]^2 r\, dr \,\,\mathbf{e_x} = \frac{\pi \rho C^2R^2}{3}\, \mathbf{e_x} \\ \int_{S_e}\, dA = \pi R^2 \\ \int_{S_e}\mathbf{V} \, dA = 2\pi C \int _0^R\left[1- \left(\frac{r}{R}\right)^2\right] r\, dr\, \mathbf{e_x} = \frac{\pi C R^2}{2}\, \mathbf{e_x} \\ \int_{S_e}\rho \mathbf{V} \cdot \mathbf{n} \, dA = 2\pi \rho C \int_0^R\left[1- \left(\frac{r}{R}\right)^2\right] r\, dr = \frac{\pi \rho CR^2}{2}$$

Hence,

$$\frac{ \int_{S_e}\rho \mathbf{V} \cdot \mathbf{n} \, dA \int_{S_e}\mathbf{V} \, dA}{ \int_{S_e}\, dA} = \frac{\pi\rho C^2 R^2}{4} \mathbf{e_x} \neq \frac{\pi\rho C^2 R^2}{3 } \mathbf{e_x} = \int_{S_e}\rho \mathbf{V} \mathbf{V} \cdot \mathbf{n} \, dA$$

Here your approximation is not particularly accurate -- deviating from the exact momentum flux by a factor of $$3/4$$.