# How can I carry out this force summation? Control volume involved

I have a stationary jet engine on a test stand, like shown in the next figure: The control volume is indicated in the figure as a dashed rectangle, that has its left side far upstream of the engine, and its bottom side far from the floor on which it rests. We consider no friction.

We know that the summation of forces over a control volume surrounding a stationary system is given by:

$\sum \textbf{F} = \int_{cs} \rho \textbf{u}\left(\textbf{u} \dot{} \textbf{n}\right)dA$

Where:

• cs stands for Control Surface.
• $\rho$ is the density of the fluid.
• $\textbf{u}$ is the velocity vector of the fluid, be it entering or exiting the cs.

The problem here is finding the expression of $\sum \textbf{F}$. I know the solution, but I am unable to find a way to reach it on my own. The solution, as given by the book, is:

$\sum \textbf{F}=T+A_ep_a-A_ep_e$

Where:

• T is the reaction to thrust shown in the above figure.
• $A_e$ is the nozzle area indicated in the figure.
• $p_a$ is the ambient pressure of air.
• $p_e$ is the exiting flow pressure.

What puts me off from finding that specific solution:

• The product $A_ep_a$ and the reaction to thrust $T$ both go in opposite directions but have the same sign in the final expresion of $\sum \textbf{F}$. Same thing happens to the product $A_ep_e$
• I am not sure I can determine whether a force is "pushing" the Control Volume (that is, entering the CV) or "pulling" (exiting) it.

Any help is greatly appreciated. Thanks.

EDIT

The hypothesis are (I think I didn't miss a single one):

• Inviscid and stationary flow.
• No fricton involved.
• Ambient pressure of air, $p_a$, doesn't appreciably change over surfaces, except possibly in the exhaust plane of the jet.
• The reaction to thrust, $T$, is considered to be a normal force.
• We do not care about any axis other than x, that is where the jet thrust exists.

## 1 Answer

The full Euler equations (so assuming inviscid) are:

$$\frac{\partial }{\partial t} \iiint \rho \vec{u} dV + \iint_{CS} \left[\rho \vec{u} \left(\vec{u}\cdot \hat{n}\right) + P \hat{n}\right] dA = 0$$

or in differential form if you prefer:

$$\frac{\partial \vec{u}}{\partial t} + \vec{u}\cdot\nabla\vec{u} + \frac{1}{\rho} \nabla P = 0$$

So your force expression is missing a term.

• Remember that the flow is stationary so the time derivative is zero. – Jose Lopez Garcia Apr 4 '15 at 12:38
• @JoseLopez You don't need to remind me of that -- that wasn't the term that is missing from your expression for force. I was just providing you the full equations. You need to identify which terms matter for your case, you already know the time derivative doesn't. – tpg2114 Apr 4 '15 at 12:39
• Thanks @tpg2114 for your answer. But still cannot figure out, how can your answer help me with the force summation? Or are you saying the book is wrong? I forgot to say we are considering no friction. Is that the missing term? – Jose Lopez Garcia Apr 4 '15 at 12:43
• @JoseLopez In your question, you wrote: $\sum \textbf{F} = \int_{cs} \rho \textbf{u}\left(\textbf{u} \dot{} \textbf{n}\right)dA$. That expression is incomplete. It is missing a term. Once you get the correct expression from what I provided you, it is just a matter of drawing your unit normals on your control surface and adding up the terms. – tpg2114 Apr 4 '15 at 12:47
• Remembering of course that the unit normals are positive when pointing outwards from the domain. – tpg2114 Apr 4 '15 at 12:48