I have a wind tunnel (subsonic flow of air) which consists of a fan inside (and ahead of) a long cylinder of cross-sectional area of $A_e$, followed by a diffuser of $A_o$ c.s. area ($A_o>A_e$). enter image description here The fan maintains a velocity of $V_e$ inside the tunnel. Assuming the flow inviscid and incompressible, I want to find the amount of force per unit area that the fan applies.

I chose the entire inside of the tunnel (fan included) as the fixed, non-deforming control volume.

Applying Conservation of Momentum for the c.v. (gage pressure used):

$$F=\rho(V_o^2A_o-V_i^2A_e)=\rho V_e^2A_e(A_e/A_o)$$

The volume integral was dropped since the flow is steady.

The last transition was done with the help of Conservation of Mass which yields:


The correct answer should be $(\rho/2)V_e^2A_e(A_e/A_o)^2$ and in the available solutions to me, Bernoulli was used on the stream line that begins after the fan and ends outside the tunnel at atm' pressure, and Momentum Conservation was applied to a control volume enclosing the fan alone.

  • $\begingroup$ You weren't able to add your image because you currently have less than 10 reputation, a limit we put on all new accounts to prevent spam (see here). However, feel free to manually upload your image to a hosting service (even imgur itself!) and then edit your question with a link to it. You can trust the community will then edit your question to embed the image itself. $\endgroup$
    – Wasabi
    Sep 13, 2018 at 23:58
  • $\begingroup$ Why do you assume incompressible flow? Air is compressible! Absent a clearer reasoning, I am reluctant to believe an answer that starts from an entirely wrong assumption. $\endgroup$ Sep 14, 2018 at 1:35
  • $\begingroup$ The question clearly states that the flow is incompressible. $\endgroup$
    – Rubenz
    Sep 14, 2018 at 5:35
  • $\begingroup$ I believe although air is compressible, under certain conditions it is possible to assume an incompressible flow. $\endgroup$
    – Rubenz
    Sep 14, 2018 at 6:42
  • $\begingroup$ @JeffreyJWeimer It is standard practice to assume air is incompressible if the flow velocities are less than about 0.3 times the speed of sound. The error is no bigger than the errors cause by leaving other things left out of the model (e.g. detailed modeling of all the boundary layers, turbulence effects, etc) and the math is very much simpler for incompressible flow. $\endgroup$
    – alephzero
    Sep 14, 2018 at 7:55

2 Answers 2


I believe I figured it out.

Since we want the force that the fan applies on the air we use Conservation of linear momentum on a CV the encapsulates the fan.

If we choose the CV, fan to outlet, excluding the tunnel itself, we must take into consideration $p{dA}\hat{n}\cdot\hat{x}$ that changes with the curve of the tunnel geometry. This we would like to avoid.

If we choose to solve this issue by expanding our CV to include the tunnel itself, we cut through matter which experiences internal forces, which we don't have.

The remaining option is to choose the CV around the fan itself. $$F+(p_i-p_e)A_e=\dot{m}(v_e-v_i)$$ Where $p_i$ and $p_e$ are the pressures just before and just after the fan, respectively, and $F$ is the force we are looking for.

Conservation of mass yields $v_i=v_e$ (incompressible flow and constant cross-section), so the RHS disappears, $$F=(p_e-p_i)A_e \tag{1}$$ Now, since the flow is steady, inviscid and incompressible, we can use Bernoulli on a streamline that stretches :

1) from "far away" before the inlet, where the pressure is atmospheric and the velocity is zero, up to the cross-section where we have defined $p_i$: $$\frac{1}{2}\rho v_e^2=p_a-p_i \tag{2}$$

2) from the point where we have defined $p_e$ up to the outlet, where the pressure is atmospheric and the velocity is determined by Conservation of mass (fan to outlet),$$v_eA_e=v_oA_o$$ $$\frac{1}{2}\rho (v_e^2-v_o^2)=p_a-p_e$$ It can shown that the last equation may be rewritten as,$$\frac{1}{2}\rho v_e^2(1-(\frac{A_e}{A_o})^2)=p_a-p_e \tag{3}$$

Now, subtract eq. (3) from (2) and we have the RHS of (1),$$p_e-p_i=\frac{1}{2}\rho v_e^2(\frac{Ae}{A_o})^2$$ Finally, the force per unit area is,$$\frac{F}{A_e}=\frac{1}{2}\rho v_e^2(\frac{Ae}{A_o})^2$$


Here is the proper approach for the starting point using air.

Take the gas as compressible and ideal. Conservation of mass gives

$$ \dot{n}_e = \dot{n}_i = \dot{n} $$

$$ v_e A_e \frac{p_e}{T_e} = v_i A_i \frac{p_i}{T_i} $$

Assume constant area and isothermal flow.

$$ v_e = \frac{p_i}{p_e} v_i $$

Under isothermal flow of an ideal gas, the force of the fan is therefore set by

$$ F = \left(p_e - p_i\right)A_e + mv_i \left(\frac{p_i}{p_e} - 1\right) $$

The two terms are the work to move the fluid (ideal gas) across a pressure difference and the work done as the gas expands/contracts across that pressure difference.

When we assume that $p_i/p_e \approx 1$, we obtain the answer found by the (incorrect) assumption of an incompressible fluid. To be blunt, a way can always be found in a physical system to validate an assumption that pressure change is small across a control volume. Nothing physically ever validates a blanket starting assumption that gases are incompressible!

This demonstrates, an analysis using a compressible fluid assumption for a gas will fail as $p_i/p_e$ deviates from unity, as mass flow increases, and as inlet velocity increases.

The remaining derivation for a compressible, ideal gas is left as an exercise for the reader.


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