I am working on an electric ducted fan rocket that has thrust vector control, see image 1.enter image description here The 90mm EDF is located a bit higher above this blue section, near 1., so I believe the blue section becomes a "thrust tube". Both the EDF tube housing and "thrust tube" are 90mm in diameter

For a given static thrust (3.62kg in my case),I am interested in the maths behind finding the velocity exiting the annular area, V1, the velocity at the exit of the "thrust tube", V2 and why the EDF thrust varies depending on the thrust tubes exit area. I had seen a formula that can find this V1 velocity enter image description here . M_dot can be expanded to densityAreaV1, where Area= annular area of EDF or FSA.

On the velocity note, another confusion arose in me. If there was no thrust tube, I could find V1 using the first equation for the given static thrust. The area would be the annular/FSA area of the EDF. With the thrust tube, I would think the same formula and values would apply, getting V1. But if the thrust tube exit area affects the overall thrust, there must be some relation to and the the 1st formula.

Edit: In short, my goal was to find the velocity at the EDF exit. I was to use this value as an initial condition for a CFD analysis( to determine the values for redirected thrust), seen attached. enter image description here. Equations for EDF thrust enter image description here involved knowing the overall thrust, made of the propeller thrust and force developed by the duct interior. The problem is that the EDF thrust (3.62kg) is specified for solely the EDF; my scenario has an additional extension, so the actual overall thrust is unknown ,leading to being unable to find the EDF exit velocity

I would appreciate any advice

  • 1
    $\begingroup$ Your overall thrust is not unkown. It is just less then the 3.62kg. With high probability, it is not much less... $\endgroup$
    – U_flow
    Commented Jul 16, 2021 at 6:56
  • $\begingroup$ @U_flow That is true. Our overall rocket weight is looking at 3.08 Kg so it honestly concerned me the extent which the thrust might vary due to the tubes presence. $\endgroup$
    – Johan M
    Commented Jul 18, 2021 at 18:55

1 Answer 1


I think you have a couple of questions, which I will try to answer in order.

Ducted Fans

As wikipedia puts it:

The duct reduces losses in thrust from the tips of the propeller blades

Normal unducted propellers produce trailing vortices, however in a duct these vortices cannot form (simply because the duct wall hinders them from doing so). This means increased efficiency, as a trailing vortex requires energy to form. It is important to note however, that these efficiency gains cannot surpass the idealized propeller assumptions/calculations. And if you get your duct design wrong, it can even hinder efficiency as it interferes with incoming or outgoing airflow.

Propeller calculations

The formula for an idealized propeller (one which sets air in motion without any losses uniformly across its cross section), the formula is:

$$ v_i = \sqrt{\frac{T}{2\rho A}}=\frac{P}{T} $$

With $v_i$ - induced velocity,
$T$ - Thrust
$\rho$ - Air density (1.2041 $\frac{kg}{m^3}$)
$P$ - Power (How much Power you need to produce the given thrust)
$A$ - Area

In order to find your velocities at point 1 and 2, you would simply plug in these numbers.

Engineering approach

Addressing your last paragraph: These formulas are for an idealized propeller. For a real propeller you would encounter all sorts of losses and real-world effects like vortices or unequal velocity distribution etc. etc.. Therefore engineers normally start with these kind of "rough" formulas in order to get a sense or idea of the quantities involved. In your case this is exactly what you would want to do. The next step would be either rather extensive CFD analysis, or real-world test (Due to time, effort and cost involved, Hobbyists normally choose the latter).

Note that as I pointed out before, the EDF is merely a way to increase the efficiency of the propeller, therefore to reduce the real-world effects in order to approach the theoretical efficiency of the propeller. Do not make the mistake to think that the duct is a magic device which will push the efficiency through the roof, because it will not! In fact if you make errors in designing your duct (for example by making it a simple straight tube), you will encounter yet other real-world effects which will hinder the airflow (e.g. airflow separation at the inlet or outlet, which will produce inefficiencies, therefore eating up power).

  • $\begingroup$ Thank you for the detailed repsonse. Regarding the formula presented, I had a slight doubt about the thrust. The EDF we intend on getting is specified at 3.62kg, however I beleive that thrust is valid only for it being mounted without an extended tube at the exits; its just the static thrust for the ducted fan only, in my scenario, the introduction of the extended blue tube will cause the thrust to differ from that specified value. So im not sure if its still valid, it seems that new thrust would just need to be experimentally measured. I edited my question with more info $\endgroup$
    – Johan M
    Commented Jul 15, 2021 at 12:38
  • 1
    $\begingroup$ @JohanM as I said: For a first guess of the velocity, this is a close as you will get with textbook formulas. The snippet you presented calculates the pressure inside the thrust tube, something which is of very little relevance in this context. The formulas presented above are as good as is necessary especially if you are using this only for a starting point of a CFD. Therefore I stand by my original point, use these formulas as a first starting point. Apart from simulation or experiment (which will present their own challenges), you will not get better results. $\endgroup$
    – U_flow
    Commented Jul 16, 2021 at 6:53
  • $\begingroup$ I see, I was honestly not sure if it was something significant to consider, thanks for the clarification. If I may ask, can you please link me to the source of that propeller equation you presented? $\endgroup$
    – Johan M
    Commented Jul 18, 2021 at 18:55
  • 1
    $\begingroup$ Unfortunately not. I have it from the lecture script: "Flugmechanik der Drehflügler I - Grundlagen" bei B. van-der Wall which I fear cannot be accessed online. However the equation still stays relevant. Also I am sure that you will find the same formula in other textbooks. $\endgroup$
    – U_flow
    Commented Jul 19, 2021 at 12:49
  • $\begingroup$ No problem. It looked similar to the static thrust equation F=1/2*ApVe^2 formula. However the 2 was at the denominator instead of the numerator so it had made me unsure if this was something different. I could only find a german book of the name but the formula you presented seems like this one :imgur.com/a/KGgMtBk $\endgroup$
    – Johan M
    Commented Jul 19, 2021 at 13:51

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