From the referenced website

There's a 195mm EDF from Schübeler that produces 25 kg of thrust at a peak power consumption of 15 kW: https://www.schuebeler-jets.de/de/produkte/hst

Now, my question is: How much thrust could you approximately get if the diameter was 400 mm instead of 200 and how much more power would you need ?

Thanks for your answers !


2 Answers 2


It can be safely assumed that thrust $L$ is a function of the input power $P$, the diameter $D$ of the gas jet and the air density $\rho$.

Thus, $L = f(P,D,\rho)$

where $f$ is a function to be determined.

From dimensional analysis, the thrust $L$ can be easily derived:

The variables are Thrust $L$, dimensions $MLT^{–2}$; Power $P$, dimensions $ML^2T^{–3}$; Gas jet diameter $D$, dimensions $L$ and air density $\rho$, dimensions $ML^{–3}$

The variables form a non-dimensional product $k$

$k = L^a\cdot P^b\cdot D^c\cdot \rho^d$ where $a,b,c,d$ are numbers to be determined.

Let’s form now a parallel product $k^*$ with the dimensions:

$k^* = (MLT^{–2})^a (ML^2T^{–3})^b (L)^c (ML^{–3})^d$

Clearly, $k^* = M^0 L^0 T^0$... We now take the exponents for each dimension:

$a + b + d = 0 \\ a + 2b + c – 3d = 0 \\ –2a – 3b = 0$

We make $a = 1$, since $L$ is the variable we’re going to solve for.

$b = –2/3 \\ d = –1/3 \\ c = –2/3$


$k = L^a\cdot P^b\cdot D^c\cdot \rho^d \rightarrow k = L\cdot P^{–2/3}\cdot D^{–2/3}\cdot \rho^{–1/3}$

Solving for $L$

$L = k\cdot P^{2/3}\cdot D^{2/3}\cdot \rho^{1/3}$

where $k$ is a constant

Hence, for gas jet diameters $D_1$ and $D_2$, and for the same power and air density, the corresponding values of thrust $L_1$ and $L_2$ are:

$L_1/L_2 = (D_1/D_2)^{2/3}$

For the case of $D_1 = 400 mm$ and $D_2 = 200 mm$, $L_1/L_2 = (400/200)^{2/3} = 1,59$

In other words, the larger (400 mm) gas jet gives you, for the same absorbed power and air density, 59% more thrust than that attained with the smaller (200 mm) jet.

Of course, this is an approximation valid for not too high aircraft speeds, based upon momentum theory, but gives you an idea... For different values of power and gas jet diameter, you can derive the constant k from the data of thrust, power and diameter you already have, $k = L\cdot P^{–2/3}\cdot D^{–2/3}\cdot \rho^{–1/3}$ and then use that constant in your calculations.

  • $\begingroup$ Probably worth noting here that a large fan can readily be designed to absorb much more than the same amount of power -- double diameter = 4x area, could be roughly 60 kW in this case. Using four motors geared together would be the simple way... $\endgroup$
    – Zeiss Ikon
    Aug 24, 2018 at 12:41
  • $\begingroup$ @xxavier wow, thanks for your great answer! An equation to roughly estimate thrust is exactly what I needed. $\endgroup$
    – Simon Henn
    Aug 24, 2018 at 14:37
  • $\begingroup$ @Zeiss Ikon Yes, that‘s a question I asked myself, too. In case I wouldn‘t just double the diameter and put in the same power but doubled or quadroupled the power either, what potential thrust could such an EDF produce ? $\endgroup$
    – Simon Henn
    Aug 24, 2018 at 14:40
  • $\begingroup$ According to the math in the answer above, doubling the power will approximately double the thrust, providing your fan can absorb the additional kW. Looks like you could get as high as roughly 6x the original, if you can supply 60 kW. That's an hour's duration on the top-end battery from a 2018 Nissan Leaf, if you can keep everything cool... $\endgroup$
    – Zeiss Ikon
    Aug 24, 2018 at 14:56
  • $\begingroup$ I dont have much to add but I would just like to say kudos for one of the clearest and easy to understand examples of Buckingham-Pi dimensional analysis I've ever seen. $\endgroup$ Aug 27, 2018 at 6:45

Simple impulse theory equations for thrust T and power P:

$$T = C_T \cdot \rho A (\Omega R)^2$$

$$P = C_P \cdot \rho A (\Omega R)^3$$

Both scale linearly with disk area A. At constant tip speed $\Omega R$ and thrust/power coefficients, the scalings are simply T = $k_T A$ and P = $k_P A$ with $k_T$ and $k_P$ constants.

So a 400 mm fan would produce (400/200)$^2$ = 4 times the thrust, at 4 times the power. Note that the larger fan has lower RPM, in order to maintain constant tip speed.

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Order Of Magnitude check: the four listed electric fans. The largest and the smallest fan have lower P/A, a matter of available electric motor size?

  • $\begingroup$ From the data table of the fans, a mean value of k = 1.08 can be calculated. A good agreement for the thrust L as a function of diameter D and power P is obtained inserting it in the dimensionally derived expression $L = k\cdot P^{2/3}\cdot D^{2/3}\cdot \rho^{1/3}$ With caution, that expression can be used for other particular cases... $\endgroup$
    – xxavier
    Aug 23, 2019 at 12:18

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