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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.

$$b = –2/3 \\ d = –1/3 \\ c = –2/3$$

Then,

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

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 based upon momentum theory, but gives you an idea...