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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.

$b = –2/3 \\ d = –1/3 \\ c = –2/3$


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

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...