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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 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.

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 valid for not too high aircraft speeds, based upon momentum theory, but gives you an idea...

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 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.

2 added 35 characters in body
source | link

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 valid for not too high aircraft speeds, based upon momentum theory, but gives you an idea...

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...

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 valid for not too high aircraft speeds, based upon momentum theory, but gives you an idea...

1
source | link

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...