Recoil drive

Recoil principle of a rocket

The recoil drive or reaction drive is a practical application of Newton's 3rd axiom . The recoil drive carries its drive medium with it; Combustion-based recoil drives carry both their fuel and their oxidizer . The propelled object , for example a rocket , is accelerated forwards by the recoil with the same force with which the propulsion medium is ejected backwards.

In space, the recoil drive is the only way to accelerate a spaceship away from massive celestial bodies and strong radiation sources.

Physical background

According to Newton's 3rd axiom (actio = reactio, also "reaction principle" or "interaction principle") two masses that exert a force on each other are accelerated. This results in a speed for both masses (after the end of the force). According to the definition for the impulse

${\displaystyle {\vec {p}}=m\cdot {\vec {v}}}$

In this case the following relations of the impulses to each other result:

${\displaystyle {\vec {p}}_{1}=-{\vec {p}}_{2}\qquad {\text{oder}}\qquad m_{1}\cdot {\vec {v}}_{1}=m_{2}\cdot -{\vec {v}}_{2}}$

(Here, for a rocket , for example, represents the momentum of the ejected combustion products, and the resulting opposite momentum of the rocket) ${\displaystyle {\vec {p}}_{1}}$ ${\displaystyle -{\vec {p}}_{2}}$

It must be taken into account that a defined energy must be available to generate these impulses, which can perform the corresponding acceleration work. If a mass has momentum, it has kinetic energy .

The following applies when calculating the proportional amounts of energy:

${\displaystyle E_{m_{1}}={\frac {m_{2}}{m_{1}+m_{2}}}\cdot E_{\mathrm {ges} }\qquad {\text{und}}\qquad E_{m_{2}}={\frac {m_{1}}{m_{1}+m_{2}}}\cdot E_{\mathrm {ges} }}$

In a continuous process, the following mathematical relationship , also known as the basic rocket equation , results :

${\displaystyle v_{n}(t)=v_{s}\cdot \ln \left({\frac {m(0)}{m(t)}}\right)}$

or:

${\displaystyle {\vec {v}}_{n}(t)=-{\vec {v}}_{s}\cdot \ln \left({\frac {m(0)}{m(t)}}\right)={\vec {v}}_{s}\cdot \ln \left({\frac {m(t)}{m(0)}}\right)}$

Where is the relative speed of the supporting mass to the actual useful mass. It must be taken into account here that the supporting mass decreases continuously as the process progresses and ultimately only the useful mass remains at its final speed (relative to the starting point). ${\displaystyle v_{s}}$ ${\displaystyle v_{n}}$

An astonishing effect occurs with a ratio of . From this point in time, the rocket and the supporting mass ejected by it move away from an observer who has remained at the launch site of the rocket in the same direction, but at different speeds. ${\displaystyle 1=\ln \left({\tfrac {m(0)}{m(t)}}\right)}$

Recoil drives that work on the basis of fluids

Outflow velocity

The pressure in the recoil chamber is higher than the ambient pressure . Due to this pressure difference, the medium in the chamber emerges from the nozzle at a certain speed . The density of the outflowing medium (inside the chamber, i.e. under pressure ) is also important. ${\displaystyle p_{i}}$ ${\displaystyle p_{a}}$ ${\displaystyle v_{s}}$ ${\displaystyle \rho }$ ${\displaystyle p_{i}}$

From the conservation of energy follows:

${\displaystyle W=(p_{i}-p_{a})\cdot dV={\frac {1}{2}}\cdot dm\cdot v_{s}^{2}={\frac {1}{2}}\cdot dV\cdot \rho \cdot v_{s}^{2}}$

${\displaystyle \Rightarrow \quad {\frac {2(p_{i}-p_{a})}{\rho }}=v_{s}^{2}}$

${\displaystyle \Rightarrow \quad v_{s}={\sqrt {\frac {2\cdot (p_{i}-p_{a})}{\rho }}}}$

This equation only applies to sufficiently small nozzles in which the chamber contents are only slightly accelerated relative to the chamber. In addition, possible friction losses were neglected.

In the case of gases, it should be noted that their density depends on the pressure and temperature. This can be (approximately) using the thermal equation of state of ideal gases ${\displaystyle \rho }$

${\displaystyle p\cdot V=m\cdot R_{s}\cdot T}$

by conversion to

${\displaystyle \rho ={\frac {m}{V}}={\frac {p}{R_{s}\cdot T}}}$

to calculate.

Since the density of gases is proportional to the pressure, an increase in the exit velocity can only be achieved by increasing the temperature.

Throughput

The throughput , which is often referred to as mass flow , can be determined according to the cross section of the nozzle, the density of the emerging medium and its exit speed . ${\displaystyle A}$ ${\displaystyle \rho }$ ${\displaystyle v_{s}}$ ${\displaystyle \mu }$

${\displaystyle \mu =A\cdot \rho \cdot v_{s}}$

thrust

The generated thrust can be calculated by multiplying the throughput by the exit speed of the medium. ${\displaystyle F_{s}}$ ${\displaystyle \mu }$ ${\displaystyle v_{s}}$

${\displaystyle F_{s}=\mu \cdot v_{s}}$

Or by replacing ${\displaystyle \mu =A\cdot \rho \cdot v_{s}}$

${\displaystyle F_{s}=A\cdot \rho \cdot v_{s}^{2}}$

and

${\displaystyle F_{s}=A\cdot \rho \cdot {\frac {2\cdot \Delta p}{\rho }}}$

one obtains the mass-independent relationship

${\displaystyle F_{s}=2\cdot \Delta p\cdot A}$

Required engine power

This does not mean the power with which such an engine would move (accelerate) a mass, but the power that is required to generate the corresponding thrust. This performance is determined from the given throughput : ${\displaystyle P_{\mathrm {Nutz} }}$ ${\displaystyle P_{\mathrm {Triebwerk} }}$ ${\displaystyle \mu }$

${\displaystyle \mu ={\frac {\Delta m}{\Delta t}}}$

To the mass of the escaping gases to the speed to accelerate, must work ${\displaystyle \Delta m}$ ${\displaystyle v_{s}}$

${\displaystyle W={\frac {1}{2}}\cdot \Delta m\cdot v_{s}^{2}}$

be performed. Thus the engine power results to ${\displaystyle P_{\mathrm {Triebwerk} }}$

${\displaystyle P_{\mathrm {Triebwerk} }={\frac {W}{\Delta t}}={\frac {1}{2}}\cdot {\frac {\Delta m}{\Delta t}}\cdot v_{s}^{2}={\frac {1}{2}}\cdot \mu \cdot v_{s}^{2}}$

or because of : ${\displaystyle F_{s}=\mu \cdot v_{s}}$

${\displaystyle P_{\mathrm {Triebwerk} }={\frac {1}{2}}\cdot v_{s}\cdot F_{s}}$

In order to generate the same thrust with a hypothetical photon engine, the engine power would have to be considerably higher than with a conventional chemical rocket engine .

Useful power

The actual performance that can be achieved by such a recoil drive results from changing the formula for the acceleration work: ${\displaystyle P_{\mathrm {Nutz} }(t)}$

${\displaystyle W_{\mathrm {Beschl.} }=m\cdot {\frac {v_{2}^{2}-v_{1}^{2}}{2}}}$

${\displaystyle P_{\mathrm {Nutz} }(t)=m(t)\cdot {\frac {v_{2}^{2}-v_{1}^{2}}{2\cdot t}}=m(t)\cdot a\cdot {\frac {v_{2}+v_{1}}{2}}=F_{s}\cdot {\frac {v_{2}+v_{1}}{2}}}$

Thereby make the initial velocity and the terminal velocity is the acceleration process. ${\displaystyle v_{1}}$ ${\displaystyle v_{2}}$

Applications

• Rocket engine

See also

• Specific impulse The influence of the outflow velocity on the fuel consumption and the thrust of a rocket engine
• Bernoulli's equation
• Overview of all types of aircraft propulsion
• Propulsion methods for space travel
• Swing-by - another method of acceleration in space

Web links

• Momentum theorem and energy law ( memento from December 17th, 2012 in the web archive archive.today )
• The concept of work (acceleration work) ( Memento from December 10, 2012 in the web archive archive.today )