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One of the consequences of the Tsiolkovsky rocket equation is that to lift a big mass you need a lot of fuel which again needs to be lifted, leading to even more fuel...

Does something similar hold for airplanes as well? In other words, all other things being constant, is there a term exponential in the cargo weight in the airplane fuel load?

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Yes there is, and it is called the Breguet Range equation. However, as one might expect, in the case of aircraft, the figure of merit is not a velocity increment $\Delta V$ (as for rockets), but a range increment $\Delta R$ or just simply $R$. To see how these equations are analogous I will write out the Tsiolkovsky rocket equation below,

$$ \Delta V = g I_{sp} \ln{\frac{m_0}{m_f}} $$

where $\Delta V$ is the ideal maximum change in velocity of the vehicle, $m_0$ is the initial mass, $m_f$ is the final mass after all of the propellant is consumed, and $I_{sp}$ is the specific impulse of the propulsion system where the product of gravity $g$ and $I_{sp}$ is simply the effective exhaust velocity. The exponential relationship you refer to in the question is then just simply,

$$ \frac{m_0}{m_f} = e^{\frac{\Delta V}{g I_{sp}}} $$

As you eluded to in the question, it becomes apparent that holding the propulsion characteristics constant, the mass ratio $m_0/m_f$ is exponentially related to the velocity increment $\Delta V$.

Now in the case of aircraft, the Breguet range equation takes the form,

$$\Delta R = I_{sp} V \frac{L}{D} \ln{\frac{m_0}{m_f}} $$

Once again, $I_{sp}$ refers to the specific impulse of the propulsion system, $V$ is the cruise velocity of the aircraft, and $L/D$ is the cruise lift to drag ratio or aerodynamic efficiency of the aircraft. In this case, we obtain,

$$ \frac{m_0}{m_f} = e^{\frac{\Delta R}{I_{sp} V \frac{L}{D}}} $$

Therefore, as indicated above, the mass ratio of the aircraft is an exponential function of its ideal cruise range, which is dependent on the efficiency of the propulsion system, aerodynamics, and cruise speed. Note, all of the above assumes steady level flight, which is a mostly adequate assumption for the cruise leg of the mission envelope.

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  • $\begingroup$ + Wikipedia has a discussion of this, with different derivations for propeller or jet propulsion. The difference is that fuel comsumption depends on different things. For propeller aircraft it depends on power. For jet aircraft it depends on thrust. $\endgroup$ – Mike Dunlavey Mar 15 '17 at 16:10

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