My graphs show that overall efficiency and thrust first increase, reach a peak and then fall in the supersonic range (M=1-5) whereas SFC first drops and then increases rapidly near M 5.

Why does this happen?

  • 3
    $\begingroup$ Welcome! A verb is missing here: "SFC first and then decreases" (increases?) $\endgroup$
    – mins
    Nov 20 '17 at 23:43

The thrust of a ramjet depends on the dynamic pressure of the airframe since this determines how high the pressure inside the ramjet can become. Thrust grows therefore initially with the square of speed.

The efficiency of heat engines depends on the temperature and pressure ratio of their cycle, and again speed helps to improve that. The specific thermodynamic process of ramjets is the Joule cycle (in the US called Brayton cycle).

Why this doesn't grow indefinitely has to do with nonlinearities at high temperature. Once gas temperatures exceed 2000 to 4000K, Oxygen starts to dissociate, depending on pressure. Nitrogen follows beyond 8000K. Now the combustion heat will not be converted into a further temperature increase but will be wasted in the dissociation of air. As this limit is approached, less margin for heating is left and the SFC drops. The optimum with current technology is at approximately Mach 2.4.

A possible remedy is to slow down the air less, so the maximum heat inside the combustion chamber can be restricted. However, the flame propagation speed of conventional hydrocarbons puts a firm and low limit to that. The higher speed simply means that mixing, evaporation and burning will happen mostly once the air-fuel mixture has left the engine. Here you need to switch to hydrogen and a supersonic ram jet, but that will bring a host of new problems.

  • $\begingroup$ May I mention the hilariously dangerous shcramjet. $\endgroup$
    – Abdullah
    Jun 29 '20 at 12:39


For a given fuel, ramjet thrust vs mach number trend may be approximately(see note 1) given by

Thrust = flight mach number(sqrt(2000 / (air temperature * (1+(7 * flight mach ^ 2) / 40))) - 1)

enter image description here


If we maintain a given air/fuel ratio, the amount of fuel consumed goes up linearly with mach, as air flow rate goes up with the same trend.

enter image description here


Note 1: The source's equation was simplified and some values rounded off to give the current equation.

  • $\begingroup$ Ooops I made a big mistake. Am fixing $\endgroup$
    – Abdullah
    Jun 29 '20 at 12:48
  • $\begingroup$ Fixed the issue $\endgroup$
    – Abdullah
    Jun 29 '20 at 15:34
  • 1
    $\begingroup$ Do you have sources or an explanation for how you derived the formula? $\endgroup$
    – ROIMaison
    Jul 2 '20 at 8:40
  • $\begingroup$ @ROIMaison Will work on it. $\endgroup$
    – Abdullah
    Jul 2 '20 at 10:01

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.