While calculating the off design conditions of the engine, we need to know the off design characteristics, such as inlet additive drag. In the book Mechanics of Flight, by Warren F. Phillips, its expression has been derived as a relation between free stream and inlet Mach no. Further while calculating the drag,in one of the example (2.6.1), the inlet Mach no. has been assumed constant at 0.6 whereas the flight is accelerating from Mach 0 to its design Mach no.

This implies that the inlet Mach no. has been somehow achieved as 0.6 even at takeoff, is it practically possible? If not, then how is the drag actually calculated during flight with varying free stream and inlet Mach numbers?

  • 2
    $\begingroup$ Remember, that the compressor sucks in the air in the static case. $\endgroup$
    – Jan Hudec
    Commented Oct 25, 2015 at 21:01

1 Answer 1


The inlets of supersonic aircraft routinely control (reduce) the Mach number of the engine (at compressor inlet) during flight. This is done using either a 2D ramp type inlet (like Concorde) or a center body (like Mig-21).

These inlets convert the incoming supersonic flow to subsonic flow by forming shock waves in the path of the airflow.

Adjustable inlet

Source: sturgeonshouse.ipbhost.com

In subsonic conditions, the inlets perform two functions:

  • At cruise condition, diffusion of the free-stream flow to the compressor inlet condition.

  • At takeoff, acceleration of static air to the compressor inlet condition.

So basically, the inlet design can be optimized such that the Mach number is constant (within some variation) over the engine operational range.

In your case, however, I think that it is a simplifying assumption made for the problem.

  • $\begingroup$ So at high speed, the concord had a non zero bypass ratio ? Or maybe the term does not apply to non turbofan engines ? $\endgroup$
    – Antzi
    Commented Oct 25, 2015 at 15:43
  • $\begingroup$ @Antzi, the air dumped overboard does not pass through the compressor and is not accelerated, so it does not count as bypass. $\endgroup$
    – Jan Hudec
    Commented Oct 25, 2015 at 20:59

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