2
$\begingroup$

For a given chamber pressure and shape,

  • thrust is proportional to nozzle area, which is proportional to the square of dimensions

  • mass is proportional to the volume, which is proportional to the cube of dimensions.

Does the s mean that smaller engines have a higher thrust to weight ratio?

Improve this question
$\endgroup$
14
  • 2
    $\begingroup$ "mass is proportional to the volume, which is proportional to the cube of dimensions.": Will scaling the nozzle area mean scaling all components of the engine as a whole equally in all three dimensions? Will the liner for an engine with twice the diameter be twice as thick? The combustion chamber twice as long? And scaling down, how thin can you make components and still have the engine be practical to build and handle? $\endgroup$
    – Christopher James Huff
    Apr 12 at 12:37
  • $\begingroup$ @ChristopherJamesHuff yes. Everything scsles. Hence I said 'for a given shape". The laws of pressure vess l scaling dictate that the liner thickness must scale with circumference to maintain a constant radius ratio. $\endgroup$
    – Abdullah
    Apr 12 at 12:44
  • 1
    $\begingroup$ The liner is not a pressure vessel, so why would it scale as one? $\endgroup$
    – Christopher James Huff
    Apr 12 at 12:51
  • 1
    $\begingroup$ @Abdullah - I think if you replaced "smaller" - with "smaller thrust output" your question would be easier to answer. For example, during WW2 Germans put aircraft engines in their tanks. It was 3 times lighter than engines in Russian tanks but with a similar thrust output. $\endgroup$
    – TheMatrix Equation-balance
    Apr 12 at 13:18
  • 1
    $\begingroup$ The inner wall of a regeneratively cooled engine is non-structural, apart from withstanding the pressure difference between the coolant and the chamber/throat/nozzle. It can be a layer of copper filled with cooling channels, with essentially no tensile strength, supported on the outside. The thickness of the inner wall is determined by the conductivity of the material, it would not be thicker on larger engines. And your assumption of the same shape is not valid in general. Compare Rocket Lab's Rutherford, SpaceX's Merlin, and Rocketdyne's F1. $\endgroup$
    – Christopher James Huff
    Apr 12 at 15:15

0

Reset to default

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Browse other questions tagged or ask your own question.