4 ms·
> You can't have it too squat or it loses structural efficiency. I'm not sure what you mean. Over "squatness," if we mean l-star to cross-sectional area, the '
by theothermkn 9y ago
> You can't have it too squat or it loses structural efficiency.
I'm not sure what you mean. Over "squatness," if we mean l-star to cross-sectional area, the 'structural efficiency' remains constant, in that we've contained (square) more fluid for (thickness x perimeter = square) more wall material, and done so for (square) more thrust.
> You can use a much shorter, and thus lighter, nozzle if you have a smaller engine.
A nozzle, again, can be modeled as a pressure vessel. (Neglecting shear stress in the first analysis.) So, if we concede that pressure vessel mass ratios are invariant under scale, then so are exhaust nozzles. What we are really worried about is the amount of fluid that is in the boundary layer for heat transfer and shear reasons, and this gets slightly worse with engine size. The area-Mach relations govern the size of the exit bell, so the exit surface for a larger engine grows linearly with the throat area, and we're back to the pressure vessel scaling laws.
> You can use a much shorter, and thus lighter, nozzle if you have a smaller engine.
The point being that, if your nozzles are 9 times smaller, they're only 9 times lighter.
It's difficult to talk to everyday SpaceX enthusiasts, who seem to have most of their information second-hand. When, for example, to I bring up frozen flow vs. equilibrium flow in this discussion? When do I point out the combustion instability limitations of larger engines? When do I point out the exorbitant research costs associated with taming those instabilities? When will any of that ever dissuade a layperson from their enthusiasm for the square-cube law?
- Robotbeat 9y agoPressure vessel equation means structural efficiency is invariant based on volume but we're talking (to zeroth order) area, thus smaller pressure vessels (nozzles, chambers, etc) are more structurally efficient (at the large size limit...). To expound: Mass ∝ Volume, Volume ∝ length^3. Thrust ∝ cross sectional area, area ∝ length^2. Thus, Mass ∝ Thrust^(3/2), or: Thrust to weight ratio ∝ sqrt(thrust) for a single engine. > The point being that, if your nozzles are 9 times smaller, they're only 9 times lighter. This is incorrect, for the reason I shared earlier. Thrust is proportional to cross section, not volume. Thus scaling laws (at some point) favor smaller engines. (Note that this is assuming we're already big enough that we have full combustion and are not experiencing minimum gauge issues, etc.) My knowledge doesn't come from being a SpaceX enthusiast. In addition to having a physics degree, years ago I was part of a very early-stage startup (never got off the ground) at one point looking at launch vehicles, and so I read Sutton's Rocket Propulsion Elements (a very nice introductory text, I highly recommend) among others and did a bunch of scaling analysis. (SpaceX would've been a competitor.)