4 ms·

> This is the linear tension in the wall of a pressure vessel(rocket engine) which varies as the square of the diameter. Twice the diameter = 4 times the pressu

> This is the linear tension in the wall of a pressure vessel(rocket engine) which varies as the square of the diameter. Twice the diameter = 4 times the pressure (hoop stress) the walls must take. This is incorrect. Cutting the cylinder in half lengthwise and taking a unit length, we see that the cross section of the walls of the chamber (unit length x 2 x wall thickness) resists the pressure force from the contained fluid (2 x radius x unit length x pressure). Since the pressure and unit length are constant under scaling, this resulting force grows linearly with the radius. Thus, wall thickness has to grow proportionally to the radius, and there are no mass penalties from the chamber wall from either a smaller or larger engine. A similar result holds for spheres. Indeed, the particular result is independent of the geometry of the pressure vessel. The larger immediate result is that pressure vessels scale just fine with size, up or down, and there's no benefit at either end of the length scale, in relation to contained volume. What you do get for a smaller engine is more manageable combustion instabilities. Look at the 5-fold symmetry (odd, rather than even) of the injector baffles in the SSMEs. This is to stop a tangential oscillation mode, an important failure mode of larger engines. What you lose for smaller engines is that you have more of the fluid "close" to the walls. The boundary layers don't grow linearly, so you get more heat transfer at the boundary in proportion to the contained fluid. (Radiant transfer in the engines complicates the picture, but convective transfer is definitely proportionately worse for smaller engines.)

Robotbeat 9y ago
It's true that pressure vessels in principle don't care about scale when it concerns mass per unit volume (at a constant pressure). But a combustion chamber's thrust is (to zeroth order) proportional to cross sectional area, not volume. The combustion chamber only needs to be a certain length (L star) to achieve efficient combustion. Any longer and you're just adding mass with no benefit. But there are practical limits to shape of the combustion chamber. You can't have it too squat or it loses structural efficiency. Thus, above a certain size, you're better off from a mass efficiency standpoint with having a bunch of smaller combustion chambers than one big huge one. And this is true even more for the nozzle. You can use a much shorter, and thus lighter, nozzle if you have a smaller engine. For the same expansion ratio, therefore, clustering a bunch of smaller engines is more mass efficient than a single big engine. (But if you go REALlY small, you have minimum gauge issues and you lose thermal and combustion efficiency.)
- 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 ago
    Pressure 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.)
aurizon 9y ago
LOL, read and weep https://en.wikipedia.org/wiki/Cylinder_stress https://en.wikipedia.org/wiki/Cylinder_stress https://en.wikipedia.org/wiki/Rocket_engine https://en.wikipedia.org/wiki/Rocket_engine the round nozzle gets bigger in two dimensions, while the walls do not, so the walls must get thicker as the nozzle gets larger in a square law manner
- theothermkn 9y ago
  I read most of that, and did not weep, 20 years ago when I was getting my degree in aero/astro engineering. Neither of your links support what you're trying to say. The walls get linearly thicker, and your intuition about why the walls need to get thicker is incorrect.
  - Retric 9y ago
    It's not quite linear. Pressure vessels like air tanks or rocket engines wall thickness scale approximately linearly with radius. But, that's not the only stress on rocket engines which need to deal with thermal expansion etc.
  - aurizon 9y ago
    The walls need to get thicker indeed, and that extra thickness gradually adds to the weight of the engine faster than adding extra small engines for the same thrust = less and less payload

Robotbeat 9y ago

It's true that pressure vessels in principle don't care about scale when it concerns mass per unit volume (at a constant pressure). But a combustion chamber's thrust is (to zeroth order) proportional to cross sectional area, not volume. The combustion chamber only needs to be a certain length (L star) to achieve efficient combustion. Any longer and you're just adding mass with no benefit. But there are practical limits to shape of the combustion chamber. You can't have it too squat or it loses structural efficiency. Thus, above a certain size, you're better off from a mass efficiency standpoint with having a bunch of smaller combustion chambers than one big huge one. And this is true even more for the nozzle. You can use a much shorter, and thus lighter, nozzle if you have a smaller engine. For the same expansion ratio, therefore, clustering a bunch of smaller engines is more mass efficient than a single big engine. (But if you go REALlY small, you have minimum gauge issues and you lose thermal and combustion efficiency.)

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 ago
  Pressure 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.)

aurizon 9y ago

LOL, read and weep https://en.wikipedia.org/wiki/Cylinder_stress https://en.wikipedia.org/wiki/Cylinder_stress https://en.wikipedia.org/wiki/Rocket_engine https://en.wikipedia.org/wiki/Rocket_engine the round nozzle gets bigger in two dimensions, while the walls do not, so the walls must get thicker as the nozzle gets larger in a square law manner

theothermkn 9y ago
I read most of that, and did not weep, 20 years ago when I was getting my degree in aero/astro engineering. Neither of your links support what you're trying to say. The walls get linearly thicker, and your intuition about why the walls need to get thicker is incorrect.
- Retric 9y ago
  It's not quite linear. Pressure vessels like air tanks or rocket engines wall thickness scale approximately linearly with radius. But, that's not the only stress on rocket engines which need to deal with thermal expansion etc.
- aurizon 9y ago
  The walls need to get thicker indeed, and that extra thickness gradually adds to the weight of the engine faster than adding extra small engines for the same thrust = less and less payload