## Tuesday, June 20, 2023

### TSTO Launch Fundamentals

I have illustrated the fundamentals of making a TSTO launch design-sizing calculation as two figures.  This is for an eastward launch at low inclination,  to a typical circular low Earth orbit (LEO).  The first figure shows the trajectory and how the delta-vees (dV’s) are estimated.  It also shows a sketched free-body diagram,  and the associated Newton’s 3rd Law solution for vehicle acceleration,  at 3 typical points coming up the trajectory.  Typical loss factor estimates are also shown.

Update 6-21-2023: made slight corrections to this first figure to make the acceleration formulas accurate,  with only the sine component of weight resisting thrust.  That sine function is of the path angle above horizontal.

The nature of the non-lifting (but thrusted) gravity turn ascent puts the vertical rise portion,  most of the steeply-inclined trajectory,  and all of the trajectory experiencing aerodynamic drag,  onto the first stage burn.   The free-body diagrams are there to emphasize this.  If the first stage thrust is not “huge” in relation to the other forces,  the acceleration capability will be small or negative (meaning the design cannot work).  The losses to be covered during this first stage burn are quite significant.

The trajectory during the second stage burn is exo-atmospheric,  and very nearly horizontal,  so that (1) there are little in the way of losses to cover,  and (2) the thrust can be much smaller and still get good acceleration in a reasonable time,  because the weight statement is different,  with much smaller numbers.  If the burn reaches the value estimated for “surface circular LEO speed”,  before reaching orbit altitude (as often happens),  then the second stage coasts onward to orbit altitude,  and makes a small circularization burn there.  That circularization burn is quite modest,  on the order of 100 m/s = 0.1 km/s.

The gravity and drag losses are percentages based on surface circular orbit speed as a measure of the total mechanical energy needed to attain orbit speed (kinetic energy) at orbit altitude (potential energy).  These losses get applied almost entirely to the first stage burn.  As a “ballpark” configuration sizing tool,  they all get applied to the first stage burn.  At most,  there is a distinct minority of the gravity loss,  that might apply to the second stage (maybe 10% of the estimated loss).

The dV required of stage 1 is thus the staging velocity Vstg plus the sum of the gravity and drag losses.  The dV required of stage 2 is the burn value dV2 plus the circularization burn dV,  where the dV2 burn value is the surface circular orbit speed less the staging speed.

Typically,  in first stages we are looking at denser propellants to reduce aerodynamic drag losses and inert tankage weights at large contained volume,  with reduced specific impulse (Isp) capability.  We are also looking at engine bells sized for sea level expansion,  which also reduces Isp capability a little further.  That,  plus shouldering all the losses,  means the staging velocity falls well short of halfway to orbit speed,  in any practical design.

Typically,  in second stages we can use lower density propellants without incurring extra drag loss and inert tankage weights,  because the total propellant mass quantities are far smaller than those in the first stage (which has to lift the entire loaded second stage as its “payload”).  Plus,  the engines can be “vacuum-optimized” (although in reality there is no such thing,  there are only engine bells that fit within the allowable room),  which makes specific impulse capability a bit better than the sea level designs.

This allows the second stage to shoulder the great majority of the total dV required to reach orbit,  as the second figure illustrates.

Once you have determined which propellants each stage will use,  and what numbers of appropriate engines,  thrust levels per engine,  and sea level or vacuum bells,  you have some idea of the average Isp capability (and thus the effective exhaust velocity Vex = gc*Isp) of each of the two stages.  From there,  the remaining significant variable to set,  is the staging velocity Vstg.  You run iterative calculations at various Vstg values to maximize your overall payload fraction at the “best” staging velocity Vstg.   As noted in the figure,  there is NO SINGLE rocket equation calculation to give you that answer!  It is an inherently iterative search,  and the result is unique to every design!

Adding recovery and re-use of stages complicates this analysis further,  and by a huge amount.  You must have at staging sufficient unburned propellant still aboard the first stage,  to support whatever kind of re-entry procedure and landing scheme that you have in mind.  That will reduce the “optimal” Vstg value,  and cause the second stage to “shoulder” more of the overall dV,  driving its required mass ratio MR higher (thus reducing stage payload fraction).  Every design is unique!  Those details are beyond scope here.

There is no way to recover and re-use second stages,  except to turn them into fully re-entry-capable spacecraft,  capable of landing by some means!  This also involves some amount of unburned propellant still aboard upon reaching orbit,  to support the return flight operations.  That reduces ignition payload fraction further,  causing such designs to be attractive only in larger sizes,  where there is an economy-of-scale effect.  This is also beyond scope here.

If you choose a liquid core with solid strap-on boosters as your first stage,  the odds are very low that the solids will burn out at the same time as the liquid first stage core.  In that case,  you break the first stage burn into two burns.  The first uses a thrust-averaged Isp for the solids and liquid burning simultaneously.  The second uses a liquid-only Isp,  and deletes the inert weights of the solids from its weight statement.  Again,  that complication is beyond scope here.

The main point of this discussion is that the notion of a single rocket equation calculation,  representing accurately all of the things that must figure into a two-stage ascent,  is utter nonsense!  The rocket equations are separate for the two stages,  with different Isp’s and different weight statements,  and these must be iteratively re-done until overall payload fraction is optimized at the “best” staging speed.

There is NO general rule-of-thumb here,  not even for simple expendable two-stage-to-orbit (TSTO) designs!

This is what is important to TSTO launch,  and when it is important.  The rocket equation always applies,  BUT,  the dV it predicts is irrelevant if your thrust is insufficient to accelerate along the path (everything is “eaten up” by gravity loss,  in that case).  Newton’s 3rd Law lets you determine whether your pathwise acceleration is adequate or not.

Staging typically occurs just exo-atmospheric,  at a rather low path angle,  and at a velocity only about 20-35% of the total velocity the launch vehicle must achieve.  It usually is about 2 km/s,  which is about 25% of orbital velocity at 8 km/s.  It is set to maximize the payload fraction out of a design concept.

The first stage shoulders virtually all the losses,  and has the highest thrust requirement,  since it must push the heaviest weight up vertically against drag.  If the first stage dV requirement is modest,  one can afford lower Isp levels in order to get the acceleration required,  with simple raw thrust,  per Newton.

The second stage has no drag loss and virtually no gravity loss.  The path is nearly horizontal,  so the path-wise weight component is very small.  And with the first stage separated,  the mass that must be accelerated is very much smaller.  The second stage has a very low thrust requirement,  but must supply the larger dV.  So the highest-possible Isp is critical to second stage designs.

Here is a way to generate realistic stage inerts from scratch,  in the absence of any actual data:

There’s basically 2 cases to consider for book-keeping payload items:  the payload within an ascent shroud,  and payload carried by a recoverable item (a capsule or an entry-capable spacecraft):

Weight statements are made up of inerts and payload,  which sum to stage burnout;  plus propellant,  which sums with burnout to ignition.  Jettisoned payload shrouds are generally counted as part of the upper stage inert.  A recoverable item is not.  The entire loaded upper stage is the “payload” for the lower stage.

If the upper stage still burns after the shroud is jettisoned,  that has to be a separately-computed rocket equation calculation,  using a separate (and different) weight statement that zeroes the shroud from the inertand only has the remaining propellant aboard the stage.  That portion of the burn before shroud jettison uses the weight statement with the shroud included in the inert,  but expels only that propellant which is used up to the jettison point

This complication is not an issue if the payload is contained inside a recoverable item.  The recoverable item plus the contained payload is the “payload” carried by the upper stage.

Here’s a summary of things that make up weight statements:

What is needed to estimate performance with the rocket equation is the effective exhaust velocity Vex of the propulsion.  This is not the actual expansion velocity coming out of the nozzle,  although it is similar in magnitude.  It is estimated from the specific impulse Isp as Vex = gc * Isp,  where gc is the standard acceleration of gravity.  There are “delta-vee” values (dV) associated with the launch trajectory.  Suitable values of dV and Vex,  and a mass ratio MR from the weight statement,  are related by the rocket equation:  dV = Vex LN(MR),  where MR = ignition mass/burnout mass.  The real “trick” is getting the right dV values to use,  from the launch trajectory,  including the effects of real-world losses:

The rocket engine produces thrust with a propellant flow rate drawn from the tanks.  It does that by expanding the hot gases through a convergent-divergent nozzle,  that has a suitable bell expansion ratio.

Where the atmospheric backpressure is high,  expansion potential is more limited,  and the thrust is less.  The same pump system and combustion chamber rig can “drive” a bell with larger expansion and thrust,  with the same flow rate,  out in vacuum.

The usual design for a first stage features “perfect expansion” to sea level-equivalent expanded pressure,  exactly matching the ambient atmospheric pressure.  That same nozzle generates somewhat more thrust as one flies out into space,  because the backpressure term that reduces delivered thrust goes to zero.

It is possible to design the nozzle (with the same chamber) at some intermediate altitude for perfect expansion.  That will raise the thrust out in vacuum a little higher than the sea level version,  but the thrust at sea level will be lower than the sea level design.  That is usually unacceptable for liftoff.

There really is no such thing as a truly “vacuum-optimized” nozzle design,  which would imply an infinite expansion to zero expanded pressure.  The size of the exit area and length of the bell would be infinite.  The max dimensions of the bell are actually set by what will fit in the allowable space at the rear of the stage.  That is what is termed a “vacuum-optimized” design,  but that is really a misnomer.  It is merely the max expansion that meets the limited space constraint into which the engine must fit.

These “vacuum-optimized” designs have significantly-higher vacuum thrust because of the greater expansion.   They would have a very low sea level thrust due to a very large backpressure term (driven by the large exit area),  except that such designs will usually suffer flow separation (and thus not produce any significant thrust).  You cannot operate with a separated bell:  the engine will self-destruct.  Clearly the compressible flow “ballistics” of the nozzle expansion are the main factor driving performance:

The specific impulse Isp = Fth/wtot,  where wtot is the total propellant flow rate drawn from the tanks.  Isp must be based on wtot to be consistent with the mass ratio calculations used in the rocket equation.  Unless there is no pump drive gas dumped overboard,  wtot and the nozzle flow rate wnoz are not the same.  If the dumped pump drive gas wbleed is nonzero,  as it is in some engine “cycles”,  then the cycle can be represented easily enough by the bleed fraction BF = wbleed/wtot.  This can be used to determine the total flow from the nozzle flow coming from the nozzle ballistics:  wtot = wnoz/(1-BF).

There are some cycles that have no pump drive gas dumped overboard.  All the flow,  including the pump drive gas,  ends up going through the nozzle.  For these cycles,  just use BF = 0.

The calculation starts with an estimate of the nozzle kinetic energy efficiency ηKE from the average bell half-angle.  This models the average of the cosine-component factors for the exiting streamlines that diverge off the centerline axis.   There is no measurable friction loss in a proper nozzle design.

For a given chamber pressure,  one chooses the appropriate expanded pressure value Pe,  and forms the pressure ratio PR,  and from that the temperature ratio TR,  as shown just above.  Expanded Mach number Me can be computed from TR,  as shown.  Using TR and Me,  find the expansion area ratio Ae/At,  as shown.  This procedure works “once-through” for sea level (or any nonzero Pa) nozzle designs.

If instead a desired Ae/At is the given requirement,  then just iterate this very same computation procedure over several values of Pe to find the one that gives you the desired Ae/At.  That iterative procedure works for all vacuum designs,  where it is the size of Ae that is your design constraint.

These computed Me,  Ae/At,  Pe,  and Pc values get used to compute the vacuum thrust coefficient CFvac.  That plus the Pa,  Pc,  and Ae/At values of the backpressure term,  combine as thrust coefficient CF at Pa.

There is some thrust requirement Fth associated with some backpressure value Pa.  Find the CF for that Pa,  and plug it,  the Pc,  and the Fth value into the thrust equation Fth = CF Pc At,  which sizes the nozzle throat area At,  and then with the Ae/At value,  also sizes the nozzle exit area Ae.  However,  if you are using the gas generating item from another calculation for this one,  you will want to keep the At and sized flow rate values of that other case.  In that case,  iteratively adjust the required Fth value at its Pa value,  until the desired At and flow rates are obtained.

So far the only thing dependent upon the propellant combination is the gas specific heat ratio γ,  and that dependence is very weak (nearly all rocket gases have γ ~ 1.20).  Once the nozzle throat area At has been sized,  the propellant combination gives you a chamber c* velocity as an empirical function of Pc,  and then the nozzle flow rate equation wnoz = Pc CD At gc/c* lets you determine wnoz.

Your engine cycle gets modeled with the BF value.  Using it,  we have wtot = wnoz/(1-BF).  Then,  the specific impulse that is consistent with the rocket equation is just Isp = Fth/wtot.

Or looked at another way,  Isp = [CF c* (1-BF)] / [gc CD].

The only thing about Isp that is truly sensitive to the propellant combination (and only the propellant combination) is the c* value.  The rest is all coming from the compressible flow ballistics of nozzles,  and Isp is very sensitive to that.

Doing the nozzle ballistics from a c* value is the only way to get a reliable Isp for your propellant combination,  as used with a specific nozzle design.

Tabulated values versus propellant combinations reflect only some Pc level (yours may be different),  with a perfect nozzle efficiency ηKE = 1.000 (about 1 or 2% too high,  usually),  and either sea level Pe expansion,  or expansion to a specified Ae/At value in vacuum (where your Ae/At in vacuum may be different).  They are therefore inherently some significant percentage off,  as a result.  Further,  a lot of reported Isp data are based on wnoz,  not wtot,  which introduces a overestimating error of the same size as the cycle BF!  While calculating corrections is possible,  just doing the scratch ballistics is just as easy,  and far more reliable.

Now,  if you know how to do this for yourself,  then you can tell if someone else’s performance prediction software is feeding you “garbage” instead of “good stuff”.  But,  if you don’t know how to do this,  you will never know “garbage” from “good stuff”.   Remember:  with computers,  it’s GIGO (“garbage in,  garbage out”).  Computers process bad numbers just as readily as good numbers.  It all looks the same.

Some typical values:

Conical bell,  15 deg half-angle..........ηKE = 0.98296

Curved bell 18 & 8 deg half angles....ηKE = 0.98719

Well-made throat with smooth profile..CD = 0.99 to 0.995

“typical” rocket gases..........................γ ~ 1.20

Modeling c* variation with pressure....

.....delivered c* = K Pcm where m ~ 0.01

.....use a K & m curve fit from real test data

.....includes c* efficiency = c*/c*o,  generally 0.95+

It is important to have a reliable estimate for the actually-delivered c* from a rocket engine chamber (true of the solids as well as the liquids,  and also the hybrids).

The c* value has the units of velocity,  and can be calculated from chamber temperature as chamber gas properties.  These are reported as theoretical values from thermochemical methods or programs (of which the NASE ODE code is considered to be the “gold standard”).

c* = [ (gc R Tc/γ)(gfactor exponent]0.5             where gfactor = (γ + 1)/2 and exponent = (γ + 1)/(γ – 1)

However,  no rocket ever delivers the theoretical value of c*,  all have some experimental c* efficiency ηc* = c*/c*o,  where c*o is the theoretical thermochemical value.  Typically,  these efficiencies are in the vicinity of 0.95+,  and they vary with pressure as a power function,  as does c*o itself,  using a function of the form

c* = K Pc m          where the exponent m is usually on the order of 0.01

As a result that function can be used to correlate actual delivered c* from real engine test data.  Once obtained,  this c* value at the chamber pressure Pc,  goes into the nozzle flow rate equation:

wnoz = Pc CD At gc / c*

In the case of a liquid engine test (or tests),  you literally measure the propellant flow rates fed to the engine,  and you measure the bleed flow rate with a metering orifice and pressure transducers,  if there is bleed.  You also have a precise before-and-after measurement of the throat diameter,  so At is known.  The gc item is just the standard Earth gravity acceleration,  so the only other item to quantify is CD.  You need a separate flow calibration result to quantify that as precisely as possible,  although well-designed nozzles have CD’s that fall in the 0.990 to 0.995 range,  usually.  (A simple drilled orifice operated choked is closer to CD = 0.80.)

You simply plug in the flow rates,  Pc values,  and your At,  your CD,  and the gc factor,  and solve for the corresponding c*.  You do that for a lot of tests at a lot of different Pc values.  Then you correlate c* versus the various levels of Pc,  and then least-squares curve fit that data set with the power function given above.  If you do all this correctly,  Pearson’s r2 parameter will always exceed 0.99.

Solids are different,  in that there is no way to directly measure the instantaneous mass flow rate.  Under the assumptions that c* and At are never far from their average values for the test,  and that the test design is such that Pc is crudely constant,  the time integral of the nozzle mass flow equation yields these results,  which can be solved for the average c* and average Pc:

∫wnoz dt = ηexp Wp = [CD At gc/(avg c*)][∫Pc dt]  and (avg Pc) = [∫Pc dt] / tB

where ηexp = expelled weight divided by the loaded propellant weight Wp (which implies a defined weighing configuration),  and CD and At are their best-estimated average values.

A test motor configuration without significant “sliver” is required for this easily.  Most of those are simple cylindrical segment internal burners.  Sizes from about 6 inch diameter on up usually get c* data that correlate very well indeed with full scale motors.  Smaller ones do not correlate very well.  2-inch burn rate motors are essentially useless for correlating c*,  although they are quite good for correlating burn rates.

Again,  you correlate avg c* vs avg Pc,  and least-squares curve-fit the data set with the power function.  Also again,  Pearson’s r2 parameter will exceed 0.99 if you do this carefully enough.

## Thursday, June 8, 2023

### What to Do About Aggressive Behavior At Sea or In the Air

Every several weeks there is another incident at sea or in the air with an adversary’s ships or planes maneuvering aggressively or dangerously,  right in front of our ships or planes.  Most of the recent ones have been Chinese misbehavior,  although we also used to get this from Putin’s Russia,  before he got so bogged down in Ukraine.

The most recent incident with the Chinese in the Taiwan Strait is pictured.  The Chinese warship cut dangerously close across the bow of a US destroyer (and a Canadian warship).  Under normal circumstances,  this is egregiously-incompetent behavior risking a collision like that.  But these circumstances are not normal:  we are being tested for signs of weakness!

Our typical measured response is to slow down or change course to avoid the collision,  which is exactly what happened in this case.  It’s a non-aggressive way to avoid the danger,  and to avoid provoking any conflict.  Any normal civilized person would not misinterpret that.

But,  these countries are run by dictators who do NOT think like that!  Both Putin in Russia and Xi in China are would-be Adolf Hitler clones,  out to secure as much world domination as they can.  You have to understand that,  in order to understand how they interpret our measured responses as a sign of weakness.  The proof of that thesis is that they continue to do these aggressions quite frequently.

The only way to stop this,  is to respond strongly in terms they actually understand:  raw naked force.  It has nothing to do with logic,  common sense,  or civilized behavior.  It has everything to do with effectively “slapping the dictator upside his head”.

You see that 5-inch gun turret on the foredeck of our destroyer in the picture?  We have to use it,  in the centuries-old unmistakable way,  of first responding with a shot across the bow of the offender.  If the offender persists,  then we hit him with it on the second shot.  If he shoots back,  we sink him.  The ship has missiles that will do that,  since only one little 5-inch gun is too small a force to do a quick sinking.

It’s called “gunboat diplomacy”,  and it has worked,  for centuries.  But,  it is really hard to do gunboat diplomacy if your boats lack guns.  Some of ours have no guns,  only missiles.  That was a mistake.

Yes,  it is conflict.  No,  it will not start a general war,  other than a war of words (about which we should not care much).  But it WILL put a stop to the offensive behavior at sea!

In the air,  it is adversary fighter jets zooming too close across,  next to,  and in front of,  our unarmed “patrol bomber” spy planes.  We have to arm them,  or else provide an armed fighter escort.  The inherent short ranges of fighters would suggest we instead just arm the spy planes.  To accomplish the analog to gunboat diplomacy at sea,  these planes need both guns and missiles,  and you need to put tracer ammo in the guns.

What you do when the adversary starts to zoom in too close,  is put a line of highly-visible tracer gunfire in front of him.  If he persists,  you then hit him with a missile.  Yes,  it is armed conflict.  No,  it will not start a general war,  only a war of words.  But it WILL put a stop to the egregious behavior!

Arming these planes does impact their mission.  From that point forward,  you must keep them in international airspace only;  you never violate sovereign territory with them again.  But,  international airspace is where they do most of their missions anyway,  so it is only a small change.

The analog with Putin’s Russia is that Ukraine MUST win its war!  They are our proxy to stop Russian armed aggression.  A draw,  or a negotiated peace without Ukrainian victory,  is just a sign of our weakness in Putin’s eyes.  He will just keep on doing these egregious things,  unless we smack him very hard to stop him.  That is just the nature of these Hitler-clone dictators.   There is no other way.

Of the two,  China is the more dangerous one,  because it has a much bigger economy than today’s Russia.  But,  there are also many other minor Hitler-wannabees out there.

Kim Jong Un in North Korea,  is one.  For him,  I would suggest that the next time he launches a rocket,  we shoot it down before it gets very far.  We have that capability.  What could he possibly do about it?  Nothing.  It starts only a war of words,  and it WILL constrain his future behavior!

The key here is that we respond to these dictators’ misbehaviors with significant force,  each time and every time!  They understand only force as a strong response!  Anything less is taken as a sign of weakness,  inviting further ever-escalating misbehavior.

And that escalating misbehavior DOES eventually lead to general war!  We’ve seen this before,  a bit over 8 decades ago.  It cost over 15 million military lives,  and over 38 million civilian lives,  even without nuclear weapons (except two right at the end).  The cost today with nuclear weapons is in the billions of lives.

A war of words,  or billions dead.  You choose.  Then tell your elected representatives what you want them to tell the military to do,  when these dictators act up.

And when they don’t do what you said,  elect somebody else.

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Some related articles:

“Russian Lies” posted 27 Dec 2022

“How To Stop the Ukraine War” posted 27 Nov 2022

“Nuclear War Warning”  posted 12 Sept 2022

“What the Ukraine Thing Is Really All About” posted 20 Feb 2022

“Bullying Demands Resistance” posted 8 June 2019  -almost the same topic,  just Russian,  not Chinese!!

“The Time Has Come to Deal With Iran and North Korea” posted 8 April 2017

“Cold War 2?” posted 18 March 2014 – this was the REAL start of the Ukraine war:  taking of Crimea.

Use the navigation tool on the left.  All you need is date and title.  Click on the year,  then the month,  then the title if more than one article was posted that month.

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Update 6-9-2023:  A close version of this article appeared on the op-ed page of the Waco "Tribune-Herald" as a column from the board of contributors,  of which I am a member.

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## Tuesday, June 6, 2023

### Frontal Thrust Density in Rockets

The design liftoff thrust performance of rocket vehicles is something I,  and many on-line correspondents,  are interested in,  and have discussed extensively.  I have long maintained that solid rockets offered the highest possible liftoff frontal thrust density,  but I never had the analysis and numbers to back that assertion up.  Now I do.

I started with a rather simple pair of point analyses,  using perfect expansion from a chamber pressure to sea level to size the nozzle expansion ratio Ae/At and sea level thrust coefficient CF.  It turns out that the variable of real interest is the ratio of the sum of nozzle exit areas to the vehicle base area Ae/Abase.  I did this for a solid design similar to a Shuttle SRB at 1000 psia,  and for a liquid operating at up to about 4000 psia,  as in SpaceX’s Raptor methane-oxygen engines.  For either,  a gas specific heat ratio γ = 1.20 is pretty close to the best available model.

The first thing that “jumped off the page at me” was the single-engine vs multiple-engine configuration trade that one must deal with,  in liquid stages.  This not only determines how far the engine bell or bells extend aft of the stage,  it also determines the effective Ae/Abase ratio,  as Figure 1 indicates.  For this initial calculation,  I just guessed Ae/Abase = 0.75 for a 3-engine configuration.  Of course Ae/Abase is 1 for the solid,  but at a Shuttle SRB chamber pressure of only 1000 psia in a 10 ft diameter size.

Be aware that extra spacing between the bells of a multi-engine installation is required of liquid engine designs that use thrust vector control (TVC).  The imposition of a TVC requirement thus lowers Ae/Abase further than the geometry of simple nested circles would suggest.

Figure 1 – Initial Point Analysis

There’s enough variable interplay going on here to confuse these results.  So I ran a little study of these results,  versus chamber pressure Pc variations all the way from 500 to 4000 psia,  expanding perfectly to a fixed sea level backpressure of PSL = 14.696 psia. Perfect expansion is when expanded pressure equals backpressure.

For these calculations,  I used γ = 1.20,  and ηKE = 0.99,  with the perfectly-expanded pressure ratio PR = Pc/Pe = Pc/PSL.  In compressible flow,  the expanded temperature ratio TR = Tc/Te = PR(γ – 1)/γ.  Expanded Mach number Me = [(TR – 1)*2/(γ – 1)]0.5.  The expansion area ratio Ae/At is then (1/Me)*[TR*2/(γ + 1)]0.5*(γ + 1)/(γ – 1).  The resulting vacuum thrust coefficient CFvac = [(Ae/At)/PR]*(1 + γ*ηKE*Me2).  The backpressure correction term is (Ae/At)*PSL/Pc,  which subtracts from CFvac to create the sea level thrust coefficient CF.  The expanded pressure term is already in CFvac.

Once you know the sea level CF,  you can easily estimate the thrust per unit exit area as Fth/Ae = CF*Pc*At/Ae.  Multiplying that by the Ae/Abase ratio gives you Fth/Abase = CF Pc (At/Ae) (Ae/Abase),  which is the frontal thrust density that the design can achieve.  Plots of this for the solid and the liquid are given in Figure 2.

Figure 2 – More Detailed Trends From Plots vs Pc

First,  notice that the frontal thrust density trend vs Pc of the solid is exactly the same as the max frontal thrust density of a 1-engine liquid,  since both have Ae/Abase = 1.  This is independent of the rocket specific impulse performance!  It only depends upon the expansion you get from Pc into a fixed sea level backpressure.  (In the real world,  the odds are against an existing engine design having an exit area exactly equal to the base area of the stage into which you want to install it.)

What I added here to the solid plot is the resulting trend of stage diameter vs Pc under the assumption that the same case material at tensile strength and thickness is used at all Pc’s.  That makes stage diameter inversely proportional to Pc,  because:  Pc ID = 2 σ t,  and ID ~ stage diameter.  The reference point is the approximately 1000 psia Pc,  at 10 ft stage diameter,  used in the Shuttle SRB’s.  2 σ t is a constant representing the best that can be done,  under these realistic assumptions.

I ran these trends for Ae/Abase = 1.0,  0.5,  and 0.25 for the liquids,  just to see the effects of Ae/Abase.  There is also a constraint here,  as to what the designer can really do.  The single-engine configuration that maximizes at Ae/Abase = 1,  leads to a very long expansion bell (of a very large engine),  protruding from the back of the stage.  Almost no designer does this in a first stage booster anymore!  Nearly all such boosters are multi-engine now,  driven to be so,  in order to get a much shorter bell protrusion length,  plus some level of engine-out capability.

One should be aware that TVC requires significant extra space around the bell exit area,  for the bell to move without hitting other engines that may vector differently,  or not at all.  That simply means that geometrically-nested circles that fit the stage base diameter are a large over-estimate of Ae/Abase.  Thus,  the value of Ae/Abase = 0.5 (versus the 0.75 I assumed initially) is actually rather realistic for a multi-engine liquid stage employing TVC.

Under those circumstances of the various design constraints driving where you are on these curves,  a 5-foot diameter SRB offers around 32 or 33 Klb/ft2 of frontal thrust density,  while a multi-engine liquid core with TVC,  at the more typical 3000 psia Pc,  only offers in the vicinity of 18 Klb/ft2 frontal thrust density at Ae/Abase = 0.5.  The solid is some factor 1.78 larger in its individual frontal thrust density!  Add only two such SRB’s to a core,  and your liftoff thrust is 2*1.78 + 1-for-the-core,  or some factor 4.56 larger than the core alone.  (That is figured for equal strap-on and core diameters.)

Add instead four of them,  for a factor 8.12 thrust increase over the liquid core alone.  That is one whopping increase in liftoff thrust!

If the solid were instead 10 feet in diameter and only 1000 psia,  its frontal thrust density is about 25 Klb/ft2.  That is still some factor 1.39 larger than the multi-engine 3000 psi liquid with TVC.  Again,  adding only two such SRB’s to the liquid core produces a liftoff thrust some factor 3.78 larger than the liquid core alone.  Adding four would still be a whopping increase at factor 6.56!

The worst case scenario for the solid under these constraints is a 10 ft diameter solid at 1000 psia,  with the 25 Klb/ft2 density,  versus a multi-engine liquid with TVC,  at the full 4000 psia Pc,  with 20 Klb/ft2 density at Ae/Abase = 0.5.  That solid is still factor 1.25 “thrustier” than the liquid.  Adding only two of them produces a total lift thrust some factor 3.5 larger than the core alone.  Add four for factor 6.

Consider also that most launch vehicles designed to use solids as SRB’s,  have SRB’s with a smaller diameter than the core liquid stage.  That means the SRB can have a larger length/diameter ratio than the liquid core stage,  up to the same length as the liquid core (but also limited to being within the ballistics limitations of the solid propellant’s burn rate).  The extra length can partially compensate for the specific impulse deficit that solids have,  with respect to the liquids.

So,  under realistic design constraints,  yes,  the solids offer significantly more frontal thrust density potential than the liquids.  Which explains their popularity for added liftoff thrust among the many launch vehicles currently flying,  reusable cores or not.  Liquid strap-ons simply cannot add as much thrust as the solids can,  because liquid frontal thrust density potential is simply lower,  in any sort of practical design.  (Strap-ons do not need TVC,  but liquid strap-ons are still likely multi-engine).

The main effect here has nothing to do with the type of propellant (liquids or solids).  It has mostly to do with the size of (the sum of) the nozzle exit areas relative to the size of the base area of the vehicle.

As it turns out,  with solids,  that area ratio is usually near 1,  because there is only 1 nozzle,  and it quite commonly has an exit area very near the base area.  That simply cannot happen with a cluster of liquid engines,  and it could only happen with a single liquid engine,  if one specifically designs it to be so.  That would actually be quite rare.

Chamber pressure driving the nozzle expansion also plays a large role,  and here the constraints upon that variable are again quite different for the solids versus the liquids.  This shows up in the thrust coefficients achievable,  which vary quite strongly with chamber pressure level.  For these purposes,  “chamber pressure” is that at the entrance to the nozzle,  not earlier in the cycle with liquids.

The solids have a trade between case diameter and design chamber pressure level,  because the steels from which the cases are made,  have a tensile strength that does not scale with size.  That is exactly why the larger sizes are associated with lower chamber pressure levels.  Liquids are more limited by the ability of the pumps to deliver adequate flow at extreme outlet pressures,  compounded by the “staged pressures” of the pump drive cycles.

As it turns out,  you get the highest frontal thrust density,  regardless of type,  when the exit diameter of the engine bell is just about equal to the base diameter of the vehicle.  That is common with solids,  and rare with liquids,  which is why the historical experience has been that solids have a higher frontal thrust density than liquids.

That frontal thrust density potential is higher at higher chamber pressures,  also quite the strong variable in the ranges where the technology is,  and has been.  Earlier in history,  when stage diameters were smaller and liquid chamber pressures were lower,  the solid typically achieved more of its potential frontal thrust density advantage over the liquid.  In recent times with larger diameters and higher liquid chamber pressures,  the disparity has lessened.

As for the effect of gas properties,  the specific heat ratio γ is pretty close to the same,  at about 1.20,  for all the solids and all the liquids.  Only γ,  and not molecular weight MW,  appears in the thrust coefficient CF,  and the engine thrust requirement defines throat area via Fth = CF Pc At,  without any gas properties.

It is the chamber c* velocity that contains MW and chamber temperature,  as well as γ.  That reflects the propellant type,  and determines propellant massflow through the nozzle.  In solids,  that massflow determines the required burning surface and burn rate.  In liquids,  it determines how much pump you must supply.

Gas properties are also in the speed of sound equation,  similar to,  but not the same as,  chamber c*.  One would need that,  to find the exit velocity from the expanded Mach number,  but that velocity does not appear in the CF equation;  only Mach number and Mach number-related ratios appear there.

We can safely conclude that gas properties (propellant type) do not determine the thrust capability achievable with a propellant,  they only determine the associated massflow (and thus specific impulse).

## Sunday, June 4, 2023

### Consolidation in Aerospace?

The figure below came from the final report of the commission on the future of the US aerospace industry.  I found it posted on LinkedIn.  It covers 2 waves of consolidation since about 1980,  with the wave since the fall of the Soviet Union in 1991 the most intense.

But,  bear in mind that this consolidation process has been going on since about the end of World War 2.  The rocket plant where I once worked closed in 1995,  early in that latest wave.  Laid off in late 1994,  I had to go do something else for a living after that,  being “on the street” along with about 1.75 million other aerospace workers.  The industry could not use hardly any of us.

This consolidation into ever-larger companies that hold ever-larger shares of the market,  has been aided and abetted by the US government,  under the excuse that “larger is more efficient”.  But that excuse has proven to be a lie.  Without competition,  there is no real innovation or efficiency.  Nor is there any incentive to produce good results,  since you get paid anyway no matter what,  in government contracting.

For just one example,  consider the case of Boeing.  In the 1930’s to 1950’s time period,  Boeing gave you the B-17 Flying Fortress,  the B-29 Superfortress,  the B-47 Stratojet,  the B-52 strategic bomber,  and the 707 airliner.  And these were deservedly famous,  marvelously-effective airplanes.  Note that Boeing had many competitors then.  Note also the innovations:  going from propellers to jets,  and from about 200 mph cruise to about 500 mph (600 mph in the case of the B-47).

After all the pictured consolidation,  in recent years Boeing is the only domestic US source of large airliners.  (It doesn’t even build bombers anymore,  that’s now Northrup-Grumman.)  Here of late Boeing has given you the 737-MAX debacle (over 300 dead),  serious assembly and quality control issues with the 787,  and serious certification issues with the 737-MAX,  the 777,  and the 787.

Consider:  every one of these modern planes flies the same high subsonic speed as the 707,  just a bit more efficiently with better engines,  and in various sizes.  I see evolution,  but not innovation,  in these recent years!  I also see no domestic competition in the airliner business at all,  and there is currently only Airbus overseas!

I think this consolidation BS has gone way too far!  Competition is demonstrably way more important than any mythical efficiency-of-large-size.

I think you might better force the breakups with a market share tax,  instead of lawsuits.  Make it suddenly very confiscatory if you exceed a certain critical market share in the relevant business niche.  That would force the divestitures that would break up these monopolies.  It would be self-enforcing.

As a first guess,  statistics says the minimum size of a credible sample is 22.  So,  as wild guess,  I would say any company then needs about 21 competitors at a minimum.  Evenly-spread,  that’s just under 5% market share each.  If the big one is 4-5 times as big as the smallest one,  then something like 15-25% market share would be about the max we should tolerate before the market share tax kicks in.

We are VERY,  VERY FAR from that competitive market right now!  And aerospace is not the only industry suffering from this,  although it may be the most expensive one.