**It is intended to give readers a means to compute accurate and realistic thrusts. This plus a knowledge of chamber characteristic velocity c* is sufficient to do very elementary rocket ballistics.**

*This article applies to anything that thrusts from a subsonic chamber through (at least) a sonic throat.*
Most (but not all) nozzles that have a sonic throat also have
a supersonic expansion bell. Scramjet is
excluded as being without a sonic throat:
the feed to the nozzle inlet is already supersonic, and there is no contraction in flow area to a
throat.

Rockets of any type are typically high pressure ratio PR chamber-to-exit, and high area ratio AR exit-to-throat. These can be ablatively cooled, or actively liquid-cooled.

Gas turbine engine nozzles are typically low pressure ratio PR
chamber-to-exit, and low area ratio AR
exit-to-throat. These are usually air-cooled, and variable geometry: anything from convergent-only to a mild
supersonic expansion bell. Lower turbine
inlet temperatures require lean mixtures and cooler flames, making air cooling possible, as long as the air itself isn’t too hot. That high speed air heat effect limits the flight
speeds achievable with gas turbine engines.

Ramjet engine nozzles are typically low pressure ratio PR chamber-to-exit, and low area ratio AR exit-to-throat. Modern missile designs are usually ablative. Some of the oldest designs were
air-cooled, similar to gas
turbines, but this approach is severely
limiting in a modern ramjet design,
which can run far richer, and at far
higher flight speeds, where the air
itself is far hotter.

**Fundamentals**

Conservation of mass: the same massflow exists throughout the nozzle
(any air cooling bleed effects or other injections or leaks are ignored, if any exist at all).

Conservation of momentum:
a

**control volume**drawn about the rocket engine is**pierced by the exit stream exactly at its exit area**, and the momentum of the propellant feeds are either inconsequential, or perpendicular to the thrust axis, or they come from tanks inside the control volume. This could be any combination of those situations, or even all three. Balancing stream momentum and the pressure forces against a restraining force, leads to evaluating the thrust.
Conservation of energy:
the drop (from chamber to exit) in enthalpy, as measured by the drop in static temperature,
equals the increase in kinetic energy of
the stream, with essentially zero
kinetic energy inside the chamber. There
is an accompanying drop in static pressure,
in an amount defined by the ideal gas assumptions and the corresponding equation
of state. See Figure 1.

__All figures are at the end.__

We use enthalpy “h” instead of internal energy “u”, because it includes the effects of pressure
change upon energy content, and internal
energy does not. Enthalpy difference Δh is
essentially the temperature difference ΔT,
multiplied by the specific heat at constant pressure c

_{p}. (Internal energy change uses the specific heat at constant volume c_{v}.)
Book-keeping: this is
done the easiest way in Mach number-pressure-temperature variables, instead of the primitive variables,

**. That last means we may use as our equation of state P = ρ R T, and we may use as the change in enthalpy Δh = c***as long as the ideal gas assumption applies*_{p}ΔT.
In this book-keeping scheme,
we make good use of total (or stagnation) pressures P

_{t}and temperatures T_{t}. Assuming no appreciable friction losses, flow is “isentropic”,__meaning total pressure and total temperature are constant through the nozzle__, a very good assumption in almost every conceivable case.
The ratio of specific heats γ = c

_{p}/c_{v}becomes a very useful value to relate totals to statics. At a location where the Mach number is M, the total/static temperature ratio TR = 1 + 0.5*(γ – 1) M^{2}, and the total/static pressure ratio is PR = TR^{exp}, where exp = γ / (γ – 1).
The streamtube area model is more complicated than the
simple mass conservation-derived relation in incompressible flow, and is based off of sonic conditions at the
throat area A

_{t}. If you know the Mach number M at another station where the area is A, you can find that area ratio AR = A/A_{t}as easily as the total/static ratios TR and PR. If you know instead the area ratio A/A_{t}, finding the Mach number M is inherently a transcendental (iterative) solution:
A/A

_{t}= (1 / M) [TR / const1]^{const2}
where TR is defined as above, const1 = 0.5 (γ + 1), and const2 = 0.5 (γ + 1) / (γ –
1)

Heat transfer:

**We***this is driven not by static temperature but by recovery temperature!*__must__do this because the supersonic flow in the nozzle is both highly compressible, and highly dissipative. At any given Mach number M, recovery temperature T_{r}is very nearly the same as total temperature T_{t}. How it varies depends upon laminar versus turbulent flow, and the gas property Prandtl number P_{r}:
T

_{r}= T + r (T_{t}– T) where the recovery factor r = P_{r}^{0.5}laminar, P_{r}^{0.33}turbulent
Only the heat transfer film coefficient h responds significantly
to the varying Mach number,
pressure, and temperature down
the nozzle profile. It does this in a
very empirical way. Multiple models
exist for this, not covered here. The local heat flux at any station is of the
form:

Q/A = h
(T

_{r}– T_{s}) where T_{s}is the material surface temperature
For heat transfer purposes,
failing real data, you can
estimate Prandtl number P

_{r}= 4 γ /(9γ – 5).**Conventional Nozzle Thrust Coefficient C**

_{F}

_{}
Your

__ideal gas model__of the gas flowing through the nozzle comprises its constant specific heat ratio γ, and its constant molecular weight MW. These can come from thermochemistry calculations, and need to reflect the high temperatures involved.
The thrust F of an idealized nozzle evaluated at its exit
plane is the momentum of the exiting gas,
plus the exit area A

_{e}times the difference in pressure between the exiting stream static pressure P_{e}and the ambient backpressure of the surroundings P_{b}. Ideally, all the streamlines are parallel to the thrust axis.
In the real
world, they are not. This
streamlines-off-angle effect is modeled with the nozzle kinetic energy
efficiency factor η

_{ke}. It applies to the momentum term in thrust,__but not the pressure term__, as long as the exit plane is perpendicular to the axis.
F = η

_{ke}m V_{e}+ (P_{e}– P_{b}) A_{e}where m = mass flow rate and V_{e}= exit velocity
To convert this to compressible flow variables, we make use of the m = ρ

_{e}A_{e}V_{e}massflow definition, the ideal gas equation of state P_{e}= ρ_{e}R T_{e}with R = R_{univ}/MW, and the exit plane speed of sound for an ideal gas c_{e}= (γ g_{c}R T_{e})^{0.5}. The variable g_{c}is the “gravity constant” to make the equation consistent with inconsistent mass and force units. If those units are consistent, g_{c}will be 1.
F = η

_{ke}(P_{e}/ R T_{e}) Ve^{2}Ae + (P_{e}– P_{b})A_{e}using massflow, then equation of state
F = γ η

_{ke}P_{e}A_{e}M_{e}^{2}+ (P_{e}– P_{b})A_{e}using speed of sound**Distribute the A**

*Note that the first term in the equation just above is the momentum term, and the second term is the static pressure difference term.*_{e}so that there are 3 separate terms, and regroup.

F = P

_{e}A_{e}[1 + γ η_{ke}M_{e}^{2}] – P_{b}A_{e}recombining terms such that P_{e}A_{e}factors out
Here, inside the
bracket, the 1 now corresponds to the exit static pressure term with P

_{e}A_{e}factored out, and the γ η_{ke}M_{e}^{2}corresponds to the momentum term with P_{e}A_{e}factored out. The backpressure effect is still a separate force term,*with the recombined bracket-containing force term really just being thrust into vacuum.*

Now we introduce the definition of thrust coefficient C

_{F}= F / P_{c}A_{t}with the understanding that the P_{c}is the total (stagnation) pressure feeding the nozzle. If the contraction from chamber to throat area is large enough, there is no measurable difference between total and static pressure at the nozzle entrance.
C

_{F}= F / P_{c}A_{t}= (P_{e}A_{e}/ P_{c}A_{t})[1 + γ η_{ke}M_{e}^{2}] – (P_{b}A_{e}/ P_{c}A_{t})
CF = (P

_{e}/P_{c})(A_{e}/A_{t})[1 + γ η_{ke}M_{e}^{2}] – (P_{b}/ P_{c})(A_{e}/ A_{t}) regrouping P’s and A’s together**CF = (AR / PR) [1 +**

**γ η**

_{ke}M_{e}^{2}] – AR / PR_{op}**(**

__the__thrust coefficient equation)
with PR

_{op }= P_{c}/P_{b}using the actual design P_{c}and P_{b}
with PR
= P

_{c}/P_{e}= (1 + 0.5 (γ – 1) M_{e}^{2})^{exp}= TR^{exp}where exp = γ / (γ – 1)
and AR
= A

_{e}/A_{t}= (1/M_{e})[ TR/const1]^{const2}
with const1 = 0.5 (γ +
1) and const2 = 0.5 (γ + 1) / (γ – 1)

This last formulation is particularly convenient when one
wants a certain exit Mach number M

_{e}, because AR = A_{e}/A_{t}and PR = P_{c}/P_{e}are easily calculated from M_{e}using the ideal gas γ. Otherwise, if conditions at a certain AR are desired, one iteratively tries M_{e}values until the desired AR obtains, then computes PR. Essentially, M_{e}and P_{c}/P_{e}are “locked in” by the AR value regardless of the value of P_{c}, although they are not most conveniently figured in that order.
The “operating pressure ratio” PR

_{op}= P_{c}/P_{b}depends directly upon your design choices for P_{c }and P_{b}. One had to choose a P_{c}to do the thermochemistry, and P_{b}is set by the altitude, or else 0 if vacuum.
Once γ, M

_{e}, PR, AR, and PR_{op}are all known, evaluating C_{F}is easy, per the above equation. If you have used a value of c* to size a throat A_{t}elsewhere in your fundamental ballistics, then the nozzle thrust is easily obtained as F = C_{F}P_{c}A_{t}. From ballistics, choked nozzle massflow w = P_{c}C_{D}A_{t}g_{c}/ c*, see Figure 2 below. C_{D}is the nozzle throat’s discharge coefficient (or efficiency).
If all the hot gas generated in the engine workings upstream
of the nozzle entrance goes through the nozzle,
then I

_{sp}= C_{F}c* / g_{c}.*If not, you must ratio down your calculated I*_{sp}, F, and A_{t}by 1 + f, where f is the fraction of generated hot gas massflow that does__not__go through the nozzle.

**Example Problem: Conventional Nozzle, Sea Level and 20 kft Designs**

I automated these calculations into a spreadsheet, and verified the numbers with hand
calculations. An image of the spreadsheet for the sea level
design is given in Figure 3. In
the spreadsheet, items highlighted
yellow are the user inputs, and items
highlighted blue are the principal outputs from the sizing calculations. These are used to generate the performance
table versus altitude, which is not
highlighted.

For this example, I
assumed Pc = 1800 psia, and a conical
nozzle of 15 degree half angle. I used
specific heat ratio γ = 1.20, and a c* =
5900 ft/sec so that specific impulse would be near 300 sec, similar to LOX-RP1. I used A

_{t}= 1.0 square inch, with a nozzle C_{D}= 0.99 to size flow rate. The resulting design is a nominal 3000 lb thrust design, completely immune to backpressure-induced separation, since it is never over-expanded. How the nozzle kinetic energy efficiency is calculated from half angle is discussed below.
Keeping all the data the same except for the design
backpressure, I ran the spreadsheet again
for perfect expansion at 20 kft instead of sea level. The effect is to increase the expansion ratio
for a higher momentum term, and then accept
the negative pressure difference term reducing thrust below 20 kft
altitude. The gas generating chamber and
throat are exactly the same. An image of the 20 kft design
spreadsheet is given as Figure 4.
The spreadsheet includes a separation backpressure estimate (see that
discussion just below), which shows the
risk starts at pressures about 9 psi larger than sea level air pressure. So,
this design is also very likely immune to backpressure-induced
separation risks.

**Flow Separation Risks**

These can only be estimated empirically. There are many correlations. My preferred one uses the inverse of PR = P

_{c}/P_{e}. P_{sep}is the estimated backpressure, at and above which nozzle flow separation is to be expected. It is empirical, and it is a rough estimate.__The designer should allow significant margin__.
P

_{sep}/ P_{c}= (1.5 P_{e}/P_{c})^{0.8333}
For the 20 kft design example just above, P

_{c}/P_{e}= 266.3, so that P_{e}/P_{c}= 0.003755. Thus P_{sep}/P_{c}= 0.013355, and for P_{c}= 1800 psia, the expected P_{sep}= 24.04 psia, quite a margin above sea level pressure. We can conclude that there is no risk of separation in the example nozzle, all the way down to sea level, where P_{b}is only 14.7 psia. The risky backpressure is even higher at about 45 psia for the sea level design.**KE-Efficiency Correlations**

Most conventional nozzles are axisymmetric. Those streamlines near the axis are aligned
along that axis, so that the cosine
factor for off-angle alignment is cos(0

^{o}) = 1.00. Those near the nozzle wall are aligned__at the angle of that wall off the axis__. For a conical nozzle, this is the half angle of the cone. The cosine factor for off-angle alignment is cos(a) where “a” is the half angle of the cone. See Figure 1 again.
Thus, there is a
distribution of local off-axis alignments for the streamlines across the exit
plane. While the “correct” way to
determine the

__effective cosine factor for the distribution__would be to integrate them for an average,**. Simply compute the arithmetic average of the centerline cosine factor value (1.00) and the wall cosine factor value cos(a), and call that the nozzle kinetic efficiency factor:***there is an easier model that is just as good***η**

_{ke}= 0.5 [1 + cos(a)] where “a” is the effective average half-angle of the nozzle wall

For a conical nozzle,
“a” is the cone’s geometric half-angle.
At 15 degrees, η

_{ke}= 0.983. For a curved bell, there is a local “a” near the throat, and a smaller local “a” at the exit lip. One simply averages the two local a’s, and uses that average as “a” in the kinetic energy efficiency formula. For most practical curved bell designs, that average “a” won’t be far from 15 degrees. See again Figure 1.**Free-Expansion Designs By “Last Point of Contact = Perpendicular Exit Plane Model”**

There are multiple techniques and geometries by which a
nozzle can be made self-compensating for perfect expansion at any altitude
backpressure. They all share two
features: (1) a free streamtube surface
unconfined by a physical shell before the “exit plane”, and (2) a point of last contact with physical
structure that is wetted by the propulsion stream that locates the exit plane.

*We want the components of the actual distribution of exhaust velocities, that are aligned with the engine axis.*
Most, if not
all, these free-expansion designs can be
analyzed for expected performance using the very same ideal gas compressible
flow techniques used for conventional nozzles.

**Note that the very same gas-generating chamber and throat area serves as the feed to the free-expansion “nozzle” at all values of P***It is just that the order in which things need to be done is revised.*_{b}.
Conceptually, we are
interested in an effective planar exit area located at the “point of last
contact” (just as the exit lip is the “last point of contact” with conventional
bell nozzles), and oriented perpendicular to the engine axis. This is shown in Figure 5.

Unlike conventional nozzles,
these are always perfectly expanded,
so that P

_{e}= P_{b},__as long as P__! Once a P_{b}is not exactly zero_{b}is known, then PR = P_{c}/P_{e}= P_{c}/P_{b}is known. One solves the PR equation for M_{e}at this value of PR, which is__not__a transcendental iteration, just a simple direct solution:
M

_{e}= { 2/(γ – 1) [PR^{(}^{γ-1)/}^{γ}– 1]}^{0.5}
With M

_{e}now known, find the area ratio from the streamtube relation, and use it with the throat area to find the effective value of the exit area A_{e}:
TR = 1+
0.5 (γ
– 1) M

_{e}^{2}
const1
= 0.5 (γ
+ 1)

const2
= 0.5 (γ
+ 1) / (γ
– 1)

AR = (1
/ M

_{e}) [TR / const1]^{const2}
A

_{e}= AR A_{t}
Referring
again to Figure 5, there is
obviously a distribution of streamline directions at the exit plane,

**.***which is different for each backpressure*__Each geometry is different__, but the idea is to find the largest half-angle off of axial and use it as “a”. This goes into the correlation for kinetic energy efficiency. That correlation is generally for “a” < 30 degrees,__so we are misusing this here!__But, it is the best I have at this time to offer.*Any such “a”-dependent model, even if flawed, is better than no model at all!*

Instead of a thrust coefficient, we estimate thrust directly from the
calculated exit plane conditions,
remembering that P

_{e}= P_{b}, and from that thrust, the thrust coefficient (to use with c* for I_{sp}):
F = η

_{ke}γ P_{e}A_{e}M_{e}^{2}
C

_{F}= F / P_{c}A_{t}
One should note that neither A

_{e}nor η_{ke}are constants here, as P_{b}changes. At high backpressures (low altitudes), “a” is small, η_{ke}is high, and A_{e}and M_{e}are lower. At low backpressures (high altitudes), “a” is quite large, η_{ke}is lower, and A_{e}and M_{e}are high.*Exactly how “a” varies is quite geometry-dependent.*
If P

_{b}= 0 (vacuum of space), PR = infinite, leading to infinite M_{e}and A_{e}. There can be no planar exit plane, and Prandtl-Meyer expansion says “a” > 90 degrees by a small amount.**, quite unlike a conventional nozzle! (Which means this free-expansion design approach is inappropriate in vacuum!)***There is no point trying to use this compressible flow analysis technique on a free-expansion nozzle in vacuum*
However, for an
axisymmetric center-spike design (aerospike nozzle), one could estimate a = tan

^{-1}[(R_{e}-R_{t})/L_{spike}]. For this, R_{e}= (A_{e}/pi)^{0.5}, and R_{t}= approximately (A_{t}/pi)^{0.5}. L_{spike}is the distance from throat plane to exit plane. Longer is lower effective “a”, but higher weight, and a tougher cooling design.
I made another worksheet in the spreadsheet for axisymmetric
aerospike nozzles, embodying the above
calculation techniques, and I verified
it with hand calculations. It lays out
differently, since the sequence is
different, and more items vary with
altitude. The same grouping of design point
data vs altitude performance is maintained,
and the same color-coding for highlighted items. However,
the volume of data is larger,
requiring two figures (vs one) to display herein.

**Example Axisymmetric Aerospike Problem**

The fairest way to compare this type of nozzle design with any
conventional nozzle design is to size both with the same P

_{c}, A_{t}, and γ. If thrust is the issue, and it usually is for launch vehicles, then the preferred performance variable to examine is thrust.
For the example problem,
we use P

_{c}= 1800 psia, A_{t}= 1.0 in^{2}, and γ = 1.20, same as the conventional nozzle examples earlier. The same c* and nozzle throat C_{D}are used.__In effect, this engine shares the very same gas generator as the two conventional examples__. The same altitude backpressures are also used, so that this design can be compared directly to the earlier examples,**.***except that vacuum performance cannot be included*
The spreadsheet results are given in Figures 6 and 7 below. The two figures together provide the image of
the spreadsheet. I have repeated the
altitude data in Figure 7 for convenience.

**Comparisons Among the Example Nozzle Designs**

How these designs compare,
especially as regards altitude performance, does not “jump off the page” from tabular
data. That takes plots, something this spreadsheet software offers. I used the same altitudes and air pressure
data for all 3 examples. Copying selected
data from each worksheet into yet another worksheet provides a way to directly
plot performance from all 3 nozzles on the same page. I did this for thrust, specific impulse, thrust coefficient, and nozzle kinetic energy efficiency.

Bear in mind that all three share the same gas generator at
P

_{c}= 1800 psia, A_{t}= 1 square inch, γ = 1.20, chamber c* = 5900 ft/sec, and nozzle throat discharge coefficient C_{D}= 0.99. All three are roughly the same 3000 lb thrust__at their design points__, within a percentage point or three.
The thrust
comparison is given in Figure 8 below.
The conventional sea level design has slightly better thrust at sea
level ( by about 82 lb out of a nominal 3000 lb) than the conventional 20 kft
design. This reflects the effects of the
negative pressure difference term at sea level, for the slightly-overexpanded 20 kft
design.

The 20 kft design has about a 107 lb thrust advantage, above 100 kft, over the sea level design. This reflects the larger expansion ratio of the
20 kft design, and the fact that the
exit momentum term dominates

__by far__over the pressure difference term, in thrust.
The axisymmetric aerospike design is “right in there” with
the other two, up to about 50 kft or 60
kft altitude. Then its performance drops
dramatically with increasing altitude,
something the free expansion is supposed to compensate! It is a little better than the conventional
sea level design at sea level, and it
remains superior all the way up to about 55 kft. It is equivalent or very slightly better to
the conventional 20 kft design at sea level,
and remains essentially equivalent to about 20 kft. Its downturn in thrust performance is quite
dramatic, and starts at about 40 kft or
50 kft.

It should not surprise anyone that the specific impulse trends in
Figure 9 tell the same tale as the thrust in Figure 8, since all three share the same gas generator
with the same propellant massflow. Nor
should it surprise anyone that the thrust coefficient trends in Figure 10 also tell exactly the
same tale, since all 3 designs share the
same gas generator operating at the same chamber pressure.

The reason for the dropoff in aerospike performance, versus the conventional designs,

**, something that in turn depends upon the effective average half-angle of the propulsion stream bondary. This is really nothing but the cosine factors of streamlines that are aligned off-axis. Kinetic energy efficiency trends are given in Figure 11.***traces directly to the trends of nozzle kinetic energy efficiency*
Remember, for the
conventional designs, half-angle is

__locked-in by the physical bell__, right up to the exit plane. Downstream of the exit lip, gas expands laterally into the vacuum, but this happens__downstream__of the “last point of contact”, where thrust is actually calculated. This is implied by how we draw the control volume about the engine and nozzle, something shown in the lower right corner of Figure 1,__touching at that last point of contact__.
For the axisymmetric aerospike free-expansion design, the last point of contact is the tip of the
spike. The free expansion surface of the
plume is

__inside__the control volume, as is the bell of the conventional nozzle. At high altitudes where the air pressure is low, the plume boundary must expand quite far laterally, between the throat, and the “exit plane” at the last point of contact. This is precisely how large AR and M_{e}are achieved, in order to match P_{e}= P_{b}. Since the length of the free-expansion zone is fixed, the boundary half-angle must be quite large at high AR. That reduces kinetic energy efficiency.
The two conventional designs share a constant kinetic energy
efficiency of 98.3%, as shown. The aerospike starts out slightly better at
99.1% (due to the choice of L

_{spike}used), but drops below conventional at about 20 kft, and falls ever more rapidly to only about 77.7% at 100 kft.__This traces directly to the effective half-angle__of the plume boundary between the throat, and the exit plane at last point of contact.
That is why I included a plot of the axisymmetric aerospike half-angle vs altitude as
Figure 12. Looking at this, please remember that half-angle is constant-with-altitude
at 15 degrees for the two conventional designs.
At 100 kft, cos(56.335

^{o}) = 0.5543. Averaging that with 1 inherently produces η_{ke}= 77.7%.**Conclusion**

I don’t
see any significant advantage to the free-expansion nozzle approach. The small performance improvement is
restricted to the lower atmosphere, and
this design approach is

__entirely inappropriate for use in vacuum__! The complications with cooling the spike outweigh any tangible performance benefits, which are low (unless you cheat by not accounting for the streamline divergence effects).Figure 2 – Modeling Nozzles with Compressible Flow

Figure 3 – Spreadsheet Image for 15 Degree Conical Nozzle As Sea Level Design

Figure 4 – Spreadsheet Image for 15 Degree Conical Nozzle As 20 Kft Design

Figure 5 – Analogous Procedure for Free-Expansion Designs

Figure 6 – Example Axisymmetric Aerospike Nozzle Results, Part A

Figure
8 – Thrust Comparison Among the 3 Designs vs Altitude

Figure 10 – Thrust Coefficient Comparison Among the 3
Designs vs Altitude

Figure 12 – Trend of Effective Half-Angle “a” for
Axisymmetric Aerospike Design