Rocket Performance and Efficiency

0.1 But first, What is a Rocket Engine? compressible isentropic flow equations become very useful
for analyzing such a beast.
In this section we are going to talk a lot about how we In this lecture we will define the key parameters that define
characterize rocket engine performance. But it is helpful to the performance for a rocket engine.
start by defining what a rocket is.
A rocket engine is a reaction engine, an engine that expels
mass to generate thrust. Rockets carry all of their own work- 0.2 Effective Exhaust Velocity and
ing fluid, called propellant, in contrast with air-breathing Specific Impulse
engines. Furthermore reaction engines are only considered
rockets when they use gasdynamic expansion of the working Now that we have covered the basic mechanics of thrust
propellant to accelerate it. This contrasts with, for instance, and the isentropic nozzle relations we will begin to show
many types of electric propulsion thrusters which use body how these apply to rockets by defining a set of convenient
forces such as the Coulomb electrostatic force to acceler- parameters.
ate gasses. And while the focus for this section will be on
Local Pressure
true rocket engines, some of the definitions and concepts
will apply when we talk about other reaction engines such
as electric thrusters.
Below is a conceptual drawing of the most-typical rocket
engine families along with an electric thruster for good mea-
sure.
Solid Bipropellant Monopropellant Hybrid Electric

Note: Station 1 is not the same as

stagnation due to non-zero velocity in
1 * e
Power chamber. For large A1/A*, stagnation is
Combustion Catalyst
Chamber still a good approximation.
Fuel

Nozzle
Let’s start by remembering that, for a rocket

T = ( Pe − P0 ) Ae + ṁ p Ue (1)
Our intuition tells us that a good parameter for the effec-
Nuclear tiveness of a rocket is the quotient of thrust (what we want)
Thermal
with propellant mass flowrate (what we have to pay):

Note the common feature of all of these engines (with the T Ae

= ( Pe − P0 ) + Ue (2)
exception of the electric thruster) is a converging / diverg- ṁ ṁ
ing nozzle De Laval nozzle. Such nozzles are the essential This parameter is called effective exhaust velocity,
feature of rocket engines as this is what enables the en-
gine to gasdynamically accelerate propellant gasses from a T
C= (3)
stagnant state to high velocity imparting momentum on the ṁ
engine. The nozzle expansion process for most gasses in and is one of the most important relations in rocketry. It
well approximated by an isentropic expansion such that the figures prominently in the Rocket Equation

1
2

to compute the mass flowrate through a choked nozzle as a

M ∗
∆V = C ln i (4) function of only nozzle throat area, A , ideal gas properties
Mf and the stangation condition temperature and pressure.
which we will be derived in a couple lectures. Let’s define a new parameter, called characteristic velocity
If we assume T and ṁ are completely independent (which and denoted c∗ using this result:
we will see is not always true), we can also interpret C as s  γ +1
Pt A∗

∗ RTt γ + 1 2(γ−1)
R
Tdt c = = (8)
I ṁ γ 2
C= R = (5)
ṁdt Mp
which is how much total impulse we get from a unit mass
of propellant. This interpretation points to a very closely
related parameter called Specific Impulse, Isp :

C T
Isp = = (6)
g0 g0 ṁ
Notice that Isp is just C normalized by Earth’s gravita-
tional acceleration with overall units of seconds. While the
unit is at first a bit non-sensical, Isp makes sense when you
consider your propellant consumption as a weight flowrate Figure 0.1: c∗ parameters of interest. ṁ, Pt and A∗ are all
rather than a mass-flowrate and so you are dividing thrust easily measureable. Tt is difficult due to the extremely high
(a force) by weight flowrate (force per time) to arrive at a temperatures of chemical rockets.
time. The etymology of Isp is rooted in the fact that for
imperial units we typically work in lbm. rather than slugs. c∗ is useful far beyond simple choked flow considerations
And while weird at first, the units of seconds do have some because it can be both measured and computed. And
intuitive utility. For a rocket with Isp = 100s a unit mass, indeed the equation above shows this directly - the LHS is
m of propellant can generate enough thrust to support its a function measure-able variables stagnation pressure Pt 1 ,
weight in Earth’s gravity for 100 seconds or 100 times its area A∗ and mass flowrate ṁ. The RHS is a function of
weight for one second. intrinsic gas properties γ, R and stagnation temperature.
Specific impulse is popularly spoken of as the "gas c∗ gives us the tools to calculate a parameter (RHS) that
mileage" for a rocket cycle and this is fairly reasonable - we can go and easily measure in the lab (LHS).
it fundamentally indicates how much bang for the buck you Given that c∗ depends on γ, R and Tt it is essentially a
get. I’ll jump the gun just a bit for the sake of intuition and function only of the working fluids thermodynamic proper-
give some typical Isp values for different types of propulsion ties and stagnation (chamber) state. This is not 100% true
in Table 0.1. - when we get to combustion we will see how a (weak) de-
This is all well and good, but all we have really done at pendency on pressure comes back into c∗ , but for conceptual
this point is some algebra. What we really are interested in purposes we should think of c∗ this way - as only a function
as engineers is how do I get the most gas mileage out of my of intrinsic thermodynamic properties of our propellants.
rocket. And for that discussion, we’ll first define another And so finally we come to a very interesting result regard-
couple useful parameters. ing the functional dependence of c∗ :

c∗ = f ( Tt , R, γ) (9)
0.3 c∗
Ru
R is the specific gas constant R = M w
which is inversely
We will go back, for a moment, to choked compressible flow. dependent on gas molecular weight, Mw . Thus
With the isentropic flow equations and the M = 1 choked s
condition, we can derive (assuming constant C p , Cv and R): ∗ Tt
c ∝ (10)
Mw
 2( γ −1)
ρ∗ Pt A∗ γ + 1 γ+1 Moreover Tt is related to the gas internal stagnation en-
   ∗ 
∗ ∗ ∗ a ∗ R Tt
ṁ = ρ a A = ρt at A = q thalpy by ∆ht = T C p dT and so we should see Tt as a
ρt at RTt 2 0
γ representation of the energy content of the rocket gasses.
(7) 1 In a finite sized rocket chamber we are not actually measuring
remembering that the t subscript denotes the total or stag-
stagnation pressure because there will be non-zero gas velocity. For
nation condition and the c subscript denotes choked (M = reasonable contraction ratios, A1 /A∗ , this is a small effect and it can
1) condition. This is a very useful relationship as it allows us be mostly corrected analytically using isentropic flow relations
0.4. C f 3

Technology Isp (s) Exhaust Velocity (m/s)

Nuclear Fusion 10,000 - 50,000+ 98,000 - 490,000+
Electric Propulsion 1,000 - 10,000 9,800 - 98,000
Nuclear Thermal (fission) 600 - 1,000 5,900 - 9,800
Beamed Thermal (microwave / laser) 600 - 1,000 5,900 - 9,800
Bipropellant Chemical Propulsion 200 - 500 2,000 - 4,900
Monopropellant Chemical Propulsion 100 - 250 980 - 2,450
Cold Gas Propulsion 10 - 120 100 - 1,150

Table 0.1: Table 1 - Representative effective exhaust velocities for different propulsion technologies

The effect of γ on c∗ is a bit more subtle. γ is fundamen- and indeed c∗ and C are closely related.
tally related to the number of vibratory degrees of freedom In order to see how, let’s define another new parameter,
a molecule has. For monatomic systems (such as helium or C f that related thrust to the nozzle throat area and rocket
atomic hydrogen) γ assymptotes to an upper limit of 1.66. total pressure:
For most simple diatoms (nitrogen, oxygen, hydrogen) it is
around 1.4 and for larger molecules or those with more com- T
Cf = ∗
(13)
plex bonding it is lower. For the conditions we are interested PtA
in within a rocket, we wouldn’t expect to find γ much lower In prose this says:
than 1.1. C f represents the amount of thrust a rocket can
produce given the stagnation pressure of its pro-
1.00 pellants and a useful characteristic fluid area - the
choked area of its nozzle.
0.98
Stated differently it is a measure of how effectively we
0.96
take the stagnation pressure we generate in the chamber
0.94 and turn it into thrust.
c∗

And since c∗ is also defined by chamber pressure and noz-

0.92
zle throat area, C f becomes the connection between C and
0.90 c∗ :
0.88 C f Pt A∗
T
0.86 C= = = C f c∗ (14)
1.1 1.2 1.3 1.4 1.5 1.6 ṁ ṁ
γ But why split C into C f and c∗ this way?
Remember that c∗ represents the potential of the rocket
Figure 0.2: c∗ shows only a weak dependence on γ propellants themselves to create thrust and essentially de-
pends exclusively on the thermodynamic characteristics of
those propellants. C f represents how well our nozzle can
The plot below shows the effect of γ on c∗ which is fairly convert the propellant’s latent utility into real thrust and
limited compared with Tt and Mw . And so if you take one thus depends almost exclusively on the physical nature of
thing away from this discussion, remember that our nozzle. And so as we go about maximizing C, we can
s divide that into two separate problems - one of picking pro-
∆ht ∗
∗
c ∝ (11) pellants (c ) and the other of chosing system pressures and
C p Mw nozzle geometry (C f ).
Since the primary role of the nozzle is to convert gasses
into thrust C f can also be seen as a measure of the "good-
0.4 C f
ness" of the nozzle. It is called thrust coefficient.
Ok, that’s interesting, but for the purposes of rocket perfor- C f can be expanded from its definition above:
∗
mance we want C not c . Take a look at the definition of
c∗ and it looks a lot like our definition of effective velocity C ( Pe − P0 ) Aṁe + Ue Pe − P0 Ae Ue
Cf = ∗ = Pt A∗
= + ∗
and specific impulse above: c Pt A∗ c
ṁ (15)
 
P t A ∗ T P 0 Pe A e U e
c∗ = ∼ =C (12) ⇒ 1− + ∗
ṁ ṁ Pe Pt A∗ c
4

It is worth noting that, like

 the thrust equation, there are
P0 Pe Ae P0 = 0.1
two pieces to C f : 1 − Pe Pt A∗ is representative of thrust
1.42 Pt 0.4

Pt A ∗
Pe Ae
created through pressure force and Uc∗e is representative of
1.28 0.3
the contribution of gas momentum to thrust. We will refer
1.15 0.2

´
to these two components when we discuss optimal nozzle
1.01 0.0

Ue
c∗

1 − PP0e
0.87 -0.1
expansion in a minute.
Beyond this things get a little messy and different people
0.74 -0.2

³
attack the derivation different ways. The next steps rely on
the fact that Uc∗e , PPet and A
Ae
∗ are all related to the isentropic
1 2
expansion of gasses through the nozzle to its exit. The exit
mach number, Me thus becomes the common parameter and 1.25
using classic isentropic relations we can derive the functional

Cf
relationship of each with Me as the independent variable:
1.20
  γ +1

1 2
Ue γMe γ+1 −2( γ −1)
= q (16)
c∗ 2
1 + γ− 1 2
2 Me

3
 −γ
Ae 2
Ac
γ−1 2

P γ −1
= 1+ M (17)
Pt 2
1
1 2
 γ +1

1+ γ −1 2
 γ +1
2( γ −1)
P0 /Pe
2 Me

Ae γ + 1 −2( γ −1)
= (18)
A∗ 2 Me
Figure 0.3: Relationship of the components of C f to each
other and to C f itself.
With some fancy algebra, we can combine these together to
arrive at
Note that the momentum component grows as exit pres-
sure decreases because the gasses are accelerated further be-
  γ +1 fore release. Conversely, the pressure component decreases
γ +1 −2( γ −1)  
2 P0
Cf = q 2
γMe + 1 − (19) as exit pressure decreases for obvious reasons. In fact the
γ −1 2 Pe pressure component of C f takes on negative values for exit
Me 1 + 2 Me
pressures below ambient - this is called over-expansion.
Ae /A∗ increases monotonically with decreasing exit pres-
Ue Pe Ae
These results shows very clearly that c∗ , Pt and A∗ are sure - this simply represents the fact that to achieve lower
not all independent (they all depend on Me ). In fact the exit pressures we must use a larger nozzle expansion ratio.
dimensionality of this set is one - picking a number for any Finally note that C f reaches a maximum where Pe /P0 =
one of these directlys sets the others. Furthermore note 1. This is called optimal expansion. It is a fundamental
∗
that these equations, unlike c , have no dependency on result in rocket theory and can be stated:
stagnation temperature or molecular weight and thus
no direct dependency on propellant properties. They do Rocket effective exhaust velocity (and therefore
depend on γ which is a property of the working fluid but as specific impulse) is maximized when the nozzle ex-
with c∗ , the dependence is not terribly strong and is really pands the exhaust gasses such that they match the
set for us by the propellant choice we made in optimizing local pressure at the nozzle exit.
c∗ .
I’d like to look at how the parameter we can control di- In general, we want to design a rocket nozzle to match
rectly, Ae /A∗ , affects the others and C f . Using the equa- the exit pressure to ambient pressure. This is difficult to do
tions above, we will sweep through Me and compute the for a rocket ascending through the atmosphere where the
other parameters directly. pressure is continuosly changing. For traditional nozzles, a
0.4. C f 5

compromise that looks at the average performance over the and the pressure distribution in the nozzle would relax to
ascent pressure profile is often chosen. a state where Pe = P0 . In supersonic (compressible flow),
pressure information travels more slowly than the fluid ve-
locity so there is no information transfer upstream. In order
Vacuum to equilibrate pressure with the ambient, a set of expansion
or compression waves form from the exit of the nozzle.

Under-expanded Case
Sea level Throat Area

Expansion Waves

Figure 0.4: Note the size difference between a sea-level

and vacuum- optimized nozzle. The rocket itself is identical
except for the nozzle expansion. Over-expanded Case
Final Area
Throat Area
And since the way we lower exit pressure is with a bigger
Ae
expansion ratio, e = A ∗ , this result explains why upper stage
"vacuum nozzles" are so much bigger than first stage "sea-
level" nozzles as can be seen in graphic above and the side-
by-side of the SpaceX Merlin 1D vacuum (left) and Merlin
Shock
1D sea-level (right) engines: (compression)
waves

Figure 0.6: In a super-sonic exhaust stream, expansion or

compression waves form to equilibrate the jet pressure to
the surrounding pressures.

Due to flow radial inhomogeneity in the flow in real noz-

zles, there will always be some equilibration wave structure
at the exit. The pattern of compression and expansion waves
and associated changes in gas temperature and luminos-
ity are what are generate Mach Diamonds in the exhaust
stream.

Figure 0.5: SpaceX’ Merlin engine has very different size

nozzles for its first and second stage even though the engine
itself is nearly identical. Figure 0.7: An example of mach diamonds in the plume of
one of Stanford’s research hybrid rockets
So what actually happens when a nozzle is under- or over-
expanded? In a subsconic (incompressivle) jet, the fluid dy- Finally, we note that if nozzles are sufficiently over-
namics allow for communication of information upstream expanded, the flow can no longer stay attached to the nozzle
 CF = 1.63 in a vacuum.

Under- and Overexpanded Nozzles

An underexpanded nozzle discharges the gases at an exit pressure greater than the
6
external pressure because its exit area is too small for an optimum expansion. Gas
expansion is therefore incomplete within the nozzle, and further expansion will take
place outside of the nozzle exit because the nozzle exit pressure is higher than the
and separates. This typically happens for PP0e < 0.4
contour pressure.
local atmospheric in the rocket thrust, effective velocity and c∗ and potentially
In an and is an undesireable
overexpanded nozzle the gascondition. Because
exits at lower pressure the
thanflow is not at-as
the atmosphere Cf .
it has a discharge
tached, itarea
maytoobe
large for optimum.generating
asymmetric, The phenomenon of overexpansion
side loads. It also Becuase many of the larger losses are associated with the
inside a supersonic nozzle is indicated
may be unstable schematically
generating unsteadyin forces
Fig. 3–8,
onfrom
thetypical early
rocket. process of combustion in the chamber, the non-ideal perfor-
pressure measurements of superheated steam along the nozzle axis and different mance is most easily and directly measured in c∗ . It is with
The diagram below shows how separate flow develops in
back pressures or pressure ratios. Curve AB shows the variation of pressure with
a nozzle as a function of pressures. this in mind that c∗ efficiency is defined:

c∗measured Pt A∗
ηc ∗ = ∗ = ∗ (20)
cideal ṁ p cideal
∗
where cideal ❦
is computed using Equation (8) for simple sub-
stances or, more commonly, using a complex chemical equi-
llibrium code as will be discussed in the next lecture.
Optimizing ηc∗ becomes a primary activity in the develop-
ment of a rocket engine, often consuming a singificant frac-
tion of the development budget and time. In liquid rockets
this might manifest as injector tuning, while in a solid rocket
it is attacked by adjusting propellant composition, ratios and
M particle sizes.

0.6 Putting it all together

The separation of C into c∗ and C f separates concerns be-
tween propellant thermodyanmics in c∗ and nozzle physical
parameters in C f .
Let’s quickly compute C, c∗ and C f to get a sense of how
we would begin to use them practically. Assume we start
FIGURE 3–8. Distribution of pressures in a converging–diverging nozzle for different with a plenum full of room-temperature, high pressure gas
FigureInlet
flow conditions. 0.8:
pressureDiagram is the same, ❦but exit the
illustrating flow
pressure separation
changes. Based on phe- early
nomenon.
Trim Size: 6.125in x 9.25in
From [1] and then vent it into vacuum through a nozzle.
Sutton c03.tex V2 - 10/06/2016 10:45am Page 65
experimental data.
Now let’s compute c∗ , C f and then C. For now we will
assume optimal expansion (P0 /Pe = 1) and calculate the
Finally we present a very nice plot of how C f varies with associated area ratio. A useful relation we will use to deter-
all of the dependencies from [1].
3.3. ISENTROPIC FLOW THROUGH NOZZLES 65 mine Me is:

k = 1.20 46
= 2.2

= ∞, ❦
2
value  1− γ
 
Line of optimum um
Maxim 500 1− γ Pe P0
 
0 γ
1.8
thrust coefficient
p2 = p3
p 1/p 3
20
Pe γ
−1 P0 Pt − 1
ion
Me2 = 2  Pt

arat 00  = 2 (21)
sep

10

1.6 t flow γ−1  γ−1 

00

ip ien tion
50

Inc ara
0

1.4 lin
e:
w sep
o
20

low fl
p1

ble
0
CF

be
/p 3

ida
10

1.2 ion vo
=

eg Un
a
e:
R

lin
50

1 lo
w Name γ Mw c∗ (m/s) Cf C (m/s)
be
n
20

0.8 N2 1.40 28 434.4 1.26 546.4

gi
Re
10

H2 1.41 2 1621.5 1.26 2039.2

5

0.6
1 3 10 30 100 300 1000 He 1.66 4 1085.2 1.25 1356.2
Area ratio = A2/At

FIGURE 3–6. Thrust coefficient CF versus nozzle area ratio for k = 1.20.
Figure 0.9: Diagram illustrating the flow separation phe- Table 0.2: c∗ , C f , C for several pure gasses at room tem-
perature, P0 /Pe = 1 and P0 /Pt = 0.1. Note how C f is
❦ ❦
nomenon. From [1]
essentially constant despite the differences in gas properties
k = 1.30
um va
lue = 1.
964
while c∗ and C vary substantially.
1.8
∞, Maxim
c∗ Efficiency
Line of optimum = 500
0.5 p /p
thrust coefficient 1
3
10
20
00
0
tion
1.6
p2 = p3
sep
ara This simple exercise confirms what we said before - c∗
00
w
flo combustion, iochamber
50

Due to imperfect mixing, ie n

t
ra t
n heat-loss depends on the nature of working fluid while C f is largely
0

cip pa
20

: In
and1.4other effects thelinerealized se
rocketflowperformance is typically fixed by the nozzle parameters. Another interesting conclu-
0
p1

10

e
ow bl
/p 3

less than that computed

el theoretically. This will be reflected sion from this is that even before we get to chemistry, we
0

da
oi
=
b

1.2 v
on

50

na
CF

gi

U
Re

e:
lin

1
w
lo
20

be
n
gio

0.8
Re
10
5
0.7. ENERGY AND EFFICIENCY 7

can see how wonderful of a working fluid Hydrogen is due to

it’s low molecular weight. Remember this because hydrogen v2 GM
e= +
will emerge again and again as we talk about rockets. 2 r
If we expand Equation (2) and simplify we arrive at a In the case of our satellite in a circular Low Earth Orbit
useful form for C: this comes out to:

s  −2( γ −1)
Pe − P0 Ae RTt γ + 1 γ+1 v2

µ µ
C= + e= + − = 29.1MJ/kg + 4.5MJ/kg = 33.6MJ/kg
Pt A∗ γ 2 2 r0 r
v (22)
Again note how much of this is due to the orbital velocity
u "  (γ−1)/γ #
u 2γ Ru Tt Pe
t 1− of the rocket. And to put in context just how much energy
γ − 1 Mw Pt
this is, it is equivalent to:

In Equation (22) we can see all the impacts of propellant • 1000x the specific energy of a Jet airliner at 650MPH
choice and nozzles on effective exhaust velocity. The term
involving Tt and Mw clearly shows the effect of propellant • 5x the specific chemical energy content of nitroglycerin
choice on C, as we noted with c∗ earlier. The term involving
Pe • 3x the required specific energy to melt aluminum
Pt demonstrates the impact of the system pressures and
nozzle geometry we choose. Figure 0.10 summarizes all of
this. It is no wonder that not much is left when object with
this much energy slams into the atmosphere!

0.8 Rocket efficiency

(s)

Clearly there is a lot of energy being liberated in rocket

engines in order to put things into space. But how efficient
are they?
There are different ways to define energy efficiency, but
the the first we’ll look at is the thermal efficiency or how
effectively a rocket takes input energy and converts it to gas
kinetic energy:

(K-kg-mol/kg) Ẇexhaust ṁC2

ηthermal = =
Pin 2Pin
Figure 0.10: How specific impulse varies
where Pin is the amount of energy being liberated in time
to power the rocket whether it be latent, chemical, nuclear
or electrical energy. In the case of chemical rockets we can
0.7 Energy and efficiency define it as the energy released by the propellants when they
combust or:
So far we have concentrated on propulsion primarily from
a conservation of momentum perspective. But there are a Pin = ṁ∆h
lot interesting observations to be made when we look at the
thermodynamics of propulsion as well. where ∆h is the amount of thermal energy going into the
With rocket propulsion and things in space we are talking rocket chamber.
about a lot of energy and to put that in to context, let’s Substituting Equation (22) gives
see how much energy our Low Earth Orbit satellite ends up   (γ−1)/γ 
with. 2γ Pe
γ −1 RT t 1 − Pt
An object moving in a conservative potential field (like ηthermal = (23)
Earth’s gravity) has total energy: 2∆h
Because
mv2 GMm ∆h
E = KE + PE = + Tt ∼
2 r Cp
or in mass-specific energy we can subsitute such that
8

" The overall efficiency is

 (γ−1)/γ #
γ R Pe
ηthermal = 1− (24) η = ηthermal η prop
γ − 1 Cp Pt

And finally recognizing that for an ideal gas A small example demonstrating how η varies over ratio of
vehicle speed to exhaust speed, Cv is shown below.
R γ−1
=
Cp γ
ηprop
1.0 η
and for a perfectly isentropic nozzle expansion process
0.8
Te Pe (γ−1)/γ
= 0.6
Tt Pt

η
we see that Equation (23) reduces to the Carnot efficiency. 0.4
Figure 0.11 shows ηthermal as a function of Pt /Pe :
0.2

0.0
0.0 0.5 1.0 1.5 2.0 2.5 3.0 3.5 4.0
0.5 V/C
ηthermal

0.4 And here we see another interesting result much like opti-
mal nozzle expansion. The rocket is most efficient when the
0.3 exhaust velocity equals the vehicle velocity. This is actually
quite intuitive - in this condition the exhaust gasses are left
0 25 50 75 100 with zero velocity in the static frame. There is no residual
kinetic energy in the exhaust so all of it’s kinetic energy have
Pt /Pe been transferred to the vehicle. Anywhere else the exhaust
is left with residual kinetic energy which is a loss.
Propulsive efficiency will come back in a big way when
Figure 0.11: How specific impulse varies discuss electrical propulsion as it is very important to system
optimization.
This is pretty good for a heat engine and a lot of early
rocket work focused on optimizing thermal efficiency. And 0.9 Energy losses
thermal efficiency will matter when we get to electric propul-
sion, but it is less useful a construct for chemical rockets Finally let’s also look at where the inefficiencies are for a
where it is largely defined by the propellant properties rather typical launch vehicle powered by a chemical engine with
than the machine they burn in. Furthermore the efficiency Isp = 340 s and moving at 5,000 m/s for intuition.
of accelerating gasses is not really what we care about - we
care about the efficiency of accelerating the vehicle. Incomplete combustionEnergy loss diagram
0.02
To understand that, we will define another, more useful Second Law loss
(unavailable thermal energy)
Heat loss 0.4
effiency metric, called propulsive efficiency as such 0.01
Friction, etc Exhaust kinetic
0.01 energy
0.1

Ẇr Tv
η prop = =
Ẇr + Ẇexhaust Tv + 12 ṁ(C − v)2 Reaction
energy
1
Useful
or the fraction of total work that actually goes to accelerat- 0.46

ing the vehicle.

Tv ṁCv
=
Tv + 1
2 ṁ (C − v )2 ṁ [Cv + (C − v)2 ] Notice that almost half of the chemical energy is unavail-
able for conversion to work due to second law of thermo-
v
C dynamics. This is expected given the Carnot efficiency we
→ η prop = 2
1 + ( Cv )2 developed earlier. The biggest contributor after second law
0.9. ENERGY LOSSES 9

loss is residual exhaust kinetic energy which we saw is some-

thing is minimized when V/C = 1. The other losses are
quite small and we can say

When a rocket operates with exhaust velocity near

vehicle velocity, it is a surprisingly efficient heat
engine
 Bibliography

[1] George P. Sutton and Oscal Biblarz. Rocket Propulsion

Elements. John Wiley & Sons, Inc., 7 edition, 2001.

11