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# A more thorough analysis of water rockets: Moist adiabats, transient flows, and inertial forces in a soda bottle

12 Feb 2010-American Journal of Physics (American Association of Physics Teachers)-Vol. 78, Iss: 3, pp 236-243

AbstractThe water rocket 1 is a popular toy that is often used in first year physics courses to illustrate Newton’s laws of motion and rocket propulsion. In its simplest version, a water rocket is made of a soda bottle, a bicycle pump, a rubber stopper, and some piping see Fig. 1. The bottle is half-filled with water, turned upside-down, and air is pushed inside the bottle via a flexible pipe that runs through the stopper. When the pressure builds up, the stopper eventually pops out of the neck. The water is then ejected and the rocket takes off. Witnesses of the launch of a water rocket cannot but be amazed that such a simple device can reach a height of tens of meters in a fraction of a second. The popularity of water rockets extends beyond physics classrooms, with many existing associations and competitions organized worldwide. 1 The more than 5000 videos posted on YouTube with the words “water rocket” in their title testify to their popularity. Some of these videos involve elaborate technical developments such as multistage water rockets, nozzles that adapt to the pressure, the replacement of water by foam or flour, underwater rocket launches, and even a water-propelled human flight. The public’s passionate explorations with water rockets contrast with the small number of articles devoted to their analysis. I found only two papers 2,3 that treat the simplest possible rocket, similar to

Topics: Water rocket (71%), Pendulum rocket fallacy (68%), Rocket (64%), Spacecraft propulsion (56%)

### A more thorough analysis of water rockets: Moist adiabats, transient flows, and inertial forces in a soda bottle

• Accurate measurements show that they outperform the usual textbook analysis at the beginning of the thrust phase.
• The bottle is half-filled with water, turned upside-down, and air is pushed inside the bottle via a flexible pipe that runs through the stopper.
• In Sec. III numerical solutions are compared with published experimental data.2.

### A. Moist air expansion

• Air expansion is the only source of energy of the rocket.
• Because this distance is much smaller than the radius of the rocket, the gas expansion has to be modeled as an adiabatic process.
• The pressure-volume relation is derived, assuming that the total entropy resulting from dry air, water vapor, and condensed water the fog is constant during the adiabatic expansion.
• The total number of water molecules NL in the liquid is the difference between the initial and the current values of NV.

### B. Water ejection

• The starting point of the analysis is to assess the importance of viscous forces.
• Water can therefore be assumed to be inviscid in this context, and ejection can be analyzed with Euler’s equation.
• The ratio in the square brackets in Eq. 10 ensures that the total flow of water is the same over any section of the rocket.
• The notation highlights that this term accounts for the inertia of the accelerated water.

### C. Rocket acceleration

• For a conventional rocket, MU is generally estimated by assuming that the fuel moves upward at the same speed as the rocket.
• This term accounts for the fact that the velocity of water in the rocket is smaller than the velocity of the rocket itself.
• The three forces on the right-hand side of Eq. 14 are estimated in the usual way.
• For the case of a uniform velocity profile, the first contribution is the product of the rate of mass loss with the exhaust velocity; the second term accounts for the fact that the thrust is not estimated in the reference frame of the rocket.
• The last two forces exerted on the rocket are the weight and the aerodynamic drag.

### A. Numerical solutions and experimental data

• These authors used a 2-l soda bottle as a rocket, the takeoff of which they measured with a high-speed camera and a “smart-pulley” system.10.
• The authors have accounted for several effects which were previously not considered and which may have an important role during the burnout.
• From dimensional analysis, the acceleration is expected to be as significant as the pressure difference between the inside and the outside of the rocket.
• For both pressures, curve 1 is slightly above the analysis of Ref. 2, but the theoretical prediction remains below experiment.

### B. Analysis of simplified equations

• A real rocket still outperforms the theoretical prediction, also known as The improvements are small.
• And thus the authors shall here assume g0=0 and CA=0.
• The time derivatives dh /dt and dZ /dt are plotted in Figs.
• For large values of / 0, the two numbers are very close to each other.
• In general, applying Bernoulli’s equation leads to an underestimation of the initial ejection velocity.

### IV. CONCLUSIONS AND PROSPECTS

• This paper was motivated by the observation2 that water rockets outperform the usual textbook analysis at the beginning of the thrust phase.
• The present analysis does answer some questions raised in Ref. 2.
• The authors have shown that this energy can be accounted for using a polytropic exponent smaller than =1.4.
• The movement of water inside the rocket contributes significantly to the total momentum of the rocket, an effect that is more pronounced for a larger hole.
• Maximum energy efficiency would require that water be ejected with a constant velocity with respect to the ground.

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A more thorough analysis of water rockets: Moist adiabats, transient ﬂows,
and inertial forces in a soda bottle
Cedric J. Gommes
Department of Chemical Engineering, University of Liège, B6A Allée du 6 août, 3, B-4000 Liège, Belgium
Received 19 May 2009; accepted 12 October 2009
Although water rockets are widely used to illustrate ﬁrst year physics principles, accurate
measurements show that they outperform the usual textbook analysis at the beginning of the thrust
phase. This paper gives a more thorough analysis of this problem. It is shown that the air expansion
in the rocket is accompanied by water vapor condensation, which provides an extra thrust; the
downward acceleration of water within the rocket also contributes to the thrust, an effect that is
negligible in other types of rockets; the apparent gravity resulting from the acceleration of the rocket
contributes as much to water ejection as does the pressure difference between the inside and outside
of the rocket; and the water ﬂow is transient, which precludes the use of Bernoulli’s equation.
Although none of these effects is negligible, they mostly cancel each other, and the overall accuracy
of the analysis is only marginally improved. There remains a difference between theory and
2010 American Association of Physics Teachers.
DOI: 10.1119/1.3257702
I. INTRODUCTION
The water rocket
1
is a popular toy that is often used in ﬁrst
year physics courses to illustrate Newton’s laws of motion
and rocket propulsion. In its simplest version, a water rocket
is made of a soda bottle, a bicycle pump, a rubber stopper,
and some piping see Fig. 1. The bottle is half-ﬁlled with
water, turned upside-down, and air is pushed inside the bottle
via a ﬂexible pipe that runs through the stopper. When the
pressure builds up, the stopper eventually pops out of the
neck. The water is then ejected and the rocket takes off.
Witnesses of the launch of a water rocket cannot but be
amazed that such a simple device can reach a height of tens
of meters in a fraction of a second.
The popularity of water rockets extends beyond physics
classrooms, with many existing associations and competi-
tions organized worldwide.
1
The more than 5000 videos
posted on YouTube with the words “water rocket” in their
title testify to their popularity. Some of these videos involve
elaborate technical developments such as multistage water
rockets, nozzles that adapt to the pressure, the replacement of
water by foam or ﬂour, underwater rocket launches, and even
a water-propelled human ﬂight. The public’s passionate ex-
plorations with water rockets contrast with the small number
of articles devoted to their analysis. I found only two
papers
2,3
that treat the simplest possible rocket, similar to
that shown in Fig. 1.
In the cited papers, the air expansion in the rocket is mod-
eled as an isothermal
3
2
process involving dry air,
which enables the authors to estimate the pressure at any
stage of water ejection. Based on Bernoulli’s steady state
equation, the pressure is then converted to a water ejection
velocity at the nozzle, from which the thrust is estimated via
the classical equation of rocket propulsion. Finally, Newton’s
laws of motion are solved numerically to predict the ﬂight of
the rocket. This analysis enables the time of ﬂight to be
predicted within the experimental uncertainties of a ﬁrst year
laboratory project.
3
However, more accurate measurements
with a high-speed camera show that the acceleration of the
rocket is substantially underestimated at the beginning of the
thrust phase.
2
Simple observations of water rockets hint at physical phe-
nomena that were not considered in Refs. 2 and 3. First, the
rocket is ﬁlled with fog at the end of a launch, as is visible in
the inset of Fig. 2b. Water condensation is an exothermic
process that is expected to contribute positively to the thrust.
Second, the acceleration of the rocket is huge, which sug-
gests that the noninertial contribution to the apparent gravity
in the rocket might help signiﬁcantly water ejection. The
rocket in Fig. 2 reaches a height of about 4 m in 0.2 s, which
implies an average acceleration of 200 m / s
2
. A more accu-
rate measurement
2
with a high-speed camera leads to an ac-
celeration of 100gs. The ﬂow of water is rapid not just at the
nozzle but also inside the rocket. The cloud in Fig. 2b
forms at the end of ejection and appears after less than 0.2 s.
A more accurate
2
ejection time is
=0.1 s. Because the
height of water in the rocket is initially H = 10 cm, this time
converts to a velocity of U =1 m/ s inside the rocket. The
latter motion carries a momentum that may also contribute
positively to the thrust.
In this paper an analysis of the water rocket is proposed,
which takes account of the phenomena we have mentioned.
A detailed physical analysis is presented in Sec. II, and the
governing equations are derived. In Sec. III numerical solu-
tions are compared with published experimental data.
2
II. PHYSICAL ANALYSIS
A. Moist air expansion
Air expansion is the only source of energy of the rocket.
As mentioned in Sec. I, it is accompanied by vapor conden-
sation. The speciﬁc issue addressed here is the relation be-
tween the pressure P and the volume V of the expanding
moist air. This relation determines the total amount of energy
released during the ejection.
The ﬁrst question is whether the expansion is adiabatic
2
or
isothermal as some authors claim.
3
For the process to be
isothermal, heat would have to diffuse into the rocket from
outside to keep the temperature constant. Generally, during a
time
, heat diffuses in air over a distance
h
given by
4,5
236 236Am. J. Phys. 78 3, March 2010 http://aapt.org/ajp © 2010 American Association of Physics Teachers

h
2
, 1
where
is the heat diffusivity of air. If we use the value
4
=0.014 cm
2
/ s and
=0.1 s, we ﬁnd
h
=0.04 cm. Be-
cause this distance is much smaller than the radius of the
rocket, the gas expansion has to be modeled as an adiabatic
process.
Adiabaticity does not necessarily imply that PV
is a con-
stant, with
=1.4 for diatomic gases.
6,7
The cooling of air
during an adiabatic expansion induces the condensation of
the water vapor it contains. Because condensation is exother-
mic, the temperature does not drop as rapidly as if the air
was dry, which keeps the pressure higher than predicted by
the PV
relation. Neglecting this effect leads to an underes-
timation of the work performed by the expanding air, and
hence of the thrust.
The pressure-volume relation is derived, assuming that the
total entropy resulting from dry air, water vapor, and con-
densed water the fog is constant during the adiabatic ex-
pansion. Assuming that air is an ideal gas, its molar entropy
s
a
is written as
6,8
s
a
T,V = s
a
T
0
,V
0
+ c
V
a
ln
T
T
0

V
V
0
−1
, 2
where c
V
a
is the speciﬁc heat of air and V is the volume. The
0 subscript refers to any arbitrarily chosen reference state.
The molar entropy of water vapor in equilibrium with liquid
water at temperature T is written as
6,8
s
V
T = s
l
T +
h
V
T
, 3
where s
l
T is the molar entropy of the liquid water and h
V
is the molar enthalpy of vaporization of water. The molar
entropy of liquid water s
l
T is
6
s
l
T = s
l
T
0
+ c
p
l
ln
T
T
0
, 4
where the speciﬁc heat of water c
P
l
has been assumed to be
temperature independent.
The number of molecules in each of the three phases—air,
water vapor, and liquid water—is obtained by assuming ther-
modynamic equilibrium. This assumption implies that the
partial pressure of water vapor at any stage of the expansion
is equal to the saturating pressure P
V
T. By using the ideal
gas law, P
V
T is converted to the total number of vapor
molecules N
V
in the volume V. The total number of water
molecules N
L
in the liquid is the difference between the ini-
tial and the current values of N
V
. The constant number of air
molecules in the rocket N
a
is determined as a function of the
initial pressure P
i
, volume V
i
, and temperature T
i
by sub-
tracting the contribution of water vapor to the total pressure.
The temperature-volume relation during the adiabatic ex-
pansion of moist air is obtained by conserving the total en-
tropy S= N
a
s
a
+N
V
s
v
+N
c
s
l
We use Eqs. 24 to ﬁnd
1−
P
V
T
i
P
i
c
V
ln
T
T
i

V
V
i
−1
+
T
i
T
V
V
i
P
V
T
P
i
h
V
T
+
P
V
T
i
P
i
c
p
l
ln
T
T
i
h
V
T
i
=0.
5
For the case of dry air, that is, for P
V
T= P
V
T
i
= 0, the
solution of Eq. 5 is TV
−1
=constant, as expected. For
moist air, however, Eq. 5 has to be solved to estimate the
temperature T reached after an adiabatic expansion. Know-
ing that temperature, the corresponding pressure is obtained
by the ideal gas law.
Solutions of Eq. 5 are plotted in Fig. 3a for three values
of P
i
and three values of T
i
=constant and isotherms PV=constant are shown for com-
parison. The numerical values used for the calculation are
c
V
=20 J/ mol K,
8
c
P
l
=75.3 J/ mol K,
6
h
V
=45 051 J/ mol,
8
and
=1.4; the saturating pressure was estimated from the
Clausius–Clapeyron equation,
68
with P
V
T
0
= 1 bar at T
0
=100 ° C.
The moist adiabats in Fig. 3a are found to be well ap-
proximated by a polytropic process
8
of the type PV
=constant dotted lines. The exponent
obtained by a least-
squares ﬁt is found to depend mostly on the initial relative
humidity P
V
T
i
/ P
i
, as shown in Fig. 3b. The dependency
is well described by the empirical function
Fig. 1. In its simplest version, a water rocket is made of a soda bottle
partially ﬁlled with water, in which air is injected with a bicycle pump.
When the pressure increases, the stopper eventually pops out, water is
ejected, and the rocket takes off.
Fig. 2. Snapshots taken at times a t =0 s, b t=0.2 s, and t = 0.4 s during
the launch of a water rocket. The insets show the rocket at the onset of
ejection and soon after the end of ejection. Note that the ejected water in the
cloud of b is moving upward.
237 237Am. J. Phys., Vol. 78, No. 3, March 2010 Cedric J. Gommes

= 1.15 + 1.4 1.15exp
−36
P
V
T
i
P
i
, 6
which reduces to
=
=1.4 for dry air.
Figure 4 shows a plot of the amount of work performed by
a given volume of moist air, initially pressurized at P
i
=3, 4,
and 7 bars, when it adiabatically doubles its volume against
atmospheric pressure. In Fig. 4, the work is normalized to the
amount of work performed by dry air in the same conditions.
From Fig. 4, we see that air compressed to 3 bars performs
20% more work when it is saturated with water at 70 ° C
than when it is dry. In the experiments of Ref. 2 with P
i
=3.4 bars and P
i
=6.8 bars, and T
i
=25 ° C, the correspond-
ing exponents are estimated to be
=1.31 and
=1.35. In
both cases, the extra work resulting from water condensation
is expected to be only a few percent. The same conclusion
applies to the launches reported in Ref. 3 with P
i
=4 bars.
B. Water ejection
The determination of the rate of water ejection is a prob-
lem in ﬂuid mechanics. The starting point of the analysis is
to assess the importance of viscous forces. Using reasoning
similar to Eq. 1, the distance
v
over which momentum
diffuses in a ﬂuid during a time
due to its viscosity is
4,5
V
2
␷␶
, 7
where
is the kinematic viscosity of the liquid. For water
4
at
20 °C,
=0.01 cm
2
/ s. Using
v
=0.3 mm. This small value means that the ﬂow of water out
of the rocket is not affected by viscosity, except for a sub-
millimeter thin boundary layer. Water can therefore be as-
sumed to be inviscid in this context, and ejection can be
analyzed with Eulers equation.
The general form of Eulers equation is
4,5
u
t
+ u
· u
=−
1
P + g
, 8
where u
is the vector velocity of water, P is the pressure,
is
the density, and g
is the gravitational ﬁeld; u
and P are gen-
erally space and time dependent. When Eq. 8 is integrated
along the axis of the rocket, from the nozzle up to the free
surface of the liquid see Fig. 5, we obtain
0
h
u
c
t
dz +
u
c
2
h u
c
2
0
2
+
P P
a
+ gh =0, 9
where ht is the time-dependent height of water on the axis
of the rocket, P
a
is atmospheric pressure, u
c
h is the nega-
tive vertical velocity of the water free surface on the axis of
the rocket, and u
c
0 is the exhaust velocity on the axis.
Bernoulli’s equation, as used in Refs. 2 and 3, results directly
Fig. 3. a Pressure-volume curves of moist air during adiabatic expansion,
starting from P
i
=2, 4, and 6 bars, and initially in equilibrium with water at
T
i
=10, 50, and 90 ° C +. The solid lines are isotherms pV=constant and
=constant. b Moist adiabats are approximated by a
polytropic process of the form pV
=constant dotted lines in a兲兴, the expo-
nent of which is a function of the initial relative humidity.
Fig. 4. Mechanical work performed by moist air upon doubling adiabati-
cally its volume against atmospheric pressure, as a function of the initial
temperature T
i
and initial pressure P
i
=3 bars +,4bars, and 7 bars
. The work is normalized by the work performed by dry air under the
same conditions.
Fig. 5. The two reference frames used in the present analysis, and the
meaning of some symbols. The free surface of water does not remain ﬂat if
the velocity proﬁle is not uniform.
238 238Am. J. Phys., Vol. 78, No. 3, March 2010 Cedric J. Gommes

from Eq. 9 by neglecting the time derivative as well as
gravity.
The order of magnitude of the ﬁrst term in Eq. 9 is
HU/
, which is equal to the magnitude of the second term
U
2
. As for the last two terms, using P P
a
=10
5
Pa,
=1000 kg/ m
3
, H= 0.1 m, and apparent gravity g
=1000 m/ s
2
, they are found to be comparable. Therefore,
Eq. 9 cannot be simpliﬁed a priori to analyze the water
rocket.
To use Eq. 9 for making predictions, assumptions have
to be made about the ﬂow pattern. We shall assume that the
velocity proﬁles at various heights are self-similar, that is,
uz,r,t =
dh
dt
h
0
f
r
Rz
, 10
where Rz is the radius of the bottle at height z see Fig. 5,
z =
R
2
z is the cross sectional area at height z, and f is
a function that characterizes the velocity proﬁle and satisﬁes
f0= 1. Uniform velocity is recovered with fx= 1; the case
where the water drops more rapidly at the center of the
rocket corresponds to a function fx with a maximum at x
=0. The ratio in the square brackets in Eq. 10 ensures that
the total ﬂow of water is the same over any section of the
rocket. Moreover, the latter ﬂow has to be equal to the rate of
air expansion. Integrating Eq. 10 over a section leads to
dV
dt
=−fh
dh
dt
11
with f=
0
1
2xfxdx.
By using Eq. 10, the various terms in Eq. 9 can be
written explicitly as a function of ht , which leads to
Ih
d
dt
h
dh
dt
+
1
2
1−
h
0
2

dh
dt
2
+
1
P
i
V
i
Vt
P
a
+ gh =0 12
with
Ih =
0
h
dz
z
. 13
The notation highlights that this term accounts for the inertia
of the accelerated water. In Eq. 12,
is the polytropic
exponent obtained in Sec. II A and
0
=0 is the area of
the nozzle.
If the rocket was forced to remain immobile, Eqs. 11 and
12 could be solved for ht and Vt. When the rocket is
free to move, however, the gravity term g includes a nonin-
ertial component resulting from the rocket’s acceleration.
The water ejection can, therefore, not be predicted indepen-
dently of the rocket acceleration: the two problems are
coupled and have to be solved simultaneously.
C. Rocket acceleration
In an inertial frame immobile with respect to the ground
conservation of vertical momentum takes the form
d
dt
MU =Th−W D, 14
where the total momentum of the rocket is written as MU,
Th is the thrust, W is the weight, and D is the aerodynamic
drag. We next discuss each of these terms.
For a conventional rocket, MU is generally estimated by
assuming that the fuel moves upward at the same speed as
the rocket. This assumption does not hold for water rockets.
More generally, the total momentum is
MU = m
dZ
dt
+
0
h
dz
0
Rz
2
rdr
uz,r,t, 15
where m is the mass of the empty rocket, Z is its vertical
position, and uz , r , t is the velocity of the water in the
rocket. Note that, contrary to Sec. II B, the water velocity
uz ,r, t is expressed here with respect to the ground. If we
correct Eq. 10 by adding dZ/ dt, Eq. 15 becomes
MU = m +
V
w
dZ
dt
+
fhh
dh
dt
, 16
where V
w
=
0
h
z dz is the volume of water in the rocket.
The second term in Eq. 16 is negative; it is speciﬁc to water
rockets, and it was not taken into account in previous
studies.
2,3
This term accounts for the fact that the velocity of
water in the rocket is smaller than the velocity of the rocket
itself.
The three forces on the right-hand side of Eq. 14 are
estimated in the usual way. The thrust is the rate of momen-
tum transfer out of the rocket resulting from water
expulsion.
9
By using the velocity proﬁle in Eq. 10, the
thrust takes the form
Th =
h
2
0
dh
dt
2
f
2
+
h
dh
dt
dZ
dt
f典共17
with f
2
=
0
1
2xf
2
xdx. For the case of a uniform velocity
proﬁle, the ﬁrst contribution is the product of the rate of
mass loss with the exhaust velocity; the second term ac-
counts for the fact that the thrust is not estimated in the
reference frame of the rocket. The last two forces exerted on
the rocket are the weight and the aerodynamic drag. The
weight is
W = m +
V
w
g
0
, 18
where g
0
is the gravity, not to be confused with g that enters
Eq. 12 and includes a noninertial contribution. The aerody-
namic drag is calculated in the usual way
4,5
as
D = C
A
max
a
2
dZ
dt
2
, 19
where
a
is the density of air,
max
is the maximum cross
section area of the rocket, and C
A
is a dimensionless drag
coefﬁcient. For a rocket with a circular cross section and no
side wings, C
A
4
but for the following analysis,
it is sufﬁcient to consider that its order of magnitude is 1.
Combining Eqs. 1419 leads to the following differen-
tial equation:
239 239Am. J. Phys., Vol. 78, No. 3, March 2010 Cedric J. Gommes

d
dt
m +
V
w
dZ
dt
+
fhh
dh
dt
=
2
h
0
dh
dt
2
f
2
+
h
dh
dt
dZ
dt
f
g
0
m +
V
w
C
A
max
a
2
dZ
dt
2
, 20
which expresses the conservation of vertical momentum.
Equations 11, 12, and 20 constitute a closed system of
three differential equations that are coupled through the term
g= g
0
+d
2
Z/ dt
2
and have to be solved simultaneously for
Zt, ht, and Vt.
III. DISCUSSION
A. Numerical solutions and experimental data
In this section the predictions of Eqs. 11, 12, and 20
are compared with the experimental data of Kagan et al.
2
which are summarized in Fig. 6. These authors used a 2-l
soda bottle as a rocket, the takeoff of which they measured
with a high-speed camera and a “smart-pulley” system.
10
Their analysis was based on a dry adiabatic expansion, on
Bernoulli’s equation, and on the estimation of the thrust as
the product mass loss rate and exhaust velocity. They also
assumed that the rocket is a cylinder with a constant section
area , ended by a sudden constriction to a smaller section
0
. The predicted height as a function of time is reproduced
in Fig. 6 as curve 0. As mentioned, the rocket outperforms
this analysis.
The discrepancy between theory and experiment is not in
contradiction with the ﬁndings of Finney,
3
according to
which the simpliﬁed analysis predicts the time of ﬂight of the
rocket with an accuracy better than 0.2 s. Figure 6 shows that
the velocity of the rocket is underestimated only at the be-
ginning of the burnout; the velocity
v
0
at the end of the
burnout is predicted fairly accurately. As a consequence, the
remaining time of ﬂight after the end of the burnout—
calculated, for example, as 2
v
0
/ g when drag is neglected—is
also accurately predicted. The main source of error for the
total time of ﬂight is the duration of the burnout, which is
overestimated by less than 0.05 s for the case in Fig. 6a.
Accuracy better than 0.2 s is therefore needed to detect a
discrepancy between theory and experiment on the basis of
time of ﬂight measurements.
We have accounted for several effects which were previ-
ously not considered and which may have an important role
during the burnout. The ﬁrst effect is water condensation
during air expansion, which is accounted for by a polytropic
exponent
smaller than
. The second effect is the rocket
acceleration, which favors water ejection through increasing
the apparent gravity in the rocket. From dimensional analy-
sis, the acceleration is expected to be as signiﬁcant as the
pressure difference between the inside and the outside of the
rocket. The third effect is the acceleration downward of the
water inside the rocket, which reduces the vertical momen-
tum of the rocket. On the other hand, water inertia is under-
estimated in the usual analysis, which makes use of Bernoul-
li’s steady equation. The transient term, proportional to Ih
in Eq. 12, is expected to reduce the overall rocket perfor-
mance.
The procedure used for the numerical integration of Eqs.
11, 12, and 20 is discussed in Appendix A. We shall ﬁrst
assume the same rocket shape as in Ref. 2. For a perfect
cylinder, we have h=
max
, and from Eq. 13, Ih
=h /
max
. The numerical solutions for P
i
=3.4 and 6.8 bars
are plotted in Fig. 6 as curve 1 for
=1.4 dry air and f
=f
2
=1 a uniform velocity proﬁle. For both pressures,
curve 1 is slightly above the analysis of Ref. 2, but the the-
oretical prediction remains below experiment.
A more realistic soda bottle has a transition from body to
neck, as presented in Appendix B together with the corre-
sponding values of h and Ih. The corresponding solu-
tions of Eqs. 11, 12, and 20 are shown in Fig. 6 as curve
2. Another reﬁnement of our analysis accounts for the con-
densation of water vapor using the exponents
=1.31 and
=1.35 as calculated in Sec. II A see Fig. 3b. The corre-
sponding predictions are shown in Fig. 6 as curve 3. For the
launch at P
i
=3.4 bars, the performances are slightly im-
proved by water vapor condensation, but the extra thrust is
negligible for P
i
=6.8 bars, as anticipated in Sec. II A.
The effect of a possible nonuniform velocity proﬁle was
also investigated. The velocity proﬁle in Eq. 10 is modeled
empirically as fx= exp−0.5x /
2
, which amounts to as-
suming that most of the ﬂow occurs at the center of the
rocket in a region that extends over a fraction
With this form of fx the coefﬁcients f and f
2
are f
=2
2
1 exp−0.5/
2

2
=
2
1 exp−1/
2

these coefﬁcients, we found that any ﬁnite value of
results
in lower performances of the rocket, as exempliﬁed in the
ﬁgure for
=0.5 curve 4.
B. Analysis of simpliﬁed equations
Although some of the effects included in the present
analysis bring theory slightly closer to experiment, the im-
provements are small: a real rocket still outperforms the the-
oretical prediction. To try and understand why, we now con-
sider the simpliﬁed cylindrical model with h=
max
for
h 0 and Ih= h /
max
. It is shown in Appendix A that grav-
ity and aerodynamic drag are negligible during the thrust
Fig. 6. Predicted and experimental heights reached by a water rocket as a
function of time for a P
i
=3.4 bars and b P
i
=6.8 bars: experimental
data from Ref. 2, 0 textbook analysis, 1 current model with simpliﬁed
geometry, 2 with realistic geometry, 3 with realistic geometry and vapor
condensation, and 4 with realistic geometry, condensation, and nonuniform
ﬂow.
240 240Am. J. Phys., Vol. 78, No. 3, March 2010 Cedric J. Gommes

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