# 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

TL;DR: The water rocket as mentioned in this paper is a popular toy that is often used in first year physics courses to illustrate Newton's laws of motion and rocket propulsion, and is made of a soda bottle, a bicycle pump, a rubber stopper, and some piping.

Abstract: 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 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

## Summary (2 min read)

Jump to: [A more thorough analysis of water rockets: Moist adiabats, transient flows, and inertial forces in a soda bottle] – [A. Moist air expansion] – [B. Water ejection] – [C. Rocket acceleration] – [A. Numerical solutions and experimental data] – [B. Analysis of simplified equations] and [IV. CONCLUSIONS AND PROSPECTS]

### 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

共Received 19 May 2009; accepted 12 October 2009兲

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

experiment with water rockets. ©

2010 American Association of Physics Teachers.

关DOI: 10.1119/1.3257702兴

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 共see Fig. 1兲. 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

or adiabatic

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. 2共b兲. 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 100g’s. The ﬂow of water is rapid not just at the

nozzle but also inside the rocket. The cloud in Fig. 2共b兲

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 共3兲, March 2010 http://aapt.org/ajp © 2010 American Association of Physics Teachers

␦

h

2

⬇

, 共1兲

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 共the fog兲 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

共T,V兲 = s

a

共T

0

,V

0

兲 + c

V

a

ln

冋

冉

T

T

0

冊冉

V

V

0

冊

␥

−1

册

, 共2兲

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

共T兲 = s

l

共T兲 +

⌬h

V

T

, 共3兲

where s

l

共T兲 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

共T兲 is

6

s

l

共T兲 = s

l

共T

0

兲 + c

p

l

ln

冉

T

T

0

冊

, 共4兲

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

共T兲. By using the ideal

gas law, P

V

共T兲 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

during the adiabatic expansion.

We use Eqs. 共2兲–共4兲 to ﬁnd

冋

1−

P

V

共T

i

兲

P

i

册

c

V

ln

冉

冉

T

T

i

冊冉

V

V

i

冊

␥

−1

冊

+

冉

T

i

T

冊

⫻

冉

V

V

i

冊

P

V

共T兲

P

i

⌬h

V

T

+

P

V

共T

i

兲

P

i

冉

c

p

l

ln

冉

T

T

i

冊

−

⌬h

V

T

i

冊

=0.

共5兲

For the case of dry air, that is, for P

V

共T兲= P

V

共T

i

兲= 0, the

solution of Eq. 共5兲 is TV

共

␥

−1兲

=constant, as expected. For

moist air, however, Eq. 共5兲 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. 共5兲 are plotted in Fig. 3共a兲 for three values

of P

i

and three values of T

i

. The dry adiabats 共PV

␥

=constant兲 and isotherms 共PV=constant兲 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,

6–8

with P

V

共T

0

兲= 1 bar at T

0

=100 ° C.

The moist adiabats in Fig. 3共a兲 are found to be well ap-

proximated by a polytropic process

8

of the type PV

=constant 共dotted lines兲. The exponent

obtained by a least-

squares ﬁt is found to depend mostly on the initial relative

humidity P

V

共T

i

兲/ P

i

, as shown in Fig. 3共b兲. 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 共a兲 t =0 s, 共b兲 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 共b兲 is moving upward.

237 237Am. J. Phys., Vol. 78, No. 3, March 2010 Cedric J. Gommes

= 1.15 + 共1.4 – 1.15兲exp

冉

−36

P

V

共T

i

兲

P

i

冊

, 共6兲

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. 共1兲, the distance

␦

v

over which momentum

diffuses in a ﬂuid during a time

due to its viscosity is

4,5

␦

V

2

⬇

, 共7兲

where

is the kinematic viscosity of the liquid. For water

4

at

20 °C,

=0.01 cm

2

/ s. Using

=0.1 s leads to

␦

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 Euler’s equation.

The general form of Euler’s equation is

4,5

u

ជ

t

+ 共u

ជ

· ⵜ兲u

ជ

=−

1

ⵜ P + g

ជ

, 共8兲

where u

ជ

is the vector velocity of water, P is the pressure,

is

the density, and g

ជ

is the gravitational ﬁeld; u

ជ

and P are gen-

erally space and time dependent. When Eq. 共8兲 is integrated

along the axis of the rocket, from the nozzle up to the free

surface of the liquid 共see Fig. 5兲, we obtain

冕

0

h

u

c

t

dz +

u

c

2

共h兲 − u

c

2

共0兲

2

+

P − P

a

+ gh =0, 共9兲

where h共t兲 is the time-dependent height of water on the axis

of the rocket, P

a

is atmospheric pressure, u

c

共h兲 is the 共nega-

tive兲 vertical velocity of the water free surface on the axis of

the rocket, and u

c

共0兲 is the exhaust velocity on the axis.

Bernoulli’s equation, as used in Refs. 2 and 3, results directly

Fig. 3. 共a兲 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 共+兲. The solid lines are isotherms 共pV=constant兲 and

dry adiabats 共pV

␥

=constant兲. 共b兲 Moist adiabats are approximated by a

polytropic process of the form pV

=constant 关dotted lines in 共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 共+兲,4bars共䊊兲, and 7 bars

共䉭兲. 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. 共9兲 by neglecting the time derivative as well as

gravity.

The order of magnitude of the ﬁrst term in Eq. 共9兲 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. 共9兲 cannot be simpliﬁed a priori to analyze the water

rocket.

To use Eq. 共9兲 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,

u共z,r,t兲 =

dh

dt

冋

⍀共h 兲

⍀共0 兲

册

f

冉

r

R共z兲

冊

, 共10兲

where R共z兲 is the radius of the bottle at height z 共see Fig. 5兲,

⍀共z 兲=

R

2

共z兲 is the cross sectional area at height z, and f is

a function that characterizes the velocity proﬁle and satisﬁes

f共0兲= 1. Uniform velocity is recovered with f共x兲= 1; the case

where the water drops more rapidly at the center of the

rocket corresponds to a function f共x兲 with a maximum at x

=0. The ratio in the square brackets in Eq. 共10兲 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. 共10兲 over a section leads to

dV

dt

=−具f典⍀共h兲

dh

dt

共11兲

with 具f典= 兰

0

1

2xf共x兲dx.

By using Eq. 共10兲, the various terms in Eq. 共9兲 can be

written explicitly as a function of h共t 兲, which leads to

I共h兲

d

dt

冉

⍀共h 兲

dh

dt

冊

+

1

2

冉

1−

冉

⍀共h 兲

⍀

0

冊

2

冊冉

dh

dt

冊

2

+

1

冉

P

i

冉

V

i

V共t兲

冊

− P

a

冊

+ gh =0 共12兲

with

I共h兲 =

冕

0

h

dz

⍀共z 兲

. 共13兲

The notation highlights that this term accounts for the inertia

of the accelerated water. In Eq. 共12兲,

is the polytropic

exponent obtained in Sec. II A and ⍀

0

=⍀共0兲 is the area of

the nozzle.

If the rocket was forced to remain immobile, Eqs. 共11兲 and

共12兲 could be solved for h共t兲 and V共t兲. 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, 共14兲

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

+

冕

0

h

dz

冕

0

R共z兲

2

rdr

u共z,r,t兲, 共15兲

where m is the mass of the empty rocket, Z is its vertical

position, and u共z , r , t兲 is the velocity of the water in the

rocket. Note that, contrary to Sec. II B, the water velocity

u共z ,r, t兲 is expressed here with respect to the ground. If we

correct Eq. 共10兲 by adding dZ/ dt, Eq. 共15兲 becomes

MU = 关m +

V

w

兴

dZ

dt

+

具f典h⍀共h兲

dh

dt

, 共16兲

where V

w

=兰

0

h

⍀共z 兲dz is the volume of water in the rocket.

The second term in Eq. 共16兲 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. 共14兲 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. 共10兲, the

thrust takes the form

Th =

⍀共h 兲

2

⍀

0

冉

dh

dt

冊

2

具f

2

典 +

⍀共h 兲

dh

dt

dZ

dt

具f典共17兲

with 具f

2

典= 兰

0

1

2xf

2

共x兲dx. 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 = 共m +

V

w

兲g

0

, 共18兲

where g

0

is the gravity, not to be confused with g that enters

Eq. 共12兲 and includes a noninertial contribution. The aerody-

namic drag is calculated in the usual way

4,5

as

D = C

A

⍀

max

a

2

冉

dZ

dt

冊

2

, 共19兲

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

is about 0.75,

4

but for the following analysis,

it is sufﬁcient to consider that its order of magnitude is 1.

Combining Eqs. 共14兲–共19兲 leads to the following differen-

tial equation:

239 239Am. J. Phys., Vol. 78, No. 3, March 2010 Cedric J. Gommes

d

dt

冋

共m +

V

w

兲

dZ

dt

+

具f典h⍀共h兲

dh

dt

册

=

⍀

2

共h兲

⍀

0

冉

dh

dt

冊

2

具f

2

典 +

⍀共h 兲

dh

dt

dZ

dt

具f典

− g

0

共m +

V

w

兲 − C

A

⍀

max

a

2

冉

dZ

dt

冊

2

, 共20兲

which expresses the conservation of vertical momentum.

Equations 共11兲, 共12兲, and 共20兲 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

Z共t兲, h共t兲, and V共t兲.

III. DISCUSSION

A. Numerical solutions and experimental data

In this section the predictions of Eqs. 共11兲, 共12兲, and 共20兲

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. 6共a兲.

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 I共h兲

in Eq. 共12兲, is expected to reduce the overall rocket perfor-

mance.

The procedure used for the numerical integration of Eqs.

共11兲, 共12兲, and 共20兲 is discussed in Appendix A. We shall ﬁrst

assume the same rocket shape as in Ref. 2. For a perfect

cylinder, we have ⍀共h兲= ⍀

max

, and from Eq. 共13兲, I共h兲

=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 共dry air兲 and 具f典

=具f

2

典=1 共a uniform velocity proﬁle兲. 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 ⍀共 h兲 and I共h兲. The corresponding solu-

tions of Eqs. 共11兲, 共12兲, and 共20兲 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 关see Fig. 3共b兲兴. 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. 共10兲 is modeled

empirically as f共x兲= exp共−0.5共x /

兲

2

兲, 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

of its radius.

With this form of f共x兲 the coefﬁcients 具f典 and 具f

2

典 are 具f典

=2

2

共1 − exp共−0.5/

2

兲兲 and 具f

2

典=

2

共1 − exp共−1/

2

兲兲. With

these coefﬁcients, we found that any ﬁnite value of

results

in lower performances of the rocket, as exempliﬁed in the

ﬁgure for

=0.5 共curve 4兲.

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 ⍀共 h兲= ⍀

max

for

h⬎ 0 and I共h兲= 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 共a兲 P

i

=3.4 bars and 共b兲 P

i

=6.8 bars: 共䊊兲 experimental

data from Ref. 2, 共0兲 textbook analysis, 共1兲 current model with simpliﬁed

geometry, 共2兲 with realistic geometry, 共3兲 with realistic geometry and vapor

condensation, and 共4兲 with realistic geometry, condensation, and nonuniform

ﬂow.

240 240Am. J. Phys., Vol. 78, No. 3, March 2010 Cedric J. Gommes

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Abstract: I Historical Roots: From Heat Engines to Cosmology
1 Basic Concepts and the Law of Gases
2 The First Law of Thermodynamics
3 The Second Law of Thermodynamics and the Arrow of Time
4 Entropy in the Realm of Chemical Reactions
II Equilibrium Thermodynamics
5 Extremum Principles and General Thermodynamics Relations
6 Basic Thermodynamics of Gases, Liquids and Solids
7 Thermodynamics of Phase Change
8 Thermodynamics of Solutions
9 Thermodynamics of Chemical Transformations
10 Fields and Internal Degrees of Freedom
11 Thermodynamics of Radiation
III Fluctuations and Stability
12 The Gibbs Stability Theory
13 Critical Phenomena and Configurational Heat Capacity
14 Entropy Productions, Fluctuations and Small Systems
IV Linear Nonequilibrium Thermodynamics
15 Nonequilibrium Thermodynamics: The Foundations
16 Nonequilibrium Thermodynamics: The Linear Regime
17 Nonequilibrium Stationary State and Their Stability: Linear Regime
V Order Through Fluctuations
18 Nonlinear Thermodynamics
19 Dissipative Structures
20 Elements of Statistical Thermodynamics
21 Self-Organization and Dissipative Structures in Nature

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