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Analysis of a water-propelled rocket: A problem in honors physics

Article in American Journal of Physics · March 2000

DOI: 10.1119/1.19415

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 Analysis of a water-propelled rocket: A problem in honors physics
G. A. Finney
Department of Physics, United States Air Force Academy, Colorado Springs, Colorado 80840
~Received 6 October 1997; accepted 16 June 1999!
The air-pumped, water-propelled rocket is a common child’s toy, yet forms a reasonably
complicated system when carefully analyzed. A lab based on this system was included as the final
laboratory project in the honors version of General Physics I at the USAF Academy. The numerical
solution for the height of the rocket is presented, as well as several analytic approximations. Five out
of six lab groups predicted the maximum height of the rocket within experimental error. © 2000
American Association of Physics Teachers.

I. INTRODUCTION height of the water and neglecting the velocity of the water at
the surface compared to the velocity at the nozzle, we obtain
The study of rocket motion has been used for decades to
excite students with the study of physics. ~See Refs. 1–4, for P5 P a 1 21 r w v 2e , ~2!
example.! Combined with the use of electronic computers,
where P is the pressure inside the rocket, P a is atmospheric
students can begin to solve many interesting, ‘‘real world’’
problems. In the honors versions of a calculus-based intro- pressure, and r w is the density of water. In addition to the
ductory mechanics course, I assigned my students the prob- assumptions listed above, we must also take as valid all the
lem of analyzing the motion of an air-pumped, water- assumptions which apply to Bernoulli’s equation ~principally
propelled rocket. The final goal was to determine the incompressible, nonviscous, irrotational flow!.
optimum amount of water to put into the rocket in order to Equation ~2! can be solved for v e and determines the ex-
achieve the maximum possible height. While we used small haust velocity as a function of internal pressure, P. The other
toy rockets, most of this analysis would also apply to the term needed to find the thrust from Eq. ~1! is the mass flow
popular demonstration using 2-l soda bottles pressurized by rate. Since the mass flow rate is just the volume flow rate
a bicycle pump. times the density of the water,
dM dV
II. MECHANICS OF ROCKET MOTION 5rw 5 r wA ev e , ~3!
dt dt
There are numerous references to the basic physics of where A e is the cross-sectional area of the exhaust nozzle.
rockets. In addition to those listed above, the reader may Combining Eqs. ~1!–~3! gives
consult almost any university physics text. The basic prob-
lem is to find the thrust, drag, and mass of the rocket as a T52 ~ P2 P a ! A e . ~4!
function of time in order to find the acceleration, velocity, Finding the thrust therefore depends on finding the pressure
and position. The following sections develop the differential within the rocket as a function of time. As the rocket expels
equations to be solved numerically, as well as some useful the water, the pressure and exhaust velocity drop, and thus
analytic approximations. the rate of pressure decrease drops. The solution begins with
A. Thrust two assumptions: ~1! the air in the rocket behaves as an ideal
gas and ~2! the air expands isothermally. ~Justification for the
The thrust, T, of a rocket due to the ejection of mass from isothermal assumption is given in Appendix A.! These as-
the nozzle is sumptions allow us to write

U
T5 v e
dM
dt
, U ~1! PV5 P 0 V 0 ,
where P and V are the pressure and volume of air inside the
~5!

where v e is the exhaust velocity of the ejected mass in the rocket at any time before all the water is ejected and P 0 and
rocket’s frame of reference and dM /dt is the rate at which V 0 are the initial pressure and volume of air. Solving for P
mass is ejected from the rocket. In our case, the mass is the and taking the derivative with respect to time
water that is pushed out as a result of the elevated air pres-
sure inside the rocket. Because v e and dM /dt both depend dP P 0 V 0 dV
52 2 . ~6!
on the pressure inside the rocket, finding the time profile of dt V dt
the thrust is nontrivial. However, it is within the capability of
Now substituting from Eqs. ~2!, ~3!, and ~5! to eliminate V,
better introductory physics students.
we get
Bernoulli’s equation ~conservation of energy! is applied at
two points along a streamline. This can be written generally
as
dP
dt
52
P2
A
P 0V 0 e
A 2 ~ P2 P a !
rw
. ~7!
P 1 1 21 r v 21 1 r gy 1 5 P 2 1 12 r v 22 1 r gy 2 . Equation ~7! can be solved to obtain P(t). The analytic so-
Figure 1 shows a schematic of the rocket. Take point 1 as the lution is presented below for comparison but, because of the
surface of the water inside the rocket and point 2 just outside complexity of the result, I had the students utilize a numeri-
the nozzle. Neglecting the pressure difference due to the cal solution for P(t). To solve ~7! analytically, separate vari-

223 Am. J. Phys. 68 ~3!, March 2000 © 2000 American Association of Physics Teachers 223
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 B. Drag
Aerodynamic drag is an important velocity-dependent
force, but not always discussed in introductory physics. Tra-
ditionally the drag force, F d , is expressed as
F d 5 12 C d A r v 2 , ~9!
where C d is the drag coefficient, A is an area corresponding
to the geometry of interest, and r is the density of air. At
moderate speeds ~see Appendix B!, the drag coefficient is
independent of the size of the object and speed of the
airflow.5 In the case of these rockets, there were two compo-
nents contributing to the drag: the rocket body and the fins.
Figure 1 shows that the body is roughly ellipsoidal in shape.
The dashed line in Fig. 1 indicates the location of a bend in
the fins to help stabilize the rocket by inducing rotation. For
these rockets, drag is a fairly small effect, so the precise
value of C d is not critical. Students could go to a number of
sources to obtain the necessary data to estimate C d . 6–8 My
estimates are C d,body50.05 and C d,fins50.1. The appropriate
area for the body is the circular cross-section normal to the
airflow. The area for the fins is the lateral area shown in Fig.
1, the surface area ~of one side of each fin! which is roughly
parallel to the airflow. Combining these into a single value,
Fig. 1. Sketch of the water rocket under consideration. and using the local average air density of r 51.05 kg/m3 ~for
an elevation of 7000 ft!, yields a total drag force
F d 5D v 2 , ~10!
ables and integrate from t50 to t and from P5 P 0 to P f , where D5231024 N/~m/s!2. This equation is incorporated
which yields into the numerical solution for the motion of the rocket.

S
Now consider the following analytic approximation for the
AP f 2 P a AP 0 2 P a
t5 A r w P 0V 0
2 P aA e Pf
2
P0
reduction in the maximum possible height due to drag. First,
drag can be neglected during the thrust phase for the follow-
ing reasons: The thrust phase lasts only about 0.1–0.2 s, or
1
1
AP a
F SA
arctan
P f2 Pa
Pa
D2arctan SA P 02 P a
Pa
DG D .
about 1.5 m out of a total altitude gain of 20 m. Furthermore,
drag is not the dominant force during the thrust phase ~or
during the coast phase, for that matter!. For a speed of 20
~8! m/s, the drag force is only about 0.08 N, compared to the
Figure 2 shows a comparison between the numerical solu- thrust of 10–20 N. However, the force of gravity on the
tion of ~7! and Eq. ~8!. Aside from the simplicity of numeri- empty rocket is about 0.4 N, so drag is a minor ~but signifi-
cally integrating ~7! compared to inverting ~8! for P(t), solv- cant! effect during the coast phase.
Therefore, it is possible to treat the drag force as a pertur-
ing it numerically also allows straightforward incorporation
bation on the kinematic solution.9 From kinematics ~i.e., ig-
of the next force, aerodynamic drag.
noring drag!, the velocity profile for an initial speed v 0 is
v~ y ! 5 Av 20 22gy. ~11!
Now calculate the work done by the force of drag using this
profile,

W nc5 E F"ds52D E 0
y
v~ y ! 2 dy52D ~ v 20 y2gy 2 ! .
~12!
Applying this value of work in conservation of energy
(DK1DU5W nc) and solving for the maximum height of
the rocket (m r is the mass of the empty rocket!,
v 20 mr
y max5 1 ~ 12 A11D 2 v 40 /m r2 g 2 ! . ~13!
2g 2D
The fraction under the radical is just the square of the ratio of
the maximum drag force to the force of gravity. For our
Fig. 2. Comparison of analytical result of Eq. ~8! with the numerical solu- rockets, this value was about 0.05. Therefore it can be ex-
tion from Eq. ~16!. Numerical solution shows only every tenth point from panded to first order. Finally, again use kinematics to replace
the numerical integration. Time of burnout is indicated by t bo . v 0 with t, the total time of flight of the rocket from launch to

224 Am. J. Phys., Vol. 68, No. 3, March 2000 G. A. Finney 224
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 impact, we obtain for the height of the rocket ~to first order
in the drag coefficient!
1 D
h5 gt 2 2 g 2t 4. ~14!
8 64m r
This expression can be used to estimate the height of the
rocket given the time of flight—a much simpler measure-
ment to make than using a sextant and trigonometry. This
result will be compared to the result of numerical integration
in a later section.
Fig. 3. Schematic diagram of the rocket pump, showing the main cylinder
with volume V c and empty space at the end of the pump with volume V e .
C. Rocket mass
Rocket ‘‘burnout’’ will be determined by one of two pos-
sible conditions: either the air expands until it forces all of limit of 4 atm ( P a ) for the pressure inside the rocket ~based
the water out of the rocket or it expands until it reaches on destructive testing of one sample that cracked at slightly
atmospheric pressure. While the latter could conceivably oc- less than 5 atm!. While it may have been possible to modify
cur if the initial volume of the air was much smaller than the the pump in order to directly measure the pressure, I wanted
total volume inside the rocket, it is of little practical interest to keep the project as simple as possible. Therefore, the pres-
and is not considered further. In the former case, once all of sure was calculated based on the volume of the rocket, the
the water is exhausted, the remainder of the air will rush out, volume of water in the rocket, and the number of ‘‘pumps.’’
but the air will contribute little to the thrust and is neglected. Figure 3 shows a schematic drawing of the pump used. Ex-
The mass of the rocket constantly decreases until all of the amining the pump, it consists of a piston moving within a
water is ejected from the rocket. For a given pressure inside cylinder. However, the pump is constructed so that there is a
the rocket, the volume of the air inside the rocket was found small amount of empty space at the end of the cylinder. As
using Eq. ~5!. The volume of the water, V w , is thus the the piston is compressed, the air in the main cylinder with
difference between the volume of the air and the total vol- volume V c ~31 mL! is forced into the small space at the end
ume of the rocket, V T . Multiplying by the density of water of the cylinder with volume V e ~8 mL!. Thus, as the rocket is
( r w 5103 kg/m3) and adding the mass of the empty rocket pressurized, the air from the main cylinder and small space at
(m r 539 g) yields the final result for the mass of the rocket the end ~with volume V c 1V e ! is compressed into the small
as a function of the internal pressure ~before ‘‘burnout’’!: space and the empty space in the rocket ~with volume V e

M ~ P !5 H r w ~ V T 2 P 0 V 0 / P ! 1m r
m r after ‘‘burnout’’.
before ‘‘burnout’’ 1V 0 !. The pressure after j11 pumps, P j11 , can be written:

P j11 5
P j V 0 1 P a ~ V c 1V e !
. ~18!
~15! V 0 1V e
This expression is used in the numerical solution.
@Note that as n→`, P n11 5 P n 5 P a (V c 1V e )/V e >5 P a .
Therefore it would appear that the rockets may have been
III. NUMERICAL SOLUTION originally designed to withstand the maximum possible pres-
Since I was working with second semester freshmen, they sure the pump could generate, but the rockets had degraded
had very little experience with numerical methods. There- with age.# Thus the students could determine the number of
fore, the students used the simple Euler method ~first order, pumps needed to achieve a pressure of 4 P a .
forward time difference! to implement the numerical solu- Finally, the students needed only to determine the opti-
tion. The students used MATHCAD® to perform the computa- mum water volume. They accomplished this task by repeat-
tions. My solution was implemented using the following set ing the calculations for a range of V 0 ’s with the MATHCAD
of equations, combining the results of Eqs. ~4!, ~7!, ~10!, and worksheet that they had developed. Figure 4 shows a plot of
~15! ~the students used similar sets of equations!: maximum height versus volume of water for P 0 54 P a , both
including and excluding the effects of drag. Interestingly, the
P 2n
P n11 5 P n 2
P 0V 0
Ae A 2~ P n2 P a !
rw
Dt,
height is not especially sensitive to the volume of water near
the maximum height ~even considering that the derivative is
zero at a maximum!. With the modeling complete, the stu-
v n11 5 v n 1a ~ P n , v n ! Dt, ~16! dents were ready to launch their rockets.
y n11 5y n 1 v n Dt,
where IV. LAUNCHING

2A e ~ P2 P a ! 2D v u v u The students launched their rockets several times during a

a ~ P, v ! 5 2g. ~17! single class period to compare the results of their model to
M~P!
the actual performance of their rockets. The height was esti-
Once the students had made various measurements of the mated in two ways. First, the students attempted to use tri-
rockets to determine the necessary physical parameters, only angulation, but the available equipment ~protractors and
two initial conditions were left to be determined: pressure plumb bobs! yielded less than satisfactory results because the
and water volume. Clearly, the higher the initial pressure, the uncertainty in the height was too large. The second technique
greater the velocity of the ejected water, and hence the was to measure the total time of flight for the rockets and use
greater the velocity of the rocket. Therefore, I set an upper Eq. ~14! to estimate the maximum height. Figure 5 shows

225 Am. J. Phys., Vol. 68, No. 3, March 2000 G. A. Finney 225
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 V. CONCLUSIONS
Most students reacted favorably to the project. The stu-
dents worked in groups of 3–4 to outline their procedures,
develop the model, and predict the maximum height. Stu-
dents did comment on the amount of time required, 10–20 h
per group, stretched over about half the semester. However,
they also commented that the ability to analyze and predict
the rocket’s motion was exciting and motivational. Although
probably beyond the ability of many introductory students,
for those willing to tackle it, they should find this a reward-
ing project.

APPENDIX A: JUSTIFICATION FOR ISOTHERMAL

EXPANSION
In deriving Eq. ~7! for the rate of change of pressure, an
approximation of isothermal expansion was used. Because of
the brief time involved, an adiabatic approximation would
Fig. 4. Predicted height as a function of initial water volume in the rocket. seem more natural. An adiabatic expansion will result in the
The solid line shows the result of numerical integration of Eq. ~16! while the
dashed line shows the result neglecting drag.
air cooling as it expands, reducing the pressure and the thrust
faster than an isothermal expansion. To test the sensitivity to
the heat gain through contact with the walls of the rocket,
one can solve the heat flow equation,
1 ]T
maximum height versus time of flight for the analytic ap- ¹ 2 T52 . ~A1!
a ]t
proximation of Eq. ~14! and numerical integration of Eq.
~16!. The time of flight from numerical integration of ~16! Here a is the ratio of the thermal conductivity to the specific
depended upon the initial volume of water—the point the heat and is taken to be approximately 231024 m2/s. For
curve doubles back indicates the time of flight corresponding simplicity, treat the rocket as an infinite cylinder. Then it is
to the optimum water volume. The error bars are based only straightforward to solve Eq. ~A1! in one dimension ~radial!
on the uncertainty in measuring the time of flight ~estimated with a fixed temperature at the walls of the rocket by expan-
to be 0.2 s!, propagated through Eq. ~14! using standard sion in Bessel functions.11 For the initial condition, use the
techniques.10 Of the six groups, five predicted the height ~as temperature change given by an adiabatic expansion, typi-
determined by time of flight! within experimental uncer- cally about 40 °C, taken uniformly across the cylinder. In 0.2
tainty. The only group which was not within uncertainty s, the temperature of the expanding air returns at least half-
elected to neglect drag in their model. The success of the way to the ambient temperature 0.8 cm from the wall of the
students as a whole is particularly significant considering the rocket. Given that ~1! the rocket is not really a cylinder and
model had no free parameters! ~2! the air has considerable volume near the surface of the
water and the front end of the rocket, then significantly less
than half the volume of the rocket deviates more than 20 °C
from the initial temperature. Since neither the isothermal nor
the adiabatic approximations rigorously hold, I used the sim-
pler isothermal approximation. This had an additional advan-
tage because my course did not include a block on thermo-
dynamics: my students had already learned about isothermal
expansions in introductory chemistry. Finally, the validity of
the approximation is confirmed by the agreement found be-
tween the model and our experimental results.

APPENDIX B: MODELING AIR DRAG

The force of drag on an object immersed in a fluid arises
as a result of two distinct processes: ~1! skin friction arising
from shear forces within the liquid ~laminar flow! and ~2!
transfer of momentum from the object to the surrounding
fluid in the form of eddy currents ~turbulent flow!. The first
process yields a force that is linearly dependent on speed of
the fluid, while the second yields a process which is depen-
dent on the square of the speed of the fluid. A qualitative
Fig. 5. Maximum altitude vs time of flight. The solid line indicates the result
explanation for each of these terms is given below.12
of numerical integration of ~16! and the dashed line indicates the analytical Consider the laminar flow of a fluid along the surface of
approximation given by Eq. ~14!. Error bars indicate uncertainty in esti- an object. The velocity profile of the fluid will appear quali-
mated height based on uncertainty in measuring time of flight. tatively as in Fig. 6, with the fluid at rest next to the object

226 Am. J. Phys., Vol. 68, No. 3, March 2000 G. A. Finney 226
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 ful in understanding the basic physics underlying the two
forms of the velocity-dependent drag force. However, the
student is still left with the question of which one to use.
This question can be answered by considering the ratio of the
two terms:

D 2 2C Dr A v 2
D1
5
1

kmlv
5 S D S DS D
1
2 CD
k
A
l
rv
m
. ~B3!

The first term in parentheses is a geometric factor that is

typically on the order of one. The second term can be re-
placed with a linear dimension d ~again, typically parallel to
the flow!. Now the ratio can be identified as
Fig. 6. Typical velocity profile of fluid in laminar flow in the vicinity of a
solid object. The dotted line is the linear approximation used in text.
D2 drv
} 5N R , ~B4!
D1 m
where N R is known as the Reynolds number. For small val-
and flowing with a speed v at some distance away. This ues of the Reynolds number (N R ,1), the flow is laminar
velocity profile results in a shear stress ~force per unit area!, and the drag is dominated by the viscous drag force, D 1 . For
s, on the rocket given by s 5 m (d v /dy), where m is the large values of the Reynolds number (N R .10 000), the flow
coefficient of viscosity, a property of the fluid. The velocity is turbulent and the drag is dominated by the inertial drag
gradient can be approximated as K v /w, where K is a dimen- force, D 2 . In between, the drag can be modeled as the sum
sionless constant based on the shape of the object and w is a of the two forces. For this study, m 5231024 P (2
distance characteristic of the object. The viscous drag will be 31025 kg/m s), r 51 kg/m3, d50.1 m, and v 520 m/s, so
proportional to the product of the shear stress and the surface N R '105 . Thus, we are justified using only the inertial drag
area of the object, or D5 s A5 m KA v /w. Since K, A, and w force.
are factors related to the geometry of the body, D can be
rewritten as 1
D. S. Gale, ‘‘Instructional uses of the computer. Rocket trajectory simu-
D 1 5 m kl v . ~B1! 2
lation,’’ Am. J. Phys. 38, 1475 ~1970!.
Robert A. Nelson and Mark E. Wilson, ‘‘Mathematical Analysis of a
Now k is a constant that depends on the shape of the object Model Rocket Trajectory,’’ Phys. Teach. 14, 150–161 ~1976!.
3
and l is a length ~typically parallel to the flow!. For example, S. K. Bose, ‘‘The rocket problem revisited,’’ Am. J. Phys. 51, 463–464
a sphere has k53 p and l is the diameter of the sphere. ~1983!.
4
R. H. Gowdy, ‘‘The physics of perfect rockets,’’ Am. J. Phys. 57, 322–
At higher speeds, the flow will not remain laminar, but 325 ~1995!.
will become turbulent. Then the object will impart momen- 5
John D. Anderson, Introduction to Flight ~McGraw–Hill, New York,
tum to the fluid by the generation of eddy currents and the 1985!, 2nd ed.
drag will be based on an inertia force. The rate of momentum 6
Handbook of Engineering Fundamentals, edited by Mott Souders and
transfer is equal to the force exerted on the object, and can be Ovid W. Eshbach ~Wiley, New York, 1975!, 3rd ed.
7
determined in the following way. Consider the object mov- Gerald M. Gregorek, ‘‘Aerodynamic Drag of Model Rockets,’’ Model
Rocket Technical Report No. TR-11, Estes Industries, Inc., Penrose, CO,
ing a distance Dx through the fluid in a time Dt. The object 1970.
will displace fluid with mass m fluid5k 1 r ADx, where k 1 is a 8
Russell A. Dodge and Milton J. Thompson, Fluid Mechanics ~McGraw–
geometrical constant, r is the fluid density, and A is the Hill, New York, 1937!, Chap. XII, Art. 171.
9
cross-sectional area. The average speed of the eddy currents James H. Head, ‘‘Faith in Physics: Building New Confidence with a Clas-
will be proportional to the speed of the object ~as long as the sic Pendulum Demonstration,’’ Phys. Teach. 33, 10–15 ~1995!.
10
Phillip R. Bevington and D. Keith Robinson, Data Reduction and Error
speed is not too high!, so momentum imparted in the time Dt Analysis for the Physical Sciences ~McGraw–Hill, New York, 1992!, 2nd
is D p5m fluid(k 2 v fluid)5k 1 k 2 r ADx v . Therefore, the rate of ed., p. 50.
momentum transfer, and hence the drag, is ~in the limit that 11
See, for example, Mary R. Boas, Mathematical Methods in the Physical
Dt approaches zero! Sciences ~Wiley, New York, 1983!, 2nd ed., pp. 558–562, or George
Arfken, Mathematical Methods for Physicists ~Academic, New York,
D 2 5D p/Dt5k 1 k 2 r A v~ Dx/Dt ! 5 21 C D r A v 2 . ~B2! 1985!, 3rd ed., pp. 448–450, 513–593.
12
See, for example, Ref. 8, Chaps. VIII and XII, or James W. Daily and
Here, C D is the familiar coefficient of drag. Donald R. F. Harleman, Fluid Dynamics ~Addison–Wesley, Reading, MA,
While these are not rigorous derivations, they prove help- 1966!, Chaps. 8 and 9.

227 Am. J. Phys., Vol. 68, No. 3, March 2000 G. A. Finney 227
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