Thermodynamic Principles on a Spud Gun Worcester Polytechnic Institute

Project Number: Term Project_PH 210X C12_01
Thermodynamic Principles on a Spud Gun
Submitted to the faculty of
Worcester Polytechnic Institute
In partial fulfillment of the requirements
for the PH210X Course
Marco Romero
George Bucşan
Approved by:
Germano S. Iannacchione, Ph.D.
Table of Contents
- What is the fuel used for the purpose of this experiment?
- How much energy can we really use?
- How much energy was lost?
- How fast will our projectile be able to go?
- Yet another question remains. How far will the projectile be able to go?
Experimental Data
Data Analysis
- Why this differences?
- What can we change to decrease discrepancies?
Concluding Data
“Thermodynamic Principles Operating on a Spud Gun”
Throughout history many powerful armies have used weapons across continents to bring
about a greater good, or to achieve an unpredictable evil. If history has taught us anything, it is
that those who possess the most technologically advanced weapons tend to overcome and
trample on those who oppose their will. One of these technologically advanced weapons was the
cannon. Although, at first sight the cannon may seem heavy and burdensome on the battlefield
(for it impedes the fast mobility of troops) a properly constructed cannon can pack a deadly blow
operating simply on basic thermodynamic principles. In principle a cannon operates with three
basic components, a cylindrical shape tube in which ballistic missiles are inserted, a friction
primer and gunpowder or other explosive-based propellants. A friction primer is essentially the
trigger used to generate a spark that will create a small explosion within the friction primer; this
explosion will instantly ignite the cartridge where the powder is and the cannon will fire.
Due to safety and liability issues we will analyze a little more in detail the processes and
the thermodynamic principles behind the cannon, by researching and conducting a series of
experiments with a spud gun. As you may or may not know all spud guns are composed of three
basic components, a chamber in which gas can reach high levels of pressure, a barrel for a
projectile, and an instrument that can ignite the propellant. As you can see, the similarities
between the two are significant, which is why studying the conversions of chemical energy to
kinetic energy will give us an insight at the thermodynamics principles governing the spud gun,
which in turn will grant us a view at the firing principles surrounding the deadly cannon.
* All sizes indicate the diameter of the pipe and not length
1-1/2” PVC Barrel 330PSI @ 23 C
1-1/2”, 3” PVC Reducer
3” PVC Pipe (Combustion Chamber) 260PSI @ 60 C
3” PVC Threaded Adapter
3” PVC End Cap into Adapter
3” PVC Threaded end Cap
PVC Primer
PVC Cement
1-1/2” Shower Drain
BBQ Igniter (any system that may generate a spark at desired
2 Metallic screws
Potatoes (any other kind of projectile)
Lighter fluid (for the purpose of this experiment, however hairspray or any other type
of combustible fuel may be used)
Hand Wood Saw
Small Knife
Drill Motor with multiple screw caps
Rag for PVC cement
One of the first things we did as we embark on this journey was to obtain the previous materials
and secure a “safe” working station with plenty of room to move around and the proper tools to
lacerate through the PVC pipe.
Step 1: Getting the materials ready for assemblage
 PVC pipe adjustments:
One of the first things to do in this process is to cut the 1-1/2” PVC pipe into the desired size, in
our case approximately 23” inches long, this will create the barrel of our gun.
You may want to leave a big enough size for the projectile to fit in and slide down, a small barrel
may increase the risk of fires and explosions for the pressure will be concentrated on a smaller
surface area.
 PVC Pipe Fittings:
Before you continue placing everything together make sure all pipes are adequately cut so that
they fit perfectly into one another. Careful not to leave any rough edges for they may cause
imperfections when bonding.
 Cleaning:
Make sure all the pipes and materials are well cleaned with no sand or other substances that can
potentially damage the outcome.
Step 2: Assembling the Chamber
Take the 3” PVC pipe desired for the chamber where combustions will occur. Drill the 2
screws on opposite sides of the pipe leaving a small breach between them in the middle.
(Make sure they do not touch inside for we need a space between then to generate a
Step 3: Assembling the Cannon
A good way to begin assembling the cannon will be from top to bottom. As a personal
choice we inserted a 1-1/2” inch shower cap between the barrel and the reducer to make
sure fragments of projectiles wont slide into our chamber. Once this was attained we
applied a coat of PVC primer to the outer edges of the barrel and quickly did the same to
the inner circumference of the reducer. Once this is done, we applied cement over the
primer and slid the barrel into the adapter with approximately 90 degree turning motion.
Hold for 30 seconds. Repeat the above process to connect the reducer to the chamber and
the chamber to the threaded adapter.
Be sure to leave the cannon alone for approximately 24 hrs. Not doing so can cause the cannon
to fail and cause serious fatal injuries.
Figure 1: The picture above represents the basic components of the spud cannon, as referred to in
this paper. If any questions should arise during construction feel free to refer to the image above.
Step 4: Wiring Electrical System
Once the cannon is solidly built, take the BBQ igniter and strip the protective plastic in
the ends. This will enable you to pass a current through a metal faster. Attach the ends to
the exposed screws on the chamber wrapping the wires clockwise. Before you proceed,
activate the BBQ igniter and make sure a spark is produced inside the chamber. If you
can see it, you have successfully placed together a potato cannon.
What is the fuel used for the purpose of this experiment?
Because it is easily available, we decided to use Ronsonol lighter fluid – a mix of light
and medium aliphatic naphtha.
How much energy can we really use?
It has been clearly established that for similar cannons the observed behavior as the
projectile leaves the barrel tends to be the same. As the projectile launches into the air one can
see a small flame shooting out and dispersing into the atmosphere. What does this mean?
Perhaps one of the most important aspects that this signifies is that the energy of the fuel is not
entirely used.
How much energy was lost?
Before we proceed with calculations about energy lost, we must generalize the problem with a
couple of assumptions.
 The burn process is very fast
 The gas inside expands adiabatically until the projectile leaves the barrel.
 The chemical energy is divided into the kinetic energy associated with the muzzle
velocity of the projectile and the remaining thermal energy in the gas that leaves the
barrel after the projectile.
We therefore can begin to approximate the temperature of our residual gas by comparing it to the
burning temperature of our fuel: 160oC or 433K. The gas is also exposed to a temperature of
about 20oC or 293K due to the surrounding environment after the projectile leaves the barrel.
To further continue of process we must note that the main ingredient (95%) of our fuel is light
aliphatic naphtha, which consists of hydrocarbons with 5-6 carbon atoms, mainly hexane and
pentane, with the chemical formulas C6H14 and C5H12. As we do not know the exact mix, we will
assume a 50% hexane 50% pentane composition.
It is noted that the burn formulas for these two compounds are as follows:
C5H12 + 8 O2 → 5 CO2 + 6 H2O
2C6H14 + 19 O2 → 12 CO2 + 14 H2O
By simple multiplication and addition we attain:
2C5H12 + 2C6H14 + 35 O2 → 22 CO2 + 26 H2O
Since we know the molar masses of these substances we can proceed to state that our reaction
yields 68% CO2 and 32% H2O.
For these complex molecules, the heat capacity at constant volume constant is
cv = 3.
Hence the number of moles of the resulting can be approximated by the above equation by the
2 moles naphtha – 48 moles of carbon dioxide and water
Given that the approximate molar mass of naphtha is 120g/mole, the above ratio can be
successively rewritten as:
120g naphtha – 24 moles of carbon dioxide and water
1g naphtha – 0.2 moles of carbon dioxide and water
Therefore, we can conclude that the thermal energy wasted for each gram of fuel used is:
ΔUf(per gram of fuel) = cv*n*R*ΔT = 3*0.2*8.314*(433-293) = 0.698 kJ/g
This in turn can be rewritten in terms of the volume of fuel in milliliters as:
ΔUf(per mL of fuel) = 698 * 0.72 = 0.502 kJ/mL
However, the gas that results from the burn of the fuel is not only the gas that is being heated but
also the initial volume of air that was already in the barrel. Considering that all the oxygen will
burn in the reaction, nitrogen will still be heated up to 518K.
The amount of nitrogen in the barrel can be calculated from the amount of air and the
concentration of nitrogen in air:
V * ρair * cN2 / µN2 = 43 moles N2
Furthermore, the energy required to increase the temperature of this amount of nitrogen is:
ΔUN = cv*n*R*ΔT = 1.5 * 43 * 8.314 *(433-293) = 75.075 kJ
Overall, the amount of energy we lose as heat as a function of the number of the number of
milliliters of fuel we use is:
ΔU = n*0.502 + 75.075 [kJ]
Averaging the heat of combustion for the two main components, hexane and pentane can
approximate the heat of combustion for our fuel:
H = 0.5 ( 48.5 kJ/g + 49.05 kJ/g) = 48.77 kJ/g
Therefore, from each gram of fuel we can use approximately 47.65kJ of energy in the kinetic
form. Considering the density of 0.72g/cm3 of naphtha in normal conditions (20oC), this
approximation becomes: 34.31kJ / mL.
*We chose to use milliliter values because liquid volume is easier to accurately measure with a
regular syringe.
If we subtract the energy lost from the total energy – as functions of the number of milliliters of
fuel added into the system – we obtain:
E = 34.31*n – 75.075 – 0.50*n = 33.81*n – 75.08 [kJ]
How fast will our projectile be able to go?
To begin our calculations regarding the speed of our projectile we first need to calculate
the mass for our object. This mass can vary a lot since our projectile for this project is a potato
with an average density of 1.1g/cm3. Trying to normalize the shape, our best guess was cylinder
with a length of 1.5 inches and a diameter of 1.6 inches. Which leads to an average calculated
mass of approximately 54g.
Using the equation for kinetic energy:
K = 0.5 * m * v2
The muzzle velocity is:
We then relied on MATLAB’s mathematical software to plot the values
(Refer to Appendix 1A for the Code)
Because we assumed that all the gas in the combustion chamber is approximately heated up to
the burn point of our fuel (for low amounts of fuel that would be impossible), our plot should be
of decent accuracy starting with values above approximately 2.5mL of fuel:
If more control over the system is desired, we can vary the length and therefore the mass of the
projectile and get different values for the kinetic energy. By arbitrarily choosing the value of 5ml
as the amount of fuel, this graph was obtained:
(Refer to Appendix 2A for the code)
Yet another question remains. How far will the projectile be able to go?
It is well known that for the maximum range, the angle of a barrel should be very close to 45
degrees; therefore we will use that into making our further calculations. Because we are trying to
accomplish a reasonable degree of accuracy in our predictions, we will need to take into
consideration the effects of drag over our projectile – the potato.
In our previous assumptions about the volume, we made an idealized potato-projectile into the
shape of a cylinder. The drag coefficient of a short cylinder is Cd=1.15. We must still remember
that the real potato will have a somewhat round front section. For a better estimation we should
decrease this number by a fraction. Given that the drag coefficient of a sphere is Cd=0.47, we can
take the average and obtain our potato drag coefficient to about Cd=0.81. This is very close to the
drag coefficient of a long cylinder (with flat ends), therefore our estimation seems to be close to
a real value.
The drag force as a function of the velocity of the projectile the equation is:
Fd = 0.5 * Cd * ρair * V2 * π * r2
(all in SI units).
Reducing the equation to the form Fd = k * V2 in SI units we get k = 6.432*10-4.
The physical way of obtaining the range equation for cannons with real drag added is very
intricate, but as engineers we found the numerical method to be far more practical and yield
accurate results.
We will use the following approach:
 The initial parameters in SI:
V0x = V0y = 59* cos(π/4) = 42 (for the 54g projectile and 5ml of fuel)
X0 = Y0 = 0
α0 = π/4
k = 6.432*10-4
m = 0.054
g = 9.81
 The time increment is dt = 0.005s – should provide enough accuracy
 The parameters after one time increment (dt):
Y1 = Y0 + V0 sin(α0) dt
X1 = X1 + V0 cos(α0) dt
V1y = [V0y – (k V02 / m)dt ] sin(α0) – g dt
V1x = [V1y - (k V02 / m)dt ] cos(α0)
α1 = tan-1(V1y / V1x)
Again we relied on MATLAB for more accurate calculations; the code used is as follows:
(Refer to Appendix 3A for the Code)
Running the script yielded the range for a 52g projectile and 5ml of fuel to be 73.46m and the
time of flight t=4.52s.
The plot of the trajectory can be seen below.
By modifying the initial velocity (muzzle velocity) and rewriting some portion of code we
plotted the range as a function of the volume of fuel for a “standard” potato with mass 52g.
(Refer to Appendix 4A the Code)
For obtaining the experimental data we decided to use the “standard sized” potato with a mass of
54g and to vary the amount of fuel – from a value smaller than we predicted possible (2ml) to a
value as big as we considered safe (6.5ml).
1. The cannon was able to fire with a smaller amount of fuel than predicted – 2ml;
2. If the amount of fuel is large, the liquid fails to fully evaporate and there can be no
further considerable increase in the range;
3. If the amount of fuel is larger or equal than 7ml, the cannon does not fire at all;
4. If the amount of fuel is above 7ml, the fuel does not explode but burns inside of the
combustion chamber, sometimes causing structural damage to the PVC piping;
5. Range difference between two almost identical setups can reach even 10m;
For data accuracy, we ran two sets of trials and averaged the data.
Data points/curve obtained:
Because our theoretical data curve had its minimum for the “fuel” axis at a value above 2ml, and
our theoretical data points ranged between 2ml and 6.5ml, we only took into account the values
ranging from 2.5ml to 6.5ml for both curves.
The average deviation of our experimental data against the theoretical data is 6.55m. Given that,
for the chosen range, the mean value of our theoretical data is 76m, the percent deviation
becomes 8.66%.
For any experiment of this kind, 8.66% would not be an unimaginably large value, but the
numbers do not reveal everything. By plotting the curves, more observations emerge:
Experimental data – red, Theoretical data - green
The above plot reveals that, although the curves have similar shapes, for low amounts of fuel the
difference is very large, whereas for larger amounts of fuel the differences become negligible.
Why this differences?
The answer lies in the way the kinetic energy is calculated. A close look at the initial reasoning
reveals some possible problems. Also, a close look at the experimental conditions can give some
1. The heat value of the fuel was obtained as an average of the theoretical heat values for
the two predominant compounds; yet we do not know what the action of the other
compounds is, how much heat they generate / use and what compounds they transform
2. The final temperature was assumed to be the burn temperature of the fuel; but it is rather
obvious that a significant percent of that number will be lost during the adiabatic
expansion of the gasses from the volume of the combustion chamber to the full internal
volume of the cannon.
3. Full vaporization of the fuel was assumed, but the small vapor pressure of this type of
fuel hints that for large amounts of fuel, a certain percentage will remain in liquid state.
4. The outside temperature was assumed to be “normal” – 293K, whereas the real
temperature, normal for winter in New England, was 273K. This change directly affects
the calculated amount of usable energy.
5. The fuel has a limited saturation amount for which it can explode, for larger amounts it
will simply burn. Initially this was not accounted for.
What can we change to decrease discrepancies?
From the ideas above, we decided to add two constants to the usable energy equation, as follows:
E = α*33.81*n – β* 75.08 [kJ]
In this case:
- α represents the adjustment to the amount of energy the fuel provides.
- β represents the adjustment to the difference between the final temperature in the
chamber and the outside temperature.
By adjusting the parameters and plotting both the new theoretical curve and the experimental
curve in parallel, a set of values that can increase the prediction accuracy was determined:
α = 0.85 & β = 0.67
The following plot was obtained:
Experimental data – red, Adjusted theoretical data - blue
For this new predicted data, the average error relative to the experimental data is 2.8m, or 3.33%.
Merging the initial theoretical calculations, with the experimentally obtained data and the
readjusted theoretical data yields:
Experimental data – red, Initial theoretical data – green, Adjusted theoretical data - blue
The initial purposes of this project were to find the useful thermodynamics equations that
describe the processes involved in firing a potato gun and relate the shooting parameters (cannon
elevation, projectile mass and shape, fuel volume) with the outcomes (muzzle velocity, range).
Through a mix of theoretical considerations and experimental based adjustments, those goals
were accomplished; therefore, we consider the project successful.
clc; clear;
for i=1:size(B)
if B(i)<0
xlabel('milliliters of fuel');
ylabel('muzzle velocity (m/s)');
grid on;
clc; clear;
xlabel('mass of projectile, grams');
ylabel('muzzle velocity (m/s)');
grid on;
clc; clear;
VX(1)=42; VY(1)=42;
X(1)=0; Y(1)=0; A(1)=pi/4; T(1)=0;
k=6.432/10000; m=0.054; g=9.81;
for i=2:1700
if Y(i)<0
plot(X,Y); grid on;
xlabel('Distance'); ylabel('Height'); title('Projectile Trajectory');
clc; clear;
for j=1:length(Vinit)
V(1)=Vinit(j); VX(1)=Vinit(j)*0.7071; VY(1)=VX(1);
X(1)=0; Y(1)=0; A(1)=pi/4; T(1)=0;
k=6.432/10000; m=0.054; g=9.81; dt=0.005;
for i=2:1700
if Y(i)<0
plot(Vol,R); grid on;
xlabel('Fuel (mL)'); ylabel('Range (m)'); title('Range vs. Fuel');
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