1

*The following is the final written report regarding my engineering team’s air motor design senior project. We were asked to come up with and develop our own design,

conduct our own analysis, and machine each part ourselves. The final product was used to power a miniature go-kart.

Engine Design

Team Muscles

Lucas Gargano

Joe Mosley

Andrew Daehn

Steven Politowitz

Michael Steiger

Yichao Ou

2

Table of Contents

Introduction

Engine Concept

Engine Detailed Engineering and Development

Thermodynamic Energy and Fluid Flow Models

Strategy, Software Coding for Valve Control, Physical Circuitry

Engine Fabrication and Manufacturing

Mechanical and Valve Timing Testing

Testing Day Results

Team

Lessons Learned

Course Improvements

Appendices

List of Figures

List of Tables

List of Equations

3

Introduction

Beginning our engine design project, we made sure to identify each of our team

members’ strengths and assign them to each team accordingly. The team was broken up into

groups of analysis, electronics, and design/manufacturing. Andrew Daehn and Lucas Gargano

were assigned to design/manufacturing portion of the project. Steven Politowitz and Joseph R.

Mosley were assigned to the electronics section of the project, which left Michael Steiger and

Yichao Ou to the analysis team. No matter which sub-group they belonged to, each team

member was encouraged to aid the other skill teams in order to help balance the work load given

to everyone.

Finding a solid direction to go when designing our engine was the first step to a

successful project. Our engine design is a bit unconventional by layman’s terms. We designed it

to be a four cylinder engine with two pistons firing in each cylinder body. These pistons are also

programmed to be in phase when firing (meaning that they pump at the same time). This gives

our four cylinder design the function of a two cylinder design. We also have two fly wheels that

are used to support the crank shaft, which is used to transfer the linear motion of the pistons into

the angular motion we desire.

Our engine is predominantly made of aluminum, while our pistons are made of bronze

and our connecting rods are made of steel. Our materials were acquired in numerous ways. We

made several to trips to a local scrap yard to obtain much of the aluminum we used to machine

the cylinders as well as the support plates. We also ordered material from a few different

suppliers online. We were able to obtain bearings to support the main shaft of our design. The

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aluminum piping used to manufacture the drive shaft was also obtained through the online

suppliers.

Nearly over one hundred hours were spent in the machine shop manufacturing our

engine’s components. Even as we were very diligent when scheduling our manufacturing

appointments, the project took longer than expected and the time slots filled up fast as well so

finishing our engine took until nearly the last minute. The pre-test was pushed back from

Monday, May 19th to Wednesday, May 21st as apparently we were not the only group going

through this. Andrew Daehn’s father also works on campus and was able to give us some spare

time on the machines he watches over. Walter Green was also very helpful in the machining

process as we are all very new to the shop. The electronics team would have never been able to

successfully complete their assignments (on time anyway) had it not been for Joe West’s, and

each of the Jasons’, constant willingness to offer help and guidance. The weekly meetings that

began in May with Professor Luscher also proved extremely helpful as it gave us reassurance to

the track we were on as well as direct access to ask any question we may have had regarding the

project, as he was always willing to clarify and suggest the best way we could go about anything.

Without these people it would have been a very rocky road and we were lucky to have

instructors who were so easy to communicate with helping us.

The learning curve throughout the project was incredible. Every single one of us was

challenged to complete assignments involving things we had little to no experience with. With

very little Solid Works experience, minimal machining experience, and no electronics

experience; each team was challenged not only by their knowledge, but their ability to learn on

the fly, as well. This design project was the strongest real life experience we have had as

5

undergraduates as far as learning to work on your own and finding ways to complete something

when all the answers may not be lying there in front of you.

6

Engine Concept

When developing our engine design we decided to keep our main focus on power. We

designed the engine keeping in mind our limited experience using CAD programs and working

in the machine shop. Our design was meant to allow as much precision error as possible when

designing and machine as we anticipated a significant amount of setbacks. We also attempted to

keep our pieces and parts designed as simply as possible so as not to bring on something that

would be potentially too difficult for such an inexperienced group. This proved vital as our

machined parts nearly never were machined precisely where we wanted them, yet our engine

was able to accommodate to these mistakes and still operate.

We developed a four cylinder engine that worked like a two cylinder engine. It had two

cylinders on each side that fired in phase with each other, while each respective side fired a half

cycle out of phase. We designed this to maximize power and in this we succeeded as our max hp

output ended up being .12. This proved vital during the power testing as we had one of the

fastest engines with a run time of 11.83 seconds. Drawings of each and every part of of our

engine design can be found later on in the appendices.

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Engine Detailed Engineering and Development

Crankshaft Stiffness Analysis

Torsional Analysis:

For the torsional analysis, an arbitrary torque was applied to the crankshaft and its deflection was

measured so that a stiffness constant, torsional, could be found. This was done using a static analysis

within the simulation tool in Solidworks. Since our crankshaft is off center at a constant radius

from the center of the flywheel, the torque was assumed to result in only an applied shear force

to one end of the crankshaft. By doing this, it is assumed that the all of the crankshaft’s material

is at the constant radius of 1”. This is not actually true, since the crankshaft’s outer-most point is

at a radius of Ro=1inch+dcrankshaft

2=1.25 , and the crankshaft’s inner-most point is at a radius of

Ri=1 inch−dcrankshaft

2=.75 . In other words, it is assumed that the crankshaft’s diameter is small

compared to the radius at which it rotates. Because the smaller torsion in the rod is neglected, the

crankshaft’s deflection will be less in the simulation than in practice all other things being equal.

A smaller deflection for the same applied torque will give a larger stiffness value. Below are the

results of the Finite Element simulation for an arbitrary applied torque of 11.24 in-lb, causing a

50 N shear force at the end of the rod.

8

Study Results

Name Type Min Max

Stress1 VON: von Mises Stress 147204 N/m^2

Node: 409

3.61192e+007 N/m^2

Node: 851

9

Name Type Min Max

cshaft-Study 1-Stress-Stress1

Name Type Min Max

Displacement1 URES: Resultant

Displacement

0 mm

Node: 1

0.220161 mm

Node: 638

10

The maximum displacement of the free end was .2202 mm, or .00867 inches. This displacement

at a radius of 1” is equal to an angular displacement of .00867 radians. The stiffness can then be

found as

k torsional=τ

θdisplacement=11.24∈−lbs

.00867 rad=1296.4 ¿−lbs

rad

Bending Model and Analysis:

The bending model was done to determine the stiffness of the crankshaft under just the loads

from the piston. For this analysis, the force of the piston was assumed to be greatest at the

piston’s top dead center position. At this position, the line of the force goes directly through the

crankshaft as well as the crankshaft’s axis of rotation. The two pistons that are in phase on our

engine were grouped as one for simplicity since they are close together, 1” apart. The force was

placed across a 1” section of the crankshaft because the two connecting rods contact the

crankshaft across a 1” portion of it. An arbitrary force of 50 N was applied for the Solidworks

Simulation. The two ends of the crankshaft were fixed, and the maximum displacement was

found. Below are the results of the Finite Element Analysis.

Study Results

Name Type Min Max

Stress1 VON: von Mises Stress 4209.9 N/m^2

Node: 9755

4.83335e+006 N/m^2

Node: 5

11

Name Type Min Max

cshaft-Study 2-Stress-Stress1

Name Type Min Max

Displacement1 URES: Resultant

Displacement

0 mm

Node: 1

0.00196683 mm

Node: 9035

12

Name Type Min Max

cshaft-Study 2-Displacement-Displacement1

For the arbitrary applied force of 11.24 lbs, the maximum displacement was .001967 mm, or

7.743x10-5 inches. This results in a bending stiffness of

k bending=F piston

δ connectingrod= 11.24 lbs

7.743 x 10−5∈¿=145163 lbs¿ ¿

For a maximum piston force of 196 lbs (two pistons at 124.7 psi), this results in a deflection of

only .001”, which should not cause any problems like a phase differences between the two sets

of pistons. From this Finite Element Analysis, our current crankshaft should be stiff enough in

bending and torsion so that if we have any problems with our engine, we can safely assume that

crankshaft flexibility is not contributing to the problem.

Stress Analysis

Rod Axial Yielding: From the basic axial yielding equation with a 3/8” diameter rod,

σ cr=FA

= F

π (38 )

2

4

=27 ksi

Fcr=2982lbs

From a Free body diagram analysis, the maximum force at the bottom of the stroke is:

Fmax=120 psi∗( π d2

4 )=94.2lbs

Therefore, the factor of safety is very high (>10).

13

Rod hole tear out: Using the approximations from a rivet-plate tear out,

τ e=F s

2 xe t=94.2 lbs

2∗xe t, so xe t ≥ .00302

For a thickness t=1/4”, xe only needs to be greater than .012”, and our connecting rod will have

at least an eighth of an inch of material surrounding the pin.

Buckling: The equation for buckling was used assuming a pin to pin connection type. Our

connecting rod is 3/8” by 3.8” long. The equation for critical buckling load is:

Pcr=π 2 Et I

Le2 =

π2 10.3E6 psi( π ( 38 )

4

64 )3.82 =6834 lbs>94 lbs

Piston: The piston cylinder should easily be able to react to the maximum pressure of 120 psi.

Since the maximum yield strength of Aluminum is 27 ksi, the piston is easily capable of

supporting the maximum 120 psi load.

Piston Pin: The piston pin must be able to withstand double shear. The maximum force of 94 lbs

is split between its supporting ends. For a factor of safety greater than or equal to 4:

τ= F2 A

= 94 lbs

2( π4 (d pin )2)

= 27000√3FOS

The pin therefore must be at least .112”, rounding up gives a nominal diameter of 1/8”.

Bearings/Bushings: We initially tried bushings for supporting the radial load between the engine

supports and the crankshaft. Assuming the crankshaft’s forces are symmetrical means that the

total radial load on the bearing/bushing is equal to one of the two max piston forces since only

two pistons fire at one time. This means that one bearing/bushing will have to support a

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Fmax=94 lbs . Using a factor of Safety of 2, this increases to 188 lbs. Using the bushing design

criteria and a bearing thickness of ¼” and diameter of ¼”,

Pmax=188 lbs

.25∗.25} =1504ps ¿

This P_max is only slightly less than the maximum Pmax for a porous bronze bushing, 2 ksi. Since

the factor of safety is low, we decided bearings would be a safer option.

We assumed a reliability of 90%, an impact factor of 1.5 (Moderate impact), and a bearing life of

120,000 revolutions, corresponding to 10 hours at 200 rpm. From the bearing design criteria,

Pe=X dF r+Y d∗Fa=1∗188 lbs=188 lbs=Pse

Therefore, the required ball bearing factor calculation is:

[Cd ( .90 ) ]req=( Ld

K R (10 )6 )1a ( IF )∗Pe=(120000 rev

1∗106 )13∗1.5∗188 lbs=139.1lbs

Most of the light duty bearings we looked at online were rated at around 600N or 134.9 lbs,

meaning we would most likely need a medium duty radial single ball bearing. This is by far the

most critical failure point in the engine due to the high required loads on the engine.

Additionally, the loads are dynamic, that is, they vary rapidly from 0 to 139.1 lbs within one

stroke of the engine. This adds in an additional element that we must make sure is covered by

using a high enough factor of safety.

Crankshaft: Our crankshaft rod lies at a radius equal to 1.5”, and since it is fixed at both ends to

the rotating flywheels, the moment provided at the ends act to decrease the maximum bending

moment in the bar. For this reason, the worst case scenario for this bar is simple supports, so we

chose our analysis based on this. From the free body diagram analysis on the rods, it was

determined that the maximum moment occurs at the points of applied force. Assuming a rod

diameter of ½” and a length of 5”, the maximum moment is:

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Mmax=53∗F p=94 lbs ¿

σ max=McI

=(156.7∈−lbs )∗.25 )/(π*.5^4/64)=12769 psi 200000psi=Ult.Tensile strength of stee≪

τ max=FA

= 94 lbs

π∗. 52

64

=7664 psi≪115470 psi=τu , steel

Since we are using a steel rod as the crankshaft, these maximum stresses are well below the

limiting strength of steel in tension.

Fasteners: To attach the engine to the provided base, screws will be used. The only major force

acting on the screw or bolt will be a shearing force on the bolt due to the acceleration of the

pistons. Assuming the engine is moving quickly at 500 rpm, and there is only one bolt holding

the assembly in place, the shear force generated is:

τ=4 pistons (ρV ω2 R )Abolt

=4∗( 5.2 slugf t 3 )¿¿

Solving for the nominal bolt diameter, d:

d ≥ .0085

All of the bolts we use will be larger than that diameter, so even one of them will be able to

withstand the engine’s shear forces.

In conclusion, the critical design components will be the bearings, the tear out from the

connecting rod hole, and the piston pins to a lesser extent. All of these components currently

have a design factor of safety of around 4 or less, so care must be taken when selecting these

particular parts.

Strength of Materials

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Reciprocating engines in a crank-slider arrangement produce unbalanced forces due to

the inertia of the piston, crankshaft, and connecting rods. While it is difficult to completely

balance many engines, a properly sized balancing mass can reduce the unbalanced force

significantly. Our engine experienced a considerable amount of shaking or unbalanced force

during a preliminary test, so to reduce this, a basic engine balancing analysis was conducted.

The engine was assumed to be operating in a constant velocity reference frame. This

assumption is close enough because the maximum acceleration of our cart is probably going to

be small compared to the acceleration experienced by the piston during the engine’s operation.

The piston’s acceleration has a primary and secondary component given by:

a p=−Rω2(cos (θ )+ RL

cos (2θ ))with the cosθ term being the primary component and the cos (2θ) term being the secondary one.

The phenomenon of dynamically equivalent bodies was used to split the mass of the connecting

rod between the crankshaft and piston so that there were only two point masses, a rotating one to

represent the crankshaft and a reciprocating one to represent the piston. Since our crankshaft was

at a radius of 1”, its mass was assumed to be centered at that radius. The overall mass of the

crankshaft was divided by four so that each of the four pistons was assumed to be connected to

an equally partitioned piece of crankshaft. The Law of Cosines was used to find the magnitude of

the vectorally added horizontal piston unbalanced force with the radially directed centripetal

force from the crankshaft equivalent mass. The net force as a function of crank angle was found:

f net=√ f p2 + f c

2+2 f p f b cos (θ )

The net force as a function of crank angle is given in the polar plot below.

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5

10

15

20

25

30

210

60

240

90

270

120

300

150

330

180 0

Net Unbalanced Force Without Balancing Mass (lbs)

To balance the engine, a mass was placed opposite the location of the crankshaft. In order

to balance this with a rotating mass, the sum of a fraction, c, of the reciprocating piston mass and

all of the rotating mass was used.

B=(meq ,crank +c meq , piston ) r

The new net unbalanced force was computed from the same reciprocating piston force

with a modified rotating force that consisted of the difference between the old rotating force and

the new centripetal force due to the rotation of the balancing mass. The balance mass that

produced the lowest total unbalanced force was found to be .9073 lbs at a distance of 1” from the

center. This mass was found by varying the fraction of reciprocating mass balanced until the

smallest net maximum unbalanced force was found. This fraction, c, was found to be .58. The

new unbalanced force is graphed with the original unbalanced force below.

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5

10

15

20

25

30

210

60

240

90

270

120

300

150

330

180 0

Net Unbalanced Force With Balancing Mass (lbs)

Original Unbalanced Force

Unbalanced Force with Balancing Mass

From the graph, the maximum unbalanced force with the balancing mass added is less

than the minimum unbalanced force without balancing. This is clearly an improvement, but since

this is the best possible revolving balancing mass, the engine cannot be perfectly balanced with a

rotating balancing mass. To completely balance the engine, some type of reciprocating mass

would need to be added. By reducing this shaking force, the engine will experience less

vibration, which could possibly extend its life and prevent any bolts from loosening. This

balancing could also help the speed and pressure sensors, whose measurements could be altered

by excessive vibration.

Thermodynamic Energy and Fluid Flow Models

Thermodynamic methods were used to estimate the power and efficiency of our engine.

The analysis is based on an engine that rotates at a constant speed of 800 rpm with a bore of 1

inch and a bore-stroke ratio of 1:2. A clearance volume of .5 in3 was used; this gives an ample

clearance distance in between .5” and 1”. Initially, we assumed that there was no pressure drop

across the line and the valve and that the mass flow rate into the cylinder was infinite. Varying

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the valve closing position, the theoretical power and efficiency curves were calculated assuming

adiabatic conditions across the cylinder boundary. The power and efficiency graphs are shown

below.

10%

15%

20%

25%

30%

35%

40%

50%

55%

60%

65%

70%

75%

80%

85%

90%

95%

100%

0

500

1000

1500

2000

2500

Power vs. Valve Position

Valve Close Percentage

Pow

er (i

n-lb

/s)

0% 20% 40% 60% 80% 100% 120%0

10

20

30

40

50

60

Efficiency vs. Valve Position

Valve Close Percentage

Efficie

ncy

(%)

20

The maximum efficiency resulted from a 20% valve closing position, while the power

reached a maximum when the valves were opened for the entire downstroke.

In order to get a more realistic graph of valve position closure versus power and efficiency, the

work at each value of valve closure needed to be calculated. To do this, Matlab was used to

calculate the work, power, and efficiency at each valve closing position in increments of 1

percent in valve position.

Once the valve position was assumed, the absolute pressure was found at each point

before and after the valve was closed. In order to do this, many assumptions were made. First,

we assumed that the pressure in the tank was at its maximum of 124.7 psi absolute, and that this

pressure was the same as the pressure in right in front of the valve. In actuality, there is a

pressure drop due to friction and flow resistance in the tube that connects the valve to the tank.

There is also a pressure drop and resistance across the valve, which causes a limitation in the

mass flow rate through the valve. An experiment was done to determine this experimental mass

flow rate, and an equation for volumetric flow rate versus tank pressure was found. Other

assumptions made were that the specific heat of air was constant at constant pressure, the air in

the tank was at room temperature, and that the volume of air in the cylinder varied sinusoidally

with time at a frequency of 800 rpm. The initial conditions were assumed to be at STP with the

mass calculated using a volume equal to the clearance volume of .5 in3. For pressure ratios of

pcylinder

ptank≥ .528, the mass flow is not at its maximum choked flow and decays rapidly towards zero

with increasing cylinder pressure, so the flow function from my Turbomachinery book was used

to find the mass flow rate at these higher cylinder pressures. The valve was assumed to be an

isentropic nozzle. In order for the this calculated mass flow rate to be less than that found in the

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electronics valve experiment, the theoretical minimum area of the valve was calculated using the

experimental mass flow rate found.

The pressures were found by first finding the mass flow rate. If the pressure ratio was

greater than .528, the flow function was used to calculate a reduced mass flow rate; otherwise,

the mass flow rate was assumed to be at the maximum value found in the experiment. From the

mass flow rate, the new mass in the cylinder was found by multiplying the mass flow rate by a

predefined time step and adding this to the previous mass in the cylinder. Then, the next pressure

was found by using the ideal gas law, with the next temperature found by using an adiabatic

assumption. Since volume as a function of time is known, the next volume can be found.

pn+1=

mn+1 R(Tn( pn+1

pn )n−1n )

V n+ 1

After rearranging the recursive formula for pn+1, the next pressure was found to be

pn+1=(mn+1 RTn

V n+ 1)n 1pn

n−1

Once the valve was closed, the remaining volume change was assumed to be an adiabatic

expansion. The pressure and volume at the intake closing position were used as the initial

condition to find the constant for adiabatic expansion. The equation for pressure was then found

to be:

pn=pintake closing∗( V intakeclosing

V displacement+V cle arance)n

where n=1.4 for air.

Since the pressure and volume at each time were known, a pressure vs. volume graph could be

made.

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Below is one of these graphs for an intake valve closing position of 75%.

0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 2 2.2 2.410

15

20

25

30

35

40

Volume (in3)

Pre

ssur

e (p

si)

Pressure vs. Volume at 75% Valve Closure

The useful work was calculated by integrating the pressure vs. volume graphs. The

trapz() function in Matlab was used to approximate this integral using a time step of .0005

seconds. This gave a good approximation since 150 intervals were used in the calculation across

a time range from 0 to .075 seconds. The work done by the atmospheric air on the other side of

the piston was subtracted from this integral so the useful work could be found. On the return part

of the stroke, the pressure on the cylinder was assumed to be atmospheric both above and below

the cylinder, meaning that no net forces acted on it and therefore no work was gained or lost.

This is a strong assumption since the same mass flow limitations for the intake valves will apply

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to the exhaust valves, so further analysis will be needed for this part of the stroke. Work will be

lost on the exhaust stroke, but because of the valve resistance, the pressure at top dead center will

be higher than atmospheric pressure and that will assist the expansion part of the next cycle so

that more work will be generated on the downstroke. It will need to be determined if this will

result in a net work gain or loss.

With the work for one cycle known, the power is computed by multiplying the work by the

number of cycles per second calculated from the speed of the engine. The efficiency was

computed by dividing the cycle work by the energy required to pressurize the tank. Power and

efficiency versus valve closing position graphs are shown below.

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1-200

-100

0

100

200

300

400

500

Valve Closing Position (Fraction of Displacement Volume)

Pow

er (i

n-lb

/s)

Power vs. Valve Closing Position

24

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1-10

-8

-6

-4

-2

0

2

4

6

8

10

Valve Closing Position (Fraction of Displacement Volume)

Effi

cien

cy (%

)

Efficiency vs. Valve Closing Position

The maximum efficiency at a valve position of .26 was 9.765%, while the maximum

power at a valve position of 1 was 429.15 in-lb/s.

The work produced for each cycle was strongly dependent on the initial pressure in the

cylinder at top dead center. For example, for an initial pressure of 124.7 psi, the max power was

1001 in-lb/s and max efficiency was 68.1%. Assuming that the pressure was atmospheric at this

point was the reason that the power and efficiency seemed much lower than those from the

infinite mass flow rate assumption. A higher initial pressure condition should give more realistic

power and efficiency values.

25

Engine Fabrication and Manufacturing

In order to abide by the DFMA techniques for low cost manufacturing we stringently

used material found at local scrap yards and designed for the least amount of machining as

possible. I believe our design used very little machining compared to other teams. We designed

for maximum adjustability to reduce the need to re-machine parts that were slightly out of

specifications.

The manufacturing plan was developed in order to expedite our machining so that testing

could be done with plenty of time to correct issues in that area. We specifically avoided all

designs that required a lot of precise machining, because our team lacked in machining expertise.

Specifically, we added an adjustment bar at the top of our engine that could expand our

frames, we added slots in the cylinder holder frame so that we could move the cylinder side to

side, clamps that could quickly be loosened so that our cylinders could move back and forth and

rod ends that could be adjusted to set our exact clearance volume.

Our cylinder to piston sealing was accomplished by setting a very precise tolerance of

0.001 between the cylinder wall and piston, and used viscous oil to help seal.

We selected a threaded rod as a crankshaft so that we could fix the crankshaft to the fly

wheels with nuts. The crankshaft is just a straight bar to reduce complex machining.

A bill of materials showing the volume of each stock purchased as well as the volume

removed of each piece of stock can be seen below. We used this to estimate the cost of the

materials we would need and just how much of this stock we would end up having to remove and

wasting.

26

Part Name Quantity Volume of Stock (in^3) Volume of Finished Piece (in^3)

Fly Wheels 2 19.24 10.394Base Plate 1 44 43.65Shaft Holder 2 19 14.23488Piston 4 7.069 3.8384Crank Shaft 1 0.567 0.442Crank Arm 4 1.787 1.54Cylinder 2 76.97 64.4End Cap 2 9.621 9.32Holder-Shaft 2 6.185 5.74Stand 2 63 19.68

Volume Removed Cost of Stock/piece ($/lb) Total Cost of Stock

($)# of faces to be

surfaced

8.846 2 3.77104 2

0.35 2 8.624 1

4.76512 2 3.724 3

3.2306 2 1.385524 2

0.125 2 0.111132 2

0.247 2 0.350252 2

12.57 2 15.08612 2

0.301 2 1.885716 1

0.445 2 1.21226 3

43.32 2 12.348 1

48.498044

27

Mechanical and Valve Timing Testing

Mechanical Testing:

After the engine was completely assembled with the valves and electronics, we tested the

motion of the engine with a basic code. We ran the engine at a low starting pressure with a

medium stroke (20 psi, 50% stroke). The engine made full revolutions; however, it was clear that

binding was slowing down the engine motion at certain locations of revolution. In addition, one

of our frame pieces was moving excessively. The following changes were made to alleviate the

problems: 1) Added an adjustable stabilizer bar connecting the two frame pieces; 2) Loosened

the crankshaft nuts to allow just enough play between the crankshafts and fly wheels. The

adjustable stabilizer bar stabilized our frame pieces and made it possible to adjust if other

problems occurred. We found that imprecise machining made our flywheels wobble about

center. By loosening the crankshaft nuts, we were able to allow the crankshaft to move freely in

the flywheels.

28

The first attempt of the 2-hour burn revealed that we had a severe balancing issue. After a

few minutes into the first attempt it was clear that our engine was not going to survive the whole

duration of the burn-in due to excessive shaking and vibrating. We alleviated this issue by adding

counterweights on the flywheels, opposite the crankshaft. Our analysis team suggested that we

need to add enough weight (approximately 20 ounces) to off-set the weight of our crankshaft and

the components attached to our crankshaft. The weight of choice was lead fishing sinkers that we

attached to the flywheels with metal putty. This addition proved vital in limiting the vibration

and shaking, and also improved our performance in other areas. In just our first attempt after

adding the counterweights we were able to complete the 2 hour burn-in seamlessly using a

simple medium-stroke code.

The dynamometer performance during the pretest provided important information about

our engine. We achieved a maximum power value of 0.059 hp. This was in middle of the pack

compared to other team’s engines. Since we designed our engine specifically for power, we were

not satisfied with that value.

In the process to make the best power-run code for the competition, we relied on the

dynamometer to find where our greatest power occurs by varying our stroke and valve delay

time. It was during this process where we found the values that we were originally expecting.

The raw data from these experiments can be found in the appendix section of this report.

Summarizing, we found our highest power at our expected engine speed (between 400-500 rpm)

to be around 0.075 hp. And our overall maximum power at start-up to be 0.12 hp.

Our analysis team found the gear ratios for each gear. We then determined that the

farthest distance traveled for one revolution of our engine was 5th gear. We assumed that we had

29

enough torque in our engine to produce the maximum rpm given the required start-up torque of

5th gear. To verify our assumption we tested our power code at every gear. Below is the time to

30 feet for each gear using our power code.

Valve Timing Testing:

In order to maximize our efficiency code we needed to find the load torque at our expected

engine speed. The first part of this process included running the engine on the cart with our

default code values in order to get the approximate engine speed. We then used the dynamometer

to match our engine speed to the power/torque. We then adjusted our valve stroke and delay until

we maximized our power and torque values. Our raw data can be found in the appendix. We then

put the engine on the cart and tested our dynamometer results.

400 425 450 475 500 525 5500.04

0.0450.05

0.0550.06

0.0650.07

0.0750.08

Effect of Valve Delay, Stroke 60%

Run 2 10.75 delayRun 2 10.5Run 1 10.25

Engine Speed, RPM

Engi

ne P

ower

, HP

30

410 420 430 440 450 460 470 480 490 5000

0.0050.01

0.0150.02

0.0250.03

0.0350.04

0.0450.05

Effect of Changing Stroke, Delay 10.25

60 % 10.2565% 10.2570% 10.25

Engine Speed, RPM

Engi

ne P

ower

, HP

Testing Day Results

Our engine functioned as well as our previous test runs predicted. Our efficiency matched

our expectations going approximately 113 feet, although it needed a few shoves to keep the

revolutions going. The power run went better than previous runs predicted by nearly a second.

Our power run went for 11.8 seconds when our previous best was 12.8 seconds. Given

additional time we would have decreased our stroke. We would have given up power but gained

in efficiency because of the amount of air we were leaking per revolution.

In conclusion of the testing day results, we were in the top four of the power run and

bottom portion of the efficiency. This is right in line with our initial team commitment to power.

We knew we would do well in power the portion of the testing in sacrifice of our efficiency run.

In terms of overall power, we recorded a value above 0.12 hp, which is the most that any team

recorded on the dynamometer.

31

Testing Day Results

Overall, the final test was a success. We achieved our best time to 30 feet in 11.83

seconds, and the cart traveled 113 feet on one 7 gallon tank of compressed air. We were able to

breakdown the engine in 2 minutes and 24 seconds. In the morning prior to the test, we made

some last minute adjustments to the valve timing and were able to run the engine noticeably

faster on one of the lab tables. The improvement in time can be attributed to the last minute

adjustments in valve position. For the efficiency run, we needed three pushes because we used a

different regulator. The regulator we had used in practice needed 11 turns to get down to a

pressure of 15 to 20 psi, which was insufficient for the one we used on the day of practice. The

engine never really had enough power at these low of pressures to shift consistently, so this was

why we needed 3 pushes in between some of our shifts. If we had some more time, we would

have tested some more on the dynamometer and worked to get the electronics more consistent.

Team

Our team consisted of six members: Joe Mosley, Lucas Gargano, Michael Steiger,

Steven Politowitz, Yichao Ou, and Andrew Daehn. We were then split up into three teams of

two according to the different aspects of the engine design project. These sub-teams were

electronics, design/manufacturing, and analysis. Joe Mosley and Steven Politowitz were

assigned to the electronics team as they had worked on this section previously throughout the

Rube Goldberg Apparatus project previously in the school year. Lucas Gargano and Andrew

Daehn were assigned to the design/manufacture team as they had both taken the ME 2900 course

32

and seemed to have the most experience in the machine shop. Yichao Ou and Michael Steiger

were assigned to the analysis team as they both had the strongest understanding of the theoretical

aspects of engine design.

Communication was paramount for effectively completing this project. Overall, our team

did a very good job of this. Whether it was through e-mail or group text messages we were

always able to get ahold of each other nearly immediately. Everyone in our group got along

extremely well, which enabled us to work long hours beside each other while enjoying one

another’s company. Joe Mosley was definitely the most important member as he unassumingly

became the team leader with his work ethic and overall knowledge. It was important having a

strong leader who had a good understanding of just about every aspect of the project and was

willing to spend as much time was necessary to complete the project no matter what. As a team

we worked very efficiently as everybody was willing and able to complete each assignment

asked of them. We all chipped in equally as far as expenses went and depending on who was

free when we were all willing to help work on whatever needed a little bit extra man power.

We believe we split up our team as best as we could have. Each member’s strengths

were utilized in the sub-team they were put into. Completing assignments and hitting deadlines

were not an issue for us. The communication and overall positive attitude of our group were

some of our best strengths and these may be seen as some of the most important qualities you

can have.

33

Lessons Learned

Design

Engineering design is a very precise, carefully planned process. No manufacturing

advancements are meant to be made until the entire structure has been carefully designed and

verified to work. This does not only mean making sure the pieces fit, this means anticipating

motions and forces that will act on the structure and verifying they will be safe. We conducted

thermodynamic analysis on our structure in order to make sure it would not overheat. We

conducted strength of materials calculations on our structure to make sure it would not fracture.

All this was done before we ever began the physical construction. This is not only a safety

precaution but a financial precaution as well seeing as if something were to go wrong with your

design post-construction you would have to repurchase all the materials used for your engine.

The designing process is easily the most important aspect of building any product.

If we could go back through the design process again what we would do differently is

conduct our analysis as we went, rather than all at once. Often times we would go through and

do a large amount of the design process without stopping to analyze each step. This can lead to

finding out something you did early on would not work with your later decisions which means

you would have to go back through everything once again. Doing your analysis as you go

provides confirmation and reassurance with each design choice you make.

Professional Skills

We became much better at communicating with each other and getting a point across the

right way. We understood we are all in this together and there was no room for anyone to act

like they were a more important member than anyone else. Even as the leader, Joe Mosley

34

would never act like he deserved any more credit than anyone else in the group no matter how

many extra hours he would spend or how much more he understood regarding the project. This

led to respect throughout the group for each member therefore there was no gap of power

between anyone allowing each of us to speak our minds freely. Our closeness as a team might

have been our biggest strength as it would give us all that extra energy and motivation we needed

to spend the tireless hours, as we all felt like we were equally responsible for the final outcome.

We also learned better ways to communicate with those above us; instructors, lab

supervisors, etc. They are the ones who can help the most, so we learned being honest with them

and being as clear as possible was the smartest route. Early on during one of our first oral

presentations we became defensive when Professor Luscher simply asked us a question

regarding a clarification on our Rube Goldberg Apparatus. Our response was rude and useless as

it helped no one and we learned this very quickly. Since then we have learned to respect and

trust our instructors and this led to a much clearer path to success as we were at an understanding

with those who would be critiquing our work. This may be one of the most important lessons we

learned throughout our entire college careers as we will always have someone above us

overseeing our work that it will always be best to respect and get on the same page with.

Manufacturing

We found out how precise of a process the actual manufacturing of our design really was.

We were not permitted to begin machining until we had physical drawings of the exact

dimensions and geometries we would need. The attentiveness that machining requires was also a

major surprise as merely one pass too many could mean disaster for your part and the necessity

35

of starting over on it beginning with repurchasing the material (which, if it wasn’t available

locally, could take days [and we didn’t have days]).

The time it took to manufacture was also a big surprise to most of us. We spent nearly

100 hours machining our engine before we could begin assembling. This amount of time was

hugely more significant then we estimated it to be in the planning phase of our project. Although

we still feel like we did a good job of getting into the shop as early and often as we could, we

would have tried even harder to start earlier during the finals week as that would allow for a

more appropriate time table.

We learned to work a variety of machines through the process of this project. We spent

significant time on the lathe, rotary table, and the mill. We learned how important it is to follow

the rules when operating this machinery as the slightest mistake could be detrimental to the

machine, or even worse, your health. This was no longer concept, nor theoretical, we were in

there physically manufacturing the parts of the engine we designed for real life use.

Programming/Electronics

36

Course Improvements

We felt this course was run very well. The toughest part is not having the proper training

for all the aspects of the engine design going into it and having to learn on the fly with a very

limited amount of time. Although, the instructors were all extremely helpful and if you were

having any sort of trouble going to them was a very good idea as they were always willing and

able to help. They were also all very knowledgeable therefore the help they gave was extremely

strong and efficient. The assignments were at a good level of difficulty as they gave you a good

idea of what they were hoping to teach you without spending too much of your time.

I would say the most improvement regarding this course would have to do with the

preparation prior. Many things we are asked to do and accomplish throughout this course are

things that we have had little to no training on. The recent addition of the Mechanical

Engineering 2900 course is a step in the right direction, but we actually had members on our

team who were not fortunate enough to be a part of that class. Between the Solid Works aspect,

the coding to run the engine, and the manufacturing itself; these were all things each of us were

very new to. This made it especially hard during the four week period we had to get everything

together during the Maymester. Luckily all the instructors were as helpful as could be whenever

we had a question.

Appendices

Budget

Expenses

arduino chip 33.63 Joe Joe Total

212.06

metal 35.08 Lucas Lucas Total

110.41

37

bearings 15 Lucasaluminmum 15.75 Joealuminum 29.24 LucasMiscellaneous` 10 Lucaspiston rod ends 24 JoeAlum Rod 31.27 Joehardware 33 Joecylinder stock 10.75 LucasBearings x2 24.75 JoeBar Stock alum 15.49 JoeOil, tool, gasket 10.34 Lucasbrass pins, brass pipe 34.17 Joe

Total 322.47Total for Each 53.745

Engine Balance Code

% All units are english system

clear all

clc

R=1/12; % Radius of rotation (half stroke)

omega=800*2*pi/60; % Engine angular speed (rad/s)

theta=0:.01:2*pi;

m_p=.566/32.2 % mass of piston (pounds to slugs)

m_r=.526/32.2 % mass of connecting rod

CG_r= 3.8/12; % Center of gravity measured from piston side in feet

l_r=5/12; % Length of rod in feet

m_c2=.106/32.2; % mass of rotating portion of crankshaft

% whose mass is centered at the crank radius

m_p1=m_r*(l_r-CG_r)/l_r % Equivalent mass added to piston from con. rod

38

m_c1=m_r-m_p1 % Equiv. mass added to crank from connecting rod

for i=1:100

c=.01*i;

b=R; % balancing mass distance

B=((m_c1+m_c2)+c*(m_p+m_p1))*R/b; % balancing mass

f_c=(m_c1+m_c2)*R*omega^2

f_b=B*omega^2*b

f_cb=f_c-f_b

f_p=(m_p+m_p1)*R*omega^2*(cos(theta)+R/l_r*cos(2*theta));

for k=1:length(theta)

f_net(i,k)=sqrt(f_p (k)^2+f_cb.^2+2*f_p(k)*f_cb.*cos(theta(k)));

end

fmax(i)=max(f_net(i,:));

end

fbest=min(fmax);

c=find(fmax==fbest)*.01

b=R; % balancing mass distance

B=((m_c1+m_c2)+c*(m_p+m_p1))*R/b % balancing mass

f_c=(m_c1+m_c2)*R*omega^2

f_b=B*omega^2*b

39

f_cb=f_c-f_b

f_p=(m_p+m_p1)*R*omega^2*(cos(theta)+R/l_r*cos(2*theta));

f_netbest=sqrt(f_p.^2+f_cb^2+2*f_p*f_cb.*cos(theta));

polar(theta,f_netbest); hold on

Thermo Code

clc

clear all

p_ 0=40:10:(110+14.7); % pressure in the tank

c_p=2241.5; % in-lb/(lbm degR)

T_ 0=529.67; %70 F in Rankine

T(1:100,1)=T_ 0; % Starting temperature

rpm=800; % speed of engine

f=rpm/60; % Frequency (Hz)

R=643.5; % in-lb/(lb,m deg R) for air

d_ 0=1.352*10^(-6); % Not needed?

V_c=.5; % cubic inches

V_D=1.571; % displacement volume

t=0:.0005:.075; % Time from 0 to period of one cycle

dt=t(2)-t(1); % Time step

V=V_c+V_D/2+V_D/2*sin(2*pi*f*t-pi/2); % Volume as function of time

40

% for h=1:4

for j=1:length(p_ 0)

p(1:100,1)=p_ 0(j)/4; % Starting pressure at TDC

m(1:100,1)=p(j)*V_c/R/T_0; % Initial mass in cylinder at TDC

% m_dotmax below calculated from electronics assignment for flow rate

m_dotmax=((.0311*p_ 0(j)^2-6.0719*p_ 0(j)+462.46)*1000/10^3/2.54^3)*p_ 0(j)/R/T_0;

% Area of valve opening below; found so that the mass flow rate

% calculated using the flow function does not exceed the mass flow rate above.

%The area is calculated at the choked condition so F=1.281

% Equations from my turbomachinery book - isentropic nozzle assumption

A_t=m_dotmax/1.281/p_0(j)*(c_p*T_ 0)^(1/2); % in^2

x=0:.01:.99; % Range of valve positions

for k=1:length(x) % For all values of valve position

i=1;

while (V(i)-V_c)<x(k)*V_D % while the valve is open

if p(i)>.528*p_ 0(j) % If back pressure (cyl. pres.) is greater than

% the pressure for choked flow

% Mach number

M(k,i)=(2/(1.4-1)*((p_ 0(j)/p(k,i))^((1.4-1)/1.4)-1))^(1/2);

% Flow function

F(k,i)=1.4*M (k,i)/.4^(1/2)*(1+.4/2*M (k,i)^2)^(-(1.4+1)/.8);

% Use flow function to find mass flow rate

41

m_dot(k,i)=F(k,i)*p_ 0(j)*A_t/(c_p*T_ 0)^(1/2);

else

m_dot(k,i)=m_dotmax; % Choked mass flow rate if cylinder pressure

% is lower than .528*tank pressure

end

m(k,i+1)=m_dot(k,i)*dt+m(k,i); % Find the new mass in cylinder

% after time step

% pressure in cylinder calculated from previous pressure, ideal

% gas law, and adiabatic assumption to find the new temperature

p(k,i+1)=m (k,i+1)^1.4*R^1.4*T (k,i)^1.4/(p (k,i)^.4*V (i)^1.4);

if p(k,i+1)>p_ 0(j) % If mass flow rate is capable of raising the cylinder

% pressure above the tank pressure, just use the tank pressure

% as the next pressure and calculate the mass based on that

p(k,i+1)=p_ 0(j);

T(k,i+1)=(T(k,i)*(p (k,i+1)/p(k,i))^(.4/1.4));

m(k,i+1)=p(k,i+1)*V (i+1)/R/T(k,i+1);

else

T(k,i+1)=(T(k,i)*(p (k,i+1)/p(k,i))^(.4/1.4));

end

i=i+1;

42

end

while V(i)<max(V) % Pressure after valve closes assuming adiabatic expansion

p(k,i+1)=p(k,i)*((V(i)+V_c)/(V(i+1)+V_c))^1.4;

i=i+1;

end

W(j,k)=trapz(V(1:length(p(k,:))),p(k,:))-14.7*V_D; % Work in in-lb

P(k)=W(j,k)*f; % Power in in-lb/s

N(j,k)=W (j,k)/(p_ 0(j)*(x(k)*V_D+V_c)*log(p_ 0(j)/14.7)-14.7*V_D)*100; % Efficiency

fillenergy(j,k)=(p_ 0(j)*(x(k)*V_D+V_c)*log(p_ 0(j)/14.7)-14.7*V_D)*100;

% if N(k)>1 % error correcting

% N(k)=0;

% end

end

maxefficiency(j)=max(N(j,:))

efficiencyposition(j)=find (N==max(N(j,:)))/j

maxpower(j)=max(P);

powerposition(j)=find(P==max(P))*.01;

end

% end

plot(p_ 0,efficiencyposition,'r'); hold on %,p_ 0,efficiencyposition(:,2),p_

0,efficiencyposition(:,3),p_ 0,efficiencyposition(:,4))

43

title('Eff. pos. vs. tank pressure')

% legend('Full TDC Pressure','Half TDC Pressure','1/3 TDC Pressure','1/4 TDC Presure')

Part Drawings

44

45

46

47

48

49

50

51

52

53

54

55

56

57

58

59

Dynamometer Raw Data

85psi 10.75Rpm Power Run 2 Rpm Power 75 psi 10

422 0.073 419 0.068 Rpm Power Run 2 Rpm Power440 0.058 444 0.068 420 0.064 430 0.05459 0.06 460 0.061 437 0.054 447 0.051473 0.055 474 0.055 454 0.056 460 0.046

500 0.054

85psi 10.5Rpm Power Run 2 Rpm Power 75 psi 9.5

407 0.073 420 0.07 Rpm Power Run 2 Rpm Power430 0.066 444 0.073 412 0.049 420 0.058451 0.07 463 0.068 428 0.047 435 0.049468 0.061 480 0.062 444 0.054 449 0.046495 0.054 495 0.059 458 0.046 462 0.043519 0.053 500 0.056

85psi 10.25Rpm Power Run 2 Rpm Power 75 psi 9

412 0.075 400 0.062 Rpm Power Run 2 Rpm Power435 0.067 424 0.059 412 0.054454 0.064 440 0.058 429 0.049470 0.055 460 0.056 445 0.051481 0.046 470 0.051 458 0.047493 0.045 484 0.042

75 9.75Rpm Power Run 2 Rpm Power 75 psi 10.25

410 0.058 Rpm Power Run 2 Rpm Power427 0.051 400 0.058 416 0.052441 0.047 418 0.05 434 0.053456 0.049 435 0.052 449 0.05468 0.042 450 0.052 462 0.046

465 0.048 472 0.036476 0.042

75 10.25 485 0.036Rpm Power Run 2 Rpm Power

418 0.057 420 0.05436 0.056 437 0.048450 0.054 450 0.048465 0.044 463 0.043476 0.042 473 0.038

483 0.038