PROBLEM NO. 5: “CAR”Build a model car powered by an engine using an elastic air-filled toy balloon as the energy source. Determine how the distance travelled by the car depends on relevant parameters and maximize the efficiency of the car.

OVERVIEW balloon observations

piston inflation/deflation model energy input and output rubber stretching losses

construction observed motion parameters

initial conditions (volume, pressure) jets, nozzles maximised travelled distance and efficiency

conclusion

PISTON MODEL ~ ENERGY piston contains all air later used for

inflation/deflation slowly pushed – balloon inflates released piston – balloon deflates

at every moment ideal gas law equation states:

V20 = Vpiston

V2f = 0V1fV10

p0 = patmpf

𝑝0(𝑉 ¿¿10+𝑉 20)=𝑛𝑅𝑇=𝑐𝑜𝑛𝑠𝑡 ¿

𝑝 (𝑉 ¿¿1+𝑉 2)=𝑛𝑅𝑇 ¿

𝑝 𝑓𝑉 1 𝑓=𝑛𝑅𝑇

𝑑𝑊=− (𝑝−𝑝0 )𝑑𝑉 2

initial state final state

integrating initial to final state

𝑅−𝑔𝑎𝑠𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡=8.314 𝐽𝑚𝑜𝑙𝐾

PISTON MODEL ~ ENERGY

maximal available energy deflation work

same equations valid

V1f V10

p0 = patmpf

𝑝0(𝑉 ¿¿10+𝑉 20)=𝑛𝑅𝑇=𝑐𝑜𝑛𝑠𝑡 ¿𝑝 𝑓𝑉 1 𝑓=𝑛𝑅𝑇

varies for inflation/deflation – determined experimentally

𝑊=∫𝑝0

𝑝 𝑓

𝑝𝑑𝑉 1+𝑝𝑓 𝑉 𝑓 ¿known conditions

V2f = 0 V2f =

PRESSURE / VOLUME RELATION INFLATION AND DEFLATION pressure

digital pressure sensor volume

photos taken at known pressureballoon edge coordinates

balloon shape determinedballoon radius/height function

numerically integratedshape rotation – body volume

height

radius

TYPICAL PRESSURE/VOLUME RELATION

deflation curve is under the inflation curve

energy partly lost due to rubber deformations

volume [cm3]

0 1000 2000 3000 4000

over

pres

sure

[kP

a]

0

2

4

6

8

10

inflationdeflation

BALOON STRETCHING – THEORETICAL CURVE pressure (force acting on a surfaca) balloon elastic energy

*

work needed to increase radius from to r+dr under pressure difference

surfaceforce

*Y. Levin, F. L. da Silveira, Two rubber balloons: Phase diagram of air transfer, PHYS. REV. E (2004)

relative extension r/r0

BALLOON STRETCHING

obtained curve has the expected maximum due to surface/force

relation

explains and confirms the obtained shape of the overpressure/strain curve

∆ 𝑃=𝛼 ( 1𝜆− 1𝜆7 )−𝑠𝑖𝑛𝑔𝑙𝑒𝑝𝑎𝑟𝑎𝑚𝑒𝑡𝑒𝑟 (𝛼𝑟𝑢𝑏𝑏𝑒𝑟 𝑝𝑟𝑜𝑝𝑒𝑟𝑡𝑦 ) 𝑓𝑖𝑡

relative extension r/r0

0,5 1,0 1,5 2,0 2,5 3,0 3,5

over

pres

sure

[kP

a]

0

1

2

3

4

5

6

INITIAL VOLUME surface under the curve enlarges with initial volume

determining part of total input and output energy efficiency is determined by the deflation and inflation

surfaces under the curve ratio between (initial + final conditions) + surface under the curve

volume [cm3]

0 1000 2000 3000 4000 5000

over

pres

sure

[kP

a]

0

1

2

3

4

5

6

inflationV1V2V3

initial volume [m3]

0,000 0,001 0,002 0,003 0,004 0,005

defla

tion/

infla

tion

effic

ienc

y

0,6

0,7

0,8

0,9

1,0

INITIAL VOLUME

𝜂=𝐸𝑑𝐸𝑖

initial volume [m3]

0,000 0,001 0,002 0,003 0,004 0,005

inpu

t ene

rgy

[J]

0

2

4

6

8

10

12

14

input energy inflation

inflation/deflation varies rubber stretching

CONSTRUCTION

cart /styrofoam

plastic tubeballoon

• plastic cart (lenght ~ 12cm)• styrofoam• plastic tube (d = 2,9 cm)• metal jets / cardboard

noozles• car mass ~120g

OBSERVED MOTION - SETUP

rails• car moving straight• position determination• lenght 5 m

video • camera 120 fps• placed 10 m from the rails

analysis time / distance coordinates

car

rails – with distance scale

OBSERVED MOTION

distance/time graph obtained from video coordinates program time-distance graph:

5 data points frame – cubic or linear curve fitted derived: one point in the v-t graph moves one point to the next frame

time [s]0 1 2 3 4 5

velo

city

[m/s

]0,0

0,2

0,4

0,6

0,8

1,0

1,2

1,4vmax

deflation accelerated motion

decceleration

time [s]0 1 2 3 4 5

dist

ance

[m]

0

1

2

3

4

5

TRAVELLED DISTANCE travelled distance:

acceleration path vmax time distance s0 determined from the video

decceleration a determined from the graph linear fit coefficient

decceleration path

time [s]0 1 2 3 4 5

velo

city

[m/s

]

0,0

0,2

0,4

0,6

0,8

1,0

1,2

1,4

BASIC WORKING PRINCIPLE input energy is used for inflation deflation transforms it into kinetic energy of the air molecules

losses due to resistance to movement (mutual collisions) momentum conservation

simplified model: air behavior mass inside the balloon velocity according to Bernoulli's principle (valid for turbulent

flows)

vcar

vair

Ffr

𝑑(𝑚¿¿𝑎𝑣𝑎)−𝐹 𝑓 𝑟 𝑑𝑡=𝑚𝑐𝑑𝑣𝑐 ¿

BASIC WORKING PRINCIPLE simplified model:

friction force main friction force source is the air drag friction force duration - flow rate

maximum car velocity vc initial volume relation obtained

vcar

vair

Ffrvcar

vair

Ffr

𝑚𝑎𝑣𝑎−𝐹 𝑓 𝑟 𝑡=𝑚𝑐𝒗𝒄

INITIAL VOLUME simplified model basic working principle fit

main friction force source is air drag

as the initial volume increases the air drag force gains significance reason of non-linear

progression simplification

𝑚𝑎𝑣𝑎−𝐹 𝑓 𝑟 𝑡=𝑚𝑐𝑣𝑐

initial volume [L]

0 1 2 3 4 5

max

imum

vel

ocity

[m/s

]

0,4

0,5

0,6

0,7

0,8

0,9

MAXIMISATION – VMAX IMPORTANCE

motion consists of acceleration and decceleration

acceleration to vmax is determined by: initial conditions

volume/pressure jets, noozles travelled distance found from

the video decceleration – same for all

parameters friction

air wheels/rails/shaft

energy input and ouput ratio energy input

work needed to inflate the balloon

piston model energy output

vmax car motion observation

travelled distance efficiency

PARAMETERS initial volume

total input energy rubber stretching losses

jet diameter volumetric flow of air Q

Reynolds number 105 tubulent flow

exit velocity deflation time

nozzle directs the flow

INITIAL VOLUME initial volume! – pressure total input energy and deflation energy

critical sizes min

needed to overcome static friction - start car movement

max radius ~ car height the balloon must not touch the floor

range ~ 0.15 dm3 - 4.5 dm3

deflated balloon ~ 0.075 dm3

air drag friction force

initial volume [m3]0,000 0,001 0,002 0,003 0,004 0,005

trave

lled

dist

ance

[m]

0

2

4

6

8

10

12

INITIAL VOLUME RESULTS

efficiency: output/input energy small

losses on the air kinetic energy maximisation smallest balloon

𝜂=𝑚𝑣𝑚𝑎𝑥 22𝐸𝑖

initial volume [m3]

0,000 0,001 0,002 0,003 0,004 0,005

effic

ienc

y [%

]

-1

0

1

2

3

4

5

travelled distance:

maximisation largest balloon

different diameters d

jet diameter changes how well the air is directed time needed for deflation

air drag duration amount of resistance the air endures in mutual collisions Reynolds number , kynematic viscosity

turbulent flow Re>4000, 105

JETS - CIRCULAR TUBE OPENINGS

d1 d2

deflation time

RESULTS (V~4DM3) different vmax for a certain diameter

both travelled distance and efficiency depend on vmax

2

max distance 47.3 m max efficiency 2.7%

best jet diameter 1.2cm-1.4cmjet r2 [cm2]

0 1 2 3 4 5

trave

lled

dist

ance

[m]

0

10

20

30

40

50

jet r2 [cm2]0 1 2 3 4 5

effic

ienc

y [%

]

0,0

0,5

1,0

1,5

2,0

2,5

3,0

NOZZLES – CONE SHAPED

different nozzles change• angle – how well the air is directed• tube lenght – amount of losses due to friction

6 cones – different angles and thus lenghts (7.5°- 30°)

effective velocity – horizontal component of rapid molecule movements

𝜷

l l

RESULTS (V~3DM3) D=1CM

max distance 20.5 m max efficiency 1.5%

best cone angle 15°-20°cone angle [°]

5 10 15 20 25 30 35

effic

ienc

y [%

]

0,0

0,2

0,4

0,6

0,8

1,0

1,2

1,4

1,6

cone angle [°]5 10 15 20 25 30 35

trave

lled

dist

ance

[m]

0

5

10

15

20

25

different vmax for a certain diameter both travelled distance and efficiency depend on

vmax2

CONCLUSION

initial V =4.5dm3 max optimal:

jet diameter1.2cm nozzle angle 20°

DISTANCE 70m

initial V=1.5dm3 min optimal:

jet diameter 1.2cm nozzle angle 20°

EFFICIENCY 6,4%

travelled distance efficiency

piston inflation/deflation model – energy evaluation stretching losses – theoretical and experimental curve basic working priciple explained found maximum conditions for jets and noozles

Thank You!

PISTON MODEL ~ ENERGYV20 = Vpiston

V2f = 0V1fV10

p0 = patmpf

𝑝0(𝑉 ¿¿10+𝑉 20)=𝑛𝑅𝑇=𝑐𝑜𝑛𝑠𝑡 ¿

𝑝 (𝑉 ¿¿1+𝑉 2)=𝑛𝑅𝑇 ¿

𝑝 𝑓𝑉 1 𝑓=𝑛𝑅𝑇

−𝑑𝑊=(𝑝−𝑝0 )𝑑𝑉 2

varies for inflation/deflation – determined from the graph

𝑊=∫𝑝0

𝑝 𝑓

𝑝𝑑𝑉 1+𝑝𝑓 𝑉 𝑓 ¿

PISTON MODEL ~ ENERGYV20 = Vpiston

V2f = 0V1fV10

p0 = patmpf

𝑝0(𝑉 ¿¿10+𝑉 20)=𝑛𝑅𝑇=𝑐𝑜𝑛𝑠𝑡 ¿ 𝑝 𝑓𝑉 1 𝑓=𝑛𝑅𝑇

−𝑑𝑊=(𝑝−𝑝0 )𝑑𝑉 2

PISTON MODEL ~ ENERGY

𝑊=∫𝑝0

𝑝 𝑓

𝑝𝑑𝑉 1+𝑝𝑓 𝑉 𝑓 ¿

𝑊=∫𝑝0

𝑝 𝑓

𝑝𝑑𝑉 1+𝑝𝑓 𝑉 𝑓 ln𝑝𝑓

𝑝0+¿𝑝0(0−𝑉 20)¿

𝑝0(𝑉 ¿¿10+𝑉 20)=𝑝 𝑓𝑉 𝑓 ¿

𝑉 20=𝑝 𝑓 𝑉 𝑓

𝑝0−𝑉 10

)

varies for inflation/deflation – determined from the graph

elastic energy

work needed to increase radius from to r+dr under pressure difference

BALOON STRETCHING – THEORETICAL CURVE

initial volum

e

BALOON STRETCHING – THEORETICAL CURVE

final expression

FRICTION

h - car-surface distances - horizontal travelled distance

s2

l𝒔=𝒔𝟏+ 𝒔𝟐

s1

F1

𝐹 1=𝐹𝑔 𝑠1𝑙Fg

F2