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Experiment 9 First Law of Thermodynamics – Bicycle Braking

Experiment 9First Law of Thermodynamics – Bicycle Braking

Object

The object of this experiment is to verify the first law of thermodynamics through the use of a bicycle brake calorimeter and to simultaneously compute Joule’s constant, J = 778.17 (ft lbf)/Btu.

Introduction

The First Law of Thermodynamics is verified yet again – this time with a bicycle. A bicycle front caliper brake is removed and replaced with a lever-mounted, copper calorimeter friction pad shown in Figure 9.1. The calorimeter friction pad rubs on the front tire, heats-up, brings the bicycle to a stop, and verifies the first law of thermodynamics. The loss in kinetic energy of the bicycle and rider is equated to the gain in internal energy of the copper calorimeter. A ratio of these two energies gives Joule’s constant. The data reduction analysis accounts for bicycle aerodynamic drag, rolling friction, and heat loss into the front tire.

thermocouple wire pivoting beam

copper calorimeter friction pad balsa wood insulation

Figure 9.1 The copper calorimeter friction pad used for braking a bicycle.

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Historical BackgroundHeating by friction has been know by man for millennia, but quantified only as late as the 19 th

century. Count Rumford [1] performed experiments involving the convertibility of mechanical work into heat as a result of friction. These experiments involved the boring of cannon with a dull tool bit on a boring bar, which was driven by two draft horses. His calorimeter was comprised of a semi-isolated end of a cannon encased in a wooden box containing water with a thermometer – see Figure 9.2.

Being engaged lately in superintending the boring of cannon in the workshops of the military arsenal at Munich, I was struck with the very considerable degree of heat which a brass gun acquires in a short time, in being bored; . . .

Benjamin Count of Rumford1798

Figure 9.2 Rumford’s cannon boring experimental equipment. The sketch is from Rumford [1].

In 1850 James Joule [2] analyzed Rumford’s 1798 data and arrived at a value of 1034 (ft lbf)/Btu for the convertibility of work into heat. This value favorably compared with the experimental results Joule himself had recently obtained. Joule's analysis on the convertibility of work into heat came from experiments reported in 1843, 1845, and 1850. This work provided the empirical basis for our contemporary first law of thermodynamics.

One of Joule’s experiments involved friction between a cast iron disk rotating in a cast iron seat all immersed in a mercury bath. The result was a reported convertibility value of 774.987 (ft

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lbf)/Btu – a value only 0.4% lower than today’s accepted Joule’s constant value of 778.17 (ft lbf)/Btu. The Experiments

The current experiment employs a bicycle with a copper calorimeter friction pad that is used as a brake on the front tire as shown in Figure 9.1. One end of the pivoting beam of this apparatus is connected to the hand-operated brake cable, the other end carries the copper calorimeter friction pad, which makes contact with the front tire. The copper pad is about 25 mm wide, 75 mm long and, 6.4 mm thick and has a type K thermocouple silver-soldered in a small hole drilled in the back surface. The copper pad is insulated from the pivoting beam with a 6.4 mm thick piece of balsa wood. Furthermore, phenolic washers thermally isolate the four mounting machine screws from the pivoting beam. Insulating balsa wood pieces, removed and not shown in Figure 9.1, fit around the exposed edges of the copper pad. A digital thermometer, audio tape recorder, and speedometer are mounted on the bicycle handlebar (also not shown).

Coast-down runs were performed with no braking to determine losses due to aerodynamic drag and rolling friction. Speeds from the bicycle speedometer were read into the audio tape recorder and later transcribed to provide speed-versus-time data. The braking runs were of short duration (~ 5 s) so only initial speed was recorded and the final speed was always zero. Temperature data at the beginning and end of the braking runs were also recorded. All of the tests were performed on a level road during early mornings when the wind was calm.

DataThe data for this experiment are given in “Experiment 9 Data.xls.” Two tables are provided in this file: Table 1 contains “Bicycle coast-down without braking” and Table 2 contains “Bicycle braking.” The mass of the bicycle and rider and the copper friction pad mass are also given.

Analysis

First law of thermodynamics analysis

We apply the first law of thermodynamics to a control volume that contains the bicycle starting at some initial velocity (state 1) and terminating at rest (state 2) – see Figure 9.3.

[Q1−2 ]rubber

= [ m2 gc

(V22−V

12)]bike+ [mg

gc(z2−z1)]bike

+(U 2−U1 )Cu + W 1−2(9.1)

(1) (2) (3) (4) (5)

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Figure 9.3 The closed system used for the first law of thermodynamics analysis.

In Equation 9.1: term 1 is the heat loss from the copper brake pad to the surroundings (we will show that

for all of the heat loss paths the rubber tire path dominates – thus the subscript “rubber” on term 1)

term 2 is the difference in kinetic energies for the bike + rider term 3 is the difference in gravitational potential energies for the bike + rider term 4 is the difference in internal energy of the copper calorimeter friction pad term 5 represents work done by the bike + rider on the surroundings.

Terms 1 and 5 will be examined in detail.

Heat loss (term 1)

During the braking process the copper friction pad heats up with a resulting heat loss to the

surroundings. (Q1−2 will be negative since heat is transferred out of the system.) Convection losses are small due to the small exposed surface area of the copper and the short braking period. The conduction losses through the balsa wood insulation and the phenolic washers that isolate the four small machine screws (see Figure 9.1) are also small. However, heat generated at the copper brake pad–rubber tire interface diffuses into each of these materials, so there is heat loss into the bicycle tire. Analysis for the heat loss into the tire is complicated by the transient generation of the heat and transient tire rotation. However, as an approximation, an analysis for the simpler case of a non-rotating tire may be used.

Our heat loss analysis begins with a solution by Carslaw and Jaeger [3] who provide the transient temperature T, as a function of time t, at the surface of a semi-infinite solid with zero initial temperature and with constant surface heat flux q / A

T=2 (q/A )

k (α t

π )1/2

(9.2)

The semi-infinite solid has thermal conductivity k, and thermal diffusivity α . For the short duration braking we may take the rubber tire and copper friction pad as semi-infinite solids with one-dimensional heat diffusion perpendicular to their contact surfaces. The initial temperature of the tire and copper brake pad are not zero as stipulated but they are the same and will cancel in the analysis. The rubbing friction provides the common surface heat flux q/A. Furthermore, the copper and rubber have the same temperature at their common surface contact location. From Equation 9.2 this means

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[2 (q/A )k

( α tπ )

1/2]Cu=[ 2 (q/A )

k ( α t

π )1/2 ]rubber

which reduces to

qCu

qrubber=( (k ρ c )Cu

(k ρ c )rubber)1/2

=51(9.3)

The heat transfer rate into the copper qCu , and the heat transfer rate into the rubber qrubber , are

integrated over time to give the net heat transferred Q 1−2 , into each material. By definition

[Q1−2 ]Cu=∫t1

t2 qCu dt (9.4)

and [Q1−2 ]rubber

=∫t1

t2 qrubber dt (9.5)

Furthermore, by the first law of thermodynamics applied to just the copper friction pad, the net heat transferred into the copper causes an increase in copper internal energy

[Q1−2 ]Cu=(U 2−U 1)Cu (9.6)

Finally, Equations 9.3-9.6 reduce to

[Q1−2 ]rubber= 1

51(U 2−U 1)Cu

(9.7)

We should note that Equation 9.7 under predicts the heat transferred into the rubber tire because of the non-rotating tire simplification.

Work due to aerodynamic drag and rolling friction (term 5)

Aerodynamic drag and rolling friction will cause the bicycle to coast-down, without braking, due to energy transferred from the moving bicycle to the surroundings. For both coast-down and for braking, speeds were read from a speedometer into an audiocassette recorder mounted on the bicycle handlebar and later transcribed with a stopwatch to give velocity-versus-time data. During coast-down the velocity-versus-time data are fitted with the curve

V=c1e−c2 t

(9.8)

and for the short duration braking the velocity-versus-time data are fitted with a linear function

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V=−c3 t+c4 (9.9)

For the decelerating bicycle we write Newton’s second law

F= 1gc

m a(9.10)

During coast-down the aerodynamic drag and rolling friction are forces acting on the combined mass of bicycle and rider that produces a negative acceleration (deceleration). This acceleration may be expressed as a=dV /dt which is obtained by taking the derivative of Equation 9.8. Using this result in Equation 9.10 we get

F=− mgc

c2 V(9.11)

Equation 9.11 expresses the force of the surroundings acting on the bicycle. For the force of the bicycle acting on the surroundings we have

F= mgc

c2 V(9.12)

This force is a function of velocity only and is present during both coast-down and braking. The work associated with this force during braking is

W 1−2=∫x1

x2 F dx =∫t 1

t 2 F V dt(9.13)

Substituting Equations 9.9 and 9.12 into 9.13 and integrating produces

W 1−2=m c2

gc[c32

t23

3−c3 c4 t

22+c42 t2]

(9.14)

Joule’s constant analysis

We now return to the first law of thermodynamics (Equation 9.1), rewritten here

[Q1−2 ]rubber

= [ m2 gc

(V22−V

12)]bike+ [mg

gc(z2−z1 )]bike

+(U 2−U1 )Cu + W 1−2 (9.1)

(1) (2) (3) (4) (5)

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The run from state 1 to state 2 was conducted over level ground thus there was no change in gravitational potential energy, eliminating term 3. The bicycle comes to rest at state 2 thus V 2=0 . We note that(U 2−U 1 )Cu= [m c (T 2−T1 )]Cu . And we substitute Equation 9.7 for term 1 and Equation 9.14 for term 5 to obtain

12gc

[m V 12 ]bike −

m c2

gc[c32

t23

3−c3 c4 t

22+c42 t2 ] = 52

51 [m c (T 2−T 1 )]Cu

(9.15)

(1) (2) (3)

In Equation 9.15 the first term represents the initial kinetic energy of the rolling bicycle; the second term is due to aerodynamic drag and rolling friction losses represented as work; and the third term represents the gain in internal energy of the copper friction pad calorimeter plus heat loss into the tire, both due to friction. Using English Engineering units* the ratio of the left-side over the right-side of this equation directly gives Joule’s constant in (ft lbf)/Btu.

Required

1. Using the data in “Experiment 9 Data.xls” Table 1, plot (with linear scales) time (s) on the

abscissa and velocity (ft/s) on the ordinate. Fit these data with the curve V=c1 e−c2 t

using TRENDLINE in EXCEL. In this same figure plot run 7 found in Table 2.

2. Construct a table with the following columns (left-to-right): run, V 1 , t 2 , (T 2−T 1 ) ,

m2gc

(V12 )

, W 1−2 , [Q1−2 ]rubber , (U2−U1 )Cu , J. Include the appropriate units for each column. Using the data in Table 2 fill in or compute the values for each column. Comment on the relative

contributions of each of the following terms:

m2 gc

(V12 )

, W 1−2 , [Q1−2 ]rubber , and (U2−U1)Cu

.

3. Plot, using linear scales, the gain in internal energy of the copper calorimeter friction pad plus heat loss into the tire (Equation 9.15, term 3) on the abscissa and the initial kinetic energy of the rolling bicycle minus the aerodynamic drag and rolling friction losses represented as work (Equation 9.15, terms 1 & 2) on the ordinate. Compute the slope of

this line, which is Joule’s constant. Hint: the origin is a datum point because, when V 1=0

, (U2−U1 )Cu = 0. Furthermore, this point is known with 100% certainty. Accordingly, in EXCEL using TRENDLINE, force your line through the origin. Also in this graph plot the known J = 778.17 (ft lbf)/Btu. Find the percent deviation between your result and the published value.

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* English units are used to distinguish the kinetic energy and work (both in ft lbf) from the internal energy and heat transfer (in Btu). SI units would give Joules (J) for all of the terms and would require conversion back to English units to obtain Joule’s constant. It is worthwhile to perform a units check on Equation 9.15. Terms 1 and 2 should each reduce to ft-lbf and term 3 should reduce to Btu. The units for , m, V, t, c, and T are given in the Nomenclature. However, the units for the constants c2 , c3 and c 4 are not immediately obvious and require examination of their origins, which are found in Equations 9.8 and 9.9. Since both Equations 9.8 and 9.9 give velocity with units (ft/s) it follows that c2 has units (1/s), c3 has units (ft/s2), and c 4 has units (ft/s).

NomenclatureEnglish

a acceleration, c1 , c2 , c3 , c4 constants c specific heat, Btu/(lbm°F)F force, lbf

proportionality constant, 32 .2 lbm ft

lbf s2

k thermal conductivity, Btu/(s ft °F)m mass, lbmQ heat loss, Btu

q/A heat flux, T temperature, °Ft time, sU internal energy, BtuV velocity, ft/sW work, ft lbfx distance, ftz elevation, ft

Greek

thermal diffusivity,

density,

SubscriptsCu copperrubber rubber tire1, 2 states

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References

1. Brown, S.C., Editor, “Collected Works of Count Rumford,” Vol. II, Practical Applications of Heat, The Belknap Press, Harvard University Press, Cambridge, Mass., 1969.

2. Joule, J. P., "The Scientific Papers of James Prescott Joule," Vol. I & II, Dawsons of Pall Mall, London, first published 1887, reprinted 1963.

3. Carslaw, H. S. & J. C. Jaeger, “Conduction of Heat in Solids,” second edition, Oxford, at the Clarendon Press, 1959. p. 75.

\ © 2005 by Ronald S. Mullisen \ Physical Experiments in Thermodynamics \ Experiment 9 \

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