Chapter 2-td

Chapter 2

Some concepts and definitions

Read BS, Chapter 2

2.1 Thermodynamic system and control volume

We take the following definitions:

• Thermodynamic system: a quantity of fixed mass under investigation,

• Surroundings: everything external to the system,

• System boundary: interface separating system and surroundings, and

• Universe: combination of system and surroundings.

The system, surroundings, system-boundary for a universe are shown for a potato-shapedsystem in Fig. 2.1. We allow two important interactions between the system and its sur-roundings:

• heat can cross into the system (our potato can get hot), and

• work can cross out of the system (our potato can expand).

Now, the system boundaries can change, for example the potato might expand on heating,but we can still distinguish the system and the surroundings. We now define an

• isolated system: a system which is not influenced by its surroundings.

Note that a potato with thick and inelastic skin will be isolated. We distinguish the system,which has constant mass, but possible variable volume, from the

• Control Volume: fixed volume over which mass can pass in and out of its boundary.

The control volume is bounded by the

29

30 CHAPTER 2. SOME CONCEPTS AND DEFINITIONS

system

surroundings

system boundaryheat in

work out

universe

Figure 2.1: Sketch of a universe composed of a system, its surroundings, and the systemboundary.

• Control Surface: boundary of the control volume.

The mass within a control volume may or may not be constant. If there is fluid flow inand out there may or may not be accumulation of mass within the control volume. We willmainly study cases in which there is no accumulation, but this need not be the case. Asketch contrasting scenarios in which a fluid is compressed in which the system approachwould be used against those where the control volume approach would be used is shown inFig. 2.2. In summary,

• system → fixed mass, closed, and

• control volume → potentially variable mass, open.

2.2 Macroscopic versus microscopic

In principle, we could solve for the forces acting on every molecule and use Newton’s lawsto determine the behavior of systems. This is difficult for even modestly sized systems.

• If we had a volume of 1 m3 of gas at atmospheric pressure and temperature, we wouldfind that it was composed of 2.4 × 1025 molecules.

• We would need six equations of motion for each molecule, three for x, y, z, position,and three for u, v, w velocity. This would require then a total of 1.4× 1026 differentialequations to solve simultaneously.

CC BY-NC-ND. 04 May 2012, J. M. Powers.

http://creativecommons.org/licenses/by-nc-nd/3.0/

2.2. MACROSCOPIC VERSUS MICROSCOPIC 31

piston-cylinder compressor

system approach control volume approach

system

!ow in !ow outcontrol

volume

Figure 2.2: Comparison of system (fixed mass) and control volume (fixed volume) approachesin thermodynamics for two common scenarios: piston-cylinder compression (left) and com-pression in a flow device whose details are not shown (right).

• Even with our largest computers, this is impossible today. Note most desktop com-puters only can store roughly 109 bytes of data in Random Access Memory (RAM).

• We can however model the average behavior of the molecules statistically.

• We can also use simple empirical relations which can be formally proved to capturethe statistical nature of the flow. This will be our approach.

• classical thermodynamics will treat macroscopic effects only and ignore individualmolecular effects. For example molecules bouncing off a wall exchange momentumwith the wall and induce pressure. We could use Newtonian mechanics for each par-ticle collision to calculate the net force on the wall. Instead our approach amounts toconsidering the average over space and time of the net effect of millions of collisions ona wall.

We will in fact assume that matter can be modelled as a

• Continuum: the limit in which discrete changes from molecule to molecule can beignored and distances and times over which we are concerned are much larger thanthose of the molecular scale. This will enable the use of calculus in our continuumthermodynamics.

The continuum theory can break down in important applications where the length and timescales are of comparable magnitude to molecular time scales. Important applications wherethe continuum assumption breaks down include

• rarefied gas dynamics of the outer atmosphere (relevant for low orbit space vehicles),and




• nano-scale heat transfer (relevant in cooling of computer chips).

To get some idea of the scales involved, we note that for air at atmospheric pressure andtemperature that the time and distance between molecular collisions provide the limits ofthe continuum. Under these conditions, we find for air

• length > 0.1 µm, and

• time > 0.1 ns,

will be sufficient to use the continuum assumption. For denser gases, these cutoff scales aresmaller. For lighter gases, these cutoff scales are larger. Details of collision theory can befound in advanced texts such as that of Vincenti and Kruger.1 They show for air that themean free path λ is well modeled by the following equation:

λ =M√

2πNρd2. (2.1)

Here, M is the molecular mass, N is Avogadro’s number, and d is the molecular diameter.

Example 2.1Find the variation of mean free path with density for air.

We turn to Vincenti and Kruger for numerical parameter values, which are seen to be M =28.9 kg/kmole, N = 6.02252× 1023 molecule/mole, d = 3.7 × 10−10 m. Thus,

λ =

(

28.9 kgkmole

) (1 kmole

1000 mole

)

√2π(6.02252× 1023 molecule

mole

)ρ (3.7 × 10−10 m)2

, (2.2)

=7.8895× 10−8 kg

molecule m2

ρ. (2.3)

Note that the unit molecule is not really a dimension, but really is literally a “unit,” which may wellbe thought of as dimensionless. Thus, we can safely say

λ =7.8895× 10−8 kg

m2

ρ. (2.4)

A plot of the variation of mean free path λ as a function of ρ is given in Fig. 2.3. Vincenti and Krugergo on to consider an atmosphere with density of ρ = 1.288 kg/m3. For this density

λ =7.8895 × 10−8 kg

m2

1.288 kgm3

, (2.5)

= 6.125× 10−8 m, (2.6)

= 6.125× 10−2µm. (2.7)

1W. G. Vincenti and C. H. Kruger, 1965, Introduction to Physical Gas Dynamics, John Wiley, New York,pp. 12-26.



2.3. PROPERTIES AND STATE OF A SUBSTANCE 33

0.01 0.1 1 10

10-7

10-6

10-5

10-4

10-8

0.001

λ (m)

ρ (kg/m3)

ρatm

∼ 1.288 kg/m3

λatm

~ 6.125 x 10-8 m

Figure 2.3: Mean free path length, λ, as a function of density, ρ, for air.

Vincenti and Kruger also show the mean molecular speed under these conditions is roughly c = 500 m/s,so the mean time between collisions, τ , is

τ ∼ λ

c=

6.125 × 10−8 m

500 ms

= 1.225× 10−10 s. (2.8)

2.3 Properties and state of a substance

We define

• Phase: a quantity of matter that is homogeneous throughout, and

• Phase Boundaries: interfaces between different phases.

An example of a single phase is ice. Another single phase is liquid water. A glass of icewater is a two-phase mixture with the phase boundaries at the edge of each ice cube.

We next define (circularly)

• State: condition described by observable macroscopic properties, and

• Property: quantity which only depends on the state of the system and is independentof the history of the system.

Examples of properties include temperature and pressure. Two states are equivalent if theyhave the same properties. So if state 1 is defined by temperature T1 and pressure P1, andstate 2 is defined is by temperature T2 and P2, state 1 is equivalent to state 2 iff (that is, ifand only if) T1 = T2 and P1 = P2.

There are two important classes of properties we consider in thermodynamics:




• Extensive Property: a property which depends on the mass (or the extent) of thesystem, example extensive properties include mass, total volume, total energy, and

• Intensive Property: a property which is independent of the mass of the system.Example intensive properties include temperature and pressure.

In general, if you cut a system in half and re-measure its properties, intensive propertiesremain unchanged, while extensive properties are cut in half. Properties are defined forsystems which are in

• Equilibrium: state in which no spontaneous changes are observed with respect totime.

We actually never totally achieve equilibrium, we only approximate it. It takes infinite timeto achieve final equilibrium. In this class we will mainly be concerned with two types ofequilibrium:

• Mechanical equilibrium: characterized by equal pressure, and

• Thermal equilibrium: characterized by equal temperature.

A third type of equilibrium is chemical equilibrium, which we will not consider here, and ischaracterized by equal chemical potentials.

A difficult conceptual challenge of thermodynamics is to reckon with two systems initiallyat their own equilibria, to bring them into contact so that they find a new equilibria. How todo this without consideration of time can be difficult. Another branch of thermodynamics,which we will consider only briefly in this course is

• Non-equilibrium thermodynamics: branch of thermodynamics which considerssystems often far from equilibrium and the time-dynamics of their path to equilibrium.

We will go to great effort to construct a thermodynamics which is generally not burdenedwith time. Occasionally we will bring time into our problems. Unfortunately, ignoring timeoccasionally requires some mental contortions, as seen in the next section.

2.4 Processes and cycles

Often systems undergo a

• Change of State: implies one or more properties of the system has changed.

How these properties would change outside of time is curiously outside the framework ofequilibrium thermodynamics! The best way to think of them is that the changes are slowrelative to the underlying molecular time scales. Fortunately, this will allow us to do a widevariety of problems of engineering relevance.

We also define a



2.5. FUNDAMENTAL VARIABLES AND UNITS 35

• Process: a succession of changes of state.

We assume our processes are all sufficiently slow such that each stage of the process is nearequilibrium. Certain common processes are given special names, based on the Greek

,ι′

σoς,isos, meaning “equal”:

• isothermal: constant temperature,

• isobaric: constant pressure, and

• isochoric: constant volume.

An important notion in thermodynamics is that of a

• cycle: series of processes which returns to the original state.

The cycle is a thermodynamic “round trip.”

2.5 Fundamental variables and units

We will mainly use the Systeme International (SI) units in this course. Occasionally, wewill use the English Engineering system of units. As found in US National Institute ofStandards and Technology (NIST) documents, the important fundamental base SI units,and corresponding English units are

• mass:

– kilogram (kg): a mass equal to the mass of the international prototype of thekilogram (a platinum-iridium bar stored in Paris), roughly equal to the mass ofone liter of water at standard temperature and pressure, and

– pound mass: (lbm),

• length:

– meter (m): the length of the path traveled by light in vacuum during a timeinterval of 1/299792458 of a second, and

– foot (ft),

• time:

– second: (s) the duration of 9192631770 periods of the radiation corresponding tothe transition between the two hyperfine levels of the ground state of the cesium133 atom, and

– second: (s) English time units are identical to those of SI,




• temperature: an equilibrium property which roughly measures how hot or cold anobject is. Note our senses are poor judges of temperature. Consider snow and air inthermal equilibrium at 20 ◦F . Usually, it is possible to keep your bare hands warmfor many hours at 20 ◦F if you are otherwise dressed warmly. However, if you placeyour bare hand in a snow bank you for a few minutes, you have a danger of frostbite.Yet both are at the same temperature. Why the difference in sense? Our bodiesactually have more sensitivity to heat fluxes instead of temperature; heat leaves ourbody more rapidly when in contact with high density objects like snow relative to thatof low density objects like air. More fundamental than common units such as ◦F areso-called absolute temperature units:

– Kelvin: (K) the fraction 1/273.16 of the thermodynamic temperature of thetriple point of water, and

– Rankine: (◦R).

2.6 Zeroth law of thermodynamics

In this class we are taking the axiomatic approach. Recall that an axiom cannot be proven.It is a statement whose truth can be ascertained only by comparison with experiment.The axiom can be disproved by a single negative experiment. The so-called zeroth law ofthermodynamics is the axiom which is probably most fundamental. It was formalized afterthe so-called first and second laws, and so it is called the zeroth law. Perhaps if a morefundamental axiom were discovered, it would be called the −1st law of thermodynamics?

• Zeroth law of thermodynamics: When two bodies have equality of temperaturewith a third body, then they have equality of temperature.

The origins of the zeroth law are murky. Sommerfeld2 attributes the notion to R. H. Fowlerin a 1931 review of a thermodynamics book. Fowler and Guggenheim explicitly introducethe term “zeroth law of thermodynamics” later.3

The equivalent statement in mathematical logic is that if x = y and x = z, then y = z.Definition of the zeroth law enables the use of a thermometer as a measurement device.A scale however needs to be defined. The old metric temperature scale, Celsius (◦C), wasdefined so that

• 0 ◦C is the freezing point of water, and

• 100 ◦C is the boiling point of water.

2A. Sommerfeld, 1956, Thermodynamics and Statistical Mechanics, Lectures on Theoretical Physics,Vol. V, Academic Press, New York, p. 1.

3R. Fowler and E. A. Guggenheim, 1939, Statistical Thermodynamics: A Version of Statistical Mechanics

for Students of Physics and Chemistry, Cambridge University Press, Cambridge, p. 56.



2.7. SECONDARY VARIABLES AND UNITS 37

These quantities varied with pressure however, so that different values would be obtainedon top of a mountain versus down in the valley, and so this is not a good standard. Themodern Celsius scale is defined to be nearly the same, but has

• 0.01 ◦C as the so-called triple point of water, and

• −273.15 ◦C as absolute zero in K.

The triple point of water is defined at the state where three phase of water (solid, liquid, andgas) are observed to co-exist. The transformation between the absolute Kelvin scale and theCelsius scale is given by

K = ◦C + 273.15. (2.9)

The English equivalents are degrees Fahrenheit (◦F ) for relative temperature and degreesRankine (◦R) for absolute temperature. The conversions are

T (◦R) = 1.8T (K), T (◦F ) = 1.8T (◦C) + 32, T (◦F ) = T (◦R) − 459.67. (2.10)

2.7 Secondary variables and units

Many units can be derived from the base units. Some important units for thermodynamicsinclude

• Force: This unit is defined from Newton’s second law, F = m d2x/dt2.

– Newton: (N), 1 N = 1 kg ms2 , and

– pound force: (lbf).

Force is straightforward in SI units. It is more confusing in English units, where theso-called gravitational constant gc is often introduced. In SI units, gc = 1. However inEnglish units, the law for force is better stated as

F =1

gcmd2x

dt2. (2.11)

Now, 1 lbf is induced by a mass of 1 lbm in places where local gravitational accelerationis g = 32.1740 ft/s2. So if we have a simple statics problem where the accelerationd2x/dt2 = g, we then get

F =1

gcmg. (2.12)

If we are at a location where g = 32.1740 ft/s2, we must have

1 lbf =(1 lbm)

(32.1740 ft

s2

)

gc

. (2.13)




Thus,

gc = 32.1740lbm ft

lbf s2. (2.14)

Example 2.2

If the local gravitational acceleration is 32.0 ft/s2, what is the weight W of an object with massof m = 1000 lbm.

W = F =1

gcmg =

1

32.1740 lbm ftlbf s2

(1000 lbm)

(

32.0ft

s2

)

= 994.59 lbf. (2.15)

• Energy: roughly speaking, the ability to do work, found from the product of forceand distance.

– Joule: (J), 1 J = 1 (N m), and

– foot-pound force: (ft lbf).

• Specific Volume: the volume per unit mass, known as v = V/m.

–(

m3

kg

)

, and

–(

ft3

lbm

)

.

• Density: the mass per unit volume, the inverse of specific volume ρ = m/V .

–(

kgm3

), and

–(

lbmft3

)

.

Note also that

v =1

ρ, ρ =

1

v. (2.16)

• Pressure: the limA→0 F/A where A is the cross-sectional area and F is the forceacting on that area. In thermodynamics, we are almost always concerned with theabsolute pressure as opposed to the gauge pressure. Most common pressure gauges donot measure the absolute pressure; instead they measure the difference between theabsolute pressure and the atmospheric pressure. The two are related via the formula

Pgauge = Pabsolute − Patm. (2.17)




We nearly always interpret P as an absolute pressure, so we could also say

Pgauge = P − Patm. (2.18)

– Pascal: (Pa), 1 Pa = 1 N/m2; note other common units are 1 bar = 105 Pa,1 atm = 1.01325 × 105 Pa = 101.325 kPa = 0.101325 MPa, and

– (psia): 1 psia = 1 lbf/in2. 1 atm = 14.696 psia. The a denotes the “absolute”pressure as opposed to the “gauge” pressure. The units psig refer to a gaugepressure.

The SI unit is named after Blaise Pascal, see Fig. 2.4, the French polymath who

Figure 2.4: Blaise Pascal (1623-1662), French scientist and philoso-pher who considered manometry, among other diverse topics; image fromhttp://www-history.mcs.st-and.ac.uk/∼history/Biographies/Pascal.html.

conducted early scientific experiments with manometers, a common measuring devicefor pressure, see Fig. 2.5. There are a variety of styles of manometers. Here, a pipecontaining fluid at pressure P and density ρ has a small tube with cross sectional areaA connecting it to the outside atmosphere at a different pressure Patm. The length H iseasily measured. The gravitational acceleration is g and is in the negative y direction.Because P > Patm, the manometer fluid is pushed up. However, it finds a mechanicalequilibrium where the weight of the manometer fluid balances the net force inducedby the pressure differential.

The figure includes a cutaway with a free body diagram. The interior fluid exerts apositive force of PA on the manometer fluid in the cutaway. The atmosphere exertsanother force of PatmA in the negative direction. The third force is the weight of thefluid: mg. Thus, Newton’s second law tells us

md2y

dt2︸︷︷︸

=0

= PA− PatmA−mg. (2.19)


http://www-history.mcs.st-and.ac.uk/~history/Biographies/Pascal.html



Patm

Hg

P A

P A atm

Fluid at P, ρ

A

mg = ρ V g = ρ A H g

y

Figure 2.5: Manometer sketch.

Now, we are concerned with cases which are static, in which case the accelerationd2y/dt2 = 0. Thus, we require a force balance, i.e. mechanical equilibrium, which isachieved when

0 = PA− PatmA−mg, (2.20)

PA = PatmA+mg. (2.21)

Now, mg = ρV g, where V is the volume of the fluid in the cutaway. Obviously fromthe geometry, we have V = AH , so mg = ρAHg. Thus,

PA = PatmA + ρAHg, (2.22)

P = Patm + ρgH. (2.23)

Or∆P = P − Patm = Pgauge = ρgH. (2.24)

Example 2.3A manometer gives a reading of H = 2 ft in a region where local g = 32.2 ft/s2. The working

fluid has specific volume v = 0.0164 ft3/lbm. The atmospheric pressure is Patm = 14.42 lbf/in2 =14.42 psia. Find the fluid pressure.

We know that in the SI system.

P = Patm + ρgH. (2.25)

In terms of specific volume, recalling that ρ = 1/v, we have

P = Patm +gH

v. (2.26)




The challenge here is really the English units. A fair way to approach English units is to replace g byg/gc in every formula. Thus, in English units, we have

P = Patm +1

v

g

gcH. (2.27)

So our fluid pressure is

P = 14.42lbf

in2+

1

0.0164 ft3

lbm

(

32.2 fts2

32.1740 lbm ftlbf s2

)

(2 ft)

(1 ft

12 in

)2

= 15.27lbf

in2= 15.27 psia. (2.28)






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