Thermodynamic Fundamentals of Refrigeration, Cryogenics

and Low Temperature Physics

Problems for colloquium

1 Zeroth law of thermodynamics – what thermodynamicparameter can be measured on the basis of the ZerothLaw?

Zeroth law of thermodynamics states that:

,,If in the case of three systems A, B and C, system A is in equilibrium C and B is inequilibrium with C, then A is in equilibrium with B”.

Consequence of this statement is defining temperature as a function, which has to beidentical in case of an arbitrary number of systems divided by heat transferring partitionsremaining in thermal equilibrium. The law permits the construction of thermometer tomeasure this property.We can calibrate the change in a thermal property, such as thelength of a column of mercury, by putting the thermometer in thermal equilibrium witha known physical system at several reference points (ex. boiling and freezing pointof water). If we then bring the thermometer into thermal equilibrium with any othersystem, we can determine the temperature of the other system by noting the change inthe thermal property.

2 First law of thermodynamics.

The existence of thermodynamic function results from the first rule of thermodynamics,stating that

,,System energy is maintained if heat is taken into account”,.

it is the internal energy of a body which may be changed as a result of work performed onthe body or heat transmitted to the body. It follows from the connected first and zerothlaws of thermodynamics, that the consequence of thermal contact between two bodies,which are not in the state of thermal equilibrium and thus they differ in temperature,is the transfer of heat between them leading to a simultaneous change of energy of thetwo bodies so that the total energy of the system was constant.First law of thermodynamics can be also expressed using equation:

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∆U = Q−W,,The change in internal energy of a system is equal to the heat added to the system

minus the work done by the system”..

Taking dU as an differential change in internal energy, one writes:

dU = δQ− δWdU = δQ− pdV

We can also express first law of thermodynamics using enthalpy instead of internalenergy: dH = δQ+ V dp

3 Second law of thermodynamics, applicability to refrig-eration.

In terms of only the first rule of thermodynamic, we cannot exclude the case of sponta-neous heat transfer from the body of lower temperature to the body of higher tempera-ture because then the energy is maintained. The fact that such spontaneous heat transferis impossible is indicated by the second rule of thermodynamics which was formulatedby Clausius in the following way:

,,It is not possible to build self-acting equipment working cyclically in such a way thatthe only effect of its activity is the transfer of heat from a colder location to a hotter

location”.

The second rule of thermodynamics indicates that lowering the temperature of a bodybelow the environment temperature requires the co-operation of at least three thermo-dynamic systems: a body whose temperature is to be lowered, environment which is thesource of heat of constant temperature and the source of energy devoid of entropy, i.e.undergoing unlimited conversion.

4 Sadi Carnot and his waterwheel analogy. What is theanalogue to water in thermodynamic systems (engines,refrigerators)?

Water/Heat will not flow spontaneously from a low level/cold object to a high level (hotobject).

The second law of thermodynamics is a general principle which places constraints uponthe direction of heat transfer and the attainable efficiencies of heat engines. In so doing, itgoes beyond the limitations imposed by the first law of thermodynamics. It’s implicationsmay be visualized in terms of the waterwheel analogy (Figure 1) proposed by SadiCarnot.

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Figure 1: Waterwheel analogy to engines and refrigerators

5 Ideal Carnot cycle of refrigeration.

Carnot cycle is a cycle with highest possible thermodynamic efficiency. It consist of twoisentropic processes and two isothermal processes. Figure 2 shows graphical representa-tion of Carnot cycle on T-s diagram. Carnot efficiency for refrigerators (COP):

COPCarnot = Q0

W = TcT−Tc

6 Transformation of Carnot cycle into vapour compressioncycle.

The cycle presented on Figure 2 (Carnot cycle in two-phase area) has some drawbacks:

• there is no sufficient signal when the evaporation should be finished in point c

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Figure 2: Carnot cycle representation on T-s diagram

• isentropic compression c-d would have to be performed in two-phase region

• also isentropic expansion a-b is difficult to be realised in two-phase region

Second version: after moving the compression process to superheated vapor area (Figure3).

Figure 3: Carnot cycle representation on T-s diagram (compression of superheated va-por)

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Advantages:

• Point c is now easy to identify using temperature measurements

• isentropic compression c-d is performed in superheated vapor area

Disadvantages:

• isentropic expansion a-b still has to be realised in two-phase region

• isothermal compression d-d’ needs additional compressor

Moving isentropic expansion to liquid area (Figure 4) moves point a to very high pressure(technical realization is impossible).

Figure 4: Carnot cycle representation on T-s diagram (expansion of liquid)

SUMMARY: real life refrigerators can not be based on Carnot cycle. Linde cycle is usedin refrigerators instead.

7 Comparison of ideal and real cycles of refrigerators.

As it was mentioned above, Linde cycle is used in refrigerators. Figure 5 shows the graph-ical representation of ideal Linde cycle. Superheated vapor is compressed in compressorin isentropic process (1-2), then it is condensed in condenser (2-3), then throttled in ex-pansion device (valve or capillary tube) - process 3-4, and then evaporated in evaporator(4-1).

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COP of Linde cycle:

COP = h1−h4h2−h1

Figure 5: a)Ideal Linde cycle representation on T-s diagram b) on lg p-h diagram

In real refrigerators there are pressure drops in pipelines and heat exchangers, alsocompression is not an isentropic process. The real Linde cycle was shown on Figure 6.

Figure 6: Real Linde cycle representation on lg p-h diagram

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8 Nernst theorem. Why the absolute zero is unattainable?Classical versus Nernst approach.

It is impossible by any procedure, no matter how idealised, to reduce any system to theabsolute zero of temperature in a finite number of operations.

The Nernst heat theorem says that as absolute zero is approached, the entropy change∆S for a chemical or physical transformation approaches 0. This can be expressedmathematically as follow:

limT→0 ∆S = 0

The entropy of all systems and of all states of a system is zero at absolute zero. It isimpossible to reach the absolute zero of temperature by any finite number of processes.

Unavailability of absolute zero can be shown using classical physics, basing on law ofequipartition of energy proposed by Maxwell, saying that energy of the body dependsonly on temperature and its is shared equally among all of its independent parts. Forevery degree of freedom of particle energy is equal:

E = 12kT ,

where k is Boltzmann constant, k = 1, 38 · 10−23JK−1.Basing only on energy equipartition law, the specific heat of a body is constant and doesnot depend on temperature. So, the entropy difference between finite temperature andabsolute zero reaches infinity:

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sT − s0 =T∫0

cT dT = c · (lnT − ln 0) =∞

In that case isobars will look like in Figure 7. The absolute zero is unavailable because itwould require infinite work: W = h1 − h0 − T (s1 − s0), where s0 = −∞. Using classical

Figure 7: Isobars in system in which energy equipartition law is valid.

physics, the absolute zero is unavailable, but it is not a reference point from which onecould calculate all the quantities necessary for thermodynamic calculations (especiallyfor entropy calculations).According to quantum mechanics the specific heat of all substances approach 0 whentemperature tend to 0K. Absolute zero can be a thermodynamic reference point. On theother hand, it may lead to situation, when isobars look like in Figure 8. Nernst approach– both the heat capacity and entropy of any substance must vanish at absolute zero (allthe substances at 0K must have to be in ideal order, which is possible only in quantumworld). Figure 9 shows the isobars according to 3rd law of thermodynamics.

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Figure 8: Isobars in system in which absolute zero can be achieved in finite number ofsteps.

Figure 9: Isobars according to 3rd law of thermodynamics.

9 Consequences of the Nernst Theorem (3. Law of ther-modynamics) with respect to heat capacity

On the terms of third law of thermodynamics, when approaching to absolute zero, allentropy changes caused by change of any parameter approaches 0: limT→0( ∂s∂z )T = 0.

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That means, the specific heat cz is vanishing, because:

cz = T ( ∂s∂T z so cz → 0 when T → 0 and (∂s/∂T )z → 0

Also thermal expansion coefficient β will be equal 0 at 0K.

10 Relation between temperature and energy. Boltzmannconstant.

The concept of temperature resulting from the Zeroth Thermodynamical Law is intuitive.The temperature is directly related to energy by the equation describing the gas moleculeenergy per one state of freedom (Maxwell equipartition):

E = 12kT ,

where k is Boltzmann constant, k = 1, 38 · 10−23JK−1.For monoatomic gas this value will be equal: E = 3

2kT .If the temperature is a measure of the energy absorbed by the physical body, there is noobjective need for a dedicated unit describing its (temperature) value. However using ofJoules [J] to desribe typical everyday temperatures would be impractical.

11 Equation of State for ideal gases and its – assumptionsallowing to derive the equation.

Lets consider a system shown on Figure 10. Assuming that mean particle velocities are

Figure 10: Velocity distribution in volume V

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the same and equal to v, we can say that 1/6 of the total particle amount N is movingwith a v velocity in the direction of As vessel wall.During a time t, the wall As[m

2] is hit by the amount of the particles n equal to:

n = v·t·As·N6V

On the assumption that the collissions of the particles with the wall are ideally elastic,and the particle energy remains constant before and after the collision, the particlemomentum is changing by the value:

∆p = 2mv

It results from the second Newton law of dynamics the the momentum change of theparticle interacting with the wall is equal to the force that the particle imposes on thewall. In time t, the total amount of N particles will collide with the surface As, creatingthe pressure P

P = 2·m·v2·N6V

Kinetic energy of particle: Ek = mv2/2, so after putting this to equation for pressureand multiplying by V :

PV = 23NEk

PV = NkTR = k ·A, A = Avogadra number A = 6, 02 · 1023

PV = nRT

We have obtain ideal gas equation of state.

12 Statistical interpretation of entropy.

Entropy of individual particle, where Ω is a number of allowable states , in a groundstate Ω = 1 and s = 0 what is in agreement with the Nernst theorem (third law ofthermodynamics):

s = k ln Ω

Entropy of one mole of substance:

s = R ln Ω

13 Zero point (state) energy

Heisenberg uncertainty principle states that certain pairs of physical properties, such asposition and momentum, cannot be simultaneously known to arbitrarily high precision(∆p∆x = h). If the position of atom is precisely defined, the uncertainty of its mo-mentum increases. Atom which position is strictly defined by strong interactions with

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lattice, is still oscillating around this position, even at 0K. At 0K the energy of atom isequal zero point energy. In higher temperatures, when kT is much higher than zero pointenergy, sum of zero point energy and thermal energy converges with energy resultingfrom energy equipartition law.

Figure 11: 1 – thermal energy resulting from classical equipartition principle 2 – thermalenergy taking into account the vanishing heat capacity 3 – thermal energy taking intoaccount zero point energy.

14 General cooling process

Figure 12: Generalized cooling. x - generalized force

Temperature drop (TA−TC) appears when generalized force x goes adiabaticaly to itsinitial value: x2 → x1. x - generalized force of conjugate variables: force - displacement.

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Example: for conjugate variables p−V : pressure p is generalize, volume V is generalizeddisplacement.

15 Examples of conjugate variables – generalized force –displacement.

Substance Force X Displacement Y Process of en-tropy decreasing(ordering)

Process of tem-perature lowering

Gas pressure, p [Pa] Volume V , [m3] Isothermal com-pression

Isentropic expan-tion

Gas pressure, p [Pa] Volume V , [m3] Isothermal com-pression

Isenthalpic ex-pantion

Paramagneticsubstance

Magnetic fluxdensity H, [A/m]

Magneric dipolmoment µ0M ,[Wb m]

Isothermal mag-netization

Adiabatic demag-netization

Dielectric sub-stance

Magnetic fluxdensity E, [V/m]

Electric dipol mo-ment P, [Cm]

Isothermal elec-trization

Adiabatic de-electrisation

Electron gas Electrical poten-tial

Electric charge Z,[C]

Electron com-pacting

Electron dilution

ε, [V]Salt Chemical poten-

tial µ[J/mol]Mole number n Drying Dissolving

Rod Mechanical forceF

Length l Compression Stress relaxation

Polymer Mechanical forceF

Length l Tension (fiberelongation)

Tension relax-ation

16 Entropy change during isothermal compression.

Isothermal expansion process is shown on Figure 12 (process A-B). Change of entropyof gas in reversible compression process can be determined using equations connecting1st and 2nd law of thermodynamics:

ds = du+pdvT

ds = dh−vdpT

Internal energy and enthalpy of ideal gas are only temperature functions. After substi-tution of: du = cv(T )dT and dh = cp(T )dT and ideal gas equation of state we get:

ds = cv(T )dTT +Rdvv

ds = cp(T )dTT −Rdpp

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After integration:

s2 − s1 =∫ T2T1cv(T )dTT +R ln v2

v1=

∫ T2T1cp(T )dTT −R ln p2

p1

If the compression is isothermal process, the work of compression is minimal, and entropychange is equal:

s2 − s1 = R ln v2v1

= −R ln p2p1

Change of entropy during isothermal gas compression can be also determined on the basisof statistical interpretation of entropy. Number of allowable states Ω is proportional tovolume Ω = a · V . As effect of compression, the volume will change from the value V1

to V2. The change of entropy will be equal to:

∆s = s2 − s1 = R(ln Ω2 − ln Ω1) = R(ln(aV2 − ln(aV1)) = R ln(V2V1 )

The result is identical with the one determined using phenomenological equations.

17 Entropy change during rubber band elongation.

The example of solid in which one can get entropy change by applying a force is rubber.In no-stress state polymer fibers can have any position. Rubber elongation causes thepolymer fibers to have identical positions, parallel to force direction. In this way entropyof rubber becomes equal 0. Assuming that before elongation each of the fibers had 3allowable positions, one can get:

∆s/R = − ln 3 ≈ −1.1

18 Isentropic expansion.

Adiabatic (isentropic) gas expansion requires no irreversible processes leading to heatrelease during expansion process, that means all energy of expanded gas has to bereceived as external work. In that way, the enthalpy and internal energy drop, as wellas temperature drop, are the highest possible. External work of expansion is equal toenthalpy difference between inlet and outlet of expander: w = h1 − h2. Temperaturedrop in isentropic expansion:

dS = ( ∂S∂T )pdT + (∂S∂p )TdpdS = 0 - isentropic process

µs = (dTdp )s = −( ∂S∂p

)T

( ∂S∂T

)p

(∂S∂p )T = ( ∂v∂T )p - Maxwell equation

( ∂S∂T )p =cpT

After substitution:

mus = (dTdp )s =T ( ∂v

∂T)p

cp= Tvβ

cp

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19 Describe and illustrate on the TS diagram a generalprocess to lower the body temperature.

Figure 13: General process to lower the body temperature

• The entropy must be a function of two parameters: temperature T and X, whereX is a generalized force (gas pressure, magnetic field ). At the room temperaturethe entropy must be high enough.

• The entropy can be lowered by external work done on the body (gas compression,paramagnetic magnetization) in isothermal process.

• By isentropic process the temperature of the body will go down.

After performing isothermally work on system, the number of energy levels decreases

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and their width is increasing. When after putting the system in order the system willbe isolated thermally from environment and the external force (pressure, magnetic field)will be removed, the width of energy levels will go back to previous one, but entropywill be the same (third picture). The temperature of the system will decrease.

20 Explain why we need some minimum storage of entropyto perform cooling?

Every system used to achieve low temperature should have ability to cool down itself:c∆T = T∆s (c is the heat capacity of system). So:

∆TT = ∆s/R

c/R

Minimum value of ∆TT does not depend on temperature and it is equal ∆T

T ≈ 0.1.In room temperature specific heat of most materials c ≈ 3R. Reduced specific heatc/R ≈ 3. After substitution to one of previous equations we get minimum entropy value∆s/R ≈ 0, 3

If the reduced entropy is lower than lower limit shown on figure above, then the processin which it may be decreased is not perspective in achieving lower temperatures.

21 Thermodynamic efficiencies of refrigerator and cryo-genic coolers.

Independently on the operation principle, the basic features of the refrigerators can becompared on basis of the Carnot cycle. Thermodynamic efficiency of Carnot cycle isdepending on temperatures:

COP = ∆Qwe

Wob= Tk

T−Tk

, where Tk is cryostating temperature and T - temperature in which the heat is rejectedto environment. In cryogenic temperatures, the thermodynamic efficiency of Carnot

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cycle is lower than 0.66.

22 Why the efficiency of cryogenic coolers is increasingwith the refrigerator capacity?

Efficiency of cryogenic cooler can be expressed as: ε = ∆Qwe

∆Qwe(T−Tk)

Tk+∆W

, where ∆Qwe

- cooling power, ∆Qwe(T−Tk)Tk

- minimum work, resulting from Carnot cycle, ∆W -additional work resulting from entropy genereation.Cooling power of refrigerator is proportional to a3 (a - characteristic dimension of cooler).Insulation losses are proportional to a2 (surface). Heat flux through the insultation:∆Qiz = k · a2 · (T − Tk)Entropy generation:∆Siz = k · a2 · (T − Tk)/TkAdditional work resulting from Gouya-Stodola law:∆Wiz = k · a2 · Tot(T − Tk)/TkSo the efficiency of cooler:ε = ∆Qwe

(T−Tk)

Tk(Qwe+ka2Tot)

Efficiency normalized versus Carnot limitation: εεC

= ∆Qwe

∆Qwe+ka2TotCooling capacity is proportional to third power of the characteristic dimension a.εεC≈ a3

a3+ka2Tot≈ a

a+kTot≈

3√Q3√Q+kTot

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23 Gouy-Stodola Law (theorem) and its applications

In real cooler there are internal sources of entropy, resulting from irreversibility of pro-cesses, that means there is a bigger amount of entropy to remove than the flux which flewinto the refrigerator with the heat flux from the refrigerated object. In case of realisationof closed thermodynamic cycle, only way of carrying away entropy to the environmentis heat exchange with the environment, real refrigereators carry away much more heatto the environment than ideal refrigerators. To satisfy the energy conservation rule, it isnecessary to perform additional work at the amount resulting from Gouy-Stodola law:∆W = ∆S · Tot

24 Entropy generation during heat transfer and fluid flow.

In cryogenic systems the most important sources of entropy generation are

• heat transferred or generated at low temperatures∆SQ = ∆Q

T

• flow with pressure dropdSdp = −V

T dp = −MρT dp

25 Please give examples of the systems that can be opti-mized with entropy minimization method.

• Heat exchangers

• Cryostats

• Thermal insulation

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• Transfer lines

• LHC superconducting magnet support

26 Heat transfer in low temperatures (convection, residualgas conduction, conduction, radiation)

26.1 Conduction heat transfer

Qs = AλsL (T2− T1), where A - material cross-section, λ - material thermal conductivity,

W/mK, L - length. Conduction heat transfer reduction methods:

• Minimum cross-section area

• Low conductivity material

• Max. material length

Gas heat conduction: Qg = Aλg1

L+l0(T2 − T1), l0 - mean free path of gas, L - distance

between walls.

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26.2 Convection

Qcon = A · hcon(Tw − Tenv),where A - material cross-section area, hcon - convective heat transfer coefficient, W/m2K,Tw - wall temperature, Tenv - environment temperature.hcon ∼ L3, so reduction of natural convection=reduction of characteristic length L.

26.3 Radiation

Qr = A · σ · ε′ · (T 41 − T 4

2 )where A - material area, σ - Stefan-Boltzmann constant, ε′ = 1

1ε1

+ 1ε2−1

- emissivity of

the system.Reduction - radiation shields:

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27 Principle of low temperature thermal insulation oper-ation (vacuum, multi-layer screens)

27.1 Gas filled close-cells foams

Foam materials: polyurethanes, polystyrenes, gums, glass. Gases: R22, R23, R134a,R407C, Carbon dioxide.Solid conduction reduction:

• Low thermal conductivity cells material

• 2% volumetric fraction of the solid phase

• Cell walls are very thin

Gas convection reduction - small cells dimensionsGas conduction reduction:

• Low thermal conductivity gases

• In cryogenic temperatures partial condensation or solidification of the gases - vac-uum in cells

Thermal radiation reduction - small temperature difference between opposite cell walls.Low cost, but the thermal conductivity is high. Application - rarely and short term usedcryogenic equipments.

27.2 Vacuum Insulation

Gas conduction reduction - high vacuumThermal radiation reduction - cold and warm boundary covered with high emissivitymetal layer.Advantages:

• Relatively easy to create - short time vacuum creation process

• Very low heat transfer in low temperature ranges

Disadvantages:

• Relatively high heat transfer in high temperature ranges

• High vacuum level sensitivity

• Very high heat flux in case of sudden vacuum loss

Applications:

• Small cryogenic equipments and experimental flasks

• Cryogenic elements where distance between boundaries is very small (no otherinsulation can be installed)

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27.3 Powder gas filled/vacuum insulation

• Perlite

• Hollow glass bubbles

• Glass fibers

• Aerogels

Solid conduction reduction

• Low thermal conductivity powders and glass fibers

• High length and very small cross-section area of the fibers

• Thin walls of the hollow glass bubbles

• High thermal contact resistance between powder grains/fibers

Gas convection reduction (gas filled insulations) - small dimension of the spaces betweenpowder grains/fibersGas conduction reduction (vacuum insulations) - high vacuum.Thermal radiation reduction

• Low temperature difference between adjacent powder grains/fibers

• Opacifier agents: Al, Cu, Au.

Advantages:

• Low thermal conductivity (in good vacuum conditions)

• Does not require HV conditions

• Relatively good insulation properties in gas filled conditions

• Mechanical support properties

• Relatively low costs of materials and production

• Sorption properties of the powders – slow vacuum degradation in case of vacuumcontainer leaking

Disadvantages:

• Very thigh vacuum pump sieves are required – long time vacuum creation process

• Powder self-compaction due thermal shrinking of the vacuum container

Applications:

• Large and very large cryogenic systems (gas liquefiers and gas separation installa-tions), tanks for low evaporated and cheap cryogens (LNG, N2, O2, Air, Ar)

• Application, where high vacuum creating and/or maintaining is very difficult orimpossible

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27.4 MLI - Multilayer Vacuum Insulation

Radiation shields - shield: single or double side metal plated Mylar, Kapton, Dracon.Pleating materials: Al, Ag, Cu, Au. Perforated (0,1-0,3% of total area). Spacers: silkyor nylon nets, glass fibers mats Solid conduction reduction

• Low thermal conductivity materials for spacers

• Special design of spacers - point contacts between solids

• Small contact area between spacers and radiation shields or between crinkled/dimpledshields

Gas conduction reduction - high vacuum.Thermal radiation reduction

• Radiation shields

• Very low emissivity of the shield surface

.Advantages:

• The best thermal insulation in very large temperature range

• Does not require HV conditions

• Weak sensitivity for slow vacuum degradation

• Good thermal behavior in case of the sudden vacuum lose

Disadvantages:

• Relatively expensive

• Long time vacuum creation process due to material outgassing

• Need additional support elements (additional heat transfer paths are creating)

Applications:

• Small, middle and some large scale cryogenic systems, cryogen storage tanks (Heand H2 tanks included), experimental equipments, cryocoolers ... .

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28 Please give the examples of engineering solutions tolimit heat flux to low temperatures regions of cryogenicdevices.

• Vacuum insulation to reduce convection

• Thermal shields to reduce radiation

• Elongation of some elements (ex. valve stems) to reduce conduction

• Avoiding thermal bridges, applying low thermal conductivity materials

29 Definition of cogeneration and trigeneration.

Cogeneration - simultaneous production of mechanical (electrical) energy and heat.

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30 Minimum work of gas mixture separation.

Entropy of gas mixture is higher than sum of entropies of components under the samepressure and temperature. From second law of thermodynamics one know that it isnecessary to perform external work to separate the gas mixture. Minimum work of gasseparation is equal to work of isothermal compression of components from partial pres-sure to pressure p.

WA =∫ pxAp

pdv = −nART lnxAWB =

∫ pxBp

pdv = −nBRT lnxBFor two component mixture: W = WA +WB = nART lnxA + nBRT lnxB

For multicomponent mixture: W =∑zi=1 niRT lnxi

31 Technologies of gas mixture separation – cryogenic, ad-sorption and membrane.

31.1 Cryogenic gas separation technology

Single or double Linde rectification column. Highest gas quality (purity), biggest capac-ity. Only technology for big industrial applications. Capacities up to 3500 tones/day.

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O2 purity up to 99%

31.2 Membrane technology

Purity of oxygen up to 40 %, low capacities (to 20 tons/day). Membrane is a continuousphase between two other phases, that allow some of one phase components to permeateto another. Advantages of this technology is that system is made of modules, whichallows to scale the process.

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Membranes can be used for nitrogen production (high purity of N2).

31.3 Adsorption technology

Purity of oxygen up to 95%, capacities of installations up to 250 tons/day. Duringoperation only one of adsorbing tanks is active, other one is being regenerated by openingthe outlet directly to the atmosphere (PSA) or by creating pressure below atmosphericin it by using vacuum pump (VPSA).

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