Lecture 12. Refrigerators. Toward

Absolute Zero (Ch. 4)

10-5

109

10-3

10-1

101

103

105

107

Center of hottest starsCenter of Sun, nuclear reactions

Electronic/chemical energySurface of Sun, hottest boiling points

Organic lifeLiquid air

Liquid 4HeUniverse

Superfluid 4He

Superfluid 3He

Lowest T of condensed matter

Tem

pera

ture

, K

Sup

erco

nduc

tivity

Ele

ctro

nic

mag

netis

mN

ucle

arm

agne

tism

More on Refrigerators

H

HH T

QS =Δ

C

CC T

QS =Δ

HQ

CQ

W

heat

work

heat

hot reservoir, TH

cold reservoir, TC

We can create a refrigerator by running a Carnot engine backwards: the gas extracts heat from the cold reservoir and deposit it in the cold reservoir.

entr

opy

TC

P

V

TH

1

2

34

rejectsheat

absorbs heat

However, it is difficult to built a practical refrigerator usingsuch an ideal gas cycles as Carnot or Stirling. Most modern refrigerators use some liquid as a working substance. These schemes are better than a reversed Carnot cycle: much more energy can be absorbed per unit mass of working substance. Thus, the fridge can be much smaller than a Carnot fridge could be.

How Low Temperatures Are Produced

Although the efficiency of an “ideal” refrigerator does not depend on the working substance, in practice the choice of working substance is very importantbecause• we can never achieve the ideal reversibility of the Carnot engine;• the rate of cooling becomes important in real low-T machines.

cooling volume“cold reservoir”

δQ from the environment

δQ to the fridge

At the lowest T, these two flows of thermal energy compensate each other.

Two important characteristics of the working substance:it should have a large entropy (δQ ~TdS) which can be reduced by practical methods at the lowest convenient starting temperature;the substance should be capable of reversible changes down to the lowest temperatures at which it is to be used.

Down to ~ 1K, gases satisfy these requirements: they can be easily compressed and transported through pipes. Liquefied gases with their relatively large latent heats form convenient temperature reservoirs.

More on Enthalpy

The notion of enthalpy simplifies consideration of many processes we’ll consider today.

The reason for such a definition of H: consider the boiling of a liquid in a cylinder closed with a piston, the outside pressure = const. When the liquid is completely converted into gas, its internal energy increases by ΔU, and the volume increases by ΔV. To achieve this transformation, we have to supply the energy

HVPU Δ=Δ+Δ

( )PdVdUdHconstPVdPPdVdUdH +=⇒=++=

In particular, the latent heat L of phase transformation (the thermal energy we need to supply/remove from a unit mass of some substance to perform phase transformation) at P = const:

( ) LHH P =− 21

⎟⎠⎞

⎜⎝⎛ =+=++=

TLdSVdPTdSVdPPdVdUdH

( ) ∫=+=T

PdTCPVTUH0

Cooling of Gases

P1 P2V1T1

V2T2

δ W

Many processes we’ll consider today are based on a flow of a working substance through a “black box”, which is driven by a constant pressure difference ΔP = P1 - P2 . Two types of “black boxes”: (a) an “expansion engine” - a mechanical device (e.g., a turbine) that “extracts” some mechanical work from the working substance (δ W ≠ 0);(b) a porous membrane or a constriction that maintains ΔP = P1 - P2. In this case, δ W = 0.

Both processes are described by the same equation based on energy conservation:

The total energy required to “fill” V1 with the gas heated up to T1 at constant P1 (to “create” the gas at a given pressure):

( ) 1111 VPTUH +=

( ) 2222 VPTUH +=Similarly, for V2 :

( ) ( ) WVPUVPUHH δ=+−+=− 22211121

( ) 1111 VPTUH +=

( ) 2222 VPTUH +=

δ W

Simple Expansion Refrigerator

Compression at T~ 300K and cooling to T~ 300K by ejecting heat into the environment

gas pre-cooling in a counterflow heat exchanger

cooling to the lowest T in the expansion engine, usually a low friction turbine

heat extraction from the cooling load.

Coolingvolume

Expansionengine

Heatexchanger

Heatejection

Compressor

The work extracted from a fixed mass of the working gas by the expansion engine:

For an ideal monatomic gas:

BB kN

WTTTkNH52

25

21 =−=

21 HHW −=δ

This process works for both ideal and real gases.

Let’s consider case (b) - the process of expansion through a cnstriction or porous plug. This is the so-called throttling or Joule-Thomson process. The JT effect is essentially irreversible (this is a disadvantage), but it does not require moving mechanical parts of the fridge at low T.

222111 VPUVPU +=+021 ==− WHH δ

The Joule-Thomson Process

21 HH =

For an ideal gas, this process won’t result in any T change:

TkNfTkNTkNfVPUH BBB 22

2+

=+=+= H = const means T = constfor an ideal gas

The JT process corresponds to an isenthalpic expansion:

The problem in implementing the expansion engine for deep cooling (e.g., gas liquefaction): the engine does not work well at low T (no good lubricants!)

Thus, we cannot cool an ideal gas by going through the Joule-Thomson process!(Recall a similar process – expansion of an ideal gas through a hole in vacuum).

Luckily, for real gases, the temperature does change in the Joule-Thomson process.

The JT Process in Real Gases

Example: Two containers of volume V0 each are separated by a closed valve. Initially, one container is empty (vacuum) and the other container is filled a van der Waals gas (UvdW=U-const/V). The valve is open to allow a free expansion of the gas. The two containers are thermally isolated from the environment. Find the expression for the change in gas temperature after the free expansion.

The internal energy of the vdW gas: V

aNTNkV

aNUUUU BidealpotkinvdW

22

23

−=−=+=

ffB

iiBvdW V

aNTNkV

aNTNkU22

25

250 −=−=ΔFree expansion (δQ = 0, δ W = 0) :

000 5211

5211

52

VkNaT

VVNa

kT

VVNa

kTT

Bi

Bi

iiBif −=⎟⎟

⎠

⎞⎜⎜⎝

⎛−−=⎟⎟

⎠

⎞⎜⎜⎝

⎛−−=

The final temperature is lower than the initial temperature: the gas molecules work against the attraction forces, and this work comes at the expense of their kinetic energy.

PVUUH potkin ++=

In real gases, molecules interact with each other. At low densities, the intermolecular forces are attractive. When the gas expands adiabatically, the average potential energy increases, at the expense of the kinetic energy. Thus, the temperature decreases because of the internal work done by the molecules during expansion.

Upot x

expansion

vdW gas

The JT Process in Real Gases (cont.)

For T < TINV, the drop in pressure (expansion) results in a temperature drop.

Gas boiling T(P=1 bar)

inversion T@ P=1 bar

CO2 195 (2050)

CH4 112 (1290)

O2 90.2 893

N2 77.4 621

H2 20.3 2054He 4.21 513He 3.19 (23)

isenthalpic curves (H =const) for

ideal and real gases

cooling

heating

All gases have two inversion temperatures: in the range between the upper and lower inversion temperatures, the JT process cools the gas, outside this range it heats the gas.

Upot x

expansion

At high densities, the effect is reversed: the free expansion results in heating, not cooling. The overall situation is complicated: the sign of ΔT depends on initial Tand P.

Liquefaction of Gases

For air, the inversion T is above RT. In 1885, Carl von Lindeliquefied air in a liquefier based solely on the JT process: thegas is recirculated and, since T is below its inversion T, it cools on expansion through the throttle. The cooled gas cools the high-pressure gas, which cools still further as it expands. Eventually liquefied gas drips from the throttle.

( ) outliqin HHH λλ −+= 1 ( ) ( )outoutoutininin PTHHPTHH ,, ==

liqout

inout

HHHH

−−

=λLiquefaction takes place if ( ) ( )inininoutoutout PTHHPTHH ,, =>=The fraction of N2 liquefied on each pass through a Linde cycle operating between Pin = 100 bar and Pout = 1 bar at Tin = 200 K:

( ) J/mole 4442K200 ,bar100 =inH( ) J/mole 5800K200 ,bar1 =outH( ) J/mole 34077K7 ,bar1 −=liqH

( ) 15.034075800

44425800=

−−−

=λ (Pr. 4.34)

Most gas liquefiers combine the expansion engines with JT process: the expansion engine helps to pre-cool the gas below the inversion T. The expansion engines are a must for He and H2 liquefiers (the inversion T is well below RT).

Linderefrigerator

Estimate of efficiency: let 1 mole of gas enter the liquefier, suppose that the fraction λ is liquefied.

Historical Development of Refrigeration

1840 1860 1880 1900 1920 1940 1960 1980 2000

10-4

10-3

10-2

10-1

100

101

102

103

02 N2H2

4He3HeLo

w T

Ultr

a-lo

w T

10-5

Magnetic refrigeration,

electronic mag. m

oments

Magnetic refrigeration,

nuclear mag. m

oments

3He -4He ,

Faraday,chlorine 1823

Kamerlingh-Onnes

Cooling by Evaporation of Liquefied GasesOnce a liquid is produced, it offers a convenient way of going to lower temperatures by reducing the pressure over the liquid under adiabatic conditions. The reduced pressure causes the liquid to evaporate: the evaporation removes the latent heat of vaporization from the system and causes the temperature to fall (only the most energetic (“hottest”) atoms will leave the liquid to replenish the vapor: the mean energy of molecules in the liquid will decrease).

[ ] vapvapliq LdtdnHH

dtdnQ =−=δ δ Q – the cooling power, dn/dt – the number of

molecules moved across the liquid/vapor interface

Usually a pump with a constant-volume pumping speed is used, and thus the mass flow dn/dt is proportional to the vapor pressure.

( ) ⎟⎠⎞

⎜⎝⎛−∝∝

TTP

dtdn

vap1exp

( ) ( ) ⎟⎠⎞

⎜⎝⎛ −∝∝

TTPLTQ vapvap

1expδ Lvap – the latent heat of evaporation

Evaporation cooling is the dominant cooling principle in everyday cooling devices such as household refrigerators and air conditioners. The main difference between the kitchen fridge and LHe fridge operating below 4 K is in the working substance.

Properties of Cryoliquids

P , torr 10-4 10-3 10-2 10-1 1 10 100

T(4He), K 0.56 0.66 0.79 0.98 1.27 1.74 2.64T(3He), K 0.23 0.28 0.36 0.47 0.66 1.03 1.79

Substance boiling T(P=1 bar)

melting T(P=1 bar)

Latent heatkJ/liter

Price$ / liter

H2O 373.15 273.15 2252

303

245

160

31.8

2.56

0.48

Xe 165.1 161.3

O2 90.2 54.4

N2 77.4 63.3 0.3

H2 20.3 14.0

4He 4.21 -- 4

3He 3.19 -- 5x104

the cooling power diminishes rapidly with decreasing T

(at T→0, δSbecomes small for

all processes)

LHe is not solidified with decreasing T at pressures less than 20 bar (large zero-point oscillations of light non-interacting atoms) – this is an important advantage, because a good thermal contact with a solid is a problem at low T. Thus, below ~4K, the simplest route to lower T is by the evaporation cooling of LHe (pumping on helium vapor above the LHe surface).

Kitchen Refrigerator

A liquid with suitable characteristics (e.g., Freon) circulates through the system. The compressor pushes the liquid through the condenser coil at a high pressure (~ 10 atm). The liquid sprays through a throttling valve into the evaporation coil which is maintained by the compressor at a low pressure (~ 2 atm).

cold reservoir(fridge interior)

T=50C

hot reservoir(fridge exterior)

T=250C

P

V

liqui

d

gas liquid+gas

1

23

4compressor

condenser

throttling valve

evaporator

( )4132

41

HHHHHH

QQQCOP

CH

C

−−−−

=−

=

processes at P = const,

δQ=dH

The enthalpies Hi can be found in tables.

Dilution Refrigerator (down to a few mK)

At sufficiently low T (< 0.87K), a 3He-4He mixture is unstable: it undergoes phase separation. The concentrated (rich) 3He phase with a lower density floats up on top of the dilute 3He phase (gravity is essential).

Cooling is achieved by transferring 3He atoms from the rich 3He phase to the dilute 3He phase (“evaporation”). The temperature dependence of the cooling power in the 3He-4He dilution process is much weaker than that of a simple evaporation process: the concentration of 3He in the dilute 3He phase never drops below ~ 6.5%. High 3He density in the dilute phase enables a high 3He molar flow rate, and, thus, a large cooling power (in other words, the 3He-4He mixture has a large entropy).

( ) ( ) 2

0

2 TTCdTHTPHTdQT

vap ∝⎥⎦

⎤⎢⎣

⎡∝Δ∝Δ∝Δ∝ ∫

ΔH – the enthalpy difference between the 3He-rich and dilute phases

(evaporation cooling with a non-exponential dependence Pvap(T)

Cooling by Adiabatic Demagnetization

iBi nTkE ln−=

the slope ~ T

TkE

iB

i

en−

∝Let’s consider a quantum system with Boltzmann distribution of population probabilities for two discrete levels:

- lnni

Ei 1

1 – 2 - Isothermal increase of B from B1 to B2 . The upper energy level rises because W has been done by external forces. If T = const, the work performed must be followed by population rearrangement, so that the red line is shifted, but its slope ~ Tremains the same: e.g., if the magnetic field is increased, the population must decrease at the highest level and increase at the lowest – S decreases!

- lnni

Ei

- lnni

Ei

2

3

B1

B2

0 2 40

0.5

1

fi

1

xikBT/ μB

S/NkB B1

B2>B1

1

23

TiTf

⎥⎦

⎤⎢⎣

⎡−⎟⎟⎠

⎞⎜⎜⎝

⎛=⎟⎟

⎠

⎞⎜⎜⎝

⎛TkB

TkB

TkBkN

TkBNS

BBBB

B

μμμμ tanhcosh2ln,

B3

2 – 3 - Adiabatic decrease of B (the specimen is thermally isolated). S = const: the population of each level must be kept constant, while its Eivaries. The red line slope decreases – T decreases!

Temperatures attained: ~ 1 mK with electronic paramagnetic systems and ~ 1 μK with nuclear paramagnetic systems.

time timebefore before new equilibriumnew equilibrium

time at whichmagnetic field

is removed

time at whichmagnetic field

is removed

entro

py

tem

pera

ture

spin

lattice

lattice

spin

During isentropic demagnetization the total entropy of the specimen is constant. The entropy can flow into the spin system only from the system of lattice vibrations. The initial entropy of the lattice should be small in comparison with the entropy of the spin system in order to obtain significant cooling of the lattice – indeed, Slattice << Sspin holds in the mK range.

μ B/kBT

C

1

NkB/2Important: the heat capacity of the spins in the two-state paramagnetic is large at low T: 1 cm3 of iron ammonium salt at T=50 mK and B ~0.05 T has a heat capacity equal to 16 tons of lead (! ) at the same T.