1. Lecture 12. Refrigerators. Toward Absolute Zero (Ch. 4) 10 -5 10 9 Temperature, K 10 -3 10 -1 10 1 10 3 10 5 10 7 Center of hottest stars Center of Sun, nuclear reactions Electronic/chemical energy Surface of Sun, hottest boiling points Organic life Liquid air Liquid 4 He Universe Superfluid 4 He Superfluid 3 He Lowest T of condensed matter Superconductivity Electronic magnetism Nuclear magnetism

2. More on Refrigerators heat work heat hot reservoir, T H cold reservoir, T C 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. entropy TCTC P V THTH 1 2 3 4 rejects heat absorbs heat However, it is difficult to built a practical refrigerator using such 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.

3. 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 important because 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”  Q from the environment  Q 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 (  Q ~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.

4. 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  U, and the volume increases by  V. To achieve this transformation, we have to supply the energy 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:

5. Cooling of Gases P1P1 P2P2 V1T1V1T1 V2T2V2T2  W 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  P = P 1 - P 2. 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 (  W  0); (b) a porous membrane or a constriction that maintains  P = P 1 - P 2. In this case,  W = 0. Both processes are described by the same equation based on energy conservation: The total energy required to “fill” V 1 with the gas heated up to T 1 at constant P 1 (to “create” the gas at a given pressure): Similarly, for V 2 :  W W

6. 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. Cooling volume Expansion engine Heat exchanger Heat ejection Compressor The work extracted from a fixed mass of the working gas by the expansion engine: For an ideal monatomic gas: This process works for both ideal and real gases.

7. 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. The Joule-Thomson Process For an ideal gas, this process won’t result in any T change: H = const means T = const for 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.

8. The JT Process in Real Gases Example: Two containers of volume V 0 each are separated by a closed valve. Initially, one container is empty (vacuum) and the other container is filled a van der Waals gas (U vdW =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: Free expansion (  Q = 0,  W = 0) : 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. 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. U pot x expansion vdW gas

9. The JT Process in Real Gases (cont.) For T < T INV, the drop in pressure (expansion) results in a temperature drop. Gasboiling T (P=1 bar) inversion T @ P=1 bar CO 2 195(2050) CH 4 112(1290) O2O2 90.2893 N2N2 77.4621 H2H2 20.3205 4 He4.2151 3 He3.19(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. U pot 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  T depends on initial T and P.

10. Liquefaction of Gases For air, the inversion T is above RT. In 1885, Carl von Linde liquefied air in a liquefier based solely on the JT process: the gas 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. Liquefaction takes place if The fraction of N 2 liquefied on each pass through a Linde cycle operating between P in = 100 bar and P out = 1 bar at T in = 200 K: (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 H 2 liquefiers (the inversion T is well below RT). Linde refrigerator Estimate of efficiency: let 1 mole of gas enter the liquefier, suppose that the fraction is liquefied.

11. Historical Development of Refrigeration 184018601880 1900192019401960 1980 2000 10 -4 10 -3 10 -2 10 -1 10 0 10 1 10 2 10 3 0202 N2N2 H2H2 4 He 3 He Low T Ultra-low T 10 -5 Magnetic refrigeration, electronic mag. moments Magnetic refrigeration, nuclear mag. moments 3 He -4 He, Faraday, chlorine 1823 Kamerlingh- Onnes

12. Cooling by Evaporation of Liquefied Gases Once 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).  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. L vap – 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.

13. Properties of Cryoliquids P, torr10 -4 10 -3 10 -2 10 -1 110100 T( 4 He), K0.560.660.790.981.271.742.64 T( 3 He), K0.230.280.360.470.661.031.79 Substanceboiling T (P=1 bar) melting T (P=1 bar) Latent heat kJ/liter Price $ / liter H2OH2O373.15273.152252 Xe165.1161.3303 O2O2 90.254.4245 N2N2 77.463.31600.3 H2H2 20.314.031.8 4 He4.21--2.564 3 He3.19--0.485x10 4 the cooling power diminishes rapidly with decreasing T (at T  0,  S becomes 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).

14. 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=5 0 C hot reservoir (fridge exterior) T=25 0 C P V liquid gas liquid+gas 1 2 3 4 compressor condenser throttling valve evaporator processes at P = const,  Q=dH The enthalpies H i can be found in tables.

15. Dilution Refrigerator (down to a few mK) At sufficiently low T (< 0.87K), a 3 He- 4 He mixture is unstable: it undergoes phase separation. The concentrated (rich) 3 He phase with a lower density floats up on top of the dilute 3 He phase (gravity is essential). Cooling is achieved by transferring 3 He atoms from the rich 3 He phase to the dilute 3 He phase (“evaporation”). The temperature dependence of the cooling power in the 3 He- 4 He dilution process is much weaker than that of a simple evaporation process: the concentration of 3 He in the dilute 3 He phase never drops below ~ 6.5%. High 3 He density in the dilute phase enables a high 3 He molar flow rate, and, thus, a large cooling power (in other words, the 3 He- 4 He mixture has a large entropy).  H – the enthalpy difference between the 3 He-rich and dilute phases (evaporation cooling with a non-exponential dependence P vap (T)

16. Cooling by Adiabatic Demagnetization the slope ~ T Let’s consider a quantum system with Boltzmann distribution of population probabilities for two discrete levels: - lnn i EiEi 1 1 – 2 - Isothermal increase of B from B 1 to B 2. 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 ~ T remains 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! - lnn i EiEi EiEi 2 3 B1B1 B2B2 k B T/  B S/Nk B B1B1 B2>B1B2>B1 1 2 3 TiTi TfTf B3B3 2 – 3 - Adiabatic decrease of B (the specimen is thermally isolated). S = const: the population of each level must be kept constant, while its E i varies. The red line slope decreases – T decreases!

17. Temperatures attained: ~ 1 mK with electronic paramagnetic systems and ~ 1  K with nuclear paramagnetic systems. time before new equilibrium time at which magnetic field is removed time at which magnetic field is removed entropy temperature spin 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, S lattice << S spin holds in the mK range.  B/k B T C 1 Nk B /2 Important: the heat capacity of the spins in the two-state paramagnetic is large at low T: 1 cm 3 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.