Lecture 12. Refrigerators. Toward Absolute Zero (Ch. 4)

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1 Center of hottest stars Center of Sun, nuclear reactions Lecture. Refrigerators. oward Absolute Zero (Ch. ) emperature, K Electronic/chemical energy Surface of Sun, hottest boiling points Organic life Liquid air Liquid e Universe Superfluid e Superfluid e Superconductivity Lowest of condensed matter Electronic magnetism Nuclear magnetism

2 More on Refrigerators hot reservoir, 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. ΔS Q Q P rejects heat entropy heat heat work W absorbs heat C Q ΔS C C C Q C cold reservoir, C owever, 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. hese schemes are better than a reversed Carnot cycle: much more energy can be absorbed per unit mass of working substance. hus, the fridge can be much smaller than a Carnot fridge could be. V

3 ow Low emperatures 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- machines. δq from the environment cooling volume cold reservoir δq to the fridge At the lowest, these two flows of thermal energy compensate each other. wo important characteristics of the working substance: it should have a large entropy (δq ~ds) 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 ~ K, 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 U ( ) + PV 0 C P d he notion of enthalpy simplifies consideration of many processes we ll consider today. he reason for such a definition of : 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. o achieve this transformation, we have to supply the energy ΔU + PΔV Δ ( d du + PdV + VdP P const d du + PdV ) 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: ( ) L P d du + PdV + VdP ds + VdP ds L

Cooling of Gases V V P P δ W ( ) P + U V 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 P - P.

5 Cooling of Gases V V P P δ W ( ) P + U V 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 P - P. wo 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 P -P. In this case, δ W 0. oth processes are described by the same equation based on energy conservation: ( ) P + U V δ W he total energy required to fill V with the gas heated up to at constant P (to create the gas at a given pressure): ( ) P + U V Similarly, for V : ( ) P + U V ( U + PV ) ( U + PV ) δ W

Simple Expansion Refrigerator Compressor eat ejection eat exchanger Compression at ~ 00K and cooling to ~ 00K by ejecting heat into the environment gas pre-cooling in a counterflow heat exchanger

6 Simple Expansion Refrigerator Compressor eat ejection eat exchanger Compression at ~ 00K and cooling to ~ 00K by ejecting heat into the environment gas pre-cooling in a counterflow heat exchanger cooling to the lowest in the expansion engine, usually a low friction turbine heat extraction from the cooling load. Expansion engine he work extracted from a fixed mass of the working gas by the expansion engine: δ W For an ideal monatomic gas: 5 N k 5 W N k Cooling volume his process works for both ideal and real gases.

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

7 he Joule-homson Process he problem in implementing the expansion engine for deep cooling (e.g., gas liquefaction): the engine does not work well at low (no good lubricants!) Let s consider case (b) - the process of expansion through a cnstriction or porous plug. his is the socalled throttling or Joule-homson process. he J effect is essentially irreversible (this is a disadvantage), but it does not require moving mechanical parts of the fridge at low. he J process corresponds to an isenthalpic expansion: δ W 0 For an ideal gas, this process won t result in any change: U + + PV U P V f f + U + PV N k + N k N k const means const for an ideal gas hus, we cannot cool an ideal gas by going through the Joule-homson 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-homson process.

8 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. hus, the temperature decreases because of the internal work done by the molecules during expansion. U kin + U pot + PV he J Process in Real Gases U pot vdw gas x expansion Example: wo 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). he valve is open to allow a free expansion of the gas. he two containers are thermally isolated from the environment. Find the expression for the change in gas temperature after the free expansion. he internal energy of the vdw gas: U vdw Free expansion (δq 0, δ W 0) : f i 5k Na Vi V i U i kin 5k + U ΔU vdw pot Na V U ideal N a V Nk i N a V 5 N a 5 0 Nki Nkf V i 0 V 0 5 he 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. Na k V 0 N V f a

9 he J Process in Real Gases (cont.) At high densities, the effect is reversed: the free expansion results in heating, not cooling. he overall situation is complicated: the sign of Δ depends on initial and P. U pot expansion x heating All gases have two inversion temperatures: in the range between the upper and lower inversion temperatures, the J process cools the gas, outside this range it heats the gas. cooling isenthalpic curves ( const) for ideal and real gases For < INV, the drop in pressure (expansion) results in a temperature drop. Gas boiling (P bar) P bar CO 95 (050) C (90) O N e. 5 e.9 ()

Liquefaction of Gases Most gas liquefiers combine the expansion engines with J process: the expansion engine helps to pre-cool the gas below the inversion.

10 Liquefaction of Gases Most gas liquefiers combine the expansion engines with J process: the expansion engine helps to pre-cool the gas below the inversion. he expansion engines are a must for e and liquefiers (the inversion is well below R). For air, the inversion is above R. In 885, Carl von Linde liquefied air in a liquefier based solely on the J process: the gas is recirculated and, since is below its inversion, it cools on expansion through the throttle. he cooled gas cools the high-pressure gas, which cools still further as it expands. Eventually liquefied gas drips from the throttle. Estimate of efficiency: let mole of gas enter the liquefier, suppose that the fraction λ is liquefied. in λ in out liq λliq + ( λ) out in ( in, Pin ) out ( out, Pout ) out out in liq ( 00 bar, 00K) J/mole ( bar, 00K) 5800 J/mole ( bar, 77K) 07 J/mole Liquefaction takes place if (, P ) > (, P ) out he fraction of N liquefied on each pass through a Linde cycle operating between P in 00 bar and P out bar at in 00 K: out 5800 λ 0.5 (Pr..) 5800 ( 07) out in Linde refrigerator in in

istorical Development of Refrigeration Faraday, chlorine 8 0 0 0 0 0 0 N e e Kamerlingh- Onnes Low 0 - e - e, Magnetic

11 istorical Development of Refrigeration Faraday, chlorine N e e Kamerlingh- Onnes Low 0 - e - e, Magnetic refrigeration, electronic mag. moments Ultra-low 0 - Magnetic refrigeration, nuclear mag. moments

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. he 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). δ dn [ liq vap ] Lvap dn Q dt dt 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. dn dt P vap ( ) exp δ Q ( ) L P ( ) exp vap vap 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. he main difference between the kitchen fridge and Le fridge operating below K is in the working substance.

13 Substance boiling melting Latent heat (P bar) (P bar) kj/liter O Price $ / liter Properties of Cryoliquids Xe O N e the cooling power diminishes rapidly with decreasing (at 0, δs becomes small for all processes) e x0 Le is not solidified with decreasing at pressures less than 0 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. hus, below ~K, the simplest route to lower is by the evaporation cooling of Le (pumping on helium vapor above the Le surface). P, torr ( e), K ( e), K

Kitchen Refrigerator A liquid with suitable characteristics (e.g., Freon) circulates through the system.

he liquid sprays through a throttling valve into the evaporation coil which is maintained by

cold reservoir (fridge interior) 5 0 C P throttling valve liquid condenser liquid+gas

14 Kitchen Refrigerator A liquid with suitable characteristics (e.g., Freon) circulates through the system. he compressor pushes the liquid through the condenser coil at a high pressure (~ 0 atm). he liquid sprays through a throttling valve into the evaporation coil which is maintained by the compressor at a low pressure (~ atm). cold reservoir (fridge interior) 5 0 C P throttling valve liquid condenser liquid+gas compressor gas processes at P const, δqd evaporator V COP Q QC Q C ( ) hot reservoir (fridge exterior) 5 0 C he enthalpies i can be found in tables.

Dilution Refrigerator (down to a few mk) (evaporation cooling with a non-exponential dependence P vap () At sufficiently low (< 0.87K), a e- e mixture is unstable: it undergoes phase separation.

15 Dilution Refrigerator (down to a few mk) (evaporation cooling with a non-exponential dependence P vap () At sufficiently low (< 0.87K), a e- e mixture is unstable: it undergoes phase separation. he concentrated (rich) e phase with a lower density floats up on top of the dilute e phase (gravity is essential). Cooling is achieved by transferring e atoms from the rich e phase to the dilute e phase ( evaporation ). he temperature dependence of the cooling power in the e- e dilution process is much weaker than that of a simple evaporation process: the concentration of e in the dilute e phase never drops below ~ 6.5%. igh e density in the dilute phase enables a high e molar flow rate, and, thus, a large cooling power (in other words, the e- e mixture has a large entropy). dq ( ) Δ P ( ) Δ ΔCd vap Δ the enthalpy difference between the e-rich and dilute phases 0

16 the slope ~ E i Cooling by Adiabatic Demagnetization Let s consider a quantum system with oltzmann distribution of population probabilities for two discrete levels: E n i i k e Ei k ln n i -lnn i -lnn i -lnn i E i E i μ S N, N k k - Isothermal increase of from to. he upper energy level rises because W has been done by external forces. If const, the work performed must be followed by population rearrangement, so that the red line is shifted, but its slope ~ 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! - Adiabatic decrease of (the specimen is thermally isolated). S const: the population of each level must be kept constant, while its E i varies. he red line slope decreases decreases! μ μ ln cosh k k f i μ tanh k S/Nk f i x i > k / μ

17 During isentropic demagnetization the total entropy of the specimen is constant. he entropy can flow into the spin system only from the system of lattice vibrations. he 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. entropy spin lattice temperature lattice time time before new equilibrium before new equilibrium spin C Nk / time at which magnetic field is removed time at which magnetic field is removed Important: the heat capacity of the spins in the two-state paramagnetic is large at low : cm of iron ammonium salt at 50 mk and ~0.05 has a heat capacity equal to 6 tons of lead (! ) at the same. μ /k emperatures attained: ~ mk with electronic paramagnetic systems and ~ μk with nuclear paramagnetic systems.

Physics Nov Cooling by Expansion

Physics Nov Cooling by Expansion Physics 301 19-Nov-2004 25-1 Cooling by Expansion Now we re going to change the subject and consider the techniques used to get really cold temperatures. Of course, the best way to learn about these techniques