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109 Center of hottest stars Lecture 12. Refrigerators. Toward Absolute Zero (Ch. 4) Center of Sun, nuclear reactions 107 Electronic/chemical energy Surface of Sun, hottest boiling points 103 101 Liquid 4He Universe Superfluid 4He 10-1 10-3 10-5 Superfluid 3He Lowest T of condensed matter Nuclear magnetism Liquid air Electronic magnetism Organic life Superconductivity Temperature, K 105 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 Q from the Q to the environment fridge cooling volume “cold reservoir” At the lowest T, these two flows of thermal energy compensate each other. T More on Enthalpy H U T PV CP dT Recall: U PV H dH dU PdV VdP P const dH dU PdV The latent heat L of phase transformation at P = const: H1 H 2 P L dH dU PdV VdP TdS VdP L dS T 0 Cooling of Gases P1 V1 T1 V2 T2 P2 constant P = P1 - P2 . (a) an “expansion engine” ( W 0); W H1 U T1 P1V1 W H 2 U T2 P2V2 (b) a porous membrane or a constriction. ( W = 0). H1 H 2 U1 P1V1 U 2 P2V2 W Compressor Heat ejection Heat exchanger Expansion engine Simple Expansion Refrigerator This process works for both ideal and real gases. W H1 H 2 For an ideal monatomic gas: Cooling volume 5 2 W H Nk BT T1 T2 2 5 Nk B The Joule-Thomson Process (b) Throttling process H1 H 2 W 0 • Irreversible!! • Real gases only Isenthalpic expansion: For an ideal gas, : H 0 H1 H 2 f 2 H Nk BT 2 T 0 The JT Process in Real Gases (low density) Upot vdW gas x expansion U T ,V Ukin T U pot V H 0, V 0 T 0 The JT Process in Real Gases (high density) At high densities, the effect is reversed: the sign of T depends on initial T and P. Upot x expansion H 0, V 0 T 0 The JT Process in Real Gases (cont.) 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. heating cooling 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 20.3 205 4He 4.21 51 3He 3.19 (23) 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: Linde refrigerator “Efficiency” of liquefaction Estimate of efficiency: let 1 mole of gas enter the liquefier, suppose that the fraction is liquefied. Hin Hliq 1 Hout Hin H Tin , Pin Hout H Tout , Pout H out H in H out H liq Liquefaction takes place if Hout H Tout , Pout Hin H Tin , Pin Example (Pr. 4.34, Pg 143) 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: H out H in H out H liq Hin 100bar, 200K 4442J/mole Hout 1bar, 200K 5800J/mole Hliq 1bar, 77K 3407J/mole 5800 4442 0.15 5800 3407 Historical Development of Refrigeration 103 Faraday, chlorine 1823 02 102 N2 101 100 Low T H2 4He KamerlinghOnnes 3He 10-2 10-3 Ultra-low T 10-1 10-4 10-5 1840 1860 1880 1900 1920 1940 1960 1980 2000 Cooling by Evaporation of Liquids dn dn H liq H vap Q Lvap dt dt 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 1 Pvap T exp dt T Pr. 5.35 Substanc e boiling T (P=1 bar) melting T (P=1 bar) Latent heat kJ/liter Price $ / liter H2O 373.15 273.15 2252 Xe 165.1 161.3 303 O2 90.2 54.4 245 N2 77.4 63.3 160 H2 20.3 14.0 31.8 4He 4.21 -- 2.56 8 3He 3.19 -- 0.48 5x104 0.3 Cryoliquids the cooling power diminishes rapidly with decreasing T (at T0, S becomes small for all processes) the evaporation cooling of Liquid Helium 10-3 10-2 10-1 1 10 100 T(4He), K 0.56 0.66 0.79 0.98 1.27 1.74 2.64 T(3He), K 0.23 0.28 0.36 0.47 0.66 1.03 1.79 P , torr 10-4 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). condenser throttling valve liquid P 2 3 compressor 1 4 cold reservoir (fridge interior) T=50C processes at P = const, Q=dH gas liquid+gas evaporator COP V QC H1 H 4 H H4 1 QH QC H 2 H 3 H1 H 4 H 2 H1 The enthalpies Hi can be found in tables. hot reservoir (fridge exterior) T=250C H 3 H 4 , H 3liquid x H 4liquid 1 x H 4gas S2 S1 T2 H 2 T2 , P2 Dilution Refrigerator (down to a few mK) evaporation cooling with a non-exponential dependence Pvap(T) T dn 2 dQ T H H CdT T T 2 dt 0 H – the enthalpy difference between the 3He-rich and dilute phases the slope ~ T 1 Ei Cooling by Adiabatic Demagnetization Let’s consider a quantum system with Boltzmann distribution of population probabilities for two discrete levels: B1 - lnni Ei 2 B2 - lnni 3 B3 - lnni Ei ni e Ei k BT Ei kBT ln ni 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 ~ 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! 2 – 3 - Adiabatic decrease of B (the specimen is thermally isolated). S = const: the population of each level must be kept constant, while its Ei varies. The red line slope decreases – T decreases! 1 S/NkB fi 0.5 0 B B B B S N, N k B ln 2 cosh k T k T tanh k T k T B B B B B1 1 3 0 2 Tf 2 B2>B1 Ti 4 1 kBT/ B xi Nobel Prize in Physics 1997 Laser cooling • Very dilute gas (avoid condensing) • Momentum transfer from photons slowing down of molecules TK.E. decrease Pphoton h , Pmolecule Pphoton , K.E. K.E. P photon min 2 Pmolecule 2 2m 2m http://www.colorado.edu/physics/2000/bec/lascool1.html