Lecture 12. Toward Absolute Zero (Ch. 4)

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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  PV  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 T0, 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
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