Lecture 06 : Refrigerators, Cooling of gases, The Joule Thomson Process.

How the domestic refrigerators operate

1. Outline
1.Refrigerators
2.Cooling of Gases
3.The Joule-Thomson Process
4.Liquefaction of Gases
5.Kitchen Refrigerator.
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2. Refrigerators
H
H
H
T
Q
S
C
C
C
T
Q
S
HQ
CQ
W
heat
work
h
e
a
t

hot reservoir, T
H
cold reservoir, T
C
The purpose of a refrigerator is to make thermal
energy flow from cold to hot. The Coefficient of
Performance (CoP) for a fridge:
e
n
t
r
o
p
y
1/
1




CHCH
CC
QQQQ
Q
W
Q
COP
C
H
C
H
T
T
Q
Q

CH
C
TT
T
COPCOP


max
COP is the largest when T
H and T
C are close
to each other!
For a typical kitchen fridge T
H
~300K, T
C
~
250K  COP ~ 6 (for each J of el. energy,
the coolant can suck as much as 6 J of heat
from the inside of the freezer).
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3. Example:
A “perfect” heat engine with e = 0.4 is used as a refrigerator (the heat
reservoirs remain the same). How much heat Q
C
can be transferred in one
cycle from the cold reservoir to the hot one if the supplied in one cycle
work is W =10 kJ?
,
H
CH
H
Q
QQ
Q
W
e


H
C
H
C
CH
cc
Q
Q
Q
Q
QQ
Q
W
Q
COP




1
,1e
Q
Q
H
C
 5.1
4.0
6.01
11
1






e
e
e
e
COP
kJ 15cycle) one(in COPWQ
C
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4. More on Refrigerators
H
H
H
T
Q
S
C
C
C
T
Q
S
HQ
CQ
W
heat
work
h
e
a
t

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 hot reservoir.
e
n
t
r
o
p
y
T
C

P
V
T
H

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.
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5. 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.
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6. 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
HVPU 
 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







T
L
dSVdPTdSVdPPdVdUdH ,
 

T
P
dTCPVTUH
0
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7. Cooling of Gases
P
1 P
2
V
1
T
1
V
2
T
2
 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):

1111 VPTUH 

2222
VPTUH Similarly, for V
2
:
  WVPUVPUHH 
22211121

1111 VPTUH 

2222 VPTUH 
 W
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8. 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:
B
B
kN
W
TTTkNH
5
2
,
2
5
21

21
HHW 
This process works for both ideal and real gases.
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9. 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 0
21
 WHH 
The Joule-Thomson
Process
21HH
For an ideal gas, this process won’t result in any T change:
TkN
f
TkNTkN
f
VPUH
BBB
2
2
2


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.
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10. 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: V
aN
TNk
V
aN
UUUU
BidealpotkinvdW
22
2
5

f
fB
i
iBvdW
V
aN
TNk
V
aN
TNkU
22
2
5
2
5
,0 
Free expansion (Q = 0,  W = 0) :
000
52
11
5
211
5
2
Vk
Na
T
VV
Na
k
T
VV
Na
k
TT
B
i
B
i
iiB
if



















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.
U
pot
x
expansion
vdW gas
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11. The JT Process in Real Gases (cont.)
For T < T
INV
, the
drop in pressure
(expansion) results
in a temperature
drop.
Gas boiling T
(P=1 bar)
inversion T
@ P=1 bar
CO
2 195 (2050)
CH
4 112 (1290)
O
2
90.2 893
N
2
77.4 621
H
2 20.3 205
4
He 4.21 51
3
He 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.
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.
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12. 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.
,1
outliqin HHH      
outoutoutininin
PTHHPTHH ,,, 
liqout
inout
HH
HH



Liquefaction takes place if    
inininoutoutout PTHHPTHH ,, 
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:
  J/mole 4442K200 ,bar100 
inH
  J/mole 5800K200 ,bar1 
outH
  J/mole 34077K7 ,bar1 
liqH
 
15.0
34075800
44425800




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.
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13. 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).
 
vapvapliq L
dt
dn
HH
dt
dn
Q 
 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.
 






T
TP
dt
dn
vap
1
exp
  






T
TPLTQ
vapvap
1
exp 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.
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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
l
i
q
u
i
d
gas
liquid+gas
1
23
4
compressor
condenser
throttling
valve
evaporator
 
4132
41
HHHH
HH
QQ
Q
COP
CH
C





processes
at P = const,
 Q=dH
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