K
Thursday, august 22, 2013
KAM office hours = T at 2 PM, or T 3-4. KAM became interested in thermoelectricity in Saclay, France. Europe is very interested in thermoelectricity. You use a temperature-difference to make a voltage. If you know something about thermoelectricity that KAM does not know, bring it to class! Old textbooks on thermoelectricity = not useful; they do no t take the nano approach. ap proach. Current application: space travel. Desired application: as common as being in your car. C. Goupil, Ch. 13; DOI 105772/12988 1057 72/12988 - SSP - thermodynamics of thermoelectricity. 2003 - Caltech - Snyder Sn yder and Ursell - Thermoelectric Efficiency and Compatibility. Compatibility. Thermodynamics of thermoelectricity
Theremoelectric engines are currently inefficient. Associated with TE engine is figure of merit. T 0 ZT [figure of merit ] ; ZT C 1 C ; T H
(1.1)
Consider an ideal gas in a box b ox of volume V. pressure of gas can be computed. However, if gas-particles are charged, you will have chemical potential. Chemical potential then becomes related to the bias across the gas. P P(V , T ); e e (eV ) eV ;
(1.2)
Problem: Qualitatively describe how a temperature-gradient (maintained by a hot and cold reservoir separated by a thermoelectric junction) elicits a voltage-gradient. Solution: You have hot vs. cold reservoir, and the thermoelectric junction, appearing as,
(1.3) The temperature-gradient elicits a density gradient. You have a compressible Fermi-gas of electrons that is hot, and a similar gas of electrons that is cold. The resulting density-gradient, in turn, elicits a number-current (diffusion).
Problem: find an order-of-magnitude estimate for the current across the junction (1.3). solution: Consider transmission of a particle of energy to be an event with probability T ( ) . Consider left-center hopping to
have a parameter L
W L2 W
, and center-right hopping to have a parameter R
for left-right transmission be a Lorentzian function,
dN d
W R2 W
dN d
. Let the density of states
f ~ 1 2 f 2 , in which f f ( L , R ) f ( R , L ) , the
license to switch the variables being a consequence of time-reversal symmetry. We realize the simplest f satisfying this symmetry is f R L . Then, the left-right transmission function (which has no dimensions), using the pi-theorem, is approximately, 1
L , R 0
T
~ deg L, R
L R dN 1 dN 2 2 ~ T 0L,R ( ); deg L, R L, R 2 2 ( L R ) d d
(1.4)
The non-interacting transmission function is subsequently computed using the series-conductor-addition-law,
eq1 11 2 1 , and this appears as, 1
1
1 1 1 1 2 2 L R LR T0 T 0 ( ) ~ L R ; 2 2 2 2 T T ( ) ( ) 0 R L R L R L 0 Finally, the thermodynamic probability is given by f L,R (e L ,R
( L ,R )
(1.5)
1)1 , which implies that we treat the
electron gas as a grand canonical canon ical ensemble. By Kirchoff’s rule, we subtract left-center and center-right currents to constitute the net current,
I N I R I L
2 L R
2W
2W 2 ( L R )2 e d
1
R ( R )
1 L ( L )
1 e
I N (T L,R , L ,R ) ; 1
(1.6)
That the number-current is a function of the indica ted thermodynamic variables allows us to establish derivative-relations, if we wanted to. Problem: Show that for L R , you have zero number current. Solution: Let the chemical potential gradient
be entirely due to a temperature gradient. Then, L R TL T R , and (1.6) immediately yields I N 0 . 1 Note, also, that L R 0 causes the following integral to vanish,
2W
1 1 0 R, L , L ,R ; 1 e 1 2W ( L R ) e
d
2
2
R
R
(1.7)
Equilibrium: when left and right chemical potentials balance out. That is when the thermoelectric effect ceases. So, you are basically making a out of a T . These are encapsulated by f FD f FD (T , ) ( e ( E ) 1) 1 .
1
This is not because the Lorentzian
1 2 2
is an even function; replacing
1 2 2
with a cosine-function yields nonzero current.
Problem: Let R and L , and similarly T R T T and T L T . Series-expand the F f R f L
about 0 and T 0 , truncating at 1 order. Solution: we shall need the derivatives, in which x st
e( )/ k T k 1T 1 1 x f f 2 x sech ; e( )/ k T 1 (e( )/ k T 1)2 4 2
k BT
,
B
B
B
B
1 f k B kB T (k BT ) e( )/ k T 1 B
e( )/ k T k( T ) 0 B
2
2
B
(e
( )/ kB T
1) 2
1 x T 4
sech 2
x 2
f T
kB x
f ;
(1.8)
st
We shall also need the fact that the 1 term of any series expansion that is between the difference of two functions always vanishes, f f f f f f ( x, y ) f ( x x, y y) f ( x, y) f x y f x y; x y x y
(1.9)
Thus, the series-expansion appears as, F
f f f f T O2 (T ) O2 ( ) k B x T T
(1.10)
f 1 x x k B x T k B x T sech 2 4 2 Plotting the prefactor/dimensionless-scaling function of f scale ( x ) 14 x sech 2 2x , we have, f scale x 0.25 x Sech x 2 2
0.2
0.1
10
5
5
10
x
k B T
0.1
0.2
(1.11) Problem: write down the units of number-flux, and energy-flux. Solution: the units of flux of [stuff] is [stuff] per unit time per unit (normal) area, so we have,
[J Q ]
J m s 2
W m
2
; [J N ]
1 m s 2
; [J E ]
W m2
;
JE
J Q J N ;
(1.12)
Problem: consider a quasi-static system. Write an expression for the entropy flux. Solution: write,
dS
dQquasi-static T
dQqs T
J s
dS dt d 2 x
d (Qqs / T ) dt d 2 x
Problem: introduce so-called thermodynamic “forces”:
F N
1 dQqs T dt d 2x
JQ
T
entropy ; (1.13) [temperature] -flux [heat-flux]
( e / T ) J N and Fe (1 / T ) J E , which
elicit the indicated currents. Let there be cou pling between the forces. Write a linear “constitutive relation”; e.g., Fick’s Law of diffusion, for each current. Solution: since there is coupling between the forces, we will have terms making each current which is due to the other elicitor of the force. Specifically, J N , E
( e / T ) LNE J N J N , N ; J N
J N ,E J N , N ;
J E,N JE
(1/ T )LEN J E J E , E ;
(1.14)
J E , N J E , E ;
Together, we have the linear relations,
J N L NN J L E EN
LNE ( e / T )
micro-
reversibility
; L NE LEN
LEE (1/ T )
(1.15)
Problem: consider J E J Q e J N , and consider the new constitutive relations,
J N L11 L12 (1/ T ) e Onsager- ; L L J 21 (1/ T ) 12 relation ; L L Q 21 22
(1.16)
Find the Lij in terms of the L NN , LNE , LEN , LEE . Solution: Writing ( e / T ) e (1/ T ) e(1/ T ) , we compute
J N
of (1.15) and then (1.16) as, J N
LNN (
e T
1
) LNE(T ) LNN eT
1
e(T 1) LNE (T 1)
(1.17)
L NE LNN e (T 1 ) LNN eT 1 L12(T 1 ) L11 eT 1; 1 From (1.17), considering the variations (T ) and e as independent, we read off the relations,
L NE LNN e L12
1 e
LNE LNN
1 e
L12 ;
LNN L11 ;
(1.18)
Carrying out the same procedure for J Q J E e J N (favouring occurrences of J Q ), we have, JQ
J E e J N LEN e T1 e( T1 ) LEE ( T1 ) e LNE LNN e ( T1 ) LNN e T1
e LNN LEN
e LEN e LEE e LNE LNN e ( T1 ) T1 eL21 ( T 1 ) L22 ;
T
From (1.19), we read off, L21 L EN LNN e T
T
L21 L EN LNN e L12 ; L22 LEE (LNN e LEN LNE ) e ;
(1.19)
(1.20)
Problem: Write the inverse-transform of the expressions (1.18) and (1.20). solution: The L NN L11 is already
solved for, so we just need to invert a 3x3 system. However, be careful to make sure all elements of the columnvector have matching-dimensions, or else the coe fficient-matrix won’t make sense,
L11 1 0 0 1 L 12 1 1 0 1 L 1 0 1 21 1 L22 1 1 1 e
0 L NN 1 0 e L EN
e
e
e
e
1 0 1 1
2
2
e
Problem: Ohm’s Law is
e
LNN 1 1 LEN 1 1 1 L NE L NE 1 L 1 L EE EE 2
e
0 0 0 L11 1 1 0 0 e L12
1 0 1 0 1 1 1 1
e
e
e
2
e
L11 1 L11 L12 1 ; (1.21) L21 L11 L12 1 1 L22 L 11 2 L12 L22 2
e
e
E J N e . Compute the isothermal ( T 0 ) electrical conductivity in the
Je
system described by (1.16) (where we have the symmetry L12 L21 ), realizing E e / e . Solution: computing
J N
from (1.17),
Problem: Fourier’s law is
JQ
Je
T 0
E
eJ N
e
e / e
eL11 eT 1 e
e 2 L11
(1.22)
T
T . Compute the iso-electric thermal conductivity2 . Solution: using
(1.19) in which e 0 ,
Problem: Fourier’s law is
the condition
JQ
J N
E 0 e / e
T
1 T
L 0 L21 ( T 1 ) L22 222 E ; T T
(1.23)
T . Compute the thermal conductivity , this time using J N 0 . Solution:
0 , using (1.17), allows us to eliminate e in favour of ( T 1 ) . Repeating (1.23),
J N 0
T
JQ
JQ
1 T
e L21 ( T 1 ) L22 T
1 T
T
L12 L11
L21 L22 ( T 1 )
T
3 Problem: the Seebeck coefficient is defined as Se
Solution: the condition
J N
1 e
L12 L21 L11L22 L11T 2
L11L22 L12 L21 L11T 2
e ; (1.24)
(e / T ) J N 0 ; re-express this in terms of the Lij .
0 allows use of (1.17) to eliminate e in favour of ( T 1 ) T 2T . Repeating
(1.24), it is trivial to write, L L 1 Seebeck- thermo- 1 L12 1 e 1 T L (T ) 1 T L S e coefficient power ; 2 e T e e T eT L11 T 12
12
11
11
Problem: the Peltier coefficient is defined as
1 e
( JQ /
J N )T 0 ;
(1.25)
re-express this in terms of the Lij .
Solution: using (1.17) and (1.19), we effect direct computation and obtain,
2 3
1 J Q e
J N
|T 0
1
JQ
e
JN
T 0 T 0
1
1 T
e L21 0 L22
e L12 0 L11 eT 1
1 L12 e e L11 e
1 L12 e L11
T
1 L12 eT L11
T S e ;
Actually, there are two different thermal conductivities, since you are holding different thermodynamic variables constant. The symbol for the Seebeck coefficient in
(1.26)
Interlude: By (1.22), (1.23), (1.24), (1.25), and (1.26), we can write, L11
T e2
1 L12
; L22 T ET ; 2
2
L11L22 L12 L21 L11T 2
T 3
1 L12
e ; Z
S e 2 T e
;
(1.27)
L11L22
1 1 Se ; TSe ; L22 2 Se2 T 2 e ; 1; ZT L11L22 ; eT L11 e L11 e L12 L21 L12L21 1 Problem: Compute the entropy flux density, re-expressing e in terms of the Lij . Solution: we write, JS
1 T
JQ
1 1 L21 e L22T T T
1 T
( e ) L21 ( T 1 ) L22 T
1 L21 eT L11
Je
1
1 L22 ( ) T T
(1.28)
number thermal entropy per thermal N Q J J S S carrier entropy ; entropy entropy
Problem: Assume a continuity equation for the entropy flux density (1.28) , and compute ; this is computed as,
. Solution: let the continuity equation appear as J S s ; we directly compute, and use the continuity for particle-number and energy, J E E 0 N J N , and we get,
JQ 1 1 T J Q T 1 J E e J N T 1 J E e J N J Q T T 0 e J N T 1 J E e J N T 1 J E e T 1 J N T 1 e J N
S JS
T
(1.29)
e 1 ( ) J N forces fluxes Onsager-hypothesis ; J E T T
S
The thermoelectric engine: consider engine: a device between a hot and cold reservoir. A load is put between the hot and cold reservoirs. It is equivalent to a battery V0, a resistor R in, and a resistor R L, all in series, 4 assuming the linear/quasi-static regime. These resistances are the internal and load resistances, respectively,
(1.30) Limits: you have Rload RL J N 0 , the broken-circuit limit, where the maximum voltage V0 is across the load. The opposite limit is Rload RL 0 I C V0 / Rin , the short-circuit limit, where the maximum current flows through the entire loop.
4
Completely-unrealistic; a battery is manifestly a non-equilibrium object.
IV (current-voltage) characteristics: We are interested in power: P VI Rin I ( I C I ) . Maximum power occurs when Rload Rin , by maximum-power-transfer. We have,
P P(I )
P 0 Rload Rin [maximum-power]; I
Problem: The thermoelectric figure of merit is defined as ZT
JQ
0
/
JQ
J N 0
(1.31)
. Write this in terms of the Lij ,
and also show that ZT ( E / e ) 1 Se 2 T / e . Solution: directly computing, we have, ZT
JQ JQ
0
J N 0
1 T
1 T
T
L12 L11
1 ) L22 0 L21 ( T
( T1 ) L21 ( T1 ) L22
L11L22
1
???
L12 L21 L11L22
L11L22 L12L21
1
ZT ;
(1.32)
Having established (1.32), we can begin from the LHS of ZT Se 2 T / e , and use (1.27) throughout, as, Se 2 T e
2
1 L12 L11e2 L11T 2 L12 2 T eT L T L L L L L L L L 11 11 22 12 21 11 22 12 21
1 L11L22 L12L21
1
ZT ;
(1.33)
We also write, E e
2
1 T L22
L11T 2 L11L22 L12 L21
1
21 1 1 L12 11L22
L L
1 1 L1211L2221 L L
1
1 L1211L2221 L L
1 L11L22 L12L21
1
ZT ;
(1.34)
Making (1.32) as large as possible, even so we’re in the linear/equilibrium-regime, is equivalent to maximizing the non-equilibrium efficiency (1.1). Result: In (1.32), we immediately see that we want a high electrical conductivity and a low thermal conductivity. Wiedemann Franz law is a problem: The problem is that we have the correlation / e ~ 13 2 T ( k B / e) 2 LT , in which we have the Lorenz number 2.4410 W K . This requires 3 possible approaches (1) use non Fermi liquids (2) use the Kondo-regime, (3) use nano-s ystems. All three of these approaches are the topic of intense research. 8
2
Thermodynamic potential: G. Snyder and T. Ursell, Phys Rev. Lett 91, 148301 (2003), Problem: compute J Q for a non-radiating closed system. Solution: that the solution is closed means we
have the following statements of mass and energy conservation,
J N N 0; J E E 0 (JQ e JN ) JQ e JN ;
Je
eJN ;
(1.35)
Poynting’s theorem (another statement of conservation) for a non-radiating electromagnetic system says P J e E , in which, for a chemical voltage
E
e / e and
Je
eJ N , the conservation laws (1.35) imply,
J E (J Q e J N ) 0 J Q e J N eE J N E ( eJ N )
E Je
;
(1.36)
Problem: compute the relative efficiency for the thermoelectric device (1.3). solution: Using
E e E J e JQ (from (1.36)),
J Q / T , Q H TS , and considering a quasi-static/reversible/linear
JS
5
thermoelectric device so that J S S , the relative efficiency is computed as, r
Ee QC
Ee QC
E e
(QH Ee )
JQ JQ E Je E Je ; (1.37) (T (S ) J Q ) J Q TS (J ST ) T J S J S T
Problem: introduce a new thermodynamic potential, u
Je
/
JQ
J N 0
J e / ( e T ) , and use this to define
another thermodynamic potential, u 1 TS e . Show that J Q J e T J S . Solution: using Fourier’s law (1.24) for J Q , we have, u
Je JQ
E 0
J e ( J Q )E0
Je J e J Q TSe J e u 1J e J e TJ S ; TS e u 1 ; eT e T
(1.38)
1 Problem: re-express the entropy current in terms of the p otentials and u , and subsequently compute
J S . Solution: using (1.38), JS
1 T
JQ
1 T
J e
1
S T
e
u 1 J e J S ( / T ) J e ;
(1.39)
Problem: compute the “relative efficiency”, r Pe / J Q , in which P e E J e (the electric power), and in
which J Q J s T . Solution: Using (1.38) in (1.37), and using J
JS
J Q / T , and
E Je
J Q J , and
J Q / , we have, r
E Je JS
T
J
(J Q / T ) T
(J Q / ) () (J Q ) (T / T )
J Q / J Q T / T
/ ; T / T
(1.40)
Problem: Re-write the relative efficiency (1.40) in terms of the thermodynamic potential u of (1.38). solution:
recalling that (u) , we solve the definition of u, (1.38), for r by inserting Je and e u
2
S e T Z
e e
( JQ J E ) (see (1.35))
(see (1.27)), yielding,
e ( J Q J E ) e
e T
Z
r
1 r Se SeT
SeT ( Z uSe ) Z
SeT ( Z uSe ) 1
Z (TS e u )
uTS e (1
uSe Z
(uTS e 1)
)
(1
uSe Z uS e Z
uSe Z
)
1 ZT
(u) ; (1.41)
Problem: Show that ZT corresponds to max C , the Carnot efficiency. Solution: it is trivial to write,
lim max C lim
ZT
5
ZT
ZT 1 1 ZT 1 1
This is the efficiency relative to the Carnot efficiency: r
C
ZT ZT
C lim max C ; ZT
/ C , in which C 1 TC / T H .
(1.42)
Not merely power and efficiency, but also “compatibility” for a good heat engine; we write, (1 Se ) ZT S e ??? 1 ZT 1 / ???? 1 ZT ( TS ) r ; ZT ( TS ) (J e / T ) T 1 ZT 2 ( TS e ) 2 J S T T / T 1 ZT S 1 ZT 1 E Je
(J e / ) J e
????
S e
e
e
(1.43)
e
max U
1 ZT 1 SeT
;
(1.44)
Tuesday, august 27, 2013 Problem: draw a schematic of a thermodynamic engine, operating between a hot and cold reservoir . Solution: just draw,
(2.1) Problem: plot the Fermi functions for low and high temperature, but with matching chemical potential. Solution: let the matching chemical potential be L R 0 . Then,
f FD
f FD
,T
1.0
Cold k B T W 0.1 0.8
Hot k B T W 8 F 0.5
0.6
0.4
0.2
2
1
0
1
2
W
(2.2) Problem: recall the order-of-magnitude estimate for current from the previous lesson, (1.6). Compute the
current resulting from T ( E ) T0 ( E E0 ) . Then, write down a thermodynamic condition for I N 0 . Solution: by direct computation, 2W
I N
e
T0 ( E E0 )d
2W
1
R ( R
T 0 ) 1 e ( ) 1 e 1
L
1
R ( E0 R
L
; ) 1 e ( E ) 1 1
L
0
L
(2.3)
The condition for the current to vanish is for the Fermi-function difference to vanish at E 0 , 1 R ( E0 R )
e
1
1 e
L ( E 0 L )
1
E0 R k BTR
E 0 L k BTL
E 0
R R L L R L
TL R T R L ; TL TR
(2.4)
Problem: consider (2.1) as a single-loop electric circuit. Write a schematic illustrating the action of an
ammeter measuring current, and another schematic illustrating a voltmeter measurement across the load R L . Solution: the current-measurement and voltage measurement respectively appear as,
vs. The R L measurement is indicative of the voltmeter’s infinite internal resistance.
(2.5)
Problem: Let the load’s resistance be constituted by a chemical potential gradient. Then, the condition of
“dead thermoelectric battery” ( I N 0 ) is characterized by 0 R L being a maximum: I N 0 0 max 0 . Note that I N 0 also happens when the condition (2.4) is satisfied for the special
case of a delta-function potential, and when an analogous condition for other transmission functions is satisfied. Plot I N vs. 0 for a fixed temperature-difference. Solution: we use the following commands, (ω-μ)/T
f[ω_,T_,μ_]:=(E
+1)-1;
(2.6)
Plot[{f[1.7,5,0]-f[1.7,1,d],0.26},{d,0,2},[formatting crap]];
This then yields, I N
0
.25
.20
.15
I N
.10
0
Fresh battery
.05
IN = 0 --> dead batter 0.0
0.5
1.0
1.5
2.0
0 W
(2.7) One can also see,
case 1 case 2 ; I 0 ; I 0 L R L L 0 N L R L L 0 N
(2.8)
Without loss of generality, we can stick to case 1 of (2.8), which is less confusing. Problem: Write down chemical potential and load-resistances for three separate cond itions: (1) fresh battery, (2) discharging/ operation, and (3) dead battery. Solution: write,
R L (disconnected) 0 R L (connected) 0 RL (connected) ; (2) 0 ; (3) (1) R L R 0 L ; (2.9) R L 0 "fresh battery" "dead battery" "operating battery"
Problem: efficiency is defined as W / Q L W / Qhot . Use the 2
nd
law of thermodynamics to derive the Carnot
nd
efficiency. Solution: the 2 law says,
S S R S L Scold S hot
Q R T R
QL TL
0
QR TR
QL
TL
QR QL
Directly computing the efficiency, we get, W Q L QR Q T 1 R 1 R C C ; Q L QL QL TL
TR ; TL
(2.10)
(2.11)
6
Interlude: For a non-interacting system out of equilibrium , you have number and thermal currents. The following equations can be heuristically derived by usin g the reasoning discussed in (1.4) through (1.6), although they are the correct expressions, f L, R ( E ) (e
( E L ,R ) L ,R
1)1 ; F (E ) f L ( E ) f R (E );
(2.12)
I N T ( E ) F ( E )dE W / 0 ; I Q ( E L )T ( E ) F ( E )dE Q; Problem: compute the power output and efficiency using (2.12). solution: write, P ( R L ) I N W 0 I N ;
Problem: compute I N and I Q for T
T ( E )
a a2 E 2
W Q
W Q
0
I N IQ
P
(2.13)
; IQ
0 and 0 , and try integrating for a “toy” transmission function
. Solution: looking at (1.10), and putting this into (2.12), 2 4t sech E a f 2 k T I N k B x T T ( E ) dE k B x T 2 dE ; 2 a E 4 k T B 0 B
4 t
I Q k B x T (E L ) 0
a a2 E 2
sech 2
E 2 k BT
4k BT
dE ;
(2.14)
(2.15)
n Problem: write the coupled linear response formulae (1.16) in terms Ln T ( E)( E L )
df dE
dE . Solution:
write,
J N At I N L11 L12 At(1/ T )e L0 J At I At(1/ T ) L L L 1 Q Q 21 22
L1
L2 T / T
;
(2.16)
Interlude: In contrast to the dimensions that may be indicated in (2.16), the transport coefficients scattered throughout (1.27), in terms of these new integral-functions, are, eI N 1 I G (2.17) e2 L0 ; R V / e e
I Q
T I
N 0
6
L
L1 1 L0
L1 L2 (T / T )
T
L0 L1T / T
L2 (T / T ) 1 LLL 1 11 1 L2 L2 L2 ; (2.18) T T T 1 ZT T 2 1
0 2
Our system (2.1) is manifestly out of equilibrium, since a disparity in temperature and chemical potential is what drives the effect.
1 S e e T I
N 0
1 L1 eT L0
L0 L1T / T
TS e
ZT
TGSe 2
1 L2 / L2 L1 eT L1 / L1 L0
1 I E e I N
1 L1
e L0
T 0
TGSe 2 e ph
ph 0
TGSe 2 e
1 L2 L12 eT L1 L0L2
1 L2
ZT
e L1 ZT 1
...
1 L2
ZT
eT L1 ZT 1
(2.19)
;
(2.20)
;
L12 L0 L2 L12
(2.21)
;
Interlude: target ZT for commercial viability, [currently] ZT ~1; [commercially viable] ZT 3;
(2.22)
Problem: compute the chemical potential difference L R 0 in the case I N 0 . Solution: the zero
current condition, from (2.16) (e.g., from (1.16)) says I N 0 L0 L1T / T ; we then write, L R
Problem: consider the case I N I N
L1 L0
T / T 0 ;
(2.23)
0 , so that there is an arbitrarily-small (but nonzero) number current.
In this case, compute the quasi-static efficiency. Solution: we can use I N 0 L0 L1T / T to simplify the definition of efficiency (2.13); using the abbreviation 0 x 0
I N IQ
0
L0 L1 T T L1 L2
T
T
L R 0
L0 0 L1 0 (1 )
1, we write,
T x(1 x) C f ( x) ; T 1 x
(2.24)
Problem: maximize the efficiency (2.24) with respect to x. State the large- and small- of the extreme-value 7 of x x( ) you find. solution: extremize (2.24) by taking a derivative of the relative efficiency;
12 18 1 O2 ( ) r (2 x 1)(1 x) ( x 2 x)(1) r ; x 1 (1 ) 2 1 C (1 x) x O 1 ( )
(2.25)
Then, the efficiency in all cases is, r
1 ( ) r
(1 ) ( (1 ) ) ; r ( 1) (1 )
1 4 2
; r ( 1) 1 ;
(2.26)
Thursday, august 29, 2013
“thermodynamic efficiency at maximum power”, C van der Broeck, PRL 95, 190602 (2005) “thermodynamic bounds on efficiency for systems with broken TRS ”, C. Benenti et al, PRL 106, 230602 (2011) “Strong bounds on onsager coefficients and efficiency for 3 terminal thermoelectric transport in magnetic field” K brandner et al., PRL 110 070603 (2013). 7
Efficiency scaled to the Carnot efficiency; this is simpler.
Dubi and DiVentura, RMP, 83, (2011) Rego and kirezenow, PRL 81, 232, (1998). Schwab et al, Nature, 404, 974, (2000). 1996 - Tennesee - Mahan and Sofo - The best thermoelectric Claim: increase efficiency if time-reversal symmetry broken? 3-terminal transport is also way to increase TE efficiency. Preliminaries: Efficiency is work done (electricity) divided by heat flow associated with this work, x(1 x) L 1 T x(1 x) ; x R ; ; c T 1 x 1 x ZT o
(3.1)
Thermodynamic efficiency at maximum power: you have, ( P max ) C
ZT ZT 2
;
(3.2)
Problem: rewrite all thermodynamic forces and fluxes in such a manner that one can write power as
proportional to J1 x
d dt
( / T ) . Solution: write the linear relations (2.16) (or (1.16)) as,
J1 L11 L12 x1 x x1 / T J L x Q ; x ( 1 ) ; L 2 2 21 22 2 T
(3.3)
W Fx J1x1T P
(3.4)
Then, the power appears as,
stop Problem: consider the condition x 0 . Compute the condition on x1 x1 resulting from this. Solution: you
have x 0 L11 x1stop L12 x2 ; this is rearranged as, x1 stop
L12 L11
x2 ; J1 J1 ( x ) J 1( x1stop ) 0 x;
(3.5)
st
stop Problem: relate x1 to the x1 which maximizes the power (found in 1 homework). solution: write,
x1max
x2 L12 2 L11
1 ZT ZT 1 x1 stop ; (P max ) C C C 2 2 ZT 2 2
1
(3.6)
You can’t reach the Carnot efficiency, even with the maximum figure of merit. Interlude - Curzon-Ahlborn limit: Not possible to reach Carnot efficiency, CA 1 TR / TL
1 (TR / TL ) . In
8
fact, if you maximize power, you do not maximize efficiency, and vice versa.
8
Recall: you can have
max P 0 . We are in nano-systems, where Wiedemann-Franz law is no longer valid
Non-linear efficiency: the following is not a very rigorous definition, but check to see if you agree. Let our engine operate between two temperatures T0 , T 1 , as,
(3.7) Now: let temperature be a function of the spatial coordinate y, T T ( y ) , which runs between the limits indicated in (3.7). Rewrite the schematic (3.7) as,
(3.8) Let an infinitesimal of heat energy produce an infinitesimal of work energy, for all time. Then, dQ dW , which in turn implies dQ dW . Series-expanding the function Q Q( y y) for y Q Q( y dy) Q( y)
dQ dy
dy O2
dy ,
Q( y) dW ( y);
(3.9)
Integrat,e y
Q( y dy) Q( y) dW ( y) 0
dW ( y) Q( y)
T ( y) T ( y dy)
(3.10)
2T ( y)
Then write, dW ( y) dy
dW ( y) 1 1 T (1) 1 d ln T ( y) Q dy ; dW ( y ) 1 ; dy 2 dy Q 0 T (0)
(3.11)
Eventually, you will show in your homework the following functional form for the mazimized efficiency, max C
ZT 1 1 ZT 1 1
C x
y 1 1 y 1 1
; y
L12 L21 onsager L122 det L
det L
ZT ; x
L21 onsager L12
1;
(3.12)
Breaking time-reversal symmetry using magnets Problem: On turning on a magnetic field, you can increase the efficiency by breaking time reversal symmetry. Draw a schematic of a thermoelectric junction with such an applied magnetic field. Solution: Both electric and heat currents flow along the horizontal axis. The system is in contact with left and right reservoirs at temperatures TL and TR and chemical potentials L and R.
(3.13) Problem: Even though fluxes are one dimensional, the motion inside the system can be two or three dimensional. Write the linear response equations for the particle and heat fluxes. Solution: write,
J J (B) L J J (B) L q q q
L (B) L q ( B) X 1L X 2L q ; Lqq X 2 Lq (B) Lqq ( B) X 1Lq X 2 Lqq
L q X 1
(3.14)
In (3.14), J and Jq are the particle and heat currents, B an applied magnetic field or any parameter breaking time reversibility (such as the Coriolis force, etc.), and, 1
X 1 ; X 2 T / T ; R L ; T ; R L ; T TR T L; 2
(3.15)
9
Under time-reversal, the Onsager reciprocity becomes broken , I i Lij f j ; Lij Lij ( B) L ji (B); nd
Problem: here, the 2
B k i A j ijk ;
(3.16)
S J X1 J q X 2 X12L X 22Lqq X 2 X 1 (L q Lqp ) 0;
(3.17)
B
law of thermodynamics is,
Write the conditions (3.17) places on the Onsager coefficients. Solution: the fluctuations and are independent, so we easily derive, for these respective cases,
0 X 2 0 S X12 L 0 L (B) 0 ; 0 X 1 0 S X 22 Lqq 0 Lqq ( B) 0 ; (3.18) You also have, 2
L (B) Lqq (B) 14 L q ( B) Lq ( B) 0
(3.19)
Problem: prove that Lij (B) L ji (B) . Solution: the proof is given in SMT 06, and the final result is in SMT
06 – 367 – [120.11].
9
Proof of this, and of the claim Lij
Lij (B) L ji (B)
in (3.16), appears in SMT 06 – 367 – [120.11].
Problem: recall ohm’s law,
Je
E J N e , the isoelectric ( E 0 / e ) Fourier law
number current ( J N 0 ) Fourier law
JQ
JQ
E T , the iso-
eT , and the relation Lij (B) L ji (B) ; these constitute four (4)
equations which can be used to find the four Lij Lij (B) in terms of the constitutive coefficients , E , e . Do this. Solution: The Onsager coefficients are related to the familiar transport coefficients , , S as follows, (B)
e2 T
L (B); (B)
1 det L(B) T 2 L (B)
; S (B)
L q (B) eTL (B)
S (B)
Lq (B) eTL (B)
;
Note that the Onsager-Casimir relations, (B) (B); (B) (B ); S (B) S ( B);
(3.20)
(3.21)
Problem: compute the power from the linear response equations (3.14) , rather than (2.16) (in other words, use Benenti’s notation). Subsequently, extremize with respet to x1. solution: write,
P I N Tx1 L11 L12
T 2 Tx1 L11 ( Tx1 ) L12 ( Tx2 ) T x1 L11x1 L12 x2 P( x1 ); (3.22) T
all of this needs to be adapted to the new notation Problem: show that the efficiency / J q at maximum power depends on two parameters, x and y,
;
(3.23)
You have,
=…
(3.24)
You write, x1
L22 L12
x
y 1 1 y 1
x2 ; max C x
y 1 1 y 1 1
(3.25)
;
Problem: show that (3.19) implies a bound on y of (3.23). solution: the result is,
Now, define h ( x)
4 x ( x 1)
2
;
x 0 h( x) y 0 x 0 0 y h( x)
; (P max ) *
x
2
2 x 1
;
(3.26)
Writing efficiency, I N T 2 x1 L11x1 L12 x2 L ; | x2 0 x1 22 1 I Q L21 T L21x1 L22 x2 x1
det L
needed physical x2 ; I Q 0 ; (3.27) fact L11L22
sgn x sgn y
x 0 0 y h( x);
h( x )
x 0 h( x) y 0;
4x
; (P max ) *
( x 1)
2
x
2
2 x 1
(3.28)
;
Plotting h(x), hx 3
hx 2 1
1
10
5
5
10
x
1
(3.29) x
C
1.0
0.8
0.6
x 0.4
M x
0.2
4
2
2
4
x
(3.30)
Extremize, and you find the following expressions for the power and efficiency,
P 1 L122 1 xy P L12 | x 0 2 x1L11 L12 x2 0 x1 x 2 ; Pmax C x 2 (P max ) max C ; (3.31) 2 L11 4 L11 IQ 2 y 2 x1 2
This is the maximum efficiency without TRS. Something provocative about (3.31): it seems you want x > 1?
Figure of merit with magnetic field: you have, 2
ZT
Se T
ZT ZT (B)
(B) Se ( B) S e ( B ) (B)
T ;
(3.32)
Three terminal devices rd
Problem: Some people are excited about having a 3 terminal in the thermoelectric junction. Draw a schematic of such a device. Solution: Sketch of a thermoelectric device in the presence of a magnetic field B. The entire dashed box represents the conductor C, which is connected to two reservoirs. For the special case of the three terminal model, the conductor consists essentially of a scattering region and an additional probe terminal.
(3.33) 10
Problem: the center-site retarded Green function for a 2 -terminal junction with a t-stub and an interaction r appears as,
Gr ;0,0
1 ( EC )
2
W0 E0
W L W
2
e
i p L
W R 2 W
e
ipR
r ( )
;
(3.34)
Write the Green function for the 3-terminal device shown in (3.33). solution: treating (3.34) as an isolated Green function, while treating the coupling to the lead P as an interaction, we just need the Green function for 11 the lead P. Indeed, this is gotten from the work of SPH ,
10 11
See 2013 - nartowt - t-stub thermoelectric junction with e ph interaction. See 1992 - OH State - Hershfield, Davies, Wilkins - Resonant tunneling through Anderson impurity – current.
g ( j, j) P r
i p P
e
j j
ep
i P(
j j )
g
2iW sin p P
g (1,1)
P r ;1,1
P r
e0 eipP 2 2iW sin pP
i pP
eipP (e
ep ) i P
2iW sin pP
e p
i P
W
(3.35)
Then, using G (1 G0 r )1 G0 ( 10 r )1 , we have, G 2
gr ,0,0 (( g r prev ) 1 r ) 1 (( EC ) 0 E 0 ;0,0 W
W L 2 W
ei L p
WR 2 W
ei
pR
1 2
( EC ) 0 E 0 W
W L W
2
e
i p L
2
WW e p i R
R
W P 2 W
e
i pP
r ( )
2 P
r () WW e p ) 1 i P
(3.36)
;
Phonons
Effect of phonons: Looking at ZT Se2T / (e ph ) , we see phonons are bad. However, in (3.33), the 3terminal device could allow phonons to carry a heat current. They could be “tuned” to actually be a good thing. Seebeck coefficient: linearized inegrals, Se
1 L1 eL L0
T (E)(E ) dE eT T (E) dE f E
1
f E
2 k B
T 0
3 e
(kBT )
d dE
ln T ( E ) |E ET ;
(3.37)
Problem: d
Now, recall Dubi and DiVentura, RMP, 83, 2011. Write Lorentzian transmission function,
2 T ( E ) 2 large efficiency, and 2 parameters to play with ; ( E E F )2
(3.38)
However, we have the unphysical limit of 0 yielding a finite value of S e . Problem: if you go to a scale where k BT , linear series-expansions break down. How to get a high ZT (i.e., good thermoelectrics): (1) large variations of T(E) near Fermi energ y. (2) violation of Wiedemann-Franz law; i.e., / e [Lorenz number] (in appropriate SI-units), (3) phonon thermal conductivity, ph
1
2 0
??? kB 2T 2 1 f B d ... ( ) ; f B ; h 3 e T 1
(3.39)
Above = Rego and kirezenow, PRL 81, 232, (1998). Schwab et al, Nature, 404, 974, (2000). Problem: when integral (3.39) is done, you sum over all phonon modes. You have big K . You must take into account ph . That must be properly tuned for a good thermoelectric.
The best thermoelectric: 1996 - Tennesee - Mahan and Sofo - The best thermoelectric. Write the figure of merit with phonon-contribution, using Ln ZT
n
( k BT ) h
L12 ( L0 L2 L ) TL0 ph 2 1
J 2
J n and J01J 2 , as, J 12
( J 0 J 2 J ) J 0 ph 2 1
J 12 J0 J 2 J 12
(1 J0 J2 )
ph
J 2
(1 )
ph
ZT ( );
(3.40)
J 2
It’s clear to see that you want 1 to extremize (3.40). The thing that does this is a delta-function T(E), T ( E ) ( E b) (E )
J 12 J 0 J 2
( xT ( x) D( x)dx)2
( T (x ) D( x)dx)( x2T (x ) D( x) dx)
b (b) D(b)
2
(b) D(b) b2 (b) D(b)
1; (3.41)
Howver, this would produce a very low power thermoelectric. Advantages of nano-systems: (1) the T ( E ) can be nano-engineered, (2) no Wiedemann-Franz law, (3) phononconribution to thermal conductivity can be made small, (4) you can have inherently sharp gradients at such a tiny length scale, which mercilessly demands one discard the quasi-static approach we just finished discussing. (5) in bulk, large efficeicny means large power, but these might become independent on the nano-scale. (6) You 23 have nano-currents and nano-powers; you can scale these up as ~ N = 6.02 x 10 .