AME 60634 Int. Heat Trans. Thermoelectric Effect &
Thermoelectric Devices
*borrowed heavily from presentation by G. Chen, MIT
D. B. Go Slide 1 AME 60634 Int. Heat Trans. Seebeck Effect
Seebeck Effect: Temperature difference generates a
voltage between two different materials
hot Conductor 1
Conductor 2
cold D. B. Go Thomas Johann Seebeck
1821, Germany
Slide 2 AME 60634 Int. Heat Trans. Seebeck Effect
electrons diffuse from hot to cold
Thot
Tcold
electric potential builds up that resists diffusion
S=−
ΔV
V −V
= − hot cold
ΔT
Thot − Tcold
Seebeck coefficient: S [V/K]
D. B. Go Slide 3 AME 60634 Int. Heat Trans. Peltier Effect
Peltier Effect: Current flow can induce a temperature
gradient depending on direction of current flow
hot Conductor 1
Conductor 2
A
Jean Charles Athanase Peltier
1834, France
D. B. Go Slide 4 AME 60634 Int. Heat Trans. Current and Heat Flow
Newton’s 2nd Law


 m*v
d
v
F = −qE −
= m*
τ
dt
drag due to collisions
Coulombic force
The
€ steady-state solution gives the average electron “drift” velocity

qτ 
v =− * E
m
µe =
qτ
≡ electron mobility
*
m
The current density is the rate of charge transport per unit area (like heat flux)

 nelec q 2τ 
j€
E = σ∇Φ
elec = −nelec qv =
*€
m
compare to Ohm’s law!
But the electrons carry heat with them!
jheat
D. B. Go  u
= nelec uv = jelec = Πjelec
q
Peltier coefficient: Π [J/A]
Slide 5 AME 60634 Int. Heat Trans. Peltier Effect
- Induced heating and cooling at the two junctions due to mismatch
- Reversible by reversing the direction of current flow
- A refrigerator! (current is “work” to drive “heat”)
1
jelec, jheat
q
q
2
jelec, jheat
q (Peltier): (Π1-Π2)×j
D. B. Go Slide 6 AME 60634 Int. Heat Trans. Thomson Effect
Thomson Effect: Current flow through a temperature gradient will
generate/absorb heat because thermoelectric properties are
temperature dependant
heat release/adsorption
Thot
Tcold
current
William Thomson, Lord Kelvin
1855, Ireland
D. B. Go Slide 7 AME 60634 Int. Heat Trans. Thomson Effect
heat release/
absorption needed
for energy balance
q(x)
electrons diffuse
from hot to cold
Thot
Tcold
current
i
Thomson coefficient: τ = (1/i)×(dq/dx)/(dT/dx)
Kelvin Relations:
Π = ST;
D. B. Go τ =T
dS
dT
Slide 8 AME 60634 Int. Heat Trans. Thermocouples & The Thermoelectric Effect
Thermocouples operate under the principle that a circuit made
by connecting two dissimilar metals produces a measurable
voltage when a temperature gradient is imposed between one
end and the other.
D. B. Go Slide 9 AME 60634 Int. Heat Trans. Thermoelectric Devices
http://www.energybandgap.com
D. B. Go Slide 10 AME 60634 Int. Heat Trans. Peltier Coolers
Tc, qc
Ideal Device:
• No conduction (hot to cold)
• No Joule heating
qc = ( Π p − Π n ) × i
Th, qh
Real Device:
• conduction (hot to cold)
• Joule heating
qc = ( Π p − Π n ) i − i 2 R 2 − σ cond (Th − Tc )
electrical resistance
Lp
L
R=
+ n
Apσ p Anσ n
D. B. Go thermal conductance
σ cond
Ap k p An kn
=
+
Lp
Ln
Slide 11 AME 60634 Int. Heat Trans. Peltier Coolers: Refrigeration Performance
Voltage Drop:
S p − Sn
V = iR +
Th − Tc
Real Device:
• conduction (hot to co
• Joule heating
Coefficient of Performance:
2
qc ( S p − Sn ) iTc − i R 2 − σ cond (Th − Tc )
COP = =
W
( S p − Sn ) i (Th − Tc ) + i 2 R
Optimal Current to Maximize COP:
1+ ZTM − Th Tc
Tc
COPmax =
(Th − Tc ) 1+ ZTM +1
D. B. Go Tm =
1
(Th + Tc )
2
Slide 12 AME 60634 Int. Heat Trans. The Z Parameter
Z=
( S p − Sn )
2
Rσ cond
2
S p − Sn )
(
=
" L
Ln %" Ap k p An kn %
p
$$
''$$
''
+
+
Ln &
# Apσ p Anσ n &# L p
To Maximize Z:
! k
kp $
p
&
+
( Rσ cond )min = ##
&
σ
σ
p
p
"
%
2
Ln Ap ! σ n kn $
&&
= ##
L p An " σ p k p %
when
1
2
Leading to Z:
Z max =
D. B. Go (S
(
p
− Sn )
2
k p σ p + kn σ n
)
2
Slide 13 AME 60634 Int. Heat Trans. Figure of Merit: ZT
For a Single Material:
σ S 2T
ZT =
k
D. B. Go increase electrical conductivity
decrease thermal losses (conduction)
http://chemgroups.northwestern.edu/kanatzidis/greatthermo.html
Slide 14 AME 60634 Int. Heat Trans. Superlattice
increase electrical conductivity
decrease thermal losses (conduction)
2
σS T
ZT =
k
constrained by
Widemann-Franz
2 2
B
2
k
π k
=
σ T 3q
2
σS T
ZT =
kelec + k phonon
D. B. Go http://lucidthoughts.com.au/
Slide 15 AME 60634 Int. Heat Trans. http://www.kickstarter.com/projects/flamestower/flamestower-charge-your-gear-with-fire
http://energyblog.nationalgeographic.com/2013/09/24/google-science-fair-winner-makes-flashlight-poweredby-body-heat/
http://www.customthermoelectric.com/
D. B. Go Slide 16