Thermoelectricity: From Atoms to Systems

Transcription

1 hermoelectricity: From Atoms to Systems Week 4: hermoelectric Systems utorial 4. Homework solutions, problems -6 By Je-Hyeong Bahk an Ali Shakouri lectrical an Computer ngineering Birck Nanotechnology Center Purue University

2 Prob.. Heat balance equation -. Secon erivative of temperature he heat generation rate per unit volume q For one-imensional problem, an assuming temperature- an position-inepenent properties, q S At steay-state, x x J x J q e Q Je J e x 0 Je J.-H. Bahk an A. Shakouri, nanohub-u Fall 03 SJ x Couple charge an heat current eq. e x J Q SJ J e e J e x S x

3 Prob.. Heat balance equation -. emperature profile with current x J e Je ( x) x Cx C ( x) K x K x C Bounary conitions: ( x 0) C ( x ) H J.-H. Bahk an A. Shakouri, nanohub-u Fall 03 3

4 Prob.. Heat balance equation -. cont ) emperature Chemical potential Voltage Comparison between with an without current flow H H V L + V - Length l RL S R R i i 0 i L By joule heating L l l l x x x H H L + V - Length l Slope= Slope= V S i 0 H l L l l l L x x x J.-H. Bahk an A. Shakouri, nanohub-u Fall 03 4

5 Prob.. Heat balance equation -3. Heat balance equation at col sie Q A out Qout A Q out A x K x0 x K From -. ( x) K K K x0 x K x C Q C S C I Q out S C I K J.-H. Bahk an A. Shakouri, nanohub-u Fall 03 5

6 Prob.. Quasi-ballistic cooler -. emperature profile for quasi-ballistic cooler q x Joule Je x exp x q Joule x x Je exp x Bounary conitions: ( x 0) C ( x ) H J.-H. Bahk an A. Shakouri, nanohub-u Fall 03 6

7 Prob.. Quasi-ballistic cooler x -. (cont ) Je exp B x x exp where B Je x B exp Bx C ( x 0) C x B B exp x Cx C B ( x ) B x B ( ) exp C C B H H K J.-H. Bahk an A. Shakouri, nanohub-u Fall 03 7

8 J.-H. Bahk an A. Shakouri, nanohub-u Fall 03 Prob.. Quasi-ballistic cooler 8 -. (cont ) H C B C B B exp B B B C C H exp B B C exp exp exp C B x B B x B x B K B C x K x K K x K x exp exp ) ( 3

9 Prob.. Quasi-ballistic cooler -. Heat issipation at col sie Qout A x x0 ( x) x K x K K 3 exp K x exp ( x) x x0 K K 3 exp K ( x) A x Q out x0 K exp exp K J.-H. Bahk an A. Shakouri, nanohub-u Fall 03 9

10 Prob.. Quasi-ballistic cooler -3. Fraction of Joule heating issipate to cathoe Q out exp 3000 nm K Fraction of 4.8 From the left figure, at x = exp 3. 0 y 4.8 y x 4.8 exp 00 nm 5 exp 0.44 J.-H. Bahk an A. Shakouri, nanohub-u Fall 03 0

11 Prob. 3. hermoelectric cooler J.-H. Bahk an A. Shakouri, nanohub-u Fall 03

12 Prob. 3. hermoelectric cooler 3-. Max cooling R SC 0 S C I K I I K K R S I K R max Z when C I J.-H. Bahk an A. Shakouri, nanohub-u Fall 03 C Z C S R C

13 Prob. 3. hermoelectric cooler 3-. Simulation ( () () J.-H. Bahk an A. Shakouri, nanohub-u Fall 03 3

14 Prob. 3. hermoelectric cooler 3-. Simulation for maximum net cooling (cont ) (3) (4) J.-H. Bahk an A. Shakouri, nanohub-u Fall 03 4

15 Prob. 3. hermoelectric cooler 3-. (cont ) Output graphs Lowest C (Maximum net cooling) J.-H. Bahk an A. Shakouri, nanohub-u Fall 03 5

16 Prob. 3. hermoelectric cooler 3-. (cont ) Click the menu an select the last one max 58.4 K, Iopt 0.97 A Check: Z C 4.6 K ( 000 V/K) (0000 m ) ( W/mK) 0.00 K - ZC (0.00 K )(4.6 K) 58.4 K max (match!) J.-H. Bahk an A. Shakouri, nanohub-u Fall 03 6

17 Prob. 3. hermoelectric cooler 3-3. With specific contact resistance= e-6 Ω cm -> Aitional Joule heating max 58.4 K 46. K With contact resistance No contact resistance I opt 0.73 A J.-H. Bahk an A. Shakouri, nanohub-u Fall 03 7

18 Prob. 3. hermoelectric cooler 3-4. With substrate with spreaing effect max 58.4 K 46. K 44.5 K With contact resistance with substrate With contact resistance No substrate I opt 0.69 A No contact resistance No substrate J.-H. Bahk an A. Shakouri, nanohub-u Fall 03 8

19 Prob. 3. hermoelectric cooler 3-5. With substrate with spreaing effect on =30.9K (ue to Peltier heating at the interface) J.-H. Bahk an A. Shakouri, nanohub-u Fall 03 9

20 Prob. 4. Segmente element In a single material: S C max Z C when I R or I R SC I (Joule)=(Peltier) I S ( S S ) I o keep this conition S I S S ) I ( S S R R Also, S S J.-H. Bahk an A. Shakouri, nanohub-u Fall 03 0

21 Prob. 4. Segmente element Material Material J.-H. Bahk an A. Shakouri, nanohub-u Fall 03

22 Prob. 4. Segmente element Single max 58.4 K 7.97 K Z C 5.0 K Segmente J.-H. Bahk an A. Shakouri, nanohub-u Fall 03

23 Prob. 5. power generator J.-H. Bahk an A. Shakouri, nanohub-u Fall 03 3

24 Prob. 5. power generator 5-. Power output < lectrical resistance network for power generation> Seebeck effect: V oc =S ( - )+S sub ( - amb ) R R c R R sub I R L V oc =S (without substrate) V oc Ohmic law: V oc =I(R+R L ) = S I S H C S R, Pout R R R R L L L J.-H. Bahk an A. Shakouri, nanohub-u Fall 03 4

25 J.-H. Bahk an A. Shakouri, nanohub-u Fall 03 Prob. 5. power generator 5 out L L R R R S P 5-. Maximum power output Differentiate P out with respect to R L R S P 4 out,max 0 4 out L L L L L R R R R R R R S R P R L R

26 Prob. 5. power generator 5-3. Maximum efficiency S H I I S IR K I Differentiate η with respect to I, an 0 (a) ( S )( K SI H / I R) I( S )( S H ) 0 I K S M S R ( ) ( ) ( ) S Z R M S R( ) ( ) (b) where Z M Z M J.-H. Bahk an A. Shakouri, nanohub-u Fall 03 6

27 Prob. 5. power generator 5-3. Maximum efficiency (cont ) Plug (b) into (a) to get max H H From I S R( ) R L R S R R L 5-4. (Carnot limit) max H H H rue only if or H Z M J.-H. Bahk an A. Shakouri, nanohub-u Fall 03 7

28 Prob. 5. power generator 5-5 & 5-6 Simulations P out η R L =0.05 Ω R L =0.04 Ω R R for η max for P out,max Z (500 M K V/K) (50000 ( W/mK) - ) J.-H. Bahk an A. Shakouri, nanohub-u Fall 03 m - Z M

29 Prob. 5. power generator 5-5 & 5-6 (cont ) Click the menu an select the last one P out,max = W at R L =0.04 Ω R η max = at R L =0.05 Ω R J.-H. Bahk an A. Shakouri, nanohub-u Fall 03 9

30 Prob. 5. power generator 5-7. Power output as a function of Click the menu an select this option η P out = 35. mw at = 300 K P out = 5.6 mw at = 00 K P out = 3.9 mw at = 00 K op surface temperature J.-H. Bahk an A. Shakouri, nanohub-u Fall 03 30

31 Prob. 6. hermoelectric system An example of topping cycle on steam turbine s = 800 K Heat source (combustion) temperature 000 x 000 elements per m h Heat transfer coefficient (W/m K) at hot sie c Fill factor=0. g = 800 K Gas temperature in the turbine J.-H. Bahk an A. Shakouri, nanohub-u Fall 03 Heat transfer coefficient (W/m K) at col sie 3

32 Prob. 6. hermoelectric system 6-. Power output as a function of material properties an heat transfer coefficients Click the menu an select Power Output vs. Z Power out vs. Z hree ifferent simulations completely overlappe:. Varying thermal conuctivity (). Varying Seebeck coefficient (S) 3. Varying electrical conuctivity () while keeping the other two constant in each simulation. J.-H. Bahk an A. Shakouri, nanohub-u Fall 03 3

33 Prob. 6. hermoelectric system 6-. (cont ) Power out vs. Z Heat transfer coefficients= otal 9 simulations 600 W/m K Varying,S, 400 W/m K 00 W/m K J.-H. Bahk an A. Shakouri, nanohub-u Fall 03 33

34 Prob. 6. hermoelectric system 6-. (cont ) which statement is correct? a. he higher the heat transfer coefficient is, the lower the power output is generate. b. Optimal power output is only a function of the heat transfer coefficients, an not of the material properties. c. Optimal power output oes not change with varying any of the three iniviual material properties an the heat transfer coefficients as long as the Z value remains the same.. Optimal power output oes not change with varying any of the three iniviual material properties as long as the Z value an the heat transfer coefficients remain the same. e. None of the above. J.-H. Bahk an A. Shakouri, nanohub-u Fall 03 34

35 Prob. 6. hermoelectric system 6-. Cost analysis Click the menu an select Power cost vs. Z Power cost ($/W) vs. Z Varying S, (power factor) Varying J.-H. Bahk an A. Shakouri, nanohub-u Fall 03 35

36 Prob. 6. hermoelectric system 6-. (Cost analysis, cont ) which statement is correct? a. Materials prouce the same power cost if their Z values are the same. b. In the high Z regime (e.g. Z higher than ~ 3.0), lowering the thermal conuctivity is more esire than enhancing the power factor in terms of power cost if Z remains the same. c. In the high Z regime (e.g. Z higher than ~ 3.0), enhancing the power factor is more esire than lowering the thermal conuctivity in terms of power cost if Z remains the same.. Power cost increases with increasing Z. e. Power cost ecreases with ecreasing efficiency. J.-H. Bahk an A. Shakouri, nanohub-u Fall 03 36