Thermoelectricrefrigerationsystem 150824020213 Lva1 App6892

Thermoelectric Refrigeration Systems

1. DEVELOPMENTS In 1821 Thomas Seebeck a German physicist, observed the production of (emf)

(electromotive force) if two dissimilar metals are joined together and their joints are kept at different temperatures. The magnitude of such emf is dependent upon the material combination and temperatures of junctions. However, he was unable to make use of his invention probably due to insignificant emf and electric power output.

Thereafter a French scientist, Jean Peltier, discovered a reversed phenomenon to that of Seebeck in 1834. He found that there is heating or cooling of a junction of a pair of dissimilar substances if direct current is passed through them. However, he was not only unable to realize the utility of the same but also failed to correlate between his invention and that of Seebeck.

In 1838 Lenz, a German scientist, applied a D.C. source to a pair of materials-bismuth and antimony connected together as shown in Fig. 1. He was astonished to see that the water droplet was frozen into ice. When the current was reversed, the ice melted. This led to the concept of thermoelectric refrigeration. However, the idea did not materialize into commercial products for about a century because of unavailability of suitable materials until 1930's when semiconductors were developed.

Fig. 1 Ice formation from thermoelectric refrigeration effect

Altenkirch analyzed qualitatively the factors which should be improved to yield reasonable thermoelectric cooling. The economical study exhibits that the thermoelectric cooling turns out to be cheaper for the smaller unit below the range of 200 300 / . Beyond this the compression system is found to be economical. The size of the thermoelectric system is reduced to2l3 of the present cooler. Thus small sized commercial units for spot cooling, laboratory tests, space exploration, etc., have been manufactured. The advances in the use of thermoelectric system are going on with the availability of better semiconductors.

2. Analysis of Thermoelectric refrigeration Systems 2.1 Thermoelectric Effects Whenever direct current passes through a pair of thermocouples with junctions

maintained at different temperatures, five effects are observed: Seebeck effect, Peltier effect, Thomson effect, Joulean effect and conduction effect.

Seebeck effect: When the two junctions of a pair of dissimilar metals are maintained at different temperatures, there is the generation of emf (electromotive force) (Fig. 2

(a)). He conducted a series of tests by varying the temperatures of the junctions of various combinations of a set of materials. The emf output was found to be:

∆ ∆ ...(1)

where ∆ and ∆ are the emf output and the temperature difference of the junctions. The phenomenon of generation of emf is called Seebeck effect.

(a) Seebeck effect.

(b) Peltier effect (c) Thomson effect

Fig. 2. Thermoelectric elements

The proportionality constant of Eq. (1) is denoted by: , ∆ ∆⁄ ...(2)

and is called Seebeck coefficient or the thermoelectric power. It is to be noted that , is the coefficient for a pair of different inetals (A and B or P and N or p

and n). Peltier effect: If direct current is passed through a pair of dissimilar metals (Fig. 2(b)), there is heating at one junction, cooling at the cither depending upon material combinations. Peltier varied the current and observed the heating and cooling rate for different sets of elements. He found that:

...(3) where q is the cooling or heating rate. The proportionality constant of Eq. (3) is called as

Peltier coefficient, , i.e., , ...(4)

where , is the coefficient for two different metals. Thomson effect: It is a reversible thermoelectric phenomenon. When a current

Passes through a single conductor having a temperature gradient as exhibited in Fig. 2c, heat transfer is given by:

...(5)

where ⁄ being Thomson coefficient and , theThomson heat transfer. using first and second laws of thermodynamics obtained the relation between

Seebeck and Peltier coefficients as: , , ...(6)

Using Eq. (6) into Eq. (4), it is found: , ...(7)

It is evident from Eq. (7) that to get the high value of cooling or heating, should be high, otherwise large current would be required. But high current will render high heat generation due to Joulean effect.

Joulean effect: When the electrical current flows through a conductor, there is dissipation of electrical energy. According to Joule it is related as:

...(8) where and are the current and electrical resistance. This effect is called Joulean

effect. Conduction effect: If the ends of any element are maintained at different

temperatures, there is heat transfer from the hot end to the cold end and is related by: . ...(9)

where being overall conductance and , are the high and low temperatures, respectively. If there is only one conductor of cross - sectional area , conductivity and length , the overall conductance is given by:

⁄ ...(10) 2.2.Thermodynamic Analysis of Thermoelectric Refrigeration System Figure 3(a) shows an enlarged view of thermoelectric elements with control volume.

In order to analyze the system to obtain refrigeration effect , etc., the following assumptions have been made:

(i) Heat transfer takes place through semi-conductors at the ends only. (ii) No energy exchange between the elements through space separating them, and (iii) Properties such as conductivity, resistance, etc., are invariant with temperature. Now cooling and heating due to the thermoelectric effect is given by (Peltier effect):

, ...(11) and

, ...(12)

(a)

(b) Heat generation (c) Heat transfer

Fig. 3 Energy transfer in thermoelectric system

Referring to Fig. 3(b), the heat generation in absence of other effects will lead to heat

transfer from both hot and cold ends equal to 2⁄ (due to symmetry and assumption (i) given earlier). The conduction effect along the element leads to heat transfer at the cold junction from the hot one. Then, from Fig. 3(a), for the control volume:

2⁄ , ...(13) and for the hot junction:

, 2⁄ ...(14) Thus the thermoelectric cooling is:

, 2⁄ ...(15) And heating,

, 2⁄ ...(16) Now using first law, one obtains the energy input to the system from outside as:

, ...(17) where negative sign indicates that energy has to be supplied to the system.

The required is:

, 2⁄ , ...(18) Where the overall conductance for both the conductors is given as:

...(19) and the total resistance for both conductors is:

...(20) where the specific resistance and 1⁄ . For above Eq. (18) one can define a term called figure of merit, as:

, ...(21) Mathematical analysis shows that the figure of merit is maximum when:

⁄ ⁄⁄ ⁄ ⁄ ...(22) Then the maximum value of the figure of merit, , is seen to be

, ...(23)

Using relations given by Eqs. (22 and 23), one can obtain the current which

corresponds to maximum from Eq. (8.18) with condition / 0. The

resulting quantity turns out to be: , 1 1⁄ ...(24)

where is the mean temperature given by 2⁄ .The maximum is given by:

⁄ ...(25a)

. . ...(25b)

where is the Carnot value for the refrigeration system operating between the given upper and lower temperature limits and

. .⁄ ...(26)

Here, in Eq. (25b) the . .stands for the effect of actual thermoelectric

behavior. From Eq. (26), it is evident that . .

1 when the figure of merit, ∞. Then, from Eq. (25b):

i.e.,

...(27) The maximum cooling is obtained from the condition / 0, which gives:

., ⁄ ...(28)

The corresponding cooling is given by:

. . ...(29) and,

. ...(30) If the heat source is removed, the . becomes zero trivially. Thus, the maximum

temperature difference is then:

. ...(31) In all the above derivations it is seen that should be as high as possible to get

reasonable amount of refrigeration effect and . Up-to-date value of figure of merit, is about 0.007/ .For 0.007/ , 310 and 260 , one gets the maximum

as: 1.0251. Figure 4 shows the achievement of maximum temperature with the year. It indicates

that there is shooting up of the achievement of temperature difference after the development of semiconductors.

Fig. 8.4 Achievement of maximum temperature rise with year

The variation in coefficient of performance with the temperature difference is shown in Fig. 5 with figure of merit as piuameters where has been kept at 40 .

Figure 6 exhibits the variation in and with current forgiven value of and other parameters. It is evident that the maxima of and , occur at different values of current. Hence a system designed to operate on least running cost is going to b expensive in the initial investment. On the other hand a system designed on the basis of maximum cooling will have high running cost.

As discussed earlier that the figure of merit should be as high as possible, it is seen from Fig. 7 that the figure of merit is small for both the insulators and conductors while the same is maximum for semiconductors for a given electron density. Here the insulators show the poor value of figure of merit because of low value of electrical conductivity. On the other hand the conductors have the low value of figure of merit because of high thermal conductivity and low thermoelectric power.

Fig. 5 Valuation in with temperature difference

There are various combinations of thermoelectric elements which are commercially fit for the production of thermoelectric refrigeration system as given in Table 1.

Table 1 Some thermoelectric elements*

P type N type

Thermo elements ⁄ 10 Thermo elements ⁄ 10 1.2 1.5 1.2 2.2 1.2 3 1.8 -

3.3 - • from table1, . stands for . ⁄ or . means: . . ⁄ . For the selection of a pair of thermoelectric elements care is taken to have and

type combination. It is usual practice to denote thermoelectric elements with if their thermoelectric power is positive. If the thermoelectric power, is negative, it is called

type thermoelectric elements.

Fig. 6 Variation in and with current

Fig. 7 Schematic representation for the ranges of insulator, semiconductor and metal

Fig. 8.S Variation of number of thermoelectric couples with 3 . ADVANTAGES OF THERMOELECTRIC REFRIGERATION SYSTEM It possesses several advantages as listed below: (i) Absence of moving parts eliminates vibration problem as well as regular

attendance. Therefore, it can be best suited for systems where vibration is undesirable. In addition there is no wear due to rubbing as such the life is expected to be almost infinite compared with other systems.

(ii) It is easy to overload if desired, just by increasing the current to a certain limit (iii) The load can be easily controlled by means of adjusting the current to meet the

situation. (iv) Very compact in size since even the system boundary may be used as the

cooling surface as exhibited in Fig. 8.9. Here one of the walls of the room forms the evaporator surface Therefore, the cost of manufacture of that wall is taken care by the evaporator surface.

Fig.9 Thermoelectric cooling element fitted on the wall the room

(v) It can be operated in any position in contrast with vapour - absorption, Vapour – compression or steam-jet refrigeration systems. As for example, if a vertical compressor is used in a refrigeration system, it cannot be operated in any other position.

(vi) It is lighter in weight for the same capacity of refrigeration. (vii) Since electric current passes through conductors, there is no problem of

leakage which is most undesirable in other refrigeration systems. Further, the leakage of refrigerant from the system causes the drastic decrease in the capacity in addition to extra cost for the refrigerant and charging operations. Thus a thermoelectric refrigeration system operates at the same capacity for long and eliminates the cost of charging and extra materials.

(viii) It is most easy to operate as a heat pump, just by reversing the terminals. Hence, a thermoelectric refrigeration system can be considered as an year round air conditioner*.

(ix) Since no refrigerant is used, there is no question of toxication etc. and can be used directly for air conditioning.

(x) Its design and manufacture are rather much simpler than the other refrigeration systems.

(xi) It is most suitable for the production of cooling suit. Main disadvantages of thermoelectric refrigeration system is the unavailability of

suitable material of high figure of merit. Presently the total cost of refrigeration system will be a few times higher than the vapour – compression or other systems for a few ton capacity. In addition, the running cost is found to be much higher compared to vapour –compression system. That is why the vapour-compression or other systems of refrigeration are most commonly employed. The overall of a thermoelectric refrigeration system is of the order of 0.1 0.2 .

4. COMPARISON BETWEEN THERMOELECTRIC REFRIGERATION AND VAPOUR-COMPRESSION SYSTEM.

The two systems are shown in Fig10 where the electron flow direction for the

thermoelectric refrigeration system and refrigerant vapour flow for the vapour compression system are exhibited. In the former the electrons are pumped by the battery, situation similar to the vapour-compression system where the compressor delivers the vapour from the evaporator to a high pressure. As the electrons reaches the junction of the dissimilar semi-conductors, its energy decreases due to heat transfer to the surroundings. This corresponds to cooling or condensation of compressed vapour. Thereafter electrons are reduced to lower potential which resembles the throttling process. The heat transfer to the cold junction imparts energy to electrons which again move to the battery similar to vapour from evaporator to the compressor to complete the cycle.

Fig. 8.10 Analogy between thermoelectric compression systems Example 1: A thermoelectric refrigeration system is to operate in an ambient

temperature of 313 . Its figure of merit is 0.0036 . Find the minimum temperature that can be achieved by it.

Solution: The maximum temperature difference is given by (Eq. 31):

2 Where 313 and 0. 0.0036

3130.0036

2 223.27

. 313 223.27 89.73 Example 2: Calculate diameter of the -type element corresponding to the maximum

value of figure of merit and from the given data for a pair of thermoelectric element.

P-type N-type

Thermoelectric power ⁄ 0.00022 - 0.00018 Specific resistivity . 0.00001 0.00002

Conductivity .⁄ 1 1 Diameter of the element 10 -

Length of the couple 12 12

Solution: Using Eq. (22), we get:

⁄ ⁄⁄ ⁄ ⁄ .. √

√ As 12 10 2 ⁄ 11.89

For maximum value of figure of merit, we have from Eq. (23): .

√ . √ .0.002745

The same may be calculated from and values i. e., , ⁄ From E q.(19) one gets:

..

..

0.0158 ⁄ From Eq. (20) one gets:

, ⁄ .. .

0.002745 Example 3: Find the value of figure of merit of a thermoelectric pair which can render

the 4 when operating within temperature limits 313 and 258 . Conclude your answer.

Solution: From Eq. (25):

1 ⁄1 1

where 313 and 258 285.5

Then, 4 √ . ⁄√ .

0.6858 Note: This value of 0.6858 is above 100 times to that of the available Thermoelectric elements. It is evident that it may take several decades when such material may become available. Example 4: Design a thermoelectric refrigeration system for an equipment to be

maintained at 258 . The ambientt emperatureis 313 The cooling load is 20 . The thermoelectric elements have the following properties:

⁄ 0.000120.00015

1 .⁄ 1000 2000

.⁄ 1 1

Thermoelectric element of type has 0.01 both the diameter and length. Both

thermoelectric elements are of the same length. Obtain: (a) Diameter of n-type couple element, (b) figure of merit, (c) , , and

corresponding current, number of thermoelectric couples and power (D.C.) needed for the same, (d) COP and current for maximum cooling, number of thermoelectric elements and D.C. power to drive the same, (e) compare the with Carnot , and (0 obtain power needed for cases (c) and (d) on the basis of overall voltage drop.

Solution: Since we know that for maximum value of figure at merit, , the relation Eq (22):

⁄⁄ = 2000 1 1000 1⁄ 1.414

4 0.01 0.00007855 1.414⁄ 1.414 0.00007855 1.414 0.00005555⁄⁄

4 0.00005555 0.0084 8.4

(b) The values and are calculated from Eqs.( 19 and 20), respectively. 0.01 1 1⁄⁄ 100⁄

0.01 1 1000 0.00007855 1 2000 0.00005555⁄⁄ 100 0.002173⁄ ⁄⁄

1 0.00007855 0.01 1 0.00005555 0.01 0.01341 ⁄⁄⁄ Therefore, the figure of merit,

, .. .

0.002502

Since , 0.00012— 0.00015 0.00027 © 285.5

⁄ . . ⁄

. .0.1953

The corresponding current is [Eq 24]: .

. . .22.095 .

The cooling produced per element is: , 2⁄

0.00027 22.095 258 22.095 0.002173 2 0.01341⁄ 313 2580.2712 ⁄

.

73.75 74 The . .power needed to run the system for maximum is:

.102.4

(d) When the design is needed for maximum cooling, we get: .

.0.1399

., .

.32.06 .

. . 0.01341 0.002502 285 313 2580.3791 ⁄

The required number of Pairs:

.52.76 53

They have to be connected in series. The power needed to run this system with . . source is :

.143

(e) The of a Carnot refrigerator is: 4.6909

Therefore relative based on Carnot value: . .

.0.0416

and for maximum cooling : ..

0.0298 (f) The overall voltage drop for maximum is:

74 , 74 0.00027 313 258 22.095 0.0021734.652

And corresponding Power 4.652 22.095 102.78

Similarly for maximum cooling: the voltage drop is: 74 , 53 0.00027 313 258 32.06 0.002173 4.479

Thus, power used is: 4.479 32.06 143.61

The negligible difference is due to arithmetical errors. This example clearly

demonstrates that the large thermoelectric refrigeration system cannot be used on commercial basis because the power consumption. would be so large that, it cannot compete even if initial investments is set a side.

Example 5: A thermoelectric-refrigeration system consists of pairs' connected in a

series. It works under a 6 solar battery. The hot and cold junctions are maintained at 40 and - 10 , respectively. The properties of the thermoelectric materials are

0.00016 ⁄ and 0.0002 ⁄ there resistance and conductance for each pair 0.003 and 0.02 ⁄ .

obtain for the cases (i) for maximum and (ii) for maximum cooling effect the following (a) number of thermoelectric couples,( b) wattage and tonnage (cooling) of the unit, (c) power consumed and (d) compare the with Carnot value.

Solution: Case (i): In order to obtain various quantities, let us obtain the figure of

merit. Now , 0.00016— 0.0002 0.00036 ⁄ Therefore,

, .. .

0.00216 (a) Now for maximum , the current is given by:

288 , .

. √ .21.9878 .

Now the overall voltage drop for couples in a series is given by: ,

,

. . .71.77 72

(b) Cooling effect is given by:

, 2⁄

72 0.00036 21.9878 263 21.98780.003

2 0.02 50 25.49 . 0.007243

(b) Power needed is given by:

6 21.8978 131.39

(c) ⁄ . ⁄.

0.194

5.26 Therefore, compared to Carnot value, the of the thermoelectric refrigeration system is:

..

0.03688 case (ii): when the cooling is to be maximum, the current is given by: (a) , .

.31.56 .

And the corresponding number of couples as above to be connected in a series:

, . . .54.3 54

(b) The maximum cooling is obtained from:

. 2⁄ 54 0.02 . 50 26.68 . 0.00762

(c) Power needed for the system 6 31.56 189.36

(c) .

.0.1389

(d) ..

0.0264 This example demonstrates that the of the refrigeration system decreases to

0.7159 ( for . .⁄ ). But the refrigeration effect is increased to 1.541( . .⁄ Example 6: A thermoelectric pair as its . twice to that of .. The

operating temperatures are 308 and 262 and 0.014 ⁄ and 0.002 . Find (a) and ., ., , corresponding to . and their ratio

and (c) power for . and . solution: Here 285 and ⁄ 308 262 1.1756⁄

⁄

√ .√

⁄ 0 0 This equation is solved using various values of as:

Z F(Z)=0 0.003 - 0.000271

0.0031 - 0.000118 0.00317 0.00000011

Required values of is then 0.00317 ,

,, 0.014 0.002 0.00317 18.049 .

. ⁄ 262√0.014 0.00317 39.028 . . 2⁄ 0.014 0.00317 262 2 308 262⁄ 0.8792

corresponding , 2⁄

0.002979 18.049 262 18.049 0.002 2 0.014 308 262 0.4389⁄ Then . 0.8792 0.4389 2.0029⁄⁄

for , 0.002979 308 262 39.0280.002 0.8989 5. OPTIMUM THERMOCOUPLE ELEMENTS It is evident from Fig. 8.6 that a refrigeration system which is designed on the basis of

maximum COP would be expensive in initial investment. On the other hand, if the above system is designed on the basis of maximum cooling, its running cost would be more compared to former. Therefore, it requires economic consideration while deciding the thermoelectric system.

Let the refrigeration capacity be TR. Then the number of pairs of thermocouple elements for this capacity is:

, ⁄ ...(32)

The power needed for these couples is: ,

,

, ⁄ ...(33)

If C is the cost of thermoelectric elements per pair of life, L,thetotal investment for L year is Nc ,.

The cost for years for the power consumption can be obtained from Eq. (32) with electrical energy charges per and operating factor oF:

. . . . ,

, ⁄ ...(34a)

Therefore, the total annual expenditure is obtained from: . . C⁄ . ,

, ⁄ ...(34)

Using Eq. (8.21) and rearranging terms in Eq. (8.34), it is found as: C⁄ . . C⁄ √ .

√ . ⁄ ⁄ ...(35) Now, using the dimensionless parameters as well as variables:

...(36a)

...(36b)

8.76 . . ...(36c) ...(36d)

⁄ ...(36e) In the dimensionless form Eq.( 35) is seen to be:

. ...(37a)

, , ,, , ...(37b) where ...(38)

For given values of parameters, , , and , the variation of and in dimensionless form has been presented with varying as given in Fig. 11. The optimum cost of thermoelectric elernents fall between the maxima of and .

Fig. 11 Variation in COP and and and optimum thermoelectric elenents If the interest is considered on the investment one can still use Eq. (39). But the parameter has to be evaluated from:

. . ...(40a) for simple interest and,

. ...(40b) for compound interest. Here /is the interest rate and the rest symbols have usual

meaning. Figure 1I shows the variation of optimum number of thermocouples for various values

of parameters (or ), , and . Thus; for known capacity and for a given location where costs of various quantities are known, it can be used to obtain the optimum number of thermoelectric elements for a given tonnage of the system.

But this will be possible only by reduces size by fig. as given above. I have looked into it & then is scope for it.

6. MUTTI.STAGE.THERMOELECTRIC SYSTEMS In order to improve the performance of a thermoelectric refrigeration system to

achieve lower temperature, the multi-staging is used since maximum temperature difference from a single-stage cannot be obtained greater than the given by Eq. (31).Figure 12(a), shows a multi-stage system which can also be reduced to an equivalent thermal system. Then, for the first stage:

1 ...(41) Similarly, for the second stage one can get:

1 ...(42)

Fig 12 Multi-stage thermoelectric system.

Substituting for from Eq. (41), one gets:

1 1⁄ 1 1⁄ ...(43) By induction one can prove that for third stage and further. stages as:

1 1⁄ 1 1⁄ 1 1⁄ ...(44) 1 1⁄ 1 1⁄ … 1 1⁄ ...(45)

Equation (8.45) can also be written in the shortened form as: ∏ 1 1⁄ ...(46)

The COP of the n-stage system can be written as:

or 1 1⁄ ...(47)

Thus Eqs. (46 and 47) yield:

∏ ⁄ ...(48)

If we consider n-stage refrigeration system with the coefficient of performance of every stage to be equal arid denoted by , one can obtain from Eq. (48) as:

⁄ ...(49) The coefficient of performance for each stage of an n-stage cascade system is given

by an approximate expression: 1 2⁄ 1 2⁄ ...(50)

Here it has been assumed that the temperature difference for each stage is equal to where and ,are the upper and lower temperature limits for the system and ⁄ . Then is given by:

.

. !

Using Eq. (5O) into Eq. (49), it is found as:

⁄ ⁄⁄ ...(51)

In case of infinite number of stages of a cascade system, the coefficient of performance can be given by:

⁄⁄ ...(52) However the infinite number of cascade system cannot be realized from the practical

point of view. Hence two-to three-stage systems can be practically a feasible choice. Figure 12(b) shows the variation of for different number of stages with varying

operating conditions. It is seen that as the number of stages increases, the coefficient of performance increases.

A brief outline for a multi-stage thermoelectric system is presented here with notations given in Fig. 13. The details of optimum number and sizes of multistage thermoelectric systems are available. The governing equations are obtained for couples as:

Heat rejection from stage 1 = heat absorbed by the second stage, i.e., ,

, , ...(53) The voltage drop through each side must be equal, i.e.,

, , ...(54)

Fig. 13 Multi-stage thermoelectric elements According to Kirchhoff' s law the total current at any junction is balanced, i.e.,

...(55) From Eqs. (54 and 55) , and are found as:

, ∆ ∆ ...(56a) , ∆ ∆ ...(56b)

where ∆ and ∆ ⁄ . Eliminating I , and I from Eq. (53) with the help of Eqs. (56a and 56b), one gets a

quadratic equation in I , as: ⁄⁄ ,

,. , ∆ ∆ . , ∆ ∆ ∆

∆ 0 ...(57) This equation was solved by Foster [3] for various values of , property values and

number of thermoelectric elernents. After calculating , from Eq. (57), cooling capacity and power are obtained

from the respective equations as: . , . . ∆ ...(58)

and, . , ∆ . . , ∆ , ∆ . ...(59)

The is then obtained as usual. The optimum results obtained have been shown in Fig. 14. It is seen that as the number of thermoelectric couples in the low stage increases, the decreases but the refrigeration effect increases. Again the maxima of

and refrigeration effect occur at different points: former at lower ampere as compared latter at the higher ampere value.

Fig. 14 Effect of changing the number of couples in the low stage of the two-stage

thermoelectric system. , , Two stage couples, , . .⁄ , Elect, conductivity / ,Seebeck coefficient ⁄ .

Example 7: The coefficient of performance of a single-stage thermoelectric system is

0.2. Obtain the for 2,3, 4 and infinite stage systems. Compare the results with respect to single-stage system.

Solution: Using Eq. (51), it is seen

⁄ ⁄⁄

. . ⁄⁄ 0.2893

. . ⁄⁄ 0.3038

. . ⁄⁄ 0.3088 In case of infinite number of stages the is given by Eq. (52),i .e.,

. .⁄ 0.3152 The required improvements due to multistaging are:

0.2893 / 0.2 1.446. 0.3038 / 0.2 1.519. 0.3088 / 0.2 1.544. 0.3152/ 0.2 1.576.

It is evident that the two-staging gives considerable improvement over the single

stage. Thereafter the improvement is very less. Even by using infinite number of rtag"r the has improved by about 3% over that of four-stages. However a three stage system is

barely 6% less efficient than infinite one. Seeing the complicacy of fabrication, it is seen that a two to four stage system is

quite reasonable choice. 7. ACTUAL THERMOELECTRIC REFRIGERATION SYSTEM The design consideration in the thermoelectric refrigeration system requires the

knowledge of optimum size, optimum number of couples, etc., of a given material. Also. one should know the contact resistance between the thermoelectric elements and the copper plates to which these elements are soldered. Similarly the increase in thermal resistances due to junction has to be incorporated. To achieve the desired performance high grade thermal insulation should be provided between the thermoelectric elements. Further the thermocouples should be insulated from the outside circuit using appropriate electric insulation having low thermal resistance. To protect the life of insulation and thermoelectric elements a good quality moisture resisting element should be provided as shown in Fig. 15. Thus an actual thermoelectric refrigeration system would perform successfully for many years to come.

Fig. S.15 An actual thermoelectric refrigeration system. Example.8: The resistance of a pair of thermoelectric elements is 0.0025 and

the conductance 0 .02 / . If the contact resistance is equivalent to 10% of the electric resistance, calculate the cooling produced by a thermoelectric system of 150 pairs for the case of maximum capacity. The cold and hot end temperatures are 310 and 260 . The donof the pair is 0.0005 ⁄ .

Also find , current and compare the aiter neglecting the effect of the contact resistance due to copper plate.

Solution: The figure of merit is obtained as:

, .. . .

0.004545 ⁄

. 2⁄ 150 0.02 0.004545 260 2 310 260⁄310.86

. ⁄.

0.2829

, ..

52 . So, without contact resistance is found to be 0.005 and therefore:

. ⁄.

0.2953 Required improvement 4%.