An earlier article in this column considered the problem

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1 --- CALC CORNER Estiating nternal Air Cooling Teperature Reduction in a Closed Box Utilizing Theroelectrically Enhanced Heat Rejection Previously published in February, 2013 Bob Sions BM Retired The following article was originally published in the February 2013 issue. t is being republished here because of its significant tutorial value. Bob served as an Associate Technical Editor of this publication fro January, 2001, to Deceber, 2011." An earlier article in this colun considered the proble of cooling electronic coponents in a closed box [1]. n outdoor applications for exaple, it ay be necessary to totally seal the box to prevent exposure to airborne particulates, water droplets or other substances in the air that could be injurious to the electronic coponents. n such an application, heat picked up by the air circulating over the electronic coponents within the box is rejected to outside air by eans of an air-to-air heat exchanger ounted in one of the walls of the box. Such a heat exchanger ay be as siple as two plate-fin heat sinks (or a heat sink with fins on both sides) ounted base to base in an opening in the wall of the box. Of course in such cases, the air that is cooling the electronics coponents will always be higher in teperature than the outside air being used for heat rejection fro the box. f this does not provide a satisfactory cooling solution, an alternative that ight be considered is augentation with theroelectric cooling (TE) odules sandwiched between the heat sinks as shown in Figure 1. TE odule [2]. For exaple in Figure 1, electric current,, ay be applied to the theroelectric odules such that the side in contact with the base of the heat sink within the box becoes cooler, thereby augenting the transfer of heat fro the air circulating within the box to the internal heat sink. The opposite side of the theroelectric odule becoes hotter and both the heat puped by the theroelectric cooler (qp)' and the heat dissipated (qt.) by the theroelectric cooler in perforing its heat puping function, is transferred to outside cooling air via the external heat sink. This article is intended to present to the reader, by exaple, a ethodology to estiate the cooling air teperatures that can be achieved by theroelectric augentation. However, to illustrate the value of theroelectric augentation, we will first consider soe equations to calculate the air teperature (Til) entering the portion of the box housing the electronic coponents without theroelectric augentation. Working fro the air teperature (To) outside the box to the teperature of the wall (T) of the box (or a base teperature coon to the external and internal heat sinks) gives, (1) where q is the heat dissipation of the electronic coponents and R2 is the theral resistance of the external heat sink. t should be noted that throughout this analysis and in the subsequent calculation exaples, heat sink resistances, R, and R 2, are assued to include any associated theral interface resistance between the heat sinks and the wall or later in the article, the TE odules. The air teperature entering the internal heat sink passages is given by, (2) - Figure 1. Closed box electronics enclosure with theroelectrically enhanced heat rejection. For those that are unfailiar with TE odules, a theroelectric cooler, soeties called a Peltier cooler, is a solid-state heat pup that transfers heat fro one side of the device to the other side depending on the direction of the applied electric current 26 Eleclronics COOLNG FALL 2017 which, substituting in eq. (1) for Tw, gives The teperature drop of the air passing through the internal heat sink is given by,, (3) (4)

2 Eleclron ics COOLNG FALL 2017 where C\ is the air heat capacity rate inside the box, given by the product of the ass flow rate,, and specific heat, c, of the inp side air cooling strea. So, the teperature, Til' of the cooling air entering the electronics copartent (without theroelectric augentation) becoes, Til = TOl +q(r +R 2 ) C 1 (5) Now, we will proceed to deterine the teperature of the air entering the electronics copartent when theroelectric cooling augentation is incorporated. To do this we ust include equations to account for theroel~ric heat puping and the heat dissipation of theroelectric odules. n an earlier article, Luo [3] presented the following equation for heat puping with a theroelectric cooling odule qp = STc-~2R -K(Th -T.) (6) and another equation for the corresponding TE heat dissipation in ters of the TE paraeters at the odule as defined in the noenclature and discussed by Luo [3]. n addition to the above two theroelectric equations we have the following heat transfer equations, qp = qte = q = Th = Tc = TO = Til = T2 = Tw =... S... K Rot - R = R2 Cl =... Table 1. Noenclature To = Ti2 -qr 1 Tb = TO +(q +qtjr 2 Heat puped by TE odules Heat dissipation of TE odules Electronics heat dissipation in box TE odule hot side teperature TE odule cold side teperature Air teperature outside box Air teperature into electronics package Air teperature out of electronics package Base teperature of heat sink(s) w/o TEs Electric current supplied to TE odule TE odule effective Seebeck coefficient TE odule theral conductance TE odule electrical resistance Theral resistance of cold side heat sink Theral resistance of cold side heat sink Air heat capacity rate within box ( ril c p ) (7) (8) (9) (10) t should be noted here that under steady-state conditions heat load, q, in Equations (8-10) is equal to the heat puped, qp' by the theroelectric cooler. ConSidering Equations (8-10) together with thero-electric Equations (6) and (7) we have five linear algebraic equations. Given that we know or can assue values for q, R\, R. ' C\, S' K' R and, we are left with five unknown variables. The unknown variables are q,., T h, Tc' Ti2 and Til (which is the variable we are really after). t is possible to solve these equations via algebraic substitution, as the author has done with the aid of software [4] capable of perforing sybolic algebraic anipulation. However, the author found that the algebraic solution obtained for Til by this ethod was so long (about two or three screen-widths) and so coplex as to be of practically no value other than deonstrating that a solution could be obtained. A ethod of practical utility is to solve these equations for nuerical values using a atrix ethod as eployed in an earlier article addressing the solution of a theral resistance network [5]. To do this we first rearrange the Equations (6-10) so that the unknown variables are on the left-hand side of the equation and the constant ters are on the right-hand side, 1 2 STcqp -KTh + KTc = q+2" R (6a) (7a) (8a) (9a) (10a) n atrix notation these sae equations can be copactly represented as [Coeffs] x [Unknowns] = [Constants] (ll) The coefficients of the unknown variables and the corresponding constant ters for each equation ay be grouped in a tabular for as shown in Table 2. So doing, the top row represents the colun vector of unknowns, the coluns beneath each unknown variable ake up the coefficient atrix and the rightost colun the constant vector. Til Ti2 Tc qte Tb Constant 0 0 S+K 0 -K q+ 2 R/2 0 0 S 1 - S 2 R - C C q qr R2 1 T O +qr 2 Table 2 - Coeffldents of Unknown Variables and Constants in Equations Electronics-COOllNG.co 27 -

3 - s - The coefficient atrix below coprises the known coefficients of the unknown variables of the equations, as shown in Table 2, 0 0 S +K 0 -K 0 0 S 1 -S Coeffs -C 1 C R2 1 the colun vector of unknown variables which we seek to solve is, Til Ti2 Unknowns = Tc and the colun vector coprising the known constants on the right hand side of each equation is, Cons tan ts = A atrix equation such as (11) ay be solved for the unknown variables by ultiplying the constant vector by the inverse of the coefficient atrix, [Unknowns] = [Coeffs]-J x [Constants] (12) Many coputational aids such as MathCad (4), Matlab (6), and EXCEL (7) can be used to obtain the atrix inverse of the coefficient atrix and ultiply by the constant vector to obtain the desired solution vector. To deonstrate the potential enhanceent with theroelectrically augented cooling, we now turn our attention to soe nuerical results obtained using the ethod described above. For purposes of illustration, a sall enclosure with an allowable opening in one side of 100 x 100 is assued. t is further assued that the base diensions of the internal and external air-cooling heat sinks are 100 x 100. Before proceeding further it is 28 Electronics COOLNG FAll2017 necessary to assign values to the theroelectric odule cooling paraeters, S' K,and R' The sae 40 x 40 TE odule considered by the author in an earlier article[81 will be considered here. n the earlier article, the author illustrated the calculation of single odule TE paraeters fro vendor data, using the ethod discussed by Luo (3). The values for the exaple TE odule were found to be: S = V/K K = W/K R ohs However, to increase the overall heat puping capacity, four of these TE odules can be sandwiched in a 2 x 2 array between the bases of the internal an external heat sink and wired electrically in series. Consequently, the paraeters for the array ofte odules will siply be four ties the value for a single TE odule or, S K R = V/K = W/K = ohs Calculations were perfored for a range of heat sink resistances to show the effects of this paraeter, coupled with the theroelectric cooling effects. The heat sink theral resistances used in the calculations were 0.075, and 0.17S)t/W and were considered to have the sae value for the heat sink within and outside the box (i.e. Rl = R 2 ). The internal air flow rate within the box was assued to be /s (20 CFM).t should be reebered that Equations (6) and (7) are in Kelvin teperature units (Le. Kelvin teperature = Centigrade teperature So the teperature of the outside cooling air, TOl, used in the calculations, ust also be in degrees Kelvin and the teperatures obtained fro Equation (12) will be in degrees Kelvin. However, for ease of understanding, the teperatures reported in the following figures have been converted to degrees Centigrade. Figure 2 illustrates both the effect of increasing electric current through the TE odules and heat sink theral resistance, with a heat load of 100 watts fro the electronics and an outside air teperature of 35 C. The solid lines represent the cooling air teperature within the box with the TE odules sandwiched between the heat sinks and the dashed lines represent the teperatures obtained without the TE odules. As can be seen, at low values of electric current, the presence of the TE odules result in higher cooling air teperatures within the box. This is because at low electric current, the Peltier heat puping effect is offset by the theral resistance across the TE odules. As current is increased the Peltier h! ~ p.uping effect becoes ore Significant and the.inside c;qqijng ~JJ teperature can be decreased Significantly. However, ~a ~lcctric current continues to be increased, the Joule heating (i.e. h9~t dissipation) within the TE odules becoes increaa(ua1y aignincant causing the inside cooling air teperature to,~*o out and if current is increased still further begin to rise. '.

4 Electronics COOLNG FALL () ~ ~ 60 :::l ~ a. 40 E... f- 20 «00 oint... end._1 vouelllc el. 10w 'll C (20 CFM) q-100w_ R1-R2 (Deg CfW) TE electric current (aps) Figure 2. Effect of increasing electric current to TE odules on internal cooling air teperature = 3.5 aps.. ' ~.., ' O L-~~~~~~~-L~-~ Electronics heat load, q (watts) Figure 3. Effect of electronic heat load on internal cooling air teperature with (solid lines) and without (dashed lines) theroelectric augentatian. -70 ~ ~.. ~ -, u.... 1= 3.5 aps ~60.., , 540 ~ 30 a. E 20 ': 10 « Outside air teperature, To1 (Oeg) Figure 4. Effect of outside air teperature on internal cooling air teperature, Til, with (solid lines) and without (dashed lines) theroelectric augentation. Figure 3 illustrates the effect of the electronics heat dissipation on cooling air teperature within the box, with a constant current of 3.5 aperes through the array of TE odules and an outdoor teperature of 35 C. As can be seen, the cooling air teperatures obtained in all the cases are lower than could be achieved without eploying TE enhanceent. t should also be noted that, depen- ding on heat load and heat sink theral resistance in any cases the cooling air teperature is even lower than could be obtained using outside air for cooling. The results in figure 4, show the effect on cooling air teperature within the box due to supplying a fixed current of 3.5 aperes to the TE odules at a fixed electronics heat load of loo watts. As in the preceding figures, it can be seen that the resulting cooling air teperatures are always below the cooling air teperatures that could be achieved without theroelectric enhanceent. The equations and solution ethodology presented in this article can provide a useful tool with which to obtain a preliinary estiate of the effectiveness of theroelectric augentation in providing cooling to electronic coponents in a closed box. Of course, when considering the use of theroelectrics for cooling applications, the designer should be cognizant of the electrical power required to drive the theroelectric eleents, which is given by Equation (7). The cooling efficiency of theroelectric devices relative to their power consuption, as with other types of heat puping devices, ay be characterized in ters of Coefficient of Perforance (COP). COP is defined as the ratio of cooling (q) provided over the electrical energy consued (qte) to produce the cooling effect. For exaple, for the results shown in Figure 3, the COPs ranged fro 0.3 to 0.4 at heat loads of 50 or 60 W to around 1 at a heat load of 150 W. The range of heat sink theral resistance values had only a little effect on the COP realized. Siilarly, considering the results shown in Figure 4, COPs ranged fro 0.6 to 0.7 with little effect due to heat sink theral resistance or outside air teperature. As one ight expect fro Equation (7) the principal effect on COP is caused by the electrical current required to drive the theroelectrics. This is aply deonstrated considering the results presented in Figure 2. At electric currents around 1.25 A, the COPs calculated ranged fro 6 to 6.4 depending on the value of heat sink theral resistance. Although these COPs ay see good, in this case the cooling air teperature within the box is no lower than could be achieved without theroelectrics. As current is increased in Figure 2, the COP realized decreases rapidly to about 0.65 at a current of 3.5 A. However, even though the COP has dropped, this coincides with the axiu reduction of air teperature within the box by as uch as 20 to 30 Centigrade degrees below those achieved without theroelectric augentation. For coparison, the COP values realized by conventional vapor copression refrigeration systes ay typically vary fro 1 to 4. Finally, it should also be noted that a nuber of vendors provide theroelectric heat exchangers for cooling air in a closed box. Electronics COOLlNG.co 29

5 11, The interested reader ay find a nuber of vendor products and exaples by conducting an internet web search using the ters "theroelectric air cooler" or "theroelectric air conditioner." REFERENCES [1] Sions, R.E., "Estiating Teperatures in an Air-Cooled Closed Box Electronics Enclosure;' ElectronicsCooling, February [2] Godfrey, S., ''An ntroduction to Theroelectric Coolers" ElectronicsCooling, Septeber [3] Luo, Z., ''A Siple Method to Estiate the PhYSical Characteristics of a Theroelectric Cooler fro Vendor Datasheets;' ElectronicsCooling, August [4] wikipedia.org/wikil Mathcad [5] Sions, R.E., "Using a Matrix nverse Method to Solve a Theral Resistance Network;' Electronics Cooling, May [6] [7] _ solve-equationsatrix-ethod-exce1.htl [8] Sions, R.E., "Using Vendor Data to Estiate Theroelectric Module Cooling Perforance in an Application Environent;' ElectronicsCooling, July ' ' /' 30 Electron ics COOLNG FALL 2017