At state 1:

occurs at the other side. This material is chosen due to being a good insulator of electricity and also being thermally conductive. The coefficient of performance of this device is defined as the ratio of the cooling or heating power to the power

**3. Theoretical analysis of thermoelectric air conditioning system**

allow for the heat from the air to be properly transferred to the heat sinks.

The temperature and relative humidity of the ambient air measured in the laboratory was 31°C and 63% respectively. In order to obtain within thermal comfort range as defined by ASHRAE, it is best to cool and dehumidify the air to 22°C with a relative humidity of 50% as shown in **Figure 6**. The cold side of the thermoelectric module would be at a temperature lower than the dew point temperature and therefore condensation is expected to take place which would decrease the

In order to pump a great amount of heat, the thermoelectric device usually consists of multiple P-type and N-type elements. A typical thermometric device contains around 250 P-type and N-type elements connected in series as shown in in **Figure 4**.

The design of thermoelectric air conditioning system is shown in **Figure 5**, a fan is mounted on top of the cover where air at ambient temperature would be sucked into the device and circulate through the heat sinks and then blow through two rectangular holes such that the direction of the air can be controlled via the flaps. This would

supplied to the module.

*Low-temperature Technologies*

*P-type and N-type elements connected in series.*

**Figure 4.**

**3.1 Determining the cooling load**

relative humidity of the air [5, 6].

**Figure 5.**

**62**

*Thermoelectric air conditioning system.*

*3.1.1 Cooling and dehumidification of the air*

31°C, 63% *relative humidity*

$$h\_1 = 76.91 \text{ kJ/kg}; \, \theta\_1 = 0.887 \text{m}^3/\text{kg}; \, W\_1 = 0.0179 \text{ kg/kg}$$

At state 2: 22°C, 50% *relative humidity*

$$h\_2 = 43.04 \text{ kJ/kg}; W\_2 = 0.0082 \text{ kg/kg}; h\_{f,2} = 83.94 \text{ kJ/kg}$$

A fan of 15 CFM was available at the lab: 15 CFM = 0.00708 m<sup>3</sup> /s

$$\begin{aligned} \dot{m}\_d &= 0.00708 \text{ m}^3/\text{s} \times \frac{1}{0.887 \text{m}^3/\text{kg}} \\ &= 0.00798 \text{ kg/s} \\ \dot{Q}\_T &= \dot{m}\_a \left[ (h\_1 - h\_2) - (W\_1 - W\_2)h\_{f,2} \right] \\ &= 0.00798 [(76.91 - 43.04) - (0.0179 - 0.0082)83.94] \\ &= 0.264 \text{ kW} \\ &= 264 \text{ W} \end{aligned} \tag{4}$$

#### *3.1.2 Sensible cooling load of the enclosure*

An enclosure shown in **Figure 7** was designed in order to test the thermoelectric air conditioning system. The enclosure was made of plywood and properly insulated with styrofoam so that the outside temperature would have minimal effect on the testing of the device.

The temperature of the ambient air is 31°C

Thermal comfort temperature is 22°C

Thermal conductivity of wood, *K*wood = 0.151 <sup>W</sup> mK

Thermal conductivity of styrofoam, *K*styrofoam = 0.033 <sup>W</sup> mK

$$R\_{\text{Total}} = \frac{L\_{\text{wood}}}{K\_{\text{wood}}} + \frac{L\_{\text{stryrofoam}}}{K\_{\text{stryrofoam}}}$$

$$= \frac{0.01}{0.151} + \frac{0.035}{0.033} \tag{5}$$

$$= 1.13 \frac{K}{W}$$

*3.1.3 Resistor cooling load*

*DOI: http://dx.doi.org/10.5772/intechopen.88664*

monitored with time.

*3.1.4 Total cooling load required*

enclosure was not properly sealed.

**3.2 Choosing the Peltier module**

each thermoelectric module.

**3.3 Sizing the heat sink**

**65**

hot side of each thermoelectric module:

A resistor/heating coil was placed in the testing enclosure to observe how long

Total cooling load ¼ cooling*=*dehumidification þ cooling load of enclosure þ resistor load

However, the device was sized at a total cooling load of 330 W to take into considerations any additional heat loads that were not accounted for, also the

One of the TEC1-12730 module is capable of producing 250 W of cooling, however in order for one module to produce 250 W of cooling, it needs a dc power supply rated at 30 amps and 18 V. Therefore, by using three modules, a power supply that is available at the lab can be used to supply each module rated at 12 V.

Cooling capacity required for each module <sup>¼</sup> <sup>330</sup>

Temperature difference*, ΔT* ¼ *Thot* � *Tcold*

The specification graphs for the TEC1-12730 module shown in **Figures 8** and **9** were used in determining the appropriate amperage and voltage needed to supply

Power consumed by the three (3) Peltier device = 3 � ð Þ¼ 12 � 21 756 W

Determining the thermal resistance of a required heat sink (**Figure 10**) for the

At *Q*\_ *<sup>c</sup>* = 110 W and *ΔT* = 32°C, amperage, I = 21 amps Using the second graph to determine the voltage to apply:

Using 21 amps and *ΔT* = 32°C, therefore V = 12 V

3

¼ 110 W

¼ 50°C � 18°C

¼ 32°C

(6)

the device takes to remove the heat and maintain a constant temperature. A resistor of 20 W rating was used and the temperature of the enclosure was

*Theoretical and Experimental Analysis of a Thermoelectric Air-conditioning System*

¼ 264 W þ 13*:*1 *W* þ 20 W

¼ 297*:*1 W

Using three Peltier TEC1-12730 modules:

Heat transfer through the wall of the box per unit area:

$$q'' = \frac{\Delta T}{R\_{total}}$$

$$= \frac{9 \, K}{1.13}$$

$$= 7.96 \, \frac{\text{W}}{\text{m}^2}$$

Area of enclosure ¼ 2 0ð Þþ *:*52 � 0*:*50 2 0ð Þþ *:*55 � 0*:*50 ð Þþ 0*:*55 � 0*:*50 ð Þ 0*:*57 � 0*:*52 m <sup>¼</sup> <sup>1</sup>*:*6414 m2

Therefore, heat transfer through the wall of the box,

$$\begin{aligned} \dot{Q} &= 7.96 \times 1.6414 \text{ m}^2 \\ &= 13.1 \text{ W} \end{aligned}$$

**Figure 7.** *Enclosure for test of thermoelectric air conditioning system.*

*Theoretical and Experimental Analysis of a Thermoelectric Air-conditioning System DOI: http://dx.doi.org/10.5772/intechopen.88664*

#### *3.1.3 Resistor cooling load*

*3.1.2 Sensible cooling load of the enclosure*

The temperature of the ambient air is 31°C Thermal comfort temperature is 22°C

Thermal conductivity of wood, *K*wood = 0.151 <sup>W</sup>

Thermal conductivity of styrofoam, *K*styrofoam = 0.033 <sup>W</sup>

Heat transfer through the wall of the box per unit area:

<sup>¼</sup> <sup>1</sup>*:*6414 m2

*Enclosure for test of thermoelectric air conditioning system.*

**Figure 7.**

**64**

Therefore, heat transfer through the wall of the box,

*<sup>R</sup>*Total <sup>¼</sup> *<sup>L</sup>*wood

*K*wood þ

<sup>¼</sup> <sup>0</sup>*:*<sup>01</sup> 0*:*151 þ

<sup>¼</sup> <sup>1</sup>*:*<sup>13</sup> *<sup>K</sup> W*

*<sup>q</sup>*″ <sup>¼</sup> *<sup>Δ</sup><sup>T</sup> Rtotal*

> <sup>¼</sup> <sup>9</sup> *<sup>K</sup>* 1*:*13

<sup>¼</sup> <sup>7</sup>*:*<sup>96</sup> <sup>W</sup> m2

Area of enclosure ¼ 2 0ð Þþ *:*52 � 0*:*50 2 0ð Þþ *:*55 � 0*:*50 ð Þþ 0*:*55 � 0*:*50 ð Þ 0*:*57 � 0*:*52 m

*<sup>Q</sup>*\_ <sup>¼</sup> <sup>7</sup>*:*<sup>96</sup> � <sup>1</sup>*:*6414 m<sup>2</sup>

¼ 13*:*1 W

testing of the device.

*Low-temperature Technologies*

An enclosure shown in **Figure 7** was designed in order to test the thermoelectric air conditioning system. The enclosure was made of plywood and properly insulated with styrofoam so that the outside temperature would have minimal effect on the

mK

*L*styrofoam *K*styrofoam

0*:*035 0*:*033 mK

(5)

A resistor/heating coil was placed in the testing enclosure to observe how long the device takes to remove the heat and maintain a constant temperature. A resistor of 20 W rating was used and the temperature of the enclosure was monitored with time.

#### *3.1.4 Total cooling load required*

Total cooling load ¼ cooling*=*dehumidification þ cooling load of enclosure þ resistor load

$$\begin{aligned} \mathbf{w} &= 264 \text{ W} + 13.1 \text{ W} + 20 \text{ W} \\\\ \mathbf{w} &= 297.1 \text{ W} \end{aligned} \tag{6}$$

However, the device was sized at a total cooling load of 330 W to take into considerations any additional heat loads that were not accounted for, also the enclosure was not properly sealed.

#### **3.2 Choosing the Peltier module**

One of the TEC1-12730 module is capable of producing 250 W of cooling, however in order for one module to produce 250 W of cooling, it needs a dc power supply rated at 30 amps and 18 V. Therefore, by using three modules, a power supply that is available at the lab can be used to supply each module rated at 12 V.

Using three Peltier TEC1-12730 modules:

$$\text{Cooling capacity required for each module} = \frac{330}{3}$$

$$= \textbf{110 W}$$

$$\text{Temperature difference, } \Delta T = T\_{hot} - T\_{cold}$$

$$= \textbf{50°C} - \textbf{18°C}$$

$$= \textbf{32°C}$$

The specification graphs for the TEC1-12730 module shown in **Figures 8** and **9** were used in determining the appropriate amperage and voltage needed to supply each thermoelectric module.

At *Q*\_ *<sup>c</sup>* = 110 W and *ΔT* = 32°C, amperage, I = 21 amps Using the second graph to determine the voltage to apply: Using 21 amps and *ΔT* = 32°C, therefore V = 12 V Power consumed by the three (3) Peltier device = 3 � ð Þ¼ 12 � 21 756 W

#### **3.3 Sizing the heat sink**

Determining the thermal resistance of a required heat sink (**Figure 10**) for the hot side of each thermoelectric module:

Maximum operating temperature of the thermoelectric module: 138°*C* [7]. Thermal resistance of the heat sink needed,

*Rhs* <sup>¼</sup> *Thot* � *Tamb <sup>Q</sup>*\_ *T,* <sup>1</sup> <sup>þ</sup> *Ppeltier*

*Theoretical and Experimental Analysis of a Thermoelectric Air-conditioning System*

*DOI: http://dx.doi.org/10.5772/intechopen.88664*

<sup>¼</sup> <sup>138</sup> � <sup>31</sup> 362

<sup>¼</sup> <sup>0</sup>*:*<sup>296</sup> *<sup>K</sup>*

*<sup>b</sup>* <sup>¼</sup> *<sup>W</sup>* � *Nfin* � *tfin Nfin* � 1

¼ 0*:*003 m

<sup>V</sup> <sup>¼</sup> *<sup>ϑ</sup>*

*Nfin* � *b* � *Hf*

20 � 0*:*003 � 0*:*057

<sup>¼</sup> <sup>0</sup>*:*<sup>142</sup>

<sup>¼</sup> <sup>41</sup>*:*52 ms�<sup>1</sup>

<sup>¼</sup> <sup>0</sup>*:*<sup>08</sup> � ð Þ <sup>20</sup> � <sup>0</sup>*:*<sup>0015</sup> 20 � 1

Air flow of fan used for heat sink, *ϑ* = 0.142 m<sup>3</sup>*s*

Width of heat sink, W = 0.08 m Width of the fin, L = 0.064 m Thickness of fin, *tfin* = 0.0015 m Number of fins, *Nfin* = 20 Height of fin, *Hf* = 0.057 m Height of heat sink, H = 0.069 m Spacing between the fins [8],

**Figure 10.** *Sizing the heat sink.*

Velocity of air between the fins,

**67**

*W*

�<sup>1</sup> (CFM of fan available at lab = 300)

**Figure 8.** *Specification graph 1 for the TEC1-12730 module.*

**Figure 9.** *Specification graph 2 for the TEC1-12730 module.*

*Theoretical and Experimental Analysis of a Thermoelectric Air-conditioning System DOI: http://dx.doi.org/10.5772/intechopen.88664*

Maximum operating temperature of the thermoelectric module: 138°*C* [7].

Thermal resistance of the heat sink needed,

*Low-temperature Technologies*

**Figure 8.**

**Figure 9.**

**66**

*Specification graph 1 for the TEC1-12730 module.*

*Specification graph 2 for the TEC1-12730 module.*

$$R\_{hs} = \frac{T\_{hot} - T\_{amb}}{\dot{Q}\_{T,1} + P\_{peltier}}$$

$$= \frac{138 - 31}{362}$$

$$= 0.296 \,\frac{K}{W}$$

Air flow of fan used for heat sink, *ϑ* = 0.142 m<sup>3</sup>*s* �<sup>1</sup> (CFM of fan available at lab = 300) Width of heat sink, W = 0.08 m Width of the fin, L = 0.064 m Thickness of fin, *tfin* = 0.0015 m Number of fins, *Nfin* = 20 Height of fin, *Hf* = 0.057 m Height of heat sink, H = 0.069 m Spacing between the fins [8],

$$\begin{split}b &= \frac{W - \left(N\_{fin} \times t\_{fin}\right)}{N\_{fin} - 1} \\ &= \frac{0.08 - \left(20 \times 0.0015\right)}{20 - 1} \\ &= 0.003\text{ m}\end{split}$$

Velocity of air between the fins,

$$\text{V} = \frac{\text{g}}{N\_{\hat{f}\hat{m}} \times b \times H\_f}$$

$$= \frac{\text{0.142}}{20 \times 0.003 \times 0.057}$$

$$= 41.52 \text{ ms}^{-1}$$

The properties of air at one (1) atmosphere and 31°C are as follows: Density, ρ = 1.15 kg m3 Specific heat capacity, Cp = 1.007 kJ kg*:*K Dynamic viscosity, *<sup>μ</sup>* = 18.718 �10�<sup>6</sup> <sup>N</sup>*:*<sup>s</sup> m2 Heat conductance, K = 26.424 �10�<sup>6</sup> kW m*:*K Prandtl number,

$$\begin{split} \text{Pr} &= \frac{\mu \times \text{C}\_p}{k} \\ &= \frac{18.718 \times 10^{-6} \times 1.007}{26.424 \times 10^{-6}} \\ &= 0.713 \end{split}$$

*m* ¼

*Theoretical and Experimental Analysis of a Thermoelectric Air-conditioning System*

¼

Surface area of the exposed base,

*DOI: http://dx.doi.org/10.5772/intechopen.88664*

Thermal resistance of the heat sink,

*Rhs* <sup>¼</sup> <sup>1</sup>

<sup>¼</sup> <sup>0</sup>*:*<sup>109</sup> <sup>K</sup>

• 12 channel thermocouple data recorder

Area of the fin,

0.109 <sup>K</sup>

**4.1 Apparatus**

<sup>W</sup> < 0.296 <sup>K</sup>

this heat sink was used.

• Thermocouples

• Power supply

• Multi-meter

• Power strip

**69**

¼ 29*:*39

*Abase* <sup>¼</sup> *Nfin* � <sup>1</sup> � � *<sup>b</sup>* � *<sup>L</sup>*

<sup>¼</sup> <sup>0</sup>*:*00365 m<sup>2</sup>

*Afin* ¼ 2 � *Hf* � *L*

<sup>¼</sup> <sup>0</sup>*:*0073 m2

*<sup>h</sup>: Abase* <sup>þ</sup> *NfinηfinAfin* � � � �

<sup>¼</sup> <sup>1</sup>

W

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 2*h kfin* � *tfin* <sup>s</sup>

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 2 *x* 108*:*16 <sup>167</sup> � <sup>0</sup>*:*<sup>0015</sup> <sup>r</sup>

¼ ð Þ 20 � 1 0*:*003 � 0*:*064

¼ 2 � 0*:*057 � 0*:*064

108*:*16*:*ð Þ 0*:*00365 þ ð Þ 20 � 0*:*556 � 0*:*0073

available is less than the required thermal resistance of the heat sink then therefore,

**4. Experimentation of the thermoelectric air conditioning system**

W, since the calculated thermal resistance of the heat sink

Reynolds number,

$$\begin{split} \text{Re} &= \frac{\rho \times V \times b}{\mu} \times \frac{b}{L} \\ &= \frac{1.15 \times 41.52 \times 0.003}{18.718 \times 10^{-6}} \times \frac{0.003}{0.064} \\ &= 358.72 \end{split}$$

Nusselt number,

$$Nu = \left[\frac{1}{\left(\frac{\text{Re}\times\text{Pr}}{2}\right)^3} + \frac{1}{\left(0.664\sqrt{\text{Re}}\text{Pr}^{0.33}\sqrt{1 + \frac{3.65}{\sqrt{\text{Re}}}}\right)^3}\right]^{\frac{1}{3}}$$

$$= \left[\frac{1}{\left(\frac{388.72 \times 0.713}{2}\right)^3} + \frac{1}{\left(0.664\sqrt{358.72}0.713^{0.33}\sqrt{1 + \frac{3.65}{\sqrt{358.72}}}\right)^3}\right]^{\frac{1}{3}}$$

$$= 12.28$$

Heat transfer coefficient,

$$\begin{aligned} \mathbf{h} &= N\boldsymbol{\mu} \times \frac{k}{b} \\ &= \mathbf{12.28} \times \frac{26.424 \times 10^{-3}}{0.003} \\ &= \mathbf{108.16} \end{aligned}$$

Efficiency of the fin,

$$\begin{aligned} \eta\_{fin} &= \frac{\tanh\left(m \times H\_f\right)}{m \times H\_f} \\ &= \frac{\tanh\left(29.39 \times 0.057\right)}{29.39 \times 0.057} \\ &= 0.556 \end{aligned}$$

*Theoretical and Experimental Analysis of a Thermoelectric Air-conditioning System DOI: http://dx.doi.org/10.5772/intechopen.88664*

$$\begin{aligned} m &= \sqrt{\frac{2h}{k\_{\hat{f}\hat{m}} \times t\_{\hat{f}\hat{m}}}} \\ &= \sqrt{\frac{2 \times 108.16}{167 \times 0.0015}} \\ &= 29.39 \end{aligned}$$

Surface area of the exposed base,

$$\begin{aligned} A\_{base} &= \left( N\_{fin} - 1 \right) b \times L \\ &= (20 - 1) 0.003 \times 0.064 \\ &= 0.00365 \text{ m}^2 \end{aligned}$$

Area of the fin,

The properties of air at one (1) atmosphere and 31°C are as follows:

Pr <sup>¼</sup> *<sup>μ</sup>* � *Cp k*

¼ 0*:*713

Re <sup>¼</sup> *<sup>ρ</sup>* � *<sup>V</sup>* � *<sup>b</sup> μ*

¼ 358*:*72

0*:*664 ffiffiffiffiffiffi Re

h ¼ *Nu* �

¼ 12*:*28 �

¼ 108*:*16

*<sup>η</sup>fin* <sup>¼</sup> tanh *<sup>m</sup>* � *Hf*

¼ 0*:*556

� �

<sup>¼</sup> tanh 29 ð Þ *:*<sup>39</sup> � <sup>0</sup>*:*<sup>057</sup> 29*:*39 � 0*:*057

*m* � *Hf*

kg*:*K

m2

m*:*K

<sup>¼</sup> <sup>18</sup>*:*<sup>718</sup> � <sup>10</sup>�<sup>6</sup> � <sup>1</sup>*:*<sup>007</sup> <sup>26</sup>*:*<sup>424</sup> � <sup>10</sup>�<sup>6</sup>

> � *b L*

> > 1

� � q <sup>3</sup>

0*:*664 ffiffiffiffiffiffiffiffiffiffiffiffiffi

*k b*

<sup>p</sup> Pr<sup>0</sup>*:*<sup>33</sup> ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi

<sup>1</sup> <sup>þ</sup> <sup>3</sup>*:*<sup>65</sup>ffiffiffiffi Re p

1

� � q <sup>3</sup>

<sup>26</sup>*:*<sup>424</sup> � <sup>10</sup>�<sup>3</sup> 0*:*003

<sup>358</sup>*:*<sup>72</sup> <sup>p</sup> <sup>0</sup>*:*713<sup>0</sup>*:*<sup>33</sup> ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi

0*:*003 0*:*064

> 3 7 5

�1 3

> <sup>1</sup> <sup>þ</sup> <sup>3</sup>*:*<sup>65</sup> ffiffiffiffiffiffiffiffiffi <sup>358</sup>*:*<sup>72</sup> <sup>p</sup>

3 7 5

�1 3

<sup>¼</sup> <sup>1</sup>*:*<sup>15</sup> � <sup>41</sup>*:*<sup>52</sup> � <sup>0</sup>*:*<sup>003</sup> <sup>18</sup>*:*<sup>718</sup> � <sup>10</sup>�<sup>6</sup> �

Density, ρ = 1.15 kg

*Low-temperature Technologies*

Prandtl number,

Reynolds number,

Nusselt number,

*Nu* <sup>¼</sup> <sup>1</sup>

2 6 4

2 6 4

¼ 12*:*28

Heat transfer coefficient,

Efficiency of the fin,

**68**

Re�Pr 2 � �<sup>3</sup> þ

358*:*72�0*:*713 2 � �<sup>3</sup> þ

<sup>¼</sup> <sup>1</sup>

m3 Specific heat capacity, Cp = 1.007 kJ

Dynamic viscosity, *<sup>μ</sup>* = 18.718 �10�<sup>6</sup> <sup>N</sup>*:*<sup>s</sup>

Heat conductance, K = 26.424 �10�<sup>6</sup> kW

$$\begin{aligned} A\_{fin} &= \mathbf{2} \times H\_f \times L \\ &= \mathbf{2} \times \mathbf{0}.057 \times \mathbf{0}.064 \\ &= \mathbf{0}.0073 \text{ m}^2 \end{aligned}$$

Thermal resistance of the heat sink,

$$\begin{split} R\_{hs} &= \frac{1}{h.\left(A\_{base} + \left(N\_{fin} \eta\_{fin} A\_{fin}\right)\right)}\\ &= \frac{1}{108.16.(0.00365 + (20 \times 0.556 \times 0.0073))}\\ &= 0.109 \left\lfloor \frac{\text{K}}{\text{W}} \right\rfloor \end{split}$$

0.109 <sup>K</sup> <sup>W</sup> < 0.296 <sup>K</sup> W, since the calculated thermal resistance of the heat sink available is less than the required thermal resistance of the heat sink then therefore, this heat sink was used.

#### **4. Experimentation of the thermoelectric air conditioning system**

#### **4.1 Apparatus**


**Time (min) Ambient temperate (°C) Temperature inside enclosure (°C)**

*Theoretical and Experimental Analysis of a Thermoelectric Air-conditioning System*

*DOI: http://dx.doi.org/10.5772/intechopen.88664*

*Temperature readings obtained from testing enclosure of thermoelectric air conditioner.*

**Table 1.**

**Figure 13.**

**71**

*Graph of temperature/°C versus time/min.*

 30.5 30.7 30.9 30.8 30.7 30.9 29.7 29.9 31.2 31.0 28.2 28.1 28.0 28.1 27.1 27.5 27.1 27.2 31.2 26.1 26.1 25.9 26.3 24.8 27.2 26.3 25.5 30.0 23.5 23.5 23.4 23.5 22.3 25.1 24.3 23.2 31.0 21.7 21.9 21.9 21.9 20.7 24.0 23.5 21.8 30.8 20.5 20.7 20.6 20.6 19.6 23.0 22.5 20.6 30.0 19.7 19.9 20.0 19.9 18.9 22.4 21.6 20.0 31.0 18.4 18.7 18.8 18.6 17.8 21.3 20.4 18.9 31.0 18.0 18.2 18.3 18.1 17.3 20.9 20.0 18.5 31.0 17.8 18.1 18.2 18.0 17.1 20.1 19.3 18.3 31.0 17.6 17.9 18.0 17.7 16.9 20.0 18.8 18.1 30.6 17.4 17.7 17.9 17.5 16.7 18.9 18.1 18.0 30.6 17.2 17.4 17.5 17.2 16.4 18.7 18.1 17.7 30.8 17.4 17.7 17.8 17.4 16.6 18.9 18.2 17.9 30.5 17.2 17.6 17.6 17.4 16.5 18.8 18.0 17.7 31.0 17.0 17.2 17.3 17.2 16.5 18.6 17.8 17.6 31.1 16.7 16.9 17.3 16.7 16.6 18.3 17.5 17.1 30.9 16.4 16.8 16.6 16.5 16.5 17.7 17.2 17.1 30.9 16.4 16.9 16.7 16.5 16.4 17.7 16.8 16.9 30.5 16.5 16.6 16.6 16.5 16.3 17.5 16.7 17.0 30.6 16.4 16.5 16.6 16.7 16.3 17.2 16.6 16.9 31.0 16.5 16.4 16.6 16.5 16.3 17.1 16.7 16.8 30.5 16.5 16.4 16.6 16.4 16.2 16.8 16.6 16.8 29.8 16.4 16.5 16.7 16.4 16.3 16.6 16.5 16.7

**T1 T2 T3 T4 T5 T6 T7 T8**

**Figure 11.**

*Testing enclosure showing points at which temperature were taken.*

**Figure 12.** *Thermoelectric air conditioner placed inside of testing enclosure.*
