Heat  Transfer  Lab    

Report  1  –  Linear  Conduction                                    Name:  Hugo  Felippe  da  Silva  Lui  Date:  10/03/2014                          

Introduction         The experiment aims to investigate the effect of length, material and diameter of piece on the temperature distribution. In addition, the influence of contact resistance on the temperature distribution will be analyzed. Conduction is the transfer of energy as heat, from the more energetic particles of a substance to the adjacent less energetic particles as a result of interaction between them. The conduction can occur in solids, liquids and gases. The rate of transfer of heat through a medium is related to the material, the geometry and thickness of this medium in addition to the temperature difference across the medium. The equation that relates all the variables above with the rate of heat transfer was described in 1822 by Fourier. That is, 𝑄 =  −𝐾𝐴

𝑑𝑇𝑑𝑥

(1)

Where:

• K - Thermal Conductivity constant, which is a measure of the ability of a material to conduct heat.

• A - Area, which is always normal to the direction of heat transfer. •

!"!"

- Temperature gradient, which is the slope of the temperature curve.

The negative sign in the equation indicates that heat is transferred in the direction of decreasing temperature.

To calculate the thermal conductivity, the Equation (1) is modified to,

𝐾 =  

−𝑄∆𝑋𝐴∆𝑇

(2)

The equation that gives the distribution of temperature through the medium is

called heat conduction equation. That is,

 𝜕𝜕𝑥 𝐾

𝜕𝑇𝜕𝑥 +  

𝜕𝜕𝑦 𝐾

𝜕𝑇𝜕𝑦 +  

𝜕𝜕𝑧 𝐾

𝜕𝑇𝜕𝑧 +  𝑔 = 𝜌𝐶!

𝜕𝑇𝜕𝑡

(3)

The first three terms of the left-hand side of this equation correspond to the net rate of heat conduction into the control volume per unit volume. The fourth term of the left-hand is the rate of energy generation per unit volume inside the control volume. The right-hand corresponds to the rate of increase in the internal energy inside the control volume per unit volume.

For the physical model of experiment and computer simulation, the assumptions

were:

- Steady state - One dimensional - Thermal conductivity constant - No heat generation

The consideration of steady state is because the experiment was done when the

temperature no longer varies with time. The experiment can be considered one dimensional, because the walls of the piece are isolated, so there is only heat transfer in one direction. For the temperature range of the experiment is considered that there is no variation of thermal conductivity with temperature. The experiment can be admitted no heat generation, because there is not any heat source inside the control volume.

After the simplifications made and integrating Equation (2) twice, one obtains:

𝑇 𝑥 =  𝐶!𝑥 + 𝐶! (4)

Where the C1 and C2 values depends on the boundary conditions of the problem. Because of the assumptions, especially without heat generation, the values of C1 and C2, will never be functions of position. Therefore the distribution of temperature is a linear function. As a result, we expect a linear temperature distribution according to equation (3) and its assumptions.

Since all procedures were performed with the rate of heat transfer constant, then the hypothesis is:

a. where only the length of the piece varies, it is expected that the difference

in temperature between the top and bottom section varies inversely with the length of the piece. However, the temperature gradient will remain constant.

b. where only the material of the piece varies, it is expected that the temperature gradient increases for a smaller thermal conductivity and reduces the temperature gradient for a greater thermal conductivity.

c. where only the diameter of piece is decreased, it is expected that the temperature gradient increases, so the slope of the line is greater.

d. when contact resistance is present, it is expected to see a jump in temperature in the gaps.

For validation of theories and their considerations, we used both experimental and

computational tools.            

Method    

                                                                                                                 Figure 1. Cross-sectional view of experiment set up    -­‐  On the top of the piece, the rate of heat transfer is 𝑄 = 8.46  𝑊 -­‐  On the bottom of the piece, T = 300 K - The piece is isolated on the side   Equipment - HT11 - Linear Heat Conduction Apparatus - Brass intermediate section: Diameter: 25 mm, Length: 30 mm - Brass intermediate section: Diameter: 13 mm, Length: 30 mm - Aluminum intermediate section: Diameter: 25 mm, Length: 30 mm - HT10X – Heat Transfer Service Unit

Procedure First connect heater to the HT10XC and set the minimum voltage. Then, connect cold water supply to the inlet of the pressure regulating valve. After that, put a thermal paste on the top and bottom section of the intermediate 25 mm brass section and clamp the piece between the heated and cooled sections of the HT11. Then, connect the eight thermocouples on the HT11. Finally, turn on the HT10XC and set selector switch to manual. Next, turn on the cooling water and adjust flow rate to 1.5 l/min and set heater voltage to 9 V. It is necessary that the system be stabilized before starting to record the temperatures, voltage and current. Repeat the previous steps for the 13 mm brass and the 25 mm aluminum intermediate sections. Results  and  Discussions

Figure 2. Temperature in function of position for cases of different materials and diameters.

In figure 2, the temperature profile has a linear trend, according to Equation (4). When the material changed to aluminum, which has a higher thermal conductivity than bronze, noticed a smaller slope of the line. The reason is   the higher the thermal

conductivity, the thermal energy conducted from a high to a low temperature region will be greater, therefore a smaller temperature gradient. By reducing the area of the piece, it is observed that the slope of the line increases, in the other words, it increased the difference of temperature between the top and bottom section. This occurs because as the rate of heat transfer is kept constant, so that the Fourier law can be obeyed, it is necessary that the temperature gradient varies inversely with the area. Table 1. Thermal conductivity of each material, experimental and theoretical values.

Material

Thermal conductivity [W/mk]

(Experimental)

Thermal conductivity [W/mk]

(Theoretical) Brass 104.6 111

Aluminum 172.34 236

Comparing the experimental and theoretical values of the thermal conductivity of table 1, for brass there is a little difference between the values, however, for aluminum, one notices a great difference between the values. One reason is that the temperature measurements were not good, because possibly the system was not stabilized when the temperature was recorded.

Figure 3. Temperature distribution for different intermediate sections.

Table 2. Maximum temperature of each intermediate section Intermediate

section (mm)

20

25

30

35

40

Maximum temperature

(℃)

46.79

47.50

48.21

48.92

49.64

The temperature distribution for each intermediate section is linear. According to Fourier’s Law, the difference in temperature between the bottom and the top section should increase. As the thermal conductivity of the material is constant, therefore, the gradient of temperature should be constant, thus the difference in temperature vary linearly with the length of the piece. In figure 3, this is observed.

Figure 4. Temperature distribution for each material

Table 3. Thermal conductivity of stainless steel and copper

Material Stainless Steel Copper Thermal Conductivity

[W/mk] 14.9 401

For figure 4, the temperature profile is linear for all the materials. In figure 4 it is

observed that the material, which experiences the greatest temperature gradient, is stainless steel and the less temperature gradient is copper. As the rate of heat transfer is kept constant, so that the Fourier law can be obeyed, it is necessary that the temperature gradient vary inversely with the thermal conductivity. In figure 4, this is noted.

Figure 5. Temperature distribution of brass with and without contact resistance In figure 5, the temperature profile without the contact resistance is linear and the temperature distribution for the situation when there is contact resistance has a linear trend.   On the contact surfaces, from Figure 5, a sudden increase in temperature is observed. The reason is that there is roughness on the surfaces of contact and these imperfections are filled with air, which has a low thermal conductivity. Thus the temperature gradient should increase with the decrease in thermal conductivity, so there is a large temperature increase in the contact surface positions.

Conclusion

The experiment aimed to determine the effects of different variables such as area, thermal conductivity and length of the piece on Fourier’s law. We determine the values of thermal conductivity for the aluminum and brass intermediate sections. It can be concluded that  the temperature difference varies directly with the length of the piece and it varies inversely with the thermal conductivity and the area. The verification of the key points was performed by experimental and computational methods, where the material, the length and the area of the piece were changed to investigate the influences of these variables on Fourier’s Law. The experiment and the computational simulation showed good validation for the confirmation of Fourier’s Law.    References   [1] Kreith, F., Manglik, R. M and BOHN, M S. Principles of heat transfer, seveth edition, 2011. [2] F. P. Incropera and D. P. DeWitt. Introduction to Heat Transfer. 4th ed. New York: John Wiley & Sons, 2002.

Appendix  1    Thermal conductivity of brass:

𝐾 =  −𝑄∆𝑋𝐴∆𝑇

For: 𝑄 = 8.46  𝑊 A = 4.91e-4 m2

∆𝑇 = −17.3  𝐾 ∆𝑥 = 0.105  𝑚

K = 104.6 W/mK Thermal conductivity of aluminum:

A sample of aluminum starts at x = 37.5 and ends at x = 67.5, by the graph is obtained, T (37.5) ≅ 37 and T(67.5) ≅ 40.

𝐾 =  −𝑄∆𝑋𝐴∆𝑇

For: 𝑄 = 8.46  𝑊 A = 4.91e-4 m2

∆𝑇 ≅ −3  𝐾 ∆𝑥 = 30𝑒 − 3  𝑚

K = 172. 34 W/m