Question

can you write a conclusion for the report?

Thermal Conductivity Lab

Introduction

The objective of this lab is to explore the thermal conductivity of various materials using Fourier’s Law and a steady state system. Certain materials are more effective heat conductors than others. The effectiveness of a given material at heat conduction is influenced by many factors, including density, molecular structure, and chemical composition. Fourier’s Law takes these factors into account to calculate Q, or the joules per second rate of energy transfer at a given location.

The simplest form of this equation is:

Q=-kATx

Where A represents the area through which the heat is conducted, k the conductivity of the medium in units of joules per seconds*meters*Kelvin, T stands for change in temperature, and x stands for the thickness of the material through which the heat must travel. This experiment generates a steady state system, so that the energy does not accumulate in the medium, but instead is transferred through.

When multiple mediums are in play, as is tested in the latter stages of the experiment, Fourier’s Law takes on a more complicated form in order to account for the varied levels of resistance. Δx is divided by the sum of the thicknesses of the materials divided by the area and the given resistance for the material.   

Experimental Apparatus and Procedure

To conduct this experiment, it is necessary to collect a variety of materials. We used stainless steel, carbon steel, aluminum, redwood, pine, and PVC for a wide range of mediums. Each substance must be cut into 3 rectangular prisms with a 3 inch2 face, and thicknesses of 0.125, 0.25, and 0.375 inches. Additionally, this experiment requires a heater, a strip of cloth to act as an insulator, and a thermometer.

Part 1

This procedure was repeated with two other materials of the same thickness of 0.125 inches which were Carbon Steel and Redwood. For the first part of this lab we started with the heater and it was set to six. Then the first rectangular piece of material, Stainless Steel which was 0.125 inches thick, and was set on top of the metal heating platform. Then we used the insulation ribbon to be wrapped around the material and platform to hold it together. The temperature was the next step. It was taken until steady state was reached. Once this was reached the temperature of the top and bottom surfaces were measures.

Part 2

Now for part 2 we used one material, which was Carbon Steel at a thickness of .125 inches, and then measured the temperatures of the top and bottom of the material. But before that, it had to wrap the the ribbon around the Carbon steel to prevent heat loss. The temperature was taken of the top and bottom of the surface once it was at a steady state. It was done an additional 2 more times for the same material.

Part 3

On the last part of this experiment, a stack was formed with two different materials but the same size. Stack one was redwood on top and carbon steel on the bottom. The two materials used were all .125 inches thick. Next, the stack was wrapped around with the ribbon to keep in the heat. Once this was completed and the temperature has been stabilized, the temperatures were recorded in a table. In addition to finding the materials temperature of one material, the temperature was read between the stacks to show the change of heat.

Results and Discussion

In the first part of the experiment, different materials were used at the same thickness to compare the thermal conductivity as a function the material heat was conducting through.

Table 1. Temperature readings through different median in Part 1 of the experiment.

Test	Material Type	Thickness (in)	Temp. of Top Surface (°C)	Temp. of Source Plate (°C)	Change in Temperature (°C)
1	Stainless Steel	0.125	48.8	54.9	6.1
2	Carbon Steel	0.125	48.7	50.2	1.5
3	Redwood	0.125	42.6	54.0	11.4

Before we discuss the meaning of these results, it is important to take note to the possible mistake made in the measurement of the temperature of the surface plate and the source plate when we calculated the change in temperature of the redwood. While we were taking these measurement the heat source was oscillating in a way that changed the temperature in our source plate dramatically.

Table 2. The rate of energy transfer through different material in Part 1 of the experiment, including the thermal conductivities to use in these calculations.

Material	Thermal Conductivity Coefficient	Heat Flux (J/s)
Stainless Steel	16	59.496
Carbon Steel	54	49.377
Redwood	0.090	0.625

The equation Q=-kATx was use as explained in the introduction in order to calculate the heat flux. The results gathered support stainless steel having the highest heat flux at 54.496 J/s. This may be the result of having a larger thermal conductivity than redwood and a higher change in temperature than carbon steel.

As described by the equation, the four factors that affect the heat flux are the thermal conductivity of the medium, the cross-sectional area heat is conducted through, the change in temperature though the object, and the thickness of an object. In general, some material conduct heat more quickly because they maximize these four variables. Material with a higher thermal conductivity, cross-sectional area, and change in temperature but a low thickness will conduct heat more quickly than object that don’t.

Table 3. Temperature readings carbon steel as a function of thickness from Part 2 of the experiment.

Test	Thickness (in)	Temp. of Top Surface (°C)	Temp. of Source Plate (°C)	Change in Temperature (°C)
1	0.125	60.2	65.5	5.3
2	0.25	41.2	48.0	6.8
3	0.375	40.8	48.1	7.3

Table 4. The rate of energy transfer through carbon steel at varying thicknesses from Part 2 of the experiment.

Thickness of Carbon Steel (in)	Heat Flux (J/s)
0.125
0.25
0.375

Figure 1.

The results of this experiment demonstrate that materials have variable conductivity, and that increasing the thickness of a given material will increase the resistance. This supports the structure of Fourier’s Law. This is important to keep in mind when proposing future experiments, or when designing a reactor in order to control the heat lost and gained during a reaction.

Answer 1

First of all, there is slight difference in the formula given, because the given Fourier formula is wrong. It should be

$Q=-\frac{kA\Delta T}{\Delta x}$

Now, calculating the heat flux from the given data of Part 2, is shown in the table using the Fourier's law given above

Sl No	K_CS	A	∆T	X	Q
	J/s-m-K	m2	K	m	J/s
1	54	1.94E-03	5.3	3.18E-03	174.42
2	54	1.94E-03	6.8	6.35E-03	111.89
3	54	1.94E-03	7.3	9.53E-03	80.08

By, plotting the data of thickness of the slab to the heat flux, we get the profile as shown in figure.

200 180 160 140 120 100 80 60 40 20 0.00E+00 2.00E-03 4.00-036.00E-03 8.00E-03 100E-02 120E-02 Thickness (m)

From the figure, it can be concluded that the heat flux reduces with increasing slab thickness where other parameters such as surface area and thermal conductivity co-efficient are same. This change was observed because the more slab thickness provide higher resistance to the heat transfer, hence the temperature difference is larger for larger slab thickness but the temperature gradient that is $\frac{\Delta T}{\Delta x}$ lesser for increasing thickness as shown in the table

∆T	X	Q	∆T/∆x
K	m	J/s	K/m
5.3	3.18E-03	1.74E+02	1.67E+03
6.8	6.35E-03	1.12E+02	1.07E+03
7.3	9.53E-03	8.01E+01	7.66E+02