Methods of Heat Transfer
Conduction
The first method through which heat is transferred from one object to another is conduction. Conduction occurs in liquids and solids through transfer of kinetic energy present in one particle to the immediate particle(s) in its surrounding. By considering the bonds existing between atoms and molecules in solids as elastic links, the vibration of one particle due to increase energy level caused by heat, the next in its line vibrate too. This process is replicated in adjacent particles thereby allowing transfer of thermal energy from a body with a higher temperature as compared to the one with low temperature. The case is different for liquids because of their ability to flow. In this case, heat transfer occurs by convection. For metals, conduction occur differently due to the presence of valence electrons, which are mobile throughout the body. From a hot region, the ‘valance’ electrons attain kinetic energy and spread rapidly over the entire metal. As the energized mobile electrons move throughout the metal, heat is distributed. In a similar manner as in solids, conduction takes place in gases, but in very slow pace. Conduction in gases depends on how frequently a gas particle with high energy level collides with cold ones. Though not effective, conduction in gases leads to transfer of heat from hot to cold parts on the container. In practical case, conduction relates to solids since other methods are far much more effective in liquids and solids.
Radiation
Heat transfer through radiation takes place through emission of electromagnetic radiation by a hot object. The actual nature of radiation emanating from a hot object depends on the temperature of such object. Whereas comparatively cool object emit near infrared radiations, hot objects radiate with high intensity to a level that can be seen by naked human eye. Extremely hot objects emit electromagnetic radiation in the range of ultraviolet. This follows that as the temperature of a material increases, there is a proportionate increase in energy emitted at various ranges of wavelengths. Notably, radiation differs from other methods of heat transfer because it does not require matter, but occurs in a vacuum. The rate of heat transfer through radiation is a factor of temperature of the source.
Convection
Convection takes place when a liquid or gas is heated leading to a notable change in its density. Upon heating, fluid expands and its density decreases. The resultant difference in density of surrounding fluids causes the fluid to flow, thereby carrying thermal energy alongside. The rate of heat transfer through convection depends on the viscosity, temperature differences involved, and thermal expansion rates the fluid involved, and shape, texture and size of a source of heat. Heat loss through convection needs a fluid with ability of changing its density when subjected to heat. In addition, gravitational field is necessary to bring about movement. Finally, the fluid should be able to move freely in the direction driven by density differences.
Poor Heat Conduction Material Case Study
The rate at which heat is transferred in material vary from one material to another depending on the ease at which atoms or molecules vibrate. Those with valence electrons, metals, are good conductors, while composites, plastics, and other carbon compounds are bad conductors because they do not have free molecules. A classic example is Tufnol, a composite material consisting of phenolic resin and an additional material, which could be cotton fabric, paper, among others. This material is desirable because of high strength and low heat conductivity making it ideal for lining in bearings. In the experiment, the material was tested for its thermal conductivity. The literature search showed that the thermal conductivity of Tufnol is 0.32W/mK whereas the value calculated through Fourier’s Law and experiment was 0.28W/mK (Tufnol Composites Limited n.p.).
Thermal Conductivity of Materials
Thermal conductivity is an attribute of material indicating its heat conduction ability. Heat conduction occurs if there is a temperature gradient existing in a stationary fluid or solid medium. Heat energy is transferred from a region with molecules with high energy to a region of less energetic molecules through collision of neighbouring molecules (Agrawal 2008, p. 136; Avallone, Baumeister, & Sadegh, 2006, p. 45). The flow of conductive heat takes place in the direction of gradually reducing temperature because a region of high temperature is related with high molecular energy (Serway 1998, p. 56). According to Fourier’s
Law, conductive heat obeys the formula below:
H=kA(T2-T1)x
Whereby;
H-the steady state heat transfer rate
k-thermal conductivity of a material
A-Cross sectional area
(T2-T1)-Difference in temperature across a thickness of ‘x’
This relationship assumes that the heat loss from the material’s sides is negligible. In this experiment, the loss from the sides was minimized using samples in the shape of a thin disk with a sufficiently large cross-sectional area as compared to the exposed area on the edges. A large rate of heat transfer was achieved by making cross sectional area very large and thickness very small. A small thickness implies that the test apparatus attains a steady state quickly. There are several ways of measuring thermal conductivity, with suitability of each method depending on the thermal properties material under investigation (Rohsenow 1998, p. 23). The commonly used techniques are Lee’s disc method and Searle’s method for bad and good heat conductors respectively.
Objective
The objective of these experiments is to investigate heat conduction through a disc and determine the heat transfer coefficient of conductivity of good and poor heat-conducting materials.
Methods
Experiment 1
Experiment 1 was executed through the formula below:
- The spacing between the output probes, disc and thickness of discs were measured
- Output probe leads were attached to numbered probes
- Once water was confirmed to be flowing, the heater was switched on to a low power
- All temperatures were recorded once T1 was confirmed to be stable
- Power was increased to 10, 20, and 40 Watts in succession as the temperature readings were recorded for each power
- Note: All readings were recorded after stability was achieved.
Figure 1 below shows the experiment setup.
Figure 1 Experimental set up for determining thermal conductivity of brass
Experiment 2
For experiment 2, Lee’s disc apparatus with a metallic disc placed on a hollow cylinder with a depth of 5cm. The hollow cylinder is a steam chamber through which water flows, thus it has inlet and outlet tubes for conveying the steam. Other than the inlet and outlet tubes, the apparatus has radial holes through which thermometers are inserted. When steam is passed through the cylinder, a steady state is attained and the readings may be collected. The entire experiment set-up relies on the fact that the heat radiated from the apparatus metallic disc is equal to the conducted across a bad conductor under investigation (Muncaster 1993, p. 2; Wright 2010, p. 103). For this experiment, the following steps were followed:
- The sample of Tufnol material was prepared in the form of a disc with the same diameter as the copper discs in the apparatus
- The thickness of the test material was measured to be 3.09mm and made flat and smooth to ensure sufficient thermal contact in its surface. The diameter and thicknesses of discs A, B, C and Specimen, S were measured.
- The heater and all discs were wiped to clean any dirt, and placed on the in the following order: - disc A, sample (S), disc B, heater, and disc C. All discs were placed in the frame in such a way that the thermometer holes point upwards before tightening the screw to ensure that all discs are held firmly as shown in Figure 2 below.
Figure 2 Experimental set up for determining thermal conductivity of Tufnol
- The terminals of the heater were connected to a 6V power supply through an ammeter and rheostat. Across the terminals, a voltmeter was connected. The fourth thermometer was placed fairly close to the apparatus to measure the ambient temperature.
- Once the experiment has been set-up as described above, the electric circuit was connected and current was allowed to flow through the heater. In order to hasten heating of the apparatus, large current was passed at first and reduced once the desired temperature is attained. The apparatus was left to go to thermal equilibrium.
- When the temperatures of all parts in the apparatus have been stable for about five minutes, the readings in the temperatures were recorded.
The results of these experiments are as recorded in the Appendix.
Discussion
The two experiments managed to estimate the heat conduction coefficients of brass and Tufnol. For brass, which is a good conductor, an experiment set-up with six temperature-measuring points was used. The calculations for tests A, B, and C showed that thermal conductivity of brass is 95.52W/mK, 119.82W/mK, and 106W/mK respectively. Notably, these values are close to the theoretical value of 109W/mK. Considering that there might have been systematical errors associated with the execution of the experiment, the values are correctly computed. It is also critical to note that the thermal conductivity value for test C, 106W/mK with 30W was so close to the theoretical value of 109W/mK (Carvill 1994, p. 17; The Engineering ToolBox n.p.).
On the other hand, Tufnol, which is the poor thermal conduction material, was investigated using Lee’s apparatus experiment setup as described above. The notable advantage of this method is its ability to estimate thermal conductivity without using huge investment. However, the fact that some heat is lost through the sides of the specimen disk compromises the accuracy of the results. Possibly, this explains the disparity between the calculated value in experimental data and theoretical value available through literature search. Just like experiment 1, the disparity may be because of errors associated with heat loss from the sides of disc specimen.
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Appendix
Experiment Results
Experiment 1: Brass
Q=2πLkTi-T0logeR0Ri
Figure 3 Thermal Gradient for tests A, B, and C
Test A
Q=2πLkTi-T0logeR0Ri
Rearranging the equation to obtain k,
k=Q2πLTi-T0logeR0Ri
k=102π×0.00328.3-13.9loge0.0550.004
k=102π×0.003(5.4961)=95.52W/mK
Test B
k=202π×0.00337.1-13.9loge0.0550.004
k=202π×0.003(8.8550)=119.82 W/mK
Test C
k=302π×0.00353.2-13.9loge0.0550.004
k=302π×0.003(15)=106.10 W/mK
Experiment 2: Tufnol
V=310 mV
I=14.5 mA
TA=29.1℃
TB=28.7℃
TC=29.4℃
TAmbient=18.8℃
TA=29.1℃-18.8℃=10.3℃
TB=28.7℃-18.8℃=9.9℃
TC=29.4℃-18.8℃=10.6℃
Still cold air gave the best results.
aA=aC=2858.762×10-6 m2
aB=1647.796×10-6 m2
aS=408.666×10-6 m2
Thickness, ts=3.09 mm
Length=3 mm
e=VIaATA+aSTA+TB2+aBTB+aCTC
Replacing the values in the equation;
e=0.310×0.01452858.762×10-6×29.1+408.666×10-629.1+28.72+1647.796×10-6×28.7+408.666×10-6×29.4
e=0.02913
Kπr2TB-TAt=e2aSTA+TB2+2aATA
Kπr228.7-29.10.00309=0.029132408.666×10-629.1+28.72+2 2858.762×10-6×29.1
129.44Kπr2=0.0145650.01181+0.16637
K=2.5953×10-3129.44πr2
Thermal Conductivity, K=0.2836W/mK
