LINEAR HEAT CONDUCTION MEC317 EXPERIMENT 1 Group #4 Matthew Stevens – First Author, Abstract, Introduction Kanchan Bhattacharyya – Discussion, Conclusions Ting Zhang – Results, Error Analysis Xie Zheng – Experimental April 29th, 2014 Abstract Fourier’s Law of Heat Conduction makes it possible to quantify not only rates and quantities of heat transferred and how temperature varies within solid objects, but also properties such as thermal conductivity of the solids that govern the manner in which heat and temperature are distributed within them. The validity of the law will be verified through Experiment, as the temperatures at various points within an insulated composite solid wrapped in a resistance heated jacket will be measured over time to quantify the temperature gradient required to determine the heat transferred through the solid. This quantity will then be used to calculate the thermal conductivity of a stainless steel sample sandwiched in between two brass samples composing the composite. In addition, the heat transfer through the solid as predicted by Fourier’s Law will be compared to the known input voltages and currents of the resistance heater. Comparison of these values, as well as the value calculated for the thermal conductivity of the steel sample with those given by engineering literature will conclude that Fourier’s Law of Heat Conduction can be used to accurately estimate rates of heat transfer and temperature distributions with relatively limited error. Introduction Since the beginning of mankind heat has been perceived as something that provides us with a sensation of warmth, but it was only in the middle of the nineteenth century and the development of the kinetic theory that we had a physical understanding of the nature of heat.[1] It was also in the nineteenth century that French mathematician Joseph Fourier presented his Théorie analytique de la chaleur (The Analytic Theory of Heat), which posed that the flow of heat between two adjacent molecules is proportional to the difference in their temperatures [2]. Since, the various heat transfer mechanisms have been developed on the premise that energy is transferred between neighboring 1 atoms and molecules, with respective definitions varying with the molecular nature we come to aspect in solids, liquids, and gases. Fourier’s Law of Heat Conduction states that the rate of heat conduction through a plane layer is proportional to the temperature difference and the heat transfer area, but inversely proportional to the thickness of the layer. [1] Heat is conducted in the direction of decreasing temperature, a phenomena easily verified by the increasing warmth we feel when we hold a cup of hot coffee in our hand for an extended period of time. The development and understanding of Fourier’s Law led to further understanding that materials store heat differently, and similarly that different materials possess varying abilities to conduct heat. The thermal conductivity of a material is defined as the rate of heat transfer through a unit thickness of the material per unit area per temperature difference, and is a measure of the ability to conduct heat. [1] In solids, heat conduction is due to two effects: the lattice vibrational waves induced by vibrational motions within molecular lattice structures and via the free flow of electrons within the solid. Fourier’s Law of Heat Conduction considers the thermal conductivity of a material, and states that in addition to being proportional to the heat transfer area and temperature difference the heat transferred through a solid is also proportional to its thermal conductivity. This explains various physical phenomena, ranging from the rapid increase in temperature metals experience when exposed to heat to the insulative nature of the wool used in clothing to keep us warm in the winter. List of Equipment The power supply provided power to the experiment and showed the voltage and currency (Model #: 11LS605, Serial #: 9800000563); A low current meter (Model #: Keithley 7700. Serial #: 1030503) read all the six channels of the thermocouples and sent the data to computer; A computer (Model #: DELL GX270) with data acquisition software to deal with the data sent from the Keithley 2 meter; The electrical caliper (manufacturer: CARRERA PRECISION) measured the inside diameter of the white cylinder and outside diameter of the metal cylinder; The experimental setup (HT11) with white cylindrical plastic insulation experienced conduction heat transfer from the top to the bottom. Experimental Theory In this experiment, we are considering the conduction in specific materials. The base of the theoretical background information is Fourier Law of conduction. In the case of a plane wall which is made of homogeneous material with certain thickness Δ x and area A, if there is a temperature difference ΔT on the two sides of the wall, then heat is transferred through the wall by conduction. In addition, if the wall temperatures are constant for a significant period of time, the phenomenon obeys Fourier’s law of conduction as described by equation 1: 𝑄 = −𝑘𝐴 𝑇2 −𝑇1 𝑥2 −𝑥1 = −𝑘𝐴 𝛥𝑇 𝛥𝑥 (Eq. 1) where k represents the thermal conductivity (W/m*k), A represents the area of the wall perpendicular to the direction of heat flow (m2), and Q represents the heat flow (W). In another case, if the walls are made of several layers with different conductivities k and there is good thermal contact between each layer, then, the Fourier’s law can be applied to each material with the assumption that the temperature of both materials is the same at every interface. In such cases, the temperature profile is linear. Fourier’s law can be simplified for the common temperatures as described in Equations 2 and 3: 𝑄 𝐴 1 𝑈 = = 𝑈(𝑇𝑎 − 𝑇𝑏 ) 𝛥𝑥1 𝑘1 + 𝛥𝑥2 𝑘2 3 + 𝛥𝑥3 𝑘3 (Eq. 2) (Eq. 3) where U is the overall heat transfer coefficient (W/m*k), and Ta and Tb represent the temperature of the hot and cold side exposed faces. The Electrical Power provided to the cylinder via a resistance heater on the top of the cylinder can be estimated via the expression 𝑃 = 𝑉𝐼 (Eq. 4) where V = voltage applied to the heating resistance (V) I = current through the heating resistance (A) If the heat losses through the insulation are negligible, the electrical power P is then equal to the heat flow Q through the cylinder. Experimental Procedure Since the experiment was already setup in the lab, we check the connections of the equipment and make sure we have all the instruments we need. First of all, we measured the inside diameter of the white cylinder and the outside diameter of the metal cylinder. After that, we turned on the power supply and Keithley meter. Then we recorded the readings for voltage and currency shown on the Keithley meter for later calculation. To keep the cooling water flow over the specimen, we opened faucet. We opened the data acquisition software to collect the readings for about 70 minutes. We stopped the software and saved the data on a disk when the steady state was reached. Once we finished, we turned off the power supply and Keithley meter, and then closed the faucet to stop the cooling water flow. 4 Results a.) Steady State Temperature Determinations as a function of Position Table 1: This table shows the 15 sets of measured temperature and their average temperature of T1 T3 T4 T5 T6 at point 1, 3, 4, 5, and 6, respectively. Table 1: Steady-State Thermocouple Temperature Readings Measurement T1( °C) T3( °C) T4( °C) T5( °C) T6( °C) 1 51.92 48.91 20.41 18.94 17.54 2 51.95 48.93 20.41 18.92 17.53 3 51.96 48.94 20.42 18.93 17.53 4 51.98 48.96 20.42 18.94 17.54 5 51.99 48.95 20.42 18.95 17.53 6 51.99 48.98 20.42 18.94 17.52 7 52.01 48.99 20.42 18.93 17.52 8 52.03 49.00 20.42 18.94 17.53 9 52.03 49.02 20.43 18.93 17.52 10 52.04 49.02 20.42 18.94 17.52 11 52.06 49.03 20.41 18.94 17.53 12 52.08 49.04 20.41 18.94 17.51 13 52.10 49.06 20.41 18.93 17.50 14 52.09 49.07 20.39 18.90 17.48 15 52.11 49.08 20.37 18.87 17.44 Average 52.02 49.00 20.41 18.93 17.52 Table 1: This table shows the 15 sets of measured temperature and their average temperature of T1 T3 T4 T5 T6 at point 1, 3, 4, 5, and 6, respectively. 5 Fig. 1. Graph plotting average steady state temperature vs. the distance along the cylinder for respective thermocouples Average Temperature (°C) Average Temperature vs. Distance Along Cylinder 60 T1 50 T2 40 30 T3 T4 T5 T6 20 10 0 0 20 40 60 80 100 120 Distance Along Cylinder (mm) Avg. Temperature vs. Position Along Cylinder Linear (Avg. Temperature vs. Position Along Cylinder) Fig. 1. This graph plots the average of the steady state temperature readings taken from each thermocouple against the position of the thermocouple within the cylinder. Table 2: Experimental Constants used in Calculations d (in) 0.983 kbrass W/(m*K) 121.00 I(amps) 0.89 V(volts) 8.50 Table 2: This table shows some constants such as thermal conductivity of brass, kbrass, diameter d, cross sectional area A, current I, voltage V, power P Table 3: Calculated Diameter, Cross-Sectional Heat Transfer Area, and Power Diameter, d (m) 0.025 Area, A (m2) 0.000490874 Power, P (W) 7.565 Table 3. Calculated values for the Diameter d of the cylindrical brass specimen, the heat transfer area A, and the supplied power P 6 b.) Determination of T(x) in Section A via Linear Regression For the expression T(x) for section A; 𝑇𝑥𝐴 = 𝑇1 + 𝑇𝑥𝐴 = 𝑇1 + 𝑇3 − 𝑇1 𝑥 ( 𝑒𝑞𝑢𝑎𝑡𝑖𝑜𝑛 5) 𝑥3 𝑇3 − 𝑇1 49 − 52.02 𝑥 = 52.02 + 𝑥 𝑥3 0.03 Using T(x) within Section A, the heat flow QA through section A can be calculated: 𝑄𝐴 = −𝑘𝐴 𝑄𝐴 = −𝑘𝐴 𝑑𝑇𝑥𝐴 𝑇3 − 𝑇1 = −𝑘𝐴 ∗ = ( 𝑒𝑞𝑢𝑎𝑡𝑖𝑜𝑛 6) 𝑑𝑥 𝑥3 𝑑𝑇𝑥𝐴 𝑇3 − 𝑇1 49 − 52.02 = −𝑘𝐴 ∗ = −121 ∗ 0.000490874 ∗ = 𝟓. 𝟗𝟕𝟗𝟏(𝑊) 𝑑𝑥 𝑥3 0.03 c.) Calculation of the Heat Flow in Section C via Linear Regression For the expression T(x) for section C, 𝑇𝑥𝐴 = 𝑇4 + 𝑇𝑥𝐶 = 𝑇4 + 𝑇6 − 𝑇4 𝑥 ( 𝑒𝑞𝑢𝑎𝑡𝑖𝑜𝑛 7) 𝑥6 − 𝑥4 𝑇6 − 𝑇4 17.52 − 20.41 𝑥 = 20.41 + 𝑥 𝑥6 − 𝑥4 0.03 Using T(x) within Section C, the heat flow Qc through section C can be calculated: 𝑄𝐶 = −𝑘𝐴 𝑄𝐶 = −𝑘𝐴 ∗ 𝑑𝑇𝑥𝐶 𝑇8 − 𝑇7 = −𝑘𝐴 ∗ = ( 𝑒𝑞𝑢𝑎𝑡𝑖𝑜𝑛 8) 𝑑𝑥 𝑥8 − 𝑥7 𝑇8 − 𝑇7 17.52 − 20.41 = −121 ∗ 0.000490874 ∗ = 𝟓. 𝟕𝟐𝟏𝟖(𝑤) 𝑥8 − 𝑥7 0.03 d.) Calculated electrical power given to the cylinder P using Equation 4 𝑃 = 𝑉𝐼 = 8.50 ∗ 0.89 = 7.565𝑤 7 e.) Comparison of the Electrical Power P to the Heat Flow through Sections A and C Part 1, calculated experimental value of ksteel First of all, we need to use the linear regression method to calculate the temperatures of both surfaces of Ta and Tb from equation 5 and equation 6 respectively. 𝑇𝑠𝑡𝑒𝑒𝑙,1 = 𝑇1 + 𝑇𝑠𝑡𝑒𝑒𝑙,2 = 𝑇4 + 𝑇3 − 𝑇1 49 − 52.02 𝑥 = 52.02 + ∗ 0.0375 = 𝟒𝟖. 𝟐𝟒𝟓 𝑥3 0.03 𝑇6 − 𝑇4 17.52 − 20.41 𝑥 = 20.41 + ∗ (−0.0075) = 𝟐𝟏. 𝟏𝟑𝟑 𝑥6 − 𝑥4 0.03 Secondly, we need to calculate the heat flow of stainless steel, Qsteel by using equation 9. Where QA and Qc can be calculated from equation 6 and Equation 8. 𝑄𝑠𝑡𝑒𝑒𝑙 = Equation 9 𝑄𝐴 +𝑄𝐶 2 = 5.9791+5.7218 2 = 5.8505 Using Fourier’s Law (Equation 1) to calculate ksteel. 𝑄𝑠𝑡𝑒𝑒𝑙 = −𝑘𝐴 ∗ 𝑇𝑠𝑡𝑒𝑒𝑙,2 − 𝑇𝑠𝑡𝑒𝑒𝑙,1 ∆𝑥 5.8505 = −𝑘𝑠𝑡𝑒𝑒𝑙 ∗ 0.000490874 ∗ Giving 21.133 − 48.245 0.03 𝒌𝒔𝒕𝒆𝒆𝒍 = 𝟏𝟑. 𝟐 Part 2, calculated theoretical value of ksteef 𝑇𝑠𝑡𝑒𝑒𝑙,𝑎𝑣𝑒𝑟𝑎𝑔𝑒 = 𝑇𝑠𝑡𝑒𝑒𝑙,2 + 𝑇𝑠𝑡𝑒𝑒𝑙,1 21.133 + 48.245 = = 34.69(𝐶) = 307.84(𝐾) 2 2 8 Using the table in the lab manual to interpolate the thermal conductivity; Table 4. Thermal Conductivities for Temperatures of 300 and 400 Kelvin for Stainless Steel Temperature (K) k T = 300K 13.400 T = 400K 15.200 Tsteel,avg = 307.84K 13.540 Table 4. The thermal conductivities of stainless steel at temperatures of 300 and 400 Kelvin. These values were used to interpolate the value of the thermal conductivity of the steel at our average temperature of 307.8 Kelvin. DISCUSSION In this experiment, the thermal conductivity k of an unknown steel is calculated by using Fourier’s Law of Conduction which holds under steady-state conditions (Equation 1). The setup involved bounding a sample cylinder of the stainless steel with two cylinders of brass of known k. At the steadystate condition, the heat transfer rate is assumed to be constant throughout and T(x) can be obtained as a piece-wise linear function with equal temperatures at interfaces between materials. To calculate k of the unknown, Equation 1 is used where ∆𝑥 and A are measured and ∆𝑇 is obtained by extrapolating the temperatures at the thermocouples in each brass cylinder to the interface. But before looking at the k obtained, let’s look at the heat transfer rates obtained at the “steady-state” time. From Table 3, Section B, and C of the results, in section A, 𝑄 = 5.9791 W, in section B, 𝑄 = 5.7218 W, and in the resistive element P = 7.565 W. The 𝑄 values are slightly different by ~0.2 and are different from the power generated in the resistive element by ~1.715. One reason that can be offered for the difference is the fact that thermal contact resistance was neglected which assumes that there is no drop in temperature over the gap. However, considering that all materials in this experiment are metals with high conductance, thermal contact resistance is not 9 enough to explain the lower 𝑄 values. It is more likely that the difference simply comes from obviously significant heat loss through the insulation. This argument accounts for how the 𝑄 in section A, which is closest to the resistive element suffers a nearly 1.715 W loss in 𝑄 from the calculated P and how 𝑄 in section C is lower still by 0.2. This is because as the bulk of the heat is flowing down from cylinder to cylinder, a large portion of heat is dissipating away, being conducted through the insulation. The fact that the experimental 𝑄 values are significantly less than the original P generated will play a role in how well k of the unknown matches accepted values. From Results Section E Part 1, the experimental k = 13.2 W/m ℃ while the accepted k is 13.540 W/m ℃. The thermal conductivity k is calculated using a rearrangement of the Fourier Equation of Equation 1. In this form, it can be noted that k is proportional to 𝑄 and thus since the 𝑄 values obtained are less than expected, it follows that k is also less than optimal. However, experimental limitations exist in all setups, the key being whether or not the discrepancy is acceptable or not. From Results Section F, the error in k is calculated to be 0.496. It’s important to note that the accepted value falls within the bounds of uncertainty for the experimental k. Thus, it can be concluded that in spite of heat loss through the insulation, statistically reasonable results are obtained from this experiment and our measured k for the unknown stainless steel is acceptable. Furthermore, the percent error is |13.540 – 13.2]/|13.540| x 100 = 2.51% which is rather remarkable for such an economical “table-top” experiment in studying conduction heat transfer. 10 Error Analysis Fig. 2. Uncertainty Analysis Tree Fig. 2. The uncertainty tree used to calculate the uncertainty in the thermal conductivity. The thermal conductivity is dependent on the heat transfer Q, the cross sectional heat transfer area A, the distance x within the cylinder, and the temperature T. Noting the dependencies of the heat transfer Q and crosssectional area A on the resistance voltage and current and spherical diameter of the sample, we therefore assume the uncertainty exists in the temperature T, distance x, diameter d, voltage V, and Current I. Given the respective uncertainties, ∆𝑙 = 0.2𝑚𝑚 ∆𝑥 = 0.2𝑚𝑚 ∆𝑉 = 0.2% ∆𝐼 = 𝐼 ± 0.2% ∆𝑇 = 1 (℃) 11 We can then calculate the uncertainty in the heat flow Q 𝜕𝑄 𝜕𝑄 ∆𝑄 = √( ∆𝑉)2 + ( ∆𝐼)2 = 0.021 𝜕𝑉 𝜕𝐼 Similarly, for the Cross-Sectional Heat Transfer Area ∆𝐴 = 𝜋 ∗ ∆𝑑4 ≅ 0(𝑐𝑎𝑛 𝑏𝑒 𝑛𝑒𝑔𝑙𝑒𝑐𝑡𝑒𝑑) 4 Subsequently, we find the uncertainty in the thermal conductivity k; 𝜕𝑘 𝜕𝑘 𝜕𝑘 𝜕𝑘 ∆𝑘 = √( ∆𝑄)2 + ( ∆𝐴)2 + ( ∆𝑥)2 + ( ∆𝑇)2 𝜕𝑄 𝜕𝐴 𝜕𝑥 𝜕𝑇 = √(− 0.03 5.8505 5.8505 ∗ 0.03 ∗ 0.021)2 + 0 + (− ∗ 0.0002)2 + ( ∗ 1)2 (−27.113)2 ∗ 0.000490874 0.000490874 ∗ (−27.113) 0.000490874 ∗ (−27.113) = 0.496 Giving the range of values for k, considering the calculated uncertainty and a confidence interval of 95% 𝒌𝒔𝒕𝒆𝒆𝒍 = 𝟏𝟑. 𝟐 ± 𝟎. 𝟒𝟗𝟔 (𝟗𝟓%) CONCLUSION To reiterate, this experiment utilized a simple setup where power is generated in a resistive element, dissipated as heat, and passed through a set of conductive sample cylinders in series. At the steady-state condition, 𝑄 is assumed constant across this series and Equation 1 is used to plot T(x) as a piece-wise linear function. From T(x), surface temperatures at the two surfaces of the unknown material can be found, and with other measurable quantities, k can be calculated. In this experiment, 𝑄 differed from P by almost 1.715 W, mainly due to heat loss through insulation and to a lesser extent possible inaccuracies in neglecting the effect of thermal contact resistance in metals. However, the k obtained in 12 spite of this issue of 13.2 W/m ℃ was in agreement with the accepted value of 13.540 W/m ℃ as the accepted fell within the bounds of uncertainty for this experiment. In light of the results achievable by this fairly economical “table-top” setup to study conductive heat transfer, there is little that can be improved other than the obvious. Results would improve or at least would have less room for error if better insulation is used such that more 𝑄 can be directed through the series of materials as intended and result in a higher experimental k given that k is proportional to 𝑄 according to Equation 1. Also, an issue that could quickly become significant if this setup was to be adapted to study any material outside of metals, is the issue of thermal conduction resistance and temperature drop at air gaps in the interface. The smoothness of the surface finish of the two surfaces in contact along with the external applied pressure are variables that can be controlled to reduce the effects of thermal contact resistance which may become significant for classes of materials with lower k’s than metals, where one can obtain fairly reasonable results even after neglecting this issue. References [1] Yunus A. Cengel and Afshin J. Ghajar, Heat and Mass Transfer Fundamentls and Applications 4th Ed., McGraw-Hill 2011, p. 3-20 [2] "Theorie Analytique De La Chaleur | Historical Mathematical Texts | Cambridge University Press." Theorie Analytique De La Chaleur | Historical Mathematical Texts | Cambridge University Press. N.p., n.d. Web. 28 Apr. 2014. 13