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EXPERIMENT C2: HEAT CONDUCTION STUDY BENCH Abstract The aim of the experiment is to study the factor which could affect linear conduction heat transfer. The results demonstrate that linear heat transfer obeys Fourier's law. Different materials (conductivity) and cross-sectional area have different flux rates. Compared with stainless steel, Brass has better heat conductivity and paper's is tiny. A large temperature drop occurs on between contact surfaces of test section and cooler section Introduction 1. Linear conduction heat transfer Conduction is a mode of heat transfer, which is the heat transfer by microscopic collisions of particles within a body. According to Fourier’s law, the heat flux is proportional to the temperature gradient and opposite to it in sign. For one-dimensional heat flow, dq/dA=-k(dT/dx) . which states that: Q kA dT dx (Eq.1) where, Q=heat flux(W) K=thermal conductivity(W/(m•K)), A=cross sectional area (m2) dT/dx=temperature gradient(K/m), Only in steady state, the temperature gradient could be replaced by temperature difference over heat flux path = 1 −2 (Eq.2) For composite systems with multiply layers, it is often convenient to work with an overall heat transfer coefficient U. Accordingly, 1 Q UAΔT (Eq.3) Where, ΔT=overall temperature difference(K) And also, 1 = 1 1 2 2 = + + 1 2 (Eq.4) 2 Where, x1, x2, x3=thickness of materials (m) k1, k2, k3=thermal conductivity for materials (W/(m•K)) 2. Heat conduction through different cross-sectional area For the continuity heat flow rate and the same conductor, Q1=Q2 and k1=k2 and from Eq.1. Following equation can be derived: ΔT2 A1 x 2 A2 ΔT1 x1 (Eq.5) Objective To demonstrate the concept of Fourier’s law that relates the rate of onedimensional heat flow to heat resistance, area and temperature gradient. Procedure Before experiment start, the water supply was turned on. Next, the computer and the power supply were switched on. Firstly, control knob was turned to 5 W, and A brass conductor installed sensors was inserted between the heater and the cooler sections. The heat power was set to 5W and the auto recorder was start. For a moment, when the reading per 5 min we got was closed, the experiment was paused and the date was saved. The brass section was taken apart and another, stainless steel section, was inserted in test area. The date saving was repeated as 2 above. Then the heat was switched off to make sure that temperature of linear conduction module below than 100 Celsius. After it was cool down, the stainless was replaced with brass section whose cross section is smaller. For a sufficient time, the temperature was recorded as above. Next, the brass section was removed, the thicker one (25mm diameter brass section) was put in linear module and a little piece of paper was insert between cool section and test section. The date saving was repeated as above. After that, paper was taken put and conducting compound was applied on the side of brass section adjant to cooler section. The date saving was repeated again as above. After that, the compound applied on interface was wiped off. In next three rounds, control knob was turned to 10W,15W,20W respectively and record the temperature as above. Last, the heater was switched off but the water supply was kept continuing until the water cooled down. Results 1. Fourier’s law study for linear conduction of heat along a homogeneous bar All data shown below (the unit of temperature is Kelvin) Power T1 T2 T3 T4 T5 T6 T7 T8 T9 5W 307.25 307.15 306.85 302.05 304.85 304.75 301.65 302.95 303.95 10W 323.95 323.25 323.35 318.15 317.55 317.75 303.15 302.95 302.45 15W 335.35 333.35 334.45 326.75 326.25 326.65 303.05 302.75 302.35 20W 329.95 328.35 327.25 320.65 319.95 319.75 303.95 303.45 302.65 temperature/k Table.1 Data of Experiment 1 345 y = -477,83x + 345,11 R² = 0,8381 340 335 330 y = -385,67x + 336,61 R² = 0,9014 325 y = -312,83x + 330,37 R² = 0,858 320 315 310 3 305 y = -55,833x + 307,4 R² = 0,5052 300 295 0 0,01 5W 0,02 0,03 0,04 10W 0,05 0,06 15W 0,07 0,08 20W 0,09 0,1 Figure.1 T-x graph of Experiment 1 dT can obtained from the graph slope. Cross section A is known which is π( 25×10−3 2 ) 2 ≈ 4.91 × 10−4 2 and from equation 1 we can get k=− . when Q=5W, then dT/dx=-55.833，k1=− 5 =182.44 25×10−3 2 ) ×(−55.833) 2 π( when Q=10W, then dT/dx=-312.83，k2=− 10 =69.34 25×10−3 2 π( ) ×(−312.83) 2 when Q=15W, then dT/dx= -477.83，k3=− when Q=20W, then dT/dx= -385.67，k4=− 15 =68.09 25×10−3 2 π( ) ×( −477.83) 2 20 =105.62 25×10−3 2 π( ) ×(−385.67) 2 so the average thermal conductivity kavg=(k1+k2+k3+k4)/4=106.37 · 2.Conduction of heat and overall heat transfer along a composite bar All data shown below (the unit of temperature is Kelvin) Power, Q T1 T2 T3 T7 T8 T9 5W 314.55 315.05 316.85 301.75 301.95 302.05 10W 327.15 327.55 329.75 302.35 302.35 302.35 15W 339.45 342.65 345.25 302.65 302.65 302.15 20W 335.45 334.05 337.15 302.95 303.25 302.35 Table.2 Data of Experiment 2 360 350 y = -611,03x + 353,02 R² = 0,8974 340 y = -505,52x + 344,48 R² = 0,9211 330 y = -395,86x + 335,04 R² = 0,9067 320 4 Figure.2 T-x graph of Experiment 2 From Eq.3: Q UAΔT From Eq.4: 1 = + + Where A=4.91 × 10−4 2 ; kbrass=106.37 · ; L=0.03m We can get thermal conductivity of the stainless steel ksteel= 1 − 2 = ∆ − 2 When Q=5W, ∆T=T1-T9=12.50K， U=814.66 ， k1=45.22 · · When Q=10W, ∆T=T1-T9=24.80K， U=821.23 ， k2=45.90 · · When Q=15W, ∆T=T1-T9=37.30K， U=819.03 ， k3=45.67 · · When Q=20W, ∆T=T1-T9=33.10K， U=1230.61 ， k4=120.71 · · So kavg=(k1+k2+k3+k4)/4=64.37 · From internet, we can get ideal thermal conductivity of brass and common Steel Kthoreticalbrass=109· , Kthoreticalsteel=50.2· From eq.4,we can get 5 1 Uthoretical= =871.03· thoretical+ thoretical+ thoretical brass brass 1 Uexperimental= =970.76· experimental+ experimental+ experimental brass brass Finally,Error error= U theoretical U exp erimental 871.03−970.76 871.03 750.03 13.9% =| |=11.4% theoretical 871.03 871.03 U 3. The effect of a change in cross-sectional area on the temperature profile along a thermal conductor All data shown below (the unit of temperature is Kelvin) Power, Q T1 T2 T3 T7 T8 T9 5 316.05 316.55 318.15 301.85 301.85 301.85 10 325.65 327.25 328.35 302.65 302.65 302.85 15 332.05 333.55 335.85 302.75 302.85 302.35 20 341.95 341.75 345.85 302.55 302.85 302.85 Table.3 Data of Experiment 3 6 360 y = -620,17x + 353,98 R² = 0,9057 350 340 y = -477,76x + 342,12 R² = 0,904 330 320 y = -373,1x + 333,56 R² = 0,9028 y = -230,17x + 320,89 R² = 0,8961 310 300 290 0 0,01 0,02 5W Линейная (5W) 0,03 10W 0,04 0,05 Линейная (10W) 0,06 15W 0,07 Линейная (15W) 0,08 20W 0,09 0,1 Distance/m Линейная (20W) Figure.3 T-x graph of Experiment 3 Cross section A is known which is π( 13×10−3 2 ) 2 when Q=5W, then dT/dx=-230.17，k1=162.11 ≈ 1.34 × 10−4 2 · when Q=10W, then dT/dx=-373.1，k2=200.02 · when Q=15W, then dT/dx= -477.76，k3=234.30 · when Q=20W, then dT/dx= -620.17，k4=240.67 · so the average thermal conductivity kavg=(k1+k2+k3+k4)/4=209.27 · 4. The influence of thermal insulation upon the conduction All data shown below (the unit of temperature is Kelvin) Power T1 T2 T3 T4 T5 T6 T7 T8 T9 5W 317.85 317.35 317.45 305.95 311.35 311.45 302.15 302.05 302.05 10W 324.55 323.55 324.15 318.65 318.15 318.35 302.65 302.35 302.25 15W 340.55 337.25 336.15 325.85 323.75 324.15 302.85 302.75 302.35 7 20W 349.85 345.75 345.25 329.65 329.15 329.05 303.05 302.85 302.75 temperature/k Table.4 Data of Experiment 4 360 350 340 y = -670,17x + 359,88 R² = 0,9092 y = -541x + 348,79 R² = 0,9136 330 320 y = -326,83x + 331,3 R² = 0,846 y = -223,67x + 320,92 R² = 0,7922 310 300 290 0 0,01 5W 0,02 0,03 10W Линейная (5W) 0,04 0,05 0,06 15W Линейная (10W) 0,07 Линейная (15W) 0,08 20W 0,09 0,1 Thickness/m Линейная (20W) Figure.4 T-x graph of Experiment 4 From Eq.3: Q UAΔT 1 From Eq.4: = + + Where A=4.91 × 10−4 2 ; kbrass=106.37 + · ; L=0.03m ; =64.37·; x paper 1mm 0.001m We can get thermal conductivity of the paper Kpaper= ∆ 2 − − When Q=5W, ∆T=T1-T9=15.8K， k1=1.92 · When Q=5W, ∆T=T1-T9=22.3K， k1=15.4 · When Q=15W, ∆T=T1-T9=38.2K， k1=4.54 · When Q=5W, ∆T=T1-T9=47.1K， k1=7.93 · 8 So kavg=(k1+k2+k3+k4)/4=7.45 · 5.The effect of surface contact on thermal conduction between adjacent slabs of material All data shown below (the unit of temperature is Kelvin) Q T1 T2 T3 T4 T5 T6 T7 T8 T9 5W 318.55 317.95 318.35 304.35 313.75 313.85 302.35 302.15 302.05 10W 320.85 320.15 320.75 305.25 315.65 315.85 302.45 302.25 302.15 15W 344.85 342.05 341.25 331.05 330.25 330.75 303.15 302.85 302.45 20W 343.05 342.25 340.95 329.25 327.05 327.05 303.95 303.15 302.45 temperature/k Table.5 Data of Experiment 5 360 y = -606,17x + 355,71 R² = 0,8745 350 340 330 320 310 y = -593,17x + 354,01 R² = 0,9063 y = -257,5x + 324,58 R² = 0,6889 y = -226,5x + 321,7 R² = 0,6857 300 290 0 0,01 5W 0,02 Линейная (5W) 0,03 10W 0,04 0,05 15W 0,06 0,07 0,08 20W 0,09 Линейная (10W) Линейная (15W) Линейная (20W) 0,1 Thickness/m Figure.5 T-x graph of Experiment 5 Sharp temperature drop occurred between the contact surfaces of three sections. But compared with contact surface of heater section and test section, the temperature drop in contact surface where the conducting compound was applied is sharper. Discussion From experiment 1,the thermal conductivity of brass is 106.37·. From wiki 9 we get, the theoretical value is 109·.. The error equal 2.41%. The experiment measured value agree well with theoretical value. Also, from experiment 2,the thermal conductivity of stainless steel is 64.37·. The error is 28.2% compared with theoretical value 50.2·. And the error in overall heat transfer coefficient is 11.4%. The reason cause the error may be the contact between test section is poor and the contact resistance could not be neglected. And if we calculate the conductivity as the date from experiment 5, the value is 54.7· which more close to theoretical value. That supports what we suppose above that the contact resistance could not be neglected in experiment 2. In experiment 3, the slope of trend line was steeper than experiment 1. It is consistent with the Eq.1. But the thermal conduction we got is far larger than theoretical value. From the graph, we could see the temperature increase in heater section. That means the date we got is not under steady state and the date is null. In experiment 4, the thermal conductivity of paper we gained is 7.45 · , which is not conform to theoretical value (0.05 watts). But once consider that the thickness should be thinned than 1mm, the error is possible. In experiment 5, temperature drop between the contact surfaces of test and heater sections is unreasonably sharp. I guess that the linear module didn’t achieve steady state. Conclusion The experiment requires us to explore the effect of material and cross-sectional area to the heat transfer and the influence of thermal insulation and conducting compounds on the thermal conduction. The thermal conductivity of brass is 106.37· and stainless steel is 64.37·. From experiment we could see the contact resistance couldn’t be ignored 10 most of time In the reality. And Recording the data until the steady state is achieved are necessary. Reference 1.FOURIER'S LAW, available from: http://www.thermopedia.com/content/781/ 2. Thermal conductivity of paper https://translate.google.com/translate?hl=zhCN&sl=en&u=https://sciencing.com/thermal-properties-paper6893512.html&prev=search 3. Properties: Stainless Steel - Grade 304 (UNS S30400) - AZoM https://www.azom.com/properties.aspx?ArticleID=965 11