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GVSU EGR 468 - LAB TWO: Determination of Thermal Resistance

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LAB TWO:PURPOSE:ABSTRACT:THEORY:APPARATUS:PROCEDURE:RESULTS/DISCUSSION:CONCLUSIONS:LAB TWO:Determination of Thermal ResistancePadnos School of EngineeringEGR 468Dr. David BlekhmanJanuary 19, 2004Brad Vander VeenPURPOSE:To measure the thermal resistance of an insulation material under steady state conditions with one-dimensional heat flow.ABSTRACT:In this lab, the thermal conductivity of a sample material was experimentally determined. An apparatus was set up that was capable of generating heat at a constant rate. Heat flowwas isolated so that it was on one direction. All the heat passed through a 6” by 6” piece of ¼” Plexiglas, and then through a piece of sample material also 6” by 6”, but 1.75” thick. Thermocouples were place on either side of the Plexiglas and on top of the sample materials. Temperature measurements were taken and charted until the system reached semi-steady state. This data was used to calculate the thermal resistance of the sample material. This thermal resistance was found to be k = 0.05 BTU/hr·ft·ºF.THEORY:Heat transfer is known to occur via three modes: conduction, convection, and radiation. This lab focuses entirely on the conduction of heat through solids. Consider the general equation for conduction:dtdTkqTgen12(1)Assuming steady state and no sources yields:02 T(2)Since heat transfer is only in one direction we get:022dxT(3)Thus we know that:CdxdT(4)-where C is a constantWe can substitute:kAqdxdTx(5)By integrating both sides from x1 = T1 to x2 = T2:)()(1221xxkAqTTx(6)-where T is the temperature, qx is the heat transfer, k is the thermal resistance, A isthe area normal to the heat flow, and x is the position in along the axis of heat transfer.APPARATUS:Apparatus Model/Type Serial Number“heat box” plexiglas k = 0.12 [BTU/hr*ft*F Sample material k = unknown ice water thermocouplesPROCEDURE:1. Measure the thickness of each of the Plexiglas and sample material.2. Assemble the apparatus as seen below in Figure 1.Figure 1 – Experiment Setup3. Turn on the light bulb4. Record temperature every minute.5. Once temperature readings have reached semi-steady state, use the steady state temperatures to calculate the thermal resistance of the sample material.RESULTS/DISCUSSION:Figure 2 below shows the temperature data after the light bulb was turned on:Figure 2 – Temperature DataThe temperature readings taken at 31 minutes were assumed to be steady state temperature readings.Table 1 below shows the temperature readings at 31 minutes at each of the thermocouples.Thermocouple Number#1 #2 #3Temperature (F) : 209.0 195.1 50.0Table 1 – Steady State Temperature ReadingsSince the dimensions and temperature of the Plexiglas are known, the heat transfer through the solid can be calculated using Equation (6).)00208.0()25.0)(12.0()1.1950.209( xq(7)Solving for qx yields:384.12xq BTU’s (7)Since the heat flux is known, the thermal resistance of the sample material can now be found using Equation (6):)0146.0()25.0)((384.12)0.501.195( k(8)Solving for k yields:08.0k BTU/hr*ft*F (9)Because temperatures had not completely reached steady state, a further analysis was alsodone. Steady state temperatures were extrapolated from Figure 2. These values can be seen below in Table 2.Thermocouple Number#1 #2 #3Temperature (F) : 235 225 50Table 2 – Extrapolated Steady State TemperaturesUsing this data and Equation (6), the thermal resistance of the sample material is found tobe only 05.0k BTU/hr*ft*FThe actual value of the thermal resistance is more likely closer to the second value rather than the first, since the second set of temperature data is a closer approximation of the steady state temperatures of the system.The results reached in this analysis seem reasonable when compared to data found in published thermal conductivity tables for certain materials. This k-value is around the k-value for Polyvinyl Chloride and some “Styrofoam-type” insulators.CONCLUSIONS:In this lab, a sample material was analyzed in order to determine is thermal resistance. Results showed that the thermal resistance was approximately k = 0.05 BTU/hr·ft·ºF. From this k-value, one could justify using it as an insulation material. When compared toStyrofoam and other low-density insulation materials, it does not insulate as well. However, it may be much cheaper to manufacture, thus making it a cost-effective alternative to a commercially available


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