Laboratory 2Determination of Thermal ResistanceBy: Brad PeirsonEGR 468 – Heat TransferInstructor: Dr. SozenSchool of EngineeringPadnos College of Engineering and ComputingGrand Valley State UniversityFebruary 28, 2008Contents1 Introduction 22 Experimental Procedure 23 Experimental Results 24 Discussion 45 Conclusions 66 References 6List of Figures1 Experimental Setup . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 22 Temperature Data for Plexiglass Sample . . . . . . . . . . . . . . . . . . . . 33 Temperature Data for Polystyrene Sample . . . . . . . . . . . . . . . . . . . 411 IntroductionThe purpose of this laboratory exercise was to experimentally determine the thermal con-ductivity, k, and the R-factor, R, of a sample of mass type insulation. In order to determineboth factors a sample of insulation was placed in an insulated box and heated. Temp e raturereadings were taken until the sample reached steady state. Then the values of k and R weredetermined using a version of the general energy balance for heat transfer in a solid.2 Experimental ProcedureTwo K-type thermocouples were attached to both sides of a 7”x7.25”x5.5mm piece of plex-iglass. The plexiglass was then placed into a heavily insulated box as shown in figure 1.The second experimental setup included a 7”x7.25”x38” piece of polystyrene insulation witha reflective barrier on top of the plexiglass.Figure 1: Experimental SetupThe thermocouples were attached to a NI 4350 Hi-Precision Temperature and Volt Meterwhich was then connected to a computer with a labview interface. When the 25 watt bulbwas switched on the labview program collected the temperature data for approximately twohours. The temperatures were recorded until the samples reached steady state and the datawas analyzed as shown in section 2.3 Experimental ResultsFigure 2 shows the data temperature data acquired from the uninsulated plexiglass sample.The equation used to calculate the thermal conductivity of the sample is shown in equa-tion 1. The derivation of this equation is given in section 4.k =QxA∆x∆T(1)2Figure 2: Temperature Data for Plexiglass SampleWhere k is the thermal conductivity, Qxis the heat generated by the light bulb, A is thearea of the sample, ∆x is the thickness of the sample and δT is the difference between thetwo surface temperatures of the sample at steady state.The heat generated by the light bulb is 25 watts, the area of the sample is 7”x7.25” or0.0327m2, the plexiglass is 0.0055m thick and the temperature difference at steady state is12.8◦C. Therefore the thermal conductivity of the plexiglass is:k =25W0.0327m20.0055m12.8◦C= 11.61Wm◦C(2)The R value of the plexiglass is then given by equation 3.R =∆xk=0.0055m11.61Wm◦C= 0.000474◦CW(3)The analysis that was carried out for the plexiglass was repeated for the insulation. Thethermal conductivity and R values for the insulation were found to bek =25W0.0327m20.009525m68◦C= 1.26Wm◦C(4)R =∆xk=0.009525m1.26Wm◦C= 0.00756◦CW(5)3Figure 3: Temperature Data for Polystyrene Sample4 DiscussionSeveral questions regarding the experiment detailed above were posed in the laboratoryhandout. The questions and answers are as follows:1. Under what conditions does the general energy balance in a solid result in equation 2QxA= qx=∆TR?The general energy balance for one dimensional heat transfer in a solid is1α∂T∂t=∂2T∂x2+˙qkWhen there are no sources inside the sample the˙qkterm drops from the equation.When the sample reaches steady state the time derivative also drops from the equa-tion, leaving:∂2T∂x2= 0.Because the second derivative of the temperature is zero,∂T∂xmust be constant.Fourier’s Law of Conduction is Qx= −kA∂T∂xTherefore∂T∂x=−QxkA= constantSeparating the equation and integrating gives:4ZT2T1dT = −QxkAZx2x1dx ⇒k∆T∆x=QxAAndQxA=∆TRWhere R =∆xkTherefore the conditions that will produce equation 2 from the laboratory handout areone-dimensional, steady state heat transfer with no internal sources.2. Consider the apparatus that we use in lab. What steps were taken to assure one-dimensional heat flow? Could a thick sample of a good ins ulator be accurately tested?The box that the light bulb was placed in in the laboratory setup was heavily insulated.Because of the insulation on the box the vast majority of heat generated by the bulbwould be lost through the top of the box, where the sample was placed. This setupprovides one-dimensional heat transfer for the plexiglass sample, however a thick pieceof insulation would not be able to be accurately tes ted. As the sample’s thickness andR value approached that of the walls the setup would tend to produce heat transfer ineach direction within the box.3. What steps did you take to assure that steady state was achieved? How would thegoverning equations change if the problem was unsteady?In order to ensure steady state conditions the samples were exposed to the heat sourcefor more than two hours apiece. The governing equation is still the general energybalance as stated in question 1. However, in an unsteady problem the time derivativeof the temperature does not drop out of the equation. This me ans that the secondderivative of temperature with respect to x does not equal zero and that the derivativeis not constant. Therefore the−QxkAterm is not constant. In most cases the area of thesample will remain constant which means that under unsteady conditions either theheat being generated is changing with time or the thermal conductivity of the materialis changing with time.4. Suppose that it takes a long time for your sample to reach steady state. Are there anypractical benefits or drawbacks to a material that reaches steady state slowly?The definition of a steady state condition is that all heat being generated is beingtransferred to the atmosphere. Before steady state is reached, the majority of heatgenerated is used in heating the enclosed volume. This is a benefit in an applicationsuch as a home where losing heat to the outside environment can mean drastic increasesin overall heating cost.5. How would you estimate experimental uncertainty for this experiment?The best way to approximate the error in this experiment is to compare the thermalconductivity and resistance values to known, published values for the sample materials.The published value of k for plexiglass is 0.2Wm◦Cand the value for polystyrene is0.14Wm◦C[1]. Both of these values make