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GVSU EGR 468 - Laboratory 5 Heat Transfer from a Circular Cylinder in Cross-Flow

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IntroductionBackgroundExperimental ProcedureExperimental ResultsDiscussionConclusionsEnergy Balance on the CylinderHand CalculationsLaboratory 5Heat Transfer from a Circular Cylinder inCross-FlowDetermination of the Heat TransferCoefficient, hOutlineBy: Brad PeirsonEGR 468 – Heat TransferInstructor: Dr. SozenSchool of EngineeringPadnos College of Engineering and ComputingGrand Valley State UniversityApril 4, 2008Contents1 Introduction 32 Background 33 Experimental Procedure 44 Experimental Results 55 Discussion 106 Conclusions 10A Energy Balance on the Cylinder 11B Hand Calculations 121List of Figures1 Experimental Setup . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 52 Plot of Natural Log of Temperature Differential with Time - Wind Speed19.5m/s . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 63 Plot of Natural Log of Temperature Differential with Time - Wind Speed26.8m/s . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 74 Plot of Natural Log of Temperature Differential with Time - Wind Speed41.8m/s . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 85 Plot of Natural Log of Temperature Differential with Time - Wind Speed50.3m/s . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 921 IntroductionThe purpose of this experiment was to experimentally determine the heat transfer coefficientfor an aluminum cylinder placed in cross-flow conditions. The cylinder was heated and thenplaced in the wind tunnel at a known wind speed. From the temperature data collected theheat transfer coefficient was determined for each of 4 wind tunnel speeds tested. This valuewas compared with the ”known” value of the heat transfer coefficient as calculated fromequation 6-17 in the text. Overall it was determined that equation 6-17 is most accurate forwind speeds that relate to Reynold’s Numbers close to the center of the ranges provided bytable 1.2 BackgroundBiot number for spherical/cylindrical geometry, uses object propertiesBi =hsk=hdk(1)Reynold’s Number for spherical/cylindrical geometry, uses fluid propertiesRe =U∞xυ=U∞dυ(2)Equation 6-17 in the text for the Nusselt Number for flow around a cylinder. This isused to determine h.Nud=hxk= C (Re)nP r1/3f(3)C and n determined from table 6-2 in text:To convert from manometer readings in in. of water to m/s. The 20.63 factor is notunitless and contains the conversion from inches of water to m/s.U = 20.63√∆z (4)3Table 1: Constants for Use With Equation 6-17 (3)RedfC n0.4-4 0.989 0.3304-40 0.911 0.38540-4000 0.683 0.4664000-40000 0.193 0.61840000-400000 0.0266 0.805Percent Error%Error =KnownV alue −ExperimentalV alueKnownV alue× 100 (5)3 Experimental Procedure• Cylinder placed in oven until significant temperature increase was measured• Wind tunnel speeds selected to represent the range of Reynold’s Numbers from table1• Wind tunnel turned to desired s peed• Cylinder placed in wind tunnel• Temperature data taken until steady state reached• Test repeated for wind speeds of 19.5m/s, 26.8m/s, 41.1m/s and 50.4m/sTable 2: Heat Transfer Coefficients Using Equation 3Pressure Difference (inH2O) Speed (m/s) Re Nu hWm2◦C0.9 19.5 16700 69.9 1441.7 26.8 228004 65.4 1344 41.1 35000 79.8 1646 50.4 42900 127 2614Figure 1: Experimental Setup4 Experimental Resultsh values taken from best fit equations using the relation given in Appendix A. Percent errorcalculated from equation 5.Table 3: Heat Transfer CoefficientsPressure (inH2O) h (3)Wm2◦Ch (best fit eq.)Wm2◦C% Error.9 144 152 5.561.7 134 116 13.44 164 246 506 261 170 34.9Hand verification of the calculations are given in Appendix B.5Figure 2: Plot of Natural Log of Temperature Differential with Time - Wind Speed 19.5m/s6Figure 3: Plot of Natural Log of Temperature Differential with Time - Wind Speed 26.8m/s7Figure 4: Plot of Natural Log of Temperature Differential with Time - Wind Speed 41.8m/s8Figure 5: Plot of Natural Log of Temperature Differential with Time - Wind Speed 50.3m/s95 Discussion• % Error is relatively small for the first two tests• The first two tests have Reynold’s Numbers close to the range for the constants intable 1• % Error is fairly large for the two higher speeds• These speeds are close to the edge of the range for the constants in table 1• This suggests that the constants in table 1 are best used when the flow falls in themiddle of a particular range in the table6 ConclusionsTemperature data was collected from a heated aluminum cylinder as it was allowed to coolin the wind tunnel set at a known speed. This temperature data was used to determine theheat transfer coefficient using the method shown in Appendix A. The ”known” value of theheat transfer coefficient was calculated using equation 6-17 in the text. The results of thiscomparison show that equation 6-17 is best used when the Reynold’s Number for the flowfalls into the middle of one of the ranges shown in table 1.10A Energy Balance on the Cylinderq = hA (T − T∞)mcdTdt= −qconvρV cdTdt= −hA (T − T∞)Let θ = T − T∞→dθdt=dTdtρV cdθdt= −hAθdθθ= −hAV cdtlnθ(t)θ0= −hAρV ctlnT − T∞T0− T∞= −hAρV ctThis equation can be solved to determine h using the best fit e quations from the plots insection 4.11B Hand CalculationsFrom Bernoulli’s Equation:U =s2 (patm− pT S)ρair=s2ρh2Og∆zinH2Oρairρair≈ 1.17kgm3U =vuuut2999kgm39.81ms2∆zinH2O1.17kgm3= 20.63q∆zinH2OU = 20.63√0.9 = 19.57m/sRe =Udυ=(19.57m/s) (0.0128m)1.5 × 10−5m2/s= 16700Nu =hdk= CRenP r1/3From table 1, for Re = 16700 → C = 0.193, n = 0.618Nu =hdk= (0.193) (16700)0.618(0.708)1/3= 70.02h =kdNu =0.02624Wm◦C0.0128m(70.02) = 143.5Wm2◦ClnhT − T∞T0−T∞i= −hAρV ctFor 0.9inH2O → −hAρV c= −0.0197h = (0.0197)ρV cA= (0.0197)(2707kg/m3)0.25π (0.0128m)2(0.263m)(896J/kg◦C)π (0.0128m) (0.263m)h = 152.55W/m2◦CNote that there is an extra1sadded to the units for the h value above. This is becauseh will be multiplied by time in the best fit equation for a value that is ultimately unitless.12%Error =|143.5 − 152.55|143.5× 100% = 6.3%The values above are not exactly the values calculated by Excel. The discrepancy is smallenough, however, that it can be attributed to differences in decimal rounding between Exceland the above


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GVSU EGR 468 - Laboratory 5 Heat Transfer from a Circular Cylinder in Cross-Flow

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