1.7 Data: Electric power (W) Specimen Material Temperature (°C) Distance (mm) 20 30 40 50 60 0 39.6 52.6 65.3 78.0 90.6

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answerhappygod
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1.7 Data: Electric power (W) Specimen Material Temperature (°C) Distance (mm) 20 30 40 50 60 0 39.6 52.6 65.3 78.0 90.6

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1 7 Data Electric Power W Specimen Material Temperature C Distance Mm 20 30 40 50 60 0 39 6 52 6 65 3 78 0 90 6 1
1 7 Data Electric Power W Specimen Material Temperature C Distance Mm 20 30 40 50 60 0 39 6 52 6 65 3 78 0 90 6 1 (174.9 KiB) Viewed 18 times
1 7 Data Electric Power W Specimen Material Temperature C Distance Mm 20 30 40 50 60 0 39 6 52 6 65 3 78 0 90 6 2
1 7 Data Electric Power W Specimen Material Temperature C Distance Mm 20 30 40 50 60 0 39 6 52 6 65 3 78 0 90 6 2 (369.96 KiB) Viewed 18 times
Please help me with the plotation of the graph and the calculations as stated
1.7 Data: Electric power (W) Specimen Material Temperature (°C) Distance (mm) 20 30 40 50 60 0 39.6 52.6 65.3 78.0 90.6 T 15 36.2 47.6 58.8 70.0 81.0 12 30 Tz 32.7 42.6 52.3 62.1 71.6 45 TA 29.6 38.2 46.6 55.1 63.3 Brass 60 Is 26.5 33.7 40.8 48.0 54.8 75 IG 22.6 28.1 33.5 39.1 443 90 17 19.6 23.9 28.1 32.5 36.4 Estimated Heat Loss (%) 9.9 11.4 12.0 12.4 12.7
1.8 Results analysis 1.8.1 Thermal conductivity I For each specimen: From the results for each power setting, plot charts of temperature against distance along the bar, with respect to the thermocouples. It should be possible to draw a good best-fit line through the results, as all parts are of the same material 3 Calculate the thermal gradient of the line for reference. Use the two readings T3 and T5, the given distance between them, the heater power, and the bar area with Equation 9, to calculate the thermal conductivity of the specimen. Compare it with the typical values given in Table 5. Indicate the differences and explain the differences. k = z OL TTL A(1,-1) 4(1,-1,)
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