Heat Transfer Essay

809 WordsNov 10, 20114 Pages
Heat Transfer Lab Report Experimental data are obtained from Pradeep as our data logger wasn’t working properly that day. Ambient Temperature used = 21°C Ambient Pressure used = 100.8k Pa Graphs of ln(θo/θ) vs t(s) were plotted for the 4 of sets of data collected. Gradients for each graph are: Set 1 = 0.0231 Set 2 = 0.0182 Set 3 = 0.0117 Set 4 = 0.007 Calculations for experimental heat transfer coefficients Set 1 Gradient, m = 0.0231 Density, ρ = 8954 kg/m³ Specific heat of copper, C = 383 J/kg°C Volume, V = π * r ² * h = π * (6.2x10-3)2 * 0.1 = 1.208x10ˉ 5 m3 Area, A = 2 * π * r * h = 2 * π * 6.2x10-3 * 0.1 = 3.896x10-3 m2 Ratio of V/A = 1.208x10ˉ 5 / 3.896x10-3 = 3.101x10-3 m Therefore, h = m * ρ * C * V/A = 0.0231 * 8954 * 383 * 3.101x10-3 = 245.65 W/m2 °C Set 2 Gradient, m = 0.0182 Density, ρ = 8954 kg/m³ Specific heat of copper, C = 383 J/kg°C Volume, V = π * r ² * h = π * (6.2x10-3)2 * 0.1 = 1.208x10ˉ 5 m3 Area, A = 2 * π * r * h = 2 * π * 6.2x10-3 * 0.1 = 3.896x10-3 m2 Ratio of V/A = 1.208x10ˉ 5 / 3.896x10-3 = 3.101x10-3 m Therefore, h = m * ρ * C * V/A = 0.0182 * 8954 * 383 * 3.101x10-3 = 193.55 W/m2 °C Set 3 Gradient, m = 0.0117 Density, ρ = 8954 kg/m³ Specific heat of copper, C = 383 J/kg°C Volume, V = π * r ² * h = π * (6.2x10-3)2 * 0.1 = 1.208x10ˉ 5 m3 Area, A = 2 * π * r * h = 2 * π * 6.2x10-3 * 0.1 = 3.896x10-3 m2 Ratio of V/A = 1.208x10ˉ 5 / 3.896x10-3 = 3.101x10-3 m Therefore, h = m * ρ * C * V/A = 0.0117 * 8954 * 383 * 3.101x10-3 = 124.42W/m2 °C Set 4 Gradient, m = 0.007 Density, ρ = 8954 kg/m³ Specific heat of copper, C = 383 J/kg°C Volume, V = π * r ² * h = π * (6.2x10-3)2 * 0.1 =

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