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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