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 = 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 =...