- #1
SalfordPhysics
- 69
- 1
1) PROBLEM:
Task is to find the thermal conductivity of steel experimentally.
I have conducted relevant experiments and I am trying to solve for the thermal conductivity (k) itself. My final however is ten times too large in size, and I am having trouble identifying where the error is coming from. Nothing seems to be adding up and its become a complete nightmare.
The process is as follows;
2) EXPERIMENT
a) Forbes Bar - long bar heated from one end in steady state. Temperature recorded at intervals with respect to distance from the source.
b) Suspend heated sample in air and record temperature drop with respect to time. (This part is for heat transfer coefficient).
c) Calorimeter to find specific heat capacity of sample of steel.
3)OUTCOME: (and relevant values + equations)...
Graph for part a) plotted using θ = θ0e-αx ⇒lnθ = lnθ0 - αx i.e. α = -gradient
graph for b) similarly from θ = θ0e-βt i.e. β = -gradient
The values I obtained are α=1.7, β=0.001
My value for Specific heat capacity ≅475
P=perimeter = 0.04m
A = area = 1x10-4
L = 0.02m
β = [(P/A + 2/L)h]/ρS
α = √(hP/KA)
where ρ = density, h = heat transfer coefficient, K = thermal conductivity,
Any help is greatly appreciated. I have the feeling my value for alpha may not be right but I can't be sure and need some advice.
Many thanks!
Task is to find the thermal conductivity of steel experimentally.
I have conducted relevant experiments and I am trying to solve for the thermal conductivity (k) itself. My final however is ten times too large in size, and I am having trouble identifying where the error is coming from. Nothing seems to be adding up and its become a complete nightmare.
The process is as follows;
2) EXPERIMENT
a) Forbes Bar - long bar heated from one end in steady state. Temperature recorded at intervals with respect to distance from the source.
b) Suspend heated sample in air and record temperature drop with respect to time. (This part is for heat transfer coefficient).
c) Calorimeter to find specific heat capacity of sample of steel.
3)OUTCOME: (and relevant values + equations)...
Graph for part a) plotted using θ = θ0e-αx ⇒lnθ = lnθ0 - αx i.e. α = -gradient
graph for b) similarly from θ = θ0e-βt i.e. β = -gradient
The values I obtained are α=1.7, β=0.001
My value for Specific heat capacity ≅475
P=perimeter = 0.04m
A = area = 1x10-4
L = 0.02m
β = [(P/A + 2/L)h]/ρS
α = √(hP/KA)
where ρ = density, h = heat transfer coefficient, K = thermal conductivity,
Any help is greatly appreciated. I have the feeling my value for alpha may not be right but I can't be sure and need some advice.
Many thanks!