Dissemination of IT for the Promotion of Materials Science (DoITPoMS)

DoITPoMS Teaching & Learning Packages Thermal Expansion and the Bi-material Strip Results: Estimating the boiling temperature of nitrogen

# Results: Estimating the boiling temperature of nitrogen

For the steel-aluminum bi-material strip the following measurements were made

 δ = 0.021 m x = 0.189 m

From equation (3) the curvature

$\kappa = \frac{{2 \times \sin \left[ {{{\tan }^{ - 1}}\left( {0.021/0.189} \right)} \right]}}{{\sqrt {{{0.021}^2} + {{0.189}^2}} }} = 1.2$

From equation (6), the misfit strain

$\Delta \varepsilon = \frac{1}{{12}}\kappa h\left( {{E_*} + 14 + \frac{1}{{{E_*}}}} \right) = \frac{1}{{12}} \times 1.2 \times 0.001 \times \left( {\frac{{210}}{{70}} + 14 + \frac{{70}}{{210}}} \right) = 1.7 \times {10^{ - 3}}$

And so, from equation (5)

$\Delta T = \frac{{\Delta \varepsilon }}{{{\alpha _{\rm{A}}} - {\alpha _{\rm{B}}}}} = \frac{{1.7 \times {{10}^{ - 3}}}}{{(1.5 - 2.3) \times {{10}^{ - 5}}}} = - 213{\rm{K}}$

Given an initial room temperatue of 20ºC, this gives the boiling temperature of nitrogen as 20 - 213 = -193ºC (compare accepted figure of -196ºC).