17 questions · timed · auto-graded
$\theta_2=50^\circ\text{C}$
$\alpha_\text{steel}=1.2\times10^{-5}\ /^\circ\text{C}$
Longitudinal stain = ?
Stain
$=\frac{\Delta\text{L}}{\text{L}}=\frac{\text{L}\alpha\Delta\theta}{\text{L}}=\alpha\Delta\theta$$=1.2\times10^{-5}\times(50-20)=3.6\times10^{-4}$
When a spherical shell is heated, its volume changes according to the equation, $\text{V}_\theta=\text{V}_0(1+\gamma\triangle\theta)$. The volume referred to here is the volume of the material used to make up the shell, as its volume expands with the rise of temperature with coefficient of expansion of volume $\gamma.$
$\Delta\text{L}=100^\circ\text{C}$
A longitudinal strain develops if and only if, there is an opposition to the expansion.
Since there is no opposition in this case, hence the longitudinal stain here = Zero.
$=373.31$
$\text{P}_\text{steam point}=\text{P}_{100^\circ}90\text{cm of Hg}$
$\text{P}_0=100\text{cm}$
$\text{t}=\frac{\text{P}-\text{P}_0}{\text{P}_{100}-\text{P}_0}\times100^\circ$$=\frac{80-100}{90-100}\times100$
$=200^\circ\text{C}$