Maximum excess pressure inside a thin-walled steel tube of radius r and thickness $\mathrm{\Delta r}(<<\mathrm{r})$, so that the tube, would not rupture would be (breaking stress of steel is ${\mathrm{\sigma }}_{max}$)

Consider a small element of the tube.

$2\mathrm{Tsin}\mathrm{\theta }=\mathrm{\Delta p}\times \mathrm{A}$

$\text{ where }\mathrm{\Delta p}={\mathrm{p}}_{\mathrm{i}}-{\mathrm{p}}_0\text{ and }\mathrm{A}\text{ is the area of element. As }\mathrm{\theta }\text{ is very }$

$\text{ small, }\sin \mathrm{\theta }\approx \mathrm{\theta }\text{ so, }2\mathrm{T}\times \mathrm{\theta }=\mathrm{\Delta p}\times \mathrm{l}\times \left(2\mathrm{r\theta }\right)$

$\Rightarrow \mathrm{\Delta p}=\frac{\mathrm{T}}{\mathrm{lr}}\mathrm{\sigma }\left(\text{ stress developed in tube }\right)=\frac{\mathrm{\Delta T}}{\mathrm{\Delta r}\times \mathrm{l}}$

$\text{ where }\mathrm{\Delta r}\times l is \text{the cross-sectional area. }$

$\mathrm{\sigma }=\frac{\mathrm{\Delta p}\times \mathrm{lr}}{\mathrm{\Delta r}\times \mathrm{l}}=\mathrm{\Delta p}\times \frac{\mathrm{r}}{\mathrm{\Delta r}}$

$\text{ For no rupturing, }\mathrm{\sigma }\le {\mathrm{\sigma }}_{max}$

$\text{ So, }\mathrm{\Delta p}\times \frac{\mathrm{r}}{\mathrm{\Delta r}}\le {\mathrm{\sigma }}_{max}$

$\mathrm{\Delta p}\left(\text{ max, value }\right)={\mathrm{\sigma }}_{max}\times \frac{\mathrm{\Delta r}}{\mathrm{r}}$