Correct Answer - Option 2 : 2
Concept:
For cylindrical pressure vessel:
Hoop stress is given by:
\({σ _{hoop}} = \frac{{Pd}}{{2t}}\)
Longitudinal stress is given by:
\({σ _{long}} = \frac{{Pd}}{{4t}}\)
For spherical pressure vessel, hoop stress and longitudinal stress is equal:
σhoop = σlong = \(\frac{{{{Pd}}}}{{4{{t}}}}\)
Calculation:
Given:
dc = ds, tc = ts, Pc = Ps
\(\frac{{{{\left( {{{\rm{\sigma }}_{\rm{h}}}} \right)}_{{\rm{Cylindrical\;}}}}}}{{{{\left( {{{\rm{\sigma }}_{\rm{h}}}} \right)}_{{\rm{spherical}}}}}} = \frac{{{{\left( {\frac{{{\rm{Pd}}}}{{2{\rm{t}}}}} \right)}_{{\rm{cyl}}}}}}{{{{\left( {\frac{{{\rm{Pd}}}}{{4{\rm{t}}}}} \right)}_{{\rm{sph}}}}}} = \frac{{{{\rm{P}}_{\rm{c}}}{{\rm{d}}_{\rm{c}}}}}{{2{{\rm{t}}_{\rm{c}}}}} \times \frac{{4{{\rm{t}}_{\rm{s}}}}}{{{{\rm{P}}_{\rm{s}}}{{\rm{d}}_{\rm{s}}}}} = 2\)
∴ the ratio of hoop stress in the cylindrical vessel to that of the spherical vessel is 2.