$150 ~ \text{psi} = 150 ~ \dfrac{\text{lb}}{\text{in}^2} \cdot \left( \dfrac{12 ~ \text{in}}{1 ~ \text{ft}} \right)^2$
$150 ~ \text{psi} = 21,600 ~ \text{lb/ft}^2$
For longitudinal joint (tangential stress):
Consider 1 ft length
$F = 2T$
$pD = 2\sigma_t t$
$\sigma_t = \dfrac{pD}{2t}$
$\dfrac{33\,000}{t} = \dfrac{21\,600D}{2t}$
$D = 3.06 \, \text{ft} = 36.67 \, \text{in.}$
For girth joint (longitudinal stress):
$F = P$
$p (\frac{1}{4} \pi D^2) = \sigma_l (\pi Dt)$
$\sigma_l = \dfrac{pD}{4t}$
$\dfrac{16\,000}{t} = \dfrac{21\,600D}{4t}$
$D = 2.96 \, \text{ft} = 35.56 \, \text{in.}$
Use the smaller diameter, $D = 35.56 \, \text{in.}$ answer
Why do you need to assume 1ft
Why do you need to assume 1ft of length when the length is left out from the formula? Also, can I ask if the "ft" from the "kips/ft" the thickness of the thin-walled vessel? I'm only assuming from the formula.