Question 10.8: Considering the pressure vessel described in Example 10.5, t...
Considering the pressure vessel described in Example 10.5, the vessel is to have one end closed by a blind flange. What is the minimum required thickness of the blind flange? Design data are as follows:
Design pressure P =2500 psi
Design temperature=250 °F
Flange material is SA-105
Bolting material is SA-325 Gr. 1
No corrosion exists.
Allowable bolt stress at gasket seating and operating conditions=19,200 psi
Allowable flange stress at gasket seating and operating conditions=17,500 psi
Gasket is spiral-wound metal, fiber-filled, stainless steel, with inside diameter 13.75 in. and width N =1 in.
Following the information calculated in Example 10.5, once the actual bolt area A_{b} is found, the design loading for the gasket seating condition W_{a} can be determined as
W_{ a }=0.5\left(A_{ m }+A_{ b }\right) S_{ a } =0.5 (36.2+36.8) (19,200)
W_{ a } = 700,800.
The moment arm is determined from
h_{ G } = 0.5(C − G) = 0.5(22.5 − 15.043) = 3.729in.
From Example 10.5, the design loading for operating condition is W_{m1} =694,700.
The minimum required thickness is determined as the greater thickness of that determined for the gasket seating load according to Eq. (10.21) and for the operating load according to Eq. (10.22). For gasket seating,
t=G \sqrt{\frac{1.9 W_{ a } h_{ G }}{S_{ a } E G^{3}}}= 15.043 \sqrt{\frac{1.9(700,800)(3.729)}{(17,500 \times 1)(15.043)^{3}}} = 4.343in.
For the operating condition,
t=G \sqrt{\frac{0.3 P}{S_{ b } E}+\frac{1.9 W_{ m 1} h_{ G }}{S_{ b } E G^{3}}}=15.043 \sqrt{\frac{0.3(2500)}{17,500 \times 1.0}+\frac{1.9(694,700)(3.729)}{(17,500 \times 1.0)(15.043)^{3}}}
= 5.329in.
Therefore, the minimum required flange thickness is t =5.329 in.