Question: A pressure vessel is sealed by a gasket as shown in Fig. 1(a) The internal fluid pressure is $650$ psi. If the gasket is made of soft steel with asbestos filling, calculate the pressure in psi necessary to seat the gasket, by tightening the bolts. cover of pressure vessel gasket seal a = $3$ in p $D_0 = 3.5$ in bolt threads w = $0.612$ in Mean diameter of effective gasket ring = $D_G$ Fig. 1 (a) Front view of pressure vessel sealed with gasket (b) Top view of gasket seal and its effective width

1. **Problem Statement:** Calculate the pressure necessary to seat the gasket in a pressure vessel sealed by a gasket, given: - Internal fluid pressure $p = 650$ psi - Outer diameter of gasket ring $D_0 = 3.5$ in - Width of gasket ring $w = 0.612$ in - Mean diameter of effective gasket ring $D_G$ 2. **Understanding the problem:** The gasket is compressed by tightening bolts to create a seating pressure $p_s$ that must counteract the internal fluid pressure $p$ to prevent leakage. 3. **Formula used:** The mean diameter of the gasket ring is: $$D_G = D_0 - w$$ The seating pressure $p_s$ needed to seat the gasket is related to the internal pressure $p$ and the geometry of the gasket. For a circular gasket ring, the force due to internal pressure is balanced by the force due to seating pressure acting on the gasket area. The effective gasket area $A_g$ is the area of the ring: $$A_g = \pi \times D_G \times w$$ The force due to internal pressure $F_p$ is: $$F_p = p \times A_i$$ where $A_i$ is the internal area enclosed by the gasket, approximated by: $$A_i = \pi \times (D_0 - 2w)^2 / 4$$ The seating pressure $p_s$ must satisfy: $$p_s \times A_g = F_p$$ 4. **Calculate mean diameter $D_G$:** $$D_G = D_0 - w = 3.5 - 0.612 = 2.888 \text{ in}$$ 5. **Calculate internal diameter $D_i$:** $$D_i = D_0 - 2w = 3.5 - 2 \times 0.612 = 3.5 - 1.224 = 2.276 \text{ in}$$ 6. **Calculate internal area $A_i$:** $$A_i = \pi \times \frac{D_i^2}{4} = \pi \times \frac{(2.276)^2}{4} = \pi \times \frac{5.181}{4} = \pi \times 1.295 = 4.068 \text{ in}^2$$ 7. **Calculate gasket area $A_g$:** $$A_g = \pi \times D_G \times w = \pi \times 2.888 \times 0.612 = \pi \times 1.767 = 5.551 \text{ in}^2$$ 8. **Calculate force due to internal pressure $F_p$:** $$F_p = p \times A_i = 650 \times 4.068 = 2644.2 \text{ lb}$$ 9. **Calculate seating pressure $p_s$:** $$p_s = \frac{F_p}{A_g} = \frac{2644.2}{5.551}$$ Show cancellation step: $$p_s = \frac{2644.2}{\cancel{5.551}}$$ Calculate: $$p_s = 476.5 \text{ psi}$$ **Final answer:** The pressure necessary to seat the gasket by tightening the bolts is approximately **476.5 psi**.