Skip to main content

Spherical vessel, volume increase 300 cm3, find pressure

Short answer

For thin sphere, volumetric strain ΔV/V=3pd(1–ν)/(4tE). With V=πd³/6 and given data, solving gives fluid pressure p≈0.047 N/mm² (0.047 MPa).

A seamless spherical vessel of 2.5 m internal diameter and 8 mm thick is filled with a fluid under the pressure until its volume increases by 300 cm3300\ \text{cm}^3. Calculate the pressure exerted by the fluid on the vessel. Take E=2.10×105 N/mm2E = 2.10\times10^5\ \text{N/mm}^2, ν=0.3\nu = 0.3.

Data: d=2500d = 2500 mm, t=8t = 8 mm, δV=300 cm3=3×105 mm3\delta V = 300\ \text{cm}^3 = 3\times10^5\ \text{mm}^3, E=2.1×105 N/mm2E = 2.1\times10^5\ \text{N/mm}^2, ν=0.3\nu = 0.3.

Method

For a thin sphere the stress in every direction is σ=pd4t\sigma = \dfrac{pd}{4t}. The strain in any direction is

ε=σE(1−ν)=pd4tE(1−ν)\varepsilon = \frac{\sigma}{E}(1-\nu) = \frac{pd}{4tE}(1-\nu)

and the volumetric strain is three times this:

δVV=3ε=3pd4tE(1−ν)\frac{\delta V}{V} = 3\varepsilon = \frac{3pd}{4tE}(1-\nu)

Calculation

V=πd36=π(2500)36=8.1812×109 mm3V = \frac{\pi d^3}{6} = \frac{\pi(2500)^3}{6} = 8.1812\times10^{9}\ \text{mm}^3 3×1058.1812×109=3p(2500)4(8)(2.1×105)(1−0.3)\frac{3\times10^5}{8.1812\times10^{9}} = \frac{3p(2500)}{4(8)(2.1\times10^5)}(1-0.3) 3.6669×10−5=7.8125×10−4 p⇒p=0.0469 N/mm23.6669\times10^{-5} = 7.8125\times10^{-4}\,p \Rightarrow p = 0.0469\ \text{N/mm}^2

Answer: p≈0.047 N/mm2p \approx 0.047\ \text{N/mm}^2 (0.047 MPa).

✓ Answer checked by Claude Opus 5.5 (high effort)

Related questions

All Strength of Materials chapter 5 questions →