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The atmospheric pressure at the surface of the earth is about 101 $kPa$.

For Carbon fiber reinforced plastic (70/30 fibre/matrix, unidirectional, along grain):

  • Young's Modulus is: $181\ GPa$
  • Density - $1.6\ g/m^3$

With these parameters in mind:

How do we calculate the minimum thickness/weight of a vacuum sealed (i.e. inside the container is vacuum) spherical container/vessel that would withstand the standard atmospheric pressure without much change in shape with its inner volume of $k\ m^3$?

And is there a general formula to calculate this, just based on the external surface area (independent of shape of the container)?

My background isn't in Physical Engineering sciences thus an apology beforehand.

Fred
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TheoryQuest1
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2 Answers2

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Typically the maximum stress for a thin walled spherical pressure vessel for a given diameter, thickness and pressure is:

$$\sigma = \frac{P\cdot R}{2\cdot t}$$

where:

  • $\sigma$ is the observed pressure (in Pa)
  • $P$ is the pressure (in Pa)
  • $R$ is the radius of the vessel (in m)
  • $t$ : is the thickness of the vessel (in m)

using this it is possible to find the minimum mass of a pressure vessel is (Wikipedia)

$$M={3 \over 2}PV{\rho \over \sigma }$$, where:

  • M is mass, (kg)
  • P is the pressure difference from ambient (the gauge pressure), (Pa) (100000Pa = 1 Atm)
  • V is volume,
  • $\rho$ is the density of the pressure vessel material, (expressed in $kg/m^3$)
  • $\sigma_{all}$ is the maximum allowable working stress that material can tolerate. ($Pa$)

However, an interesting parameter in the question which is without changing much its shape. That can be interpreted as having a very small deformation. Usually a common value which is used to denote small deformation within the elastic range is 0.2% or (0.002). So the value that you could use for the allowable stress is $\sigma_{all}= E\cdot 0.002 = 181 GPa \cdot 0.002= 362 MPa$

Compression buckling

An very important additional check is that of buckling (because the shell will be under compressive loads).

The buckling pressure $q_C$ of a elastic thin spherical shell was obtained by Zoelly (1915) and Van der Neut (1932):

$$q_c = \frac{2 E}{[3(1-\nu^2)]^{1/2}} \left(\frac{h}{R}\right)^2 \tag{eq:2}$$

where

  • E is the elastic modulus
  • $\nu$ is the Poisson Ratio
  • $h$ is the shell thickness
  • $R$ is the shell radius. Assuming a volume V, the radius of the sphere will be $R= \sqrt[3]{\frac{3V}{4\pi}}$

Therefore, given that the critical pressure needs to be 1 atm ($q_C= 1Atm = 100000Pa$), the minimum thickness can be estimated:

$$q_c = \frac{2 E}{[3(1-\nu^2)]^{1/2}} \left(\frac{h}{R}\right)^2 \tag{eq:2}$$

$$ h = \sqrt[3]{\frac{3V}{4\pi}} \sqrt{\frac{q_c}{2 E}\sqrt{3(1-\nu^2)}} $$

After having calculated the minimum thickness, it is possible to obtain the total required volume of material $V_m$ and weight $M$ by:

$$V_m = 4\pi R^2 h $$ $$M = V_m\cdot \rho = 4\pi R^2 h \cdot \rho $$

NMech
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In a sphere container with the volume K

$$r= (\frac{3K}{4\pi} )^{1/3}$$

Assuming the strength, which you have not provided, $\sigma_c = 2.02×10^5 psi = 1.39 GPa$

The pressure transferred by the atmosphere is equivalent to the pressure over the surface of the circle as compression with a factor of safety. And in a thin-walled sphere, thickness<< r:

$$C_{stress/m}= 101kPa *\pi r^2/2\pi r=101kPa* r/2= 101kPa*(\frac{3K}{8\pi} )^{1/3}$$

The thikness is $thicknes =\frac{1.39GPa}{C_{stress}}$

kamran
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