Pressure vessel

Apr 23, 2026 - 11:52

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Adding unit to the spherical minimum mass formula

← Previous revision		Revision as of 06:20, 23 April 2026
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	* $M$ is mass, (kg)		* $M$ is mass, (kg)
	* $P$ is the pressure difference from ambient (the [[gauge pressure]]), (Pa)		* $P$ is the pressure difference from ambient (the [[gauge pressure]]), (Pa)
	* $V$ is volume,		* $V$ is volume, (m³)
	* $\rho$ is the density of the pressure vessel material, (kg/m³)		* $\rho$ is the density of the pressure vessel material, (kg/m³)
	* $\sigma$ is the maximum working [[stress (physics)\|stress]] that material can tolerate. (Pa)For a sphere the thickness d = rP/2σ, where r is the radius of the tank. The volume of the spherical surface then is 4πr²d = 4πr³P/2σ. The mass is determined by multiplying by the density of the material that makes up the walls of the spherical vessel. Further the volume of the gas is (4πr³)/3. Combining these equations give the above results. The equations for the other geometries are derived in a similar manner		* $\sigma$ is the maximum working [[stress (physics)\|stress]] that material can tolerate. (Pa)For a sphere the thickness d = rP/2σ, where r is the radius of the tank. The volume of the spherical surface then is 4πr²d = 4πr³P/2σ. The mass is determined by multiplying by the density of the material that makes up the walls of the spherical vessel. Further the volume of the gas is (4πr³)/3. Combining these equations give the above results. The equations for the other geometries are derived in a similar manner