A large spherical tank (see figure) contains gas at a pressure of 3.5 MPa. The tank is 20 m in d

step 1 Hello and welcome to this video solution of numerate. So here it's given that a large spherical

A large spherical tank (see figure) contains gas at a...

A large spherical tank (see figure) contains gas at a pressure of $3.5 \mathrm{MPa}$. The tank is $20 \mathrm{m}$ in diameter and is constructed of high-strength steel having a yield stress in tension of $550 \mathrm{MPa}$ (a) Determine the required thickness of the wall of the tank if a factor of safety of 3.5 with respect to yielding is required. (b) If the tank wall thickness is $100 \mathrm{mm}$, what is the maximum permissible internal pressure?

A large spherical tank (see figure) contains gas at a pressure of $3.5 \mathrm{MPa}$. The tank is $20 \mathrm{m}$ in diameter and is constructed of high-strength steel having a yield stress in tension of $550 \mathrm{MPa}$
(a) Determine the required thickness of the wall of the tank if a factor of safety of 3.5 with respect to yielding is required.
(b) If the tank wall thickness is $100 \mathrm{mm}$, what is the maximum permissible internal pressure?

Key Concepts

Spherical Pressure Vessels

Spherical pressure vessels are containers designed to hold fluids at a pressure substantially different from the ambient pressure. Their geometry leads to uniform stress distribution, which makes them efficient in resisting internal pressures. The analysis of stress in a spherical vessel revolves around the balance of forces due to the internal pressure acting over the curved surface.

Stress Analysis in Spherical Pressure Vessels

The internal pressure in a spherical vessel induces tensile stresses in the wall. One of the key formulas employed is that the hoop (or membrane) stress is given by the internal pressure multiplied by the radius, divided by twice the wall thickness. This relationship is crucial for determining whether the material will deform or yield under the applied load.

Material Yield Strength

Yield strength is a material property that indicates the stress level at which a material begins to deform plastically. In the context of pressure vessels, knowing the yield strength of the construction material is vital because it defines the upper limit of stress that the vessel's wall can sustain before permanent deformation occurs.

Factor of Safety

A factor of safety is an engineering parameter used to provide a design margin over the theoretical design capacity. It accounts for uncertainties in material properties, loading conditions, and potential flaws in manufacturing. When applied to the yield strength, the permissible working stress is the yield strength divided by the factor of safety, ensuring that the vessel operates well within its safe limits.

Wall Thickness Determination

Determining the required wall thickness of a pressure vessel involves rearranging the stress formula to solve for the thickness. This process incorporates the internal pressure, the vessel's radius, and the allowable stress (which is the material yield strength divided by the factor of safety). Calculating the wall thickness ensures that the vessel can withstand the internal pressure without experiencing yielding.

Maximum Permissible Internal Pressure

Calculating the maximum permissible internal pressure is essential when the wall thickness is predetermined. Using the formula for hoop stress, one can rearrange to solve for the internal pressure, considering the material's yield strength and the used factor of safety. This calculation ensures that the vessel operates within safe limits and confirms that the chosen wall thickness is adequate for the intended pressure conditions.

Transcript

00:01 Hello and welcome to this video solution of numerate.

00:04 So here it's given that a large spherical tank contains gas at a pressure of 3 .5 mega pascal right.

00:12 So i can write pressure p as 3 .5 mega pascal right and the gas is having the container which is a tank whose diameter is d let's say that is equal to 20 meters right and this is constructed of high strength steel having tensile or yield stress sigma y we can write that is equal to 550 mega pascal right.

00:47 Now you have to determine the thickness of the wall of the tank if a factor of safety n i'm taking of 3 .5 is provided with respect to the yielding right.

01:00 So the net force will be equal to the factor of safety or the net stress that is that needs to withstand will be equal to sigma sorry n which is the factor of safety times sigma y right and this is equal to 3 .5 times of 550 mega pascal right and this is equal to 195 mega pascal right.

01:37 Now what you can do is you can simply apply the expression of a maximum stress sigma that is equal to pd by 4t right.

01:52 So from here you can easily calculate the thickness t or t will be equal to pd by 4 sigma right the pressure is here 550 mega pascal times the diameter is we have got 20 meters by 4 times the sigma is 1925 and it will be in meters itself right.

02:17 So let's see how much we're getting here so this will be around 1 .43 meters in right.

02:39 So this is the thickness that is required sorry i just made a small error because here the pressure is not 550 it's 3 .5 that's why i'm feeling that why it's not coming in millimeters right...

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