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A pressurized spherical tank is made of 0.5 -in.-thick steel. If it is subjected to an internal pressure of \(p=200 \mathrm{psi},\) determine its outer radius if the maximum normal stress is not to exceed 15 ksi.

Short Answer

Expert verified
The outer radius of the pressurized spherical tank should be approximately 10.20 inches.

Step by step solution

01

Understand the Given Parameters

The exercise provides the thickness of the steel as 0.5 inch, the internal pressure applied to the tank as \(p=200 \mathrm{psi}=\frac{200}{0.0069} \mathrm{ksi}\) (since 1 ksi = 1000 psi), and the maximum normal stress as 15 ksi.
02

Apply Stress Formula for Thin Spherical Shells

The formula for normal stress \( \sigma \) in thin-walled spherical pressure vessels is given by: \(\sigma = \frac{p \cdot r}{t}\), where \(p\) is the internal pressure, \(r\) is the radius of the shell, and \(t\) is the thickness of the shell.
03

Solve for the Outer Radius

Substituting the given values into the stress formula, we have \(15 = \frac{(200/0.0069) x r}{0.5}\). Solving for \(r\), we find: \(r = \frac{15 x 0.5 x 0.0069}{200} = \frac{0.05175}{200} km\).
04

Convert to Appropriate Units

The required radius is currently in kilometers, which isn't appropriate for the context of the problem. Let's convert it to a more suitable unit, inches, using the conversion factor \(1 km = 39370.1 inches\). So, \(r = \frac{0.05175}{200} x 39370.1 = 10.1989 inches \approx 10.20 inches\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Thin-Walled Pressure Vessels
Understanding the concept of thin-walled pressure vessels is crucial for mechanical engineering, especially when dealing with the design of tanks and containers that hold gases or liquids at a pressure different from the ambient pressure. These vessels are characterized by having a wall thickness that is small relative to their radius.

For the thin-walled assumption to be valid, the wall thickness should generally be less than one-tenth of the vessel's radius. This simplification allows us to assume that the stress distribution across the wall's thickness is uniform. The significance here lies in the analysis and design of pressure vessels, as it influences safety, efficiency, and material economy. When vessels are thin-walled, two significant stresses need to be considered: the circumferential or hoop stress, and the longitudinal stress. In the case of spherical pressure vessels, such as the tank in our problem, the stresses are uniformly distributed and only the hoop stress needs to be considered.
Spherical Shell Stress Formula
The spherical shell stress formula is a fundamental piece of knowledge in mechanical engineering. It's derived from the principles of stress analysis on thin-walled pressure vessels. For a spherical shell, the hoop stress, which is the stress exerted circumferentially as a result of the internal pressure, is given as:\[\sigma = \frac{p \cdot r}{t}\]where \(\sigma\) represents the normal stress, \(p\) is the internal pressure, \(r\) is the internal radius of the shell, and \(t\) is the wall thickness of the vessel.

In practical applications, this formula allows engineers to calculate how much stress a vessel can withstand before yielding. The design must ensure that the material used can support the calculated stress. This is done by setting the maximum normal stress below the yield strength of the material to avoid failure under the given pressure.
Mechanical Engineering Principles
Mechanical engineering principles cover a broad spectrum of concepts ranging from the design, analysis, manufacturing, and maintenance of mechanical systems. These principles are founded on laws of physics and material science. As demonstrated in the step-by-step solution for the pressure vessel problem, applying these principles involves understanding material properties such as yield strength, which signifies the maximum stress that can be applied to a material without causing plastic deformation.

Moreover, the principles guide the engineer to choose appropriate units for calculations, address real-world constraints, and consider safety factors. In our exercise, converting units is exemplified by changing the computed radius from kilometers to inches for practical use. This level of attention to detail is critical and reflects the rigorous nature of mechanical engineering design processes.

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Most popular questions from this chapter

The coiled spring is subjected to a force \(P\). If we assume the shear stress caused by the shear force at any vertical section of the coil wire to be uniform, show that the maximum shear stress in the coil is \(\tau_{\max }=P / A+P R r / J\) where \(J\) is the polar moment of inertia of the coil wire and \(A\) is its cross-sectional area.

The inner ring \(A\) has an inner radius \(r_{1}\) and outer radius \(r_{2}\). The outer ring \(B\) has an inner radius \(r_{3}\) and an outer radius \(r_{4},\) and \(r_{2}>r_{3} .\) If the outer ring is heated and then fitted over the inner ring, determine the pressure between the two rings when ring \(B\) reaches the temperature of the inner ring. The material has a modulus of elasticity of \(E\) and a coefficient of thermal expansion of \(\alpha\).

A spherical gas tank has an inner radius of \(r=1.5 \mathrm{m}\) If it is subjected to an internal pressure of \(p=300 \mathrm{kPa}\) determine its required thickness if the maximum normal stress is not to exceed \(12\) \(\mathrm{MPa}\).

The steel bracket is used to connect the ends of two cables If the applied force \(P=1.50\) kip, determine the maximum normal stress in the bracket. Assume the bracket is a rod having a diameter of 1.5 in.

The steel bracket is used to connect the ends of two cables. If the allowable normal stress for the steel is \(\sigma_{\text {allow }}=30 \mathrm{ksi}\), determine the largest tensile force \(P\) that can be applied to the cables. Assume the bracket is a rod having a diameter of 1.5 in.

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