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Andreas93 [3]
4 years ago
15

A large spherical tank contains gas at a pressure of 400 psi. The tank is constructed of high-strength steel having a yield stre

ss in tension of 80 ksi. For volume requirements, the inner diameter of the tank is 45-ft. Determine the required thickness (to the nearest ¼ inch) of the wall of the tank if a factor of safety of 3.0 with respect to yielding is required.
Engineering
1 answer:
jok3333 [9.3K]4 years ago
5 0

Answer:

t=2.025 inches

Explanation:

Given that

P = 400 Psi

Yield stress ,σ = 80 ksi

Diameter ,d= 45 ft

We know that

1 ft = 12 inches

d= 540 inches

Factor of safety ,K= 3

The required thickness given as

\dfrac {Pd}{4t}=\dfrac{\sigma}{K}

t=thickness

\dfrac {PdK}{4\sigma}=t

\dfrac {400\times 540\times 3}{4\times 80\times 1000}=t

t=2.025 inches

Therefore thickness will be 2.025 inches.

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Answer:

78 MPa

Explanation:

Given that the critical resolved shear stress for a metal is 39 MPa, the maximum possible yield strength for a single crystal of this metal is twice the critical resolved shear stress for the metal. The maximum yield yield strength for a single crystal of this metal that is pulled in tension (\sigma_y) is given as:

\sigma_y=2*critical\ resolved\ shear\ stress(\tau_{css})\\\\\sigma_y=2*\tau_{css}\\\\\sigma_y=2*39\\\\\sigma_y=78\ MPa

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Answer:

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Explanation:

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Answer:

The rate of entropy change of the air is -0.10067kW/K

Explanation:

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