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PolarNik [594]
3 years ago
9

A thin-walled cylindrical pressure vessel is subjected to an internal gauge pressure, p=75 psip=75 psi. It had a wall thickness

of 0.25 inches and an inner diameter of 8 inches. Use Mohr’s Circle to determine the absolute maximum shear stress in the pressure vessel when it is subjected to this pressure.

Physics
1 answer:
Mekhanik [1.2K]3 years ago
6 0

To solve this problem we must apply the concept related to the longitudinal effort and the effort of the hoop. The effort of the hoop is given as

\sigma_h = \frac{Pd}{2t}

Here,

P = Pressure

d = Diameter

t = Thickness

At the same time the longitudinal stress is given as,

\sigma_l = \frac{Pd}{4t}

The letters have the same meaning as before.

Then he hoop stress would be,

\sigma_h = \frac{Pd}{2t}

\sigma_h = \frac{75 \times 8}{2\times 0.25}

\sigma_h = 1200psi

And the longitudinal stress would be

\sigma_l = \frac{Pd}{4t}

\sigma_l = \frac{75\times 8}{4\times 0.25}

\sigma_l = 600Psi

The Mohr's circle is attached in a image to find the maximum shear stress, which is given as

\tau_{max} = \frac{\sigma_h}{2}

\tau_{max} = \frac{1200}{2}

\tau_{max} = 600Psi

Therefore the maximum shear stress in the pressure vessel when it is subjected to this pressure is 600Psi

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Answer:

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Explanation:

The expression to use here is the following:

ΔS = Q/T

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Now, in order to do this, we need to gather the data. We know that the temperature in part a) is above the freezing temperature of water, which is 0 ° C or 273 K. and the mass of the ice cube is 14.6 g.

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Solving for Q first we have:

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b) In this part, we follow the same procedure than in part a) but using the water heat of boiling (Lv = 2,256,000 J/kg), the temperature of boiling which is 100 °C (or 373 K) and the mass of 5.51 g (0.00551 kg)

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Now the entropy change:

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8 0
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What is the relationship between the angle of an incline and the acceleration of an object moving down the incline? How would yo
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Answer:

See Explanation

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As the angle of an incline increases, so does the acceleration of the body moving down the incline increases, resolving the force acting on an inclined object

Parallel force = mgsin, perpendicular = mgcosΘ

With th weigh component 'mg' of the parallel force accounting for the acceleration of the body down the incline.

mgsinΘ = ma

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895522 times faster.

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From the question given above, the following data were obtained:

Speed of sound in air (v) = 335 m/s

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How many times faster =.?

To obtain how many times faster light travels in air than sound, do the following

c : v => 3×10⁸ : 335

c/v = 3×10⁸ / 335

c/v = 895522

Cross multiply

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Answer:

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question 1:

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Question 2

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