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3 years ago

For a brass alloy, the following engineering stresses produce the corresponding plastic engineering strains prior to necking:

Engineering

1 answer:

tatuchka [14]3 years ago

4 0

Answer:

<h2><em> </em>The engineering stress necessary to produce an engineering strain of 0.28 is 339 Mpa</h2>

Explanation:

<em>On the basis of this information, we are going to compute the engineering stress necessary to produce an engineering strain of 0.28.</em>

<h2>Engineering Engineering </h2><h2> Stress (MPa) Strain</h2><h2>315 0.105 </h2><h2>340 0.220 </h2>

For clarity and you will find the solving in the attached file for your reference.

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Water vapor at 1.0 MPa, 300°C enters a turbine operating at steady state and expands to 15 kPa. The work developed by the turbin

Andre45 [30]

Answer:

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b) rate of entropy generation = .341 kj/(kg.k)

Please kindly see explaination and attachment.

Explanation:

a) isentropic efficiency = 84.905%

b) rate of entropy generation = .341 kj/(kg.k)

The Isentropic efficiency of a turbine is a comparison of the actual power output with the Isentropic case.

Entropy can be defined as the thermodynamic quantity representing the unavailability of a system's thermal energy for conversion into mechanical work, often interpreted as the degree of disorder or randomness in the system.

Please refer to attachment for step by step solution of the question.

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3 years ago

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denis-greek [22]

Answer:

275 MPa

Explanation:

Regardless of what it is holding, the stiffness of a bolt depends on its own material properties and geometry.

The stiffness is:

$k = E * \frac{A}{l}$

I assume this one is made of steel, because regular bolts are steel.

The Young's modulus for steel is E = 210 GPa

The longitude is given. (But note that in a real application you have to consider the length up to the nut.)

The section is (using the nominal diameter of 10 mm)

$A = \frac{\pi * d^2}{4} = \frac{\pi * 0.01^2}{4} = 7.85e-5 m^2$

Then:

$k = 2.1e11 * \frac{7.85e-5}{0.06} = 275e6 Pa = 275 MPa$

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3 years ago

In a voltage-divider biased pnp transistor, there is no base current, but the base voltage is approximately correct. The most li

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Answer:

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Explanation:

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3 years ago

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Answer:

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3 years ago