The material properties of a machine part are Sut 5 85 kpsi, f 5 0.86, and a fully corrected endurance limit of Se 5 45 kpsi. The part is to be cycled at sa 5 35 kpsi and sm 5 30 kpsi for 12 (103 ) cycles. Using the Gerber criterion, estimate the new endurance limit after cycling

Answers

Answer 1

Answer:

according to Miners method the new endurance limit S[tex]_e[/tex]" is 45 kpsi

Explanation:

Given the data in the question;

the amplitude stress σₐ = 35 kpsi

the mean stress for 12(10³) cycles σ[tex]_m[/tex] = 30 kpsi

the ultimate tensile stress S[tex]_{ut[/tex] = 85 kpsi

endurance limit of S[tex]_e[/tex] = 45 kpsi

so, we calculate the reversing stress for Gerber criteria

σ[tex]_{rev[/tex] = σₐ / [ 1 - ( σ[tex]_m[/tex]/S[tex]_{ut[/tex])²

we substitute

σ[tex]_{rev[/tex] = 35 / [ 1 - ( 30/85)²]

σ[tex]_{rev[/tex] = 35 / [ 1 - 0.124567]

σ[tex]_{rev[/tex] = 35 / 0.875433

σ[tex]_{rev[/tex] = 39.98 kpsi

Using Miners Method;

the reversing stress ( 39.98 kpsi ) is less than endurance limit ( 45 kpsi )

σ[tex]_{rev[/tex]  < S[tex]_e[/tex]

According to Miners method, if reversing stress is less than endurance limit, then the endurance limit has not reduced and new endurance limit remains the same.

Hence, according to Miners method the new endurance limit S[tex]_e[/tex]" is 45 kpsi


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Answer:

A

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Consider a layer of insulation installed around a steam pipe. The radius of the pipe is R1 and the thickness of the insulation is R3-R1.The stream is maintained at a temperature TW and the air surrounding the insulation is at a temperature To and flows cross wise over the pipe. The air is flowed over the steam pipe at high enough velocity so a thermal boundary layer develops over the surface described by a heat transfer coefficient h.

Required:
Beginning with the heat equation, calculate the total quantity of heat being conducted per unit time (heat flow) through the insulation, Q (units: energy/time).

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2.3
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Answers

Answer:

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4.In a hydroelectric power plant, 100 m3/s of water flows from an elevation of 120 m to a turbine, where electric power is generated. The electric power you get out from the 80% efficiency turbine is known to be 50 MW, what is the rate of irreversible loss in the piping system (in MW unit)

Answers

Answer:

The rate of irreversible loss will be "55.22 MW".

Explanation:

The given values are:

Elevation,

h = 120 m

Flow of water,

Q = 100 m³/s

Efficiency,

= 80%

i.e,

= 0.8

Efficiency turbine,

= 50 MW

Now,

Without any loss,

The power generated by turbine will be:

⇒ [tex]P=\delta gQh[/tex]

On substituting the values, we get

⇒     [tex]=1000\times 9.8\times 100\times 120[/tex]

⇒     [tex]=117.72 \ MW[/tex]

Power generated in actual will be:

= [tex]\frac{50}{0.8}[/tex]

= [tex]62.5 \ MW[/tex]

Hence,

Throughout the piping system,

The rate of irreversible loss is:

= [tex]Power \ generated \ by \ turbine-Power \ generated \ in \ actual[/tex]

= [tex]117.72-62.5[/tex]

= [tex]55.22 \ MW[/tex]

Please I need help with this

Photosynthesis energy is stored in the cells of green plants through
process called ______?

Answers

light energy is captured and used to convert water, carbon dioxide, and minerals into oxygen and energy-rich organic compounds.
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