The Clausius inequality expresses which of the following laws? i. Law of Conservation of Mass ii. Law of Conservation of Energy iii. First Law of Thermodynamics iv. Second Law of Thermodynamics

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Answer 1

Answer:

Answer:

(iv) second law of thermodynamics

Explanation:

The Clausius inequality expresses the second law of thermodynamics it applies to the real engine cycle.It is defined as the cycle integral of change in entropy of a reversible system is zero. It is nothing but mathematical form of second law of thermodynamics . It also states that for irreversible process the cyclic integral of change in entropy is less than zero

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Air is compressed in a piston-cylinder device. List three examples of irreversibilities that could occur

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Answer:

While air is compressed in a piston cylinder there are following types of irreversibilities

1.Due to finite temperature difference heat transfer take place between cylinder and surrounding.

2.Due friction force between cylinder and piston .

3.Compression process is so fast due to this ,it leads in the irreversibility of system.

When looking at a chain of processes with a low yield (high defective rate), what is a good place to start investigating the source of variation?

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The most upstream process with issues would be a good location to start exploring the cause of the variance.

Chain processes

A manufacturing technique would be a specific procedure for generating a commodity.

Throughout manufacturing, a six sigma process has been utilized just to generate a product throughout which 99.99966 percent among all possibilities to produce certain aspects of a part seem to be likely toward being defect-free.

Thus the response above is correct.

Find out more information about chain processes here:

brainly.com/question/25646504

Answer: The furthest upstream process that has problems.

A process in manufacturing is a particular method used for producing a product.

A six sigma process is used in processing to produce a product that is 99.99966% of all opportunities to produce some feature of a part are statistically expected to be free of defects.

According to the rules of the six sigma process, when there's a defect, the best thing to do is investigate the furthest upstream process that has problems.

Use Euler’s Method: ????????????????=???? ????????????????=−????????−????3+????cos(????) ????(0)=1.0 ????(0)=1.0 ????=0.4 ????=20.0 ℎ=0.01 ????=10000 ▪ Write the data (y1, y2) to a file named "LASTNAME_Prob1.dat" Example: If your name is John Doe – file name would be "DOE_Prob1.dat" ▪ Plot the result with lines using GNUPLOT (Hint: see lecture 08) ▪ Submit full code (copy and paste). Plot must be on a separate page. ▪ Run the code again and plot the result for: ????=0.1 ????=11.0

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Answer:

Too many question marks

Explanation:

Shear modulus is analogous to what material property that is determined in tensile testing? (a)- Percent reduction of area (b) Yield strength (c)- Elastic modulus (d)- Poisson's ratio

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Answer:

(c)- Elastic modulus

Explanation:

We know that in tensile test we measure the properties of the material like yield strength,ultimate tensile strength ,Poisson ratio.

In tensile test

σ = ε E

Where σ is the stress

ε is the strain.

E is the elastic modulus.

Now for shear tress

τ = Φ G

Where τ the shear stress

Φ is the shear strain.

G is the shear modulus.

So we can say that Shear modulus is analogous to Elastic modulus.

Considering only (110), (1 1 0), (101), and (10 1 ) as the possible slip planes, calculate the stress at which a BCC single crystal will yield if the critical resolved shear stress is 50 MPa and the load is applied in the [100] direction.

Answers

Solution :

i. Slip plane (1 1 0)

Slip direction -- [1 1 1]

Applied stress direction = ( 1 0 0 ]

τ = 50 MPa ( Here slip direction must be perpendicular to slip plane)

τ = σ cos Φ cos λ

$\cos \phi = ((1,0,0) \cdot (1,1,0))/(1 * \sqrt2)$

$=(1)/(\sqrt2 )$

$\cos \lambda = ((1,0,0) \cdot (1,-1,1))/(1 * \sqrt3)$

$=(1)/(\sqrt3 )$

τ = σ cos Φ cos λ

∴ $50= \sigma * (1)/(\sqrt2) * (1)/(\sqrt3)$

σ = 122.47 MPa

ii. Slip plane --- (1 1 0)

Slip direction -- [1 1 1]

$\cos \phi = ((1, 0, 0) \cdot (1, -1, 0))/(1 * \sqrt2) =(1)/(\sqrt2)$

$\cos \lambda = ((1, 0, 0) \cdot (1, 1, -1))/(1 * \sqrt3) =(1)/(\sqrt3)$

τ = σ cos Φ cos λ

∴ $50= \sigma * (1)/(\sqrt2) * (1)/(\sqrt3)$

σ = 122.47 MPa

iii. Slip plane --- (1 0 1)

Slip direction --- [1 1 1]

$\cos \phi = ((1, 0, 0) \cdot (1, 0, 1))/(1 * \sqrt2) =(1)/(\sqrt2)$

$\cos \lambda = ((1, 0, 0) \cdot (1, 1, -1))/(1 * \sqrt3) =(1)/(\sqrt3)$

τ = σ cos Φ cos λ

∴ $50= \sigma * (1)/(\sqrt2) * (1)/(\sqrt3)$

σ = 122.47 MPa

iv. Slip plane -- (1 0 1)

Slip direction ---- [1 1 1]

$\cos \phi = ((1, 0, 0) \cdot (1, 0, -1))/(1 * \sqrt2)=(1)/(\sqrt2)$

$\cos \lambda = ((1, 0, 0) \cdot (1, -1, 1))/(1 * \sqrt3) =(1)/(\sqrt3)$

τ = σ cos Φ cos λ

∴ $50= \sigma * (1)/(\sqrt2) * (1)/(\sqrt3)$

σ = 122.47 MPa

∴ (1, 0, -1). (1, -1, 1) = 1 + 0 - 1 = 0

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