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3 years ago

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A bar of 75 mm diameter is reduced to 73mm by a cutting tool while cutting orthogonally. If the mean length of the cut chip is 7

3.5 mm, find the cutting ratio. If the rake angle is 15 deg, what is the shear angle?

1 answer:

barxatty [35]3 years ago

4 0

Answer:

r=0.31

Ф=18.03°

Explanation:

Given that

Diameter of bar before cutting = 75 mm

Diameter of bar after cutting = 73 mm

Mean diameter of bar d= (75+73)/2=74 mm

Mean length of uncut chip = πd

Mean length of uncut chip = π x 74 =232.45 mm

So cutting ratio r

$Cutting\ ratio=\dfrac{Mean\ length\ of cut\ chip}{Mean\ length\ of uncut\ chip}$

$r=\dfrac{73.5}{232.45}$

r=0.31

So the cutting ratio is 0.31.

As we know that shear angle given as

$tan\phi =\dfrac{rcos\alpha }{1-rsin\alpha }$

Now by putting the values

$tan\phi =\dfrac{rcos\alpha }{1-rsin\alpha }$

$tan\phi =\dfrac{0.31cos15 }{1-0.31sin15 }$ \

Ф=18.03°

So the shear angle is 18.03°.

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The diffusion coefficients for species A in metal B are given at two temperatures:

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Answer:

a) 149 kJ/mol, b) 6.11*10^-11 m^2/s ,c) 2.76*10^-16 m^2/s

Explanation:

Diffusion is governed by Arrhenius equation

$D = D_0e^{\frac{-Q_d}{RT} }$

I will be using R in the equation instead of k_b as the problem asks for molar activation energy

I will be using

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°C + 273 = K

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Since we got 2 difusion coefficients at 2 temperatures alredy, we can simply turn these into 2 linear equations to solve for a) and b) simply by taking logarithm

So:

$ln(6.69*10^{-17})=ln(D_0) -\frac{Q_d}{R*(1030+273)}$

and

$ln(6.56*10^{-16}) = ln(D_0) -\frac{Q_d}{R*(1290+273)}$

You might notice that these equations have the form of

$d=y-ax$

You can solve this equation system easily using calculator, and you will eventually get

$D_0 =6.11*10^{-11}\ m^2/s\\ Q_d=1.49 *10^3\ J/mol$

After you got those 2 parameters, the rest is easy, you can just plug them all including the given temperature of 1180°C into the Arrhenius equation

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A 500-km, 500-kV, 60-Hz, uncompensated three-phase line has a positivesequence series impedance. z = 5 0.03 1 + j 0.35 V/km and

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Answer:

A) 282.34 - j 12.08 Ω

B) 0.0266 + j 0.621 / unit

C)

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B = 164.6 < 85.42°Ω

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Z ( impedance ) = 0.03 i + j 0.35 Ω/km

positive sequence shunt admittance ( Y ) = j4.4*10^-6 S/km

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Zc = $\sqrt{\frac{z}{y} }$ = $\sqrt{\frac{0.03 i + j 0.35}{j4.4*10^-6 } }$

= $\sqrt{79837.128< 4.899^o}$ = 282.6 < -2.45°

hence Zc = 282.34 - j 12.08 Ω

B) Calculate gl

gl = $\sqrt{zy} * d$

d = 500

z = 0.03 i + j 0.35

y = j4.4*10^-6 S/km

gl = $\sqrt{0.03 i + j 0.35* j4.4*10^-6} * 500$

= $\sqrt{1.5456*10^{-6} < 175.1^0} * 500$

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gl = 0.0266 + j 0.621 / unit

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A = cos h (gl) . per unit = 0.812 < 1.09° per unit ( as calculated )

B = Zc sin h (gl) Ω = 164.6 < 85.42°Ω ( as calculated )

C = 1/Zc sin h (gl) s = 2.061 * 10^-3 < 90.32° s ( as calculated )

D = cos h (gl) . per unit = 0.812 < 1.09° per unit ( as calculated )

where : cos h (gl) = $\frac{e^{gl} + e^{-gl} }{2}$

sin h (gl) = $\frac{e^{gl}-e^{-gl} }{2}$

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For a steel alloy it has been determined that a carburizing heat treatment of 11.3 h duration will raise the carbon concentratio

diamong [38]

This question is incomplete, the complete question is;

For a steel alloy it has been determined that a carburizing heat treatment of 11.3 h duration at Temperature T1 will raise the carbon concentration to 0.44 wt% at a point 1.8 mm from the surface. A separate experiment is performed at T2 that doubles the diffusion coefficient for carbon in steel.

Estimate the time necessary to achieve the same concentration at a 4.9 mm position for an identical steel and at the same carburizing temperature T2.

Answer:

the required time to achieve the same concentration at a 4.9 is 83.733 hrs

Explanation:

Given the data in the question;

treatment time t₁ = 11.3 hours

Carbon concentration = 0.444 wt%

thickness at surface x₁ = 1.8 mm = 0.0018 m

thickness at identical steel x₂ = 4.9 mm = 0.0049 m

Now, Using Fick's second law inform of diffusion

$x^2$ / Dt = constant

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then

$x^2$ / t = constant

$x^2_1$ / t₁ = $x^2_2$ / t₂

$x^2_1$ t₂ = t₁ $x^2_2$

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t₂ = 83.733 hrs

Therefore, the required time to achieve the same concentration at a 4.9 is 83.733 hrs

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