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laiz [17]
3 years ago
14

Rank the magnitudes of the diffusion coefficients from greatest to least for the following systems:(a) Cr in Fe at 600°C; (b) C

in Fe at 900°C; (c) Cr in Fe at 900°C. Now justify this ranking. You MUST write a short paragraph to explain your ranking
Engineering
2 answers:
lisabon 2012 [21]3 years ago
8 0

Answer:

The rank of the magnitude of diffusion coefficients from greatest to least is as follows:

1) "C" in "Fe" at 900°C

2) "Cr" in "Fe" at 900°C

3) "Cr" in "Fe" at 600°C

Explanation:

"C" in "Fe" is an interstitial impurity while "Cr" in "Fe" is a substitutional impurity".

With that being said, interstitial diffusion occurs in "C" in "Fe" systems while substitutional diffusion occurs in "Cr" in "Fe" systems.

Also, Interstitial diffusion is much faster than substitutional diffusion, hence the order we got in the answer.

Also, with increasing temperature, magnitude of diffusion coefficient increases due to the relation;

D =D° Exp (-Qd/RT)

Where;

D° = temperature independent pre-exponential

Qd = the activation energy for diffusion

R = universal gas constant

T = absolute temperature

jek_recluse [69]3 years ago
3 0

Answer:

Explanation:

The rank of the magnitude of the diffusion coefficient from greatest to least is as follows:

C in Fe at 900°C > Cr in Fe at 900°C > Cr in Fe at 600°C

Reason

C in Fe is an interstitial impurity while Cr in Fe is a substutional impurity.Therefore interstitial impurity occurs in C in Fe systems,while substutitional diffusion occurs in Cr in Fe system.Interstitial is much faster than substitutional diffusion hence the order

Also with increasing temperature magnitude of diffusion coefficient increases,due to the relation.

     D = D₀exp(-Qd/RT)

Where D₀=Temperature independent per exponential

           Qd= The activation energy for diffusion

             R= Universal gas constant

              T=absolute temperature

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A steel bar 100 mm long and having a square cross section 20 mm x 20 mm is pulled in
Ierofanga [76]

Answer:

222.5 Gpa

Explanation:

From definition of engineering stress, \sigma=\frac {F}{A}

where F is applied force and A is original area

Also, engineering strain, \epsilon=\frac {\triangle l}{l} where l is original area and \triangle l is elongation

We also know that Hooke's law states that E=\frac {\sigma}{\epsilon}=\frac {\frac {F}{A}}{\frac {\triangle l}{l}}=\frac {Fl}{A\triangle l}

Since A=20 mm* 20 mm= 0.02 m*0.02 m

F= 89000 N

l= 100 mm= 0.1 m

\triangle l= 0.1 mm= 0.1\times 10^{-3} m

By substitution we obtain

E=\frac {89000\times 0.1}{0.02^{2}\times 0.1\times 10^{-3}}=2.225\times 10^{11}= 225.5 Gpa

5 0
3 years ago
Which one of the following activities is not an example of incident coordination
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Directing, ordering, or controlling
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3 years ago
Consider a plane composite wall that is composed of two materials of thermal conductivities kA = 0.1 W/m*K and kB = 0.04 W/m*K a
nadya68 [22]

Answer:

q=39.15 W/m²

Explanation:

We know that

Thermal resistance due to conductivity given as

R=L/KA

Thermal resistance due to heat transfer coefficient given as

R=1/hA

Total thermal resistance

R_{th}=\dfrac{L_A}{AK_A}+\dfrac{L_B}{AK_B}+\dfrac{1}{Ah_1}+\dfrac{1}{Ah_2}+\dfrac{1}{Ah_3}

Now by putting the values

R_{th}=\dfrac{0.01}{0.1A}+\dfrac{0.02}{0.04A}+\dfrac{1}{10A}+\dfrac{1}{20A}+\dfrac{1}{0.3A}

R_{th}=4.083/A\ K/W

We know that

Q=ΔT/R

Q=\dfrac{\Delta T}{R_{th}}

Q=A\times \dfrac{200-40}{4.086}

So heat transfer per unit volume is 39.15 W/m²

q=39.15 W/m²

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3 years ago
Consider a very long, slender rod. One end of the rod is attached to a base surface maintained at Tb, while the surface of the r
9966 [12]

Answer:

(a) Calculate the rod base temperature (°C). = 299.86°C

(b) Determine the rod length (mm) for the case where the ratio of the heat transfer from a finite length fin to the heat transfer from a very long fin under the same conditions is 99 percent.  = 0.4325m

Explanation:

see attached file below

3 0
3 years ago
Water flows through a horizontal plastic pipe with a diameter of 0.15 m at a velocity of 15 cm/s. Determine the pressure drop pe
Sonja [21]

Answer:0.1898 Pa/m

Explanation:

Given data

Diameter of Pipe\left ( D\right )=0.15m

Velocity of water in pipe\left ( V\right )=15cm/s

We know viscosity of water is\left (\mu\right )=8.90\times10^{-4}pa-s

Pressure drop is given by hagen poiseuille equation

\Delta P=\frac{128\mu \L Q}{\pi D^4}

We have asked pressure Drop per unit length i.e.

\frac{\Delta P}{L} =\frac{128\mu \ Q}{\pi D^4}

Substituting Values

\frac{\Delta P}{L}=\frac{128\times8.90\times10^{-4}\times\pi \times\left ( 0.15^{3}\right )}{\pi\times 4 \times\left ( 0.15^{2}\right )}

\frac{\Delta P}{L}=0.1898 Pa/m

4 0
3 years ago
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