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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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Answer:

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A cyclic tensile load ranging from 0 kN to 55 kN force is applied along the length of a 100 mm long bar with a 15 mm x 15 mm squ
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Answer:

square cross section. The bar is made of a 7075-T6 aluminum alloy which has a yield strength of 500 MPa, a tensile strength of 575 MPa, and a fracture toughness of 27.5 MPaâm.

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3 years ago
Five kilograms of air at 427°C and 600 kPa are contained in a piston–cylinder device. The air expands adiabatically until the pr
son4ous [18]

Answer:

The entropy change of the air is 0.240kJ/kgK

Explanation:

T_{1} =427+273K,T_{1} =700K\\P_{1} =600kPa\\P_{2} =100kPa

T_{2}  is unknown

we can apply the following expression to find T_{2}

-w_{out} =mc_{v} (T_{2} -T_{1} )

T_{2} =T_{1} -\frac{w_{out } }{mc_{v} }

now substitute

T_{2} =700K-\frac{600kJ}{5kg*0.718kJ/kgK} \\T_{2}=533K

To find entropy change of the air we can apply the ideal gas relationship

Δs_{air}=c_{p} ln\frac{T_{2} }{T_{1} } -Rln\frac{P_{2} }{P_{1} }

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Δs_{air} =0.240kJ/kgK

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A hair dryer is basically a duct in which a few layers of electric resistors are placed. A small fan pulls the air in and forces
nika2105 [10]

Answer:

a) volume flow rate of air at the inlet is 0.0471 m³/s

b) the velocity of the air at the exit is  8.517 m/s

Explanation:

Given that;

The electrical power Input W_elec = -1400 W = -1.4 kW

Inlet temperature of air T_in = 22°C

Inlet pressure of air p_in = 100 kPa

Exit temperature T_out = 47°C

Exit area of the dyer is A_out = 60 cm²= 0.006 m²

cp = 1.007 kJ/kg·K

R = 0.287 kPa·m3/kg·K

Using mass balance

m_in = m_out = m_air

W _elec = m_air ( h_in - h_out)

we know that h = CpT

so

W _elec = m_air.Cp ( T_in - T_out)

we substitute

-1.4 = m_air.1.007 ( 22 - 47 )

-1.4 =  - m_air.25.175

m_air = -1.4 / - 25.175

m_ air = 0.0556 kg/s

a) volume flow rate of air at the inlet

we know that

m_air = P_in × V_in

now from the ideal gas equation

P_in = p_in / RT_in

we substitute our values

= (100×10³) / ((0.287×10³)(22+273))

= 100000 / 84665

P_in = 1.18 kg/m³

therefore inlet volume flowrate will be;

V_in = m_air / P_in

= 0.0556 / 1.18

= 0.0471 m³/s

the volume flow rate of air at the inlet is 0.0471 m³/s

b) velocity of the air at the exit

the mass flow rate remains unchanged across the duct

m_ air = P_in.A_in.V_in = P_out.A_out.V_out

still from the ideal gas equation

P_out = p_out/ RT_out   ( assume p_in = p_out)

P_out = (100×10³) / ((0.287×10³)(47+273))

P_out  = 1.088 kg/m³

so the exit velocity will be;

V_out = m_air / P_out.A_out

we substitute our values

V_out = 0.0556 / ( 1.088 × 0.006)

= 0.0556 / 0.006528

= 8.517 m/s

 Therefore the velocity of the air at the exit is  8.517 m/s

6 0
3 years ago
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