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insens350 [35]
3 years ago
13

For a steel alloy it has been determined that a carburizing heat treatment of 14 h duration at 809°C will raise the carbon conce

ntration to 0.54 wt% at a point 3.6 mm from the surface. Estimate the time necessary to achieve the same concentration at a 6.0 mm position for an identical steel and at a carburizing temperature of 1100°C. Assume that D0 is 2.5 × 10-5 m2/s and Qd is 120 kJ/mol.
Engineering
1 answer:
Yakvenalex [24]3 years ago
3 0

Answer:

t_2 = 27.7 hr

Explanation:

Given data:

carbon concentration  =  0.54%

from the relation given below calculate the time required to achieve concentration at 6.00 mm from surface

\frac{x^2}{Dt} = constant

D considered constant

\frac{x^2}{t} =  constant

here, x POSITION FROM SURFACE, t is time required to achieve concentration

\frac{x_1^2}{t_1} = \frac{x_2^2}{t_2}

\frac{3.6^2}{14} = \frac{6^2}{t_2}

t_2 = 27.7 hr

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Air is compressed by a 40-kW compressor from P1 to P2. The air temperature is maintained constant at 25°C during this process a
AlexFokin [52]

Answer:

the rate of entropy change of the air is -0.1342 kW/K

the assumptions made in solving this problem

- Air is an ideal gas.

- the process is isothermal ( internally reversible process ). the change in internal energy is 0.

- It is a steady flow process

- Potential and Kinetic energy changes are negligible.

Explanation:

Given the data in the question;

From the first law of thermodynamics;

dQ = dU + dW ------ let this be equation 1

where dQ is the heat transfer, dU is internal energy and dW is the work done.

from the question, the process is isothermal ( internally reversible process )

Thus, the change in internal energy is 0

dU = 0

given that; Air is compressed by a 40-kW compressor from P1 to P2

since it is compressed, dW = -40 kW

we substitute into equation 1

dQ = 0 + ( -40 kW )

dQ = -40 kW

Now, change in entropy of air is;

ΔS_{air = dQ / T

given that T = 25 °C = ( 25 + 273.15 ) K = 298.15 K

so we substitute

ΔS_{air =  -40 kW / 298.15 K

ΔS_{air =  -0.13416 ≈ -0.1342 kW/K

Therefore, the rate of entropy change of the air is -0.1342 kW/K

the assumptions made in solving this problem

- Air is an ideal gas.

- the process is isothermal ( internally reversible process ). the change in internal energy is 0.

- It is a steady flow process

- Potential and Kinetic energy changes are negligible.

7 0
3 years ago
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