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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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Flow and Pressure Drop of Gases in Packed Bed. Air at 394.3 K flows through a packed bed of cylinders having a diameter of 0.012
devlian [24]

The pressure drop of air in the bed is  14.5 kPa.

<u>Explanation:</u>

To calculate Re:

R e=\frac{1}{1-\varepsilon} \frac{\rho q d_{p}}{\mu}

From the tables air property

\mu_{394 k}=2.27 \times 10^{-5}

Ideal gas law is used to calculate the density:

ρ = \frac{2.2}{2.83 \times 10^{-3} \times 394.3}

ρ = 1.97 Kg / m^{3}

ρ = \frac{P}{RT}

R = \frac{R_{c} }{M} = 8.2 × 10^{-5} / 28.97×10^{-3}

R = 2.83 × 10^{-3} m^{3} atm / K Kg

q is expressed in the unit m/s

q=\frac{2.45}{1.97}

q = 1.24 m/s

Re = \frac{1}{1-0.4} \frac{1.97 \times 1.24 \times 0.0127}{2.27 \times 10^{-5}}

Re = 2278

The Ergun equation is used when Re > 10,

\frac{\Delta P}{L}=\frac{180 \mu}{d_{p}^{2}} \frac{(1-\varepsilon)^{2}}{\varepsilon^{3}} q+\frac{7}{4} \frac{\rho}{d_{p}} \frac{(1-\varepsilon)}{\varepsilon^{3}} q^{2}

\frac{\Delta P}{L}=\frac{180 \times 2.27 \times 10^{-5}}{0.0127^{2}} \frac{(1-0.4)^{2}}{0.4^{3}} 1.24 +\frac{7}{4} \frac{1.97}{0.0127} \frac{(1-0.4)}{0.4^{3}} 1.24^{2}

= 4089.748 Pa/m

ΔP = 4089.748 × 3.66

ΔP = 14.5 kPa

4 0
3 years ago
A steel rod, which is free to move, has a length of 200 mm and a diameter of 20 mm at a temperature of 15oC. If the rod is heate
kherson [118]

Explanation:

thermal expansion ∝L = (δL/δT)÷L ----(1)

δL = L∝L + δT ----(2)

we have δL = 12.5x10⁻⁶

length l = 200mm

δT = 115°c - 15°c = 100°c

putting these values into equation 1, we have

δL = 200*12.5X10⁻⁶x100

= 0.25 MM

L₂ = L + δ L

= 200 + 0.25

L₂ = 200.25mm

12.5X10⁻⁶ *115-15 * 20

= 0.025

20 +0.025

D₂ = 20.025

as this rod undergoes free expansion at 115°c, the stress on this rod would be = 0

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