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Ostrovityanka [42]
3 years ago
9

A sheet of steel 1.4 mm thick has nitrogen atmospheres on both sides at 1200°C and is permitted to achieve a steady-state diffus

ion condition. The diffusion coefficient for nitrogen in steel at this temperature is 7.0 × 10-11 m2/s, and the diffusion flux is found to be 1.0 × 10-7 kg/m2-s. Also, it is known that the concentration of nitrogen in the steel at the high-pressure surface is 4.9 kg/m3. How far into the sheet from this high-pressure side will the concentration be 3.6 kg/m3? Assume a linear concentration profile.
Physics
1 answer:
marta [7]3 years ago
7 0

Answer:

9.1 × 10^-4m

Explanation:

From Fick's first law of diffusion:

J = -D dc/dx

Where J ,= diffusion flux

D = diffusion coefficient and dc/dx = concerntration gradient

Since we can assume a linear concentration profile, we rewrite the equation above as:

J = - D ◇c/◇x = -D (Cs - Cx)/(0 - x)

Where Cs = surface concerntration

C = concerntration of depth

X = target variable

J = diffusion flux = 1.0×10^-7kg/m^2/s

D ,= diffusion coefficient = 7.0×10^-12m^2/s

Substituting into the equation

1.0 ×10^-7 = (-7.0×10^-11)(4.9 -3.6) / ( 0 - X)

1.0 × 10^-7 = (-9.1 ×10^-11)/(0 - X)

Cross multiply

(1.0 ×10^-7) ×(0-X) = (-9.1 ×10^-11)

X = (-9.1 × 10^-11)/(-1.0 × 10^-7)

X = 9.1 × 10^-4m

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Problem 3: Thermal expansionThe steel rod has the length 2 m and cross-section area 200 cm2at the room temperature 20◦C. Weapply
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Answer:

a) 2.00024 m

b) 0.036%

c) 436.67°C

Explanation:

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Initial cross sectional Area = A₀ = 200 cm² = 0.02 m²

We can obtain initial volume = V₀ = A₀L₀ = 0.02 × 2 = 0.04 m³

Initial Temperature = T₀ = 20°C

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ΔL = L₀ × α ×ΔT

ΔL = 2 × 2 × 10⁻⁶ × (80 - 20) = 0.00024 m = 0.24 mm

New length = old length + expansion = 2 + 0.00024 = 2.00024 m

b) The percentage of the volume change of the rod.

Volume expansion is given by

ΔV = V₀ × (3α) × ΔT

Volume expansivity ≈ 3 × (linear expansivity)

ΔV = 0.04 × (3×2×10⁻⁶) × (80 - 20) = 0.0000144 m³

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c) The maximal temperature we can allow if the volume should not increase by more than half percent.

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ΔV = 0.0002 m³

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ΔV = V₀ × (3α) × ΔT

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Maximal temperature = T₀ + ΔT = 20 + 416.67 = 436.67°C

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2 years ago
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