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svetoff [14.1K]

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lol what

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For some transformation having kinetics that obey the Avrami equation, the parameter n is known to have a value of 1.7. If, afte

luda_lava [24]

Answer:

It would take approximately 305 s to go to 99% completion

Explanation:

Given that:

y = 50% = 0.5

n = 1.7

t = 100 s

We need to first find the parameter k from the equation below.

$exp(-kt^n)=1-y$

taking the natural logarithm of both sides:

$-kt^n=ln(1-y)\\kt^n=-ln(1-y)\\k=-\frac{ln(1-y)}{t^n}$

Substituting values:

$k=-\frac{ln(1-y)}{t^n}= -\frac{ln(1-0.5)}{100^1.7} = 2.76*10^{-4}$

Also

$t^n=-\frac{ln(1-y)}{k}\\t=\sqrt[n]{-\frac{ln(1-y)}{k}}$

Substituting values and y = 99% = 0.99

$t=\sqrt[n]{-\frac{ln(1-y)}{k}}=\sqrt[1.7]{-\frac{ln(1-0.99)}{2.76*10^{-4}}}=304.6s$

∴ t ≅ 305 s

It would take approximately 305 s to go to 99% completion

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A steel pipe of 400-mm outer diameter is fabricated from 10-mm-thick plate by welding along a helix that forms an angle of 20° w

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Explanation:

Outer di ameter $d_{0}=400 \mathrm{mm}[tex] Thickness of the cylinder [tex]t=10 \mathrm{mm}$

$\therefore[tex] Inner diam eter [tex]d_{i}=d_{0}-2 t=400-2 \times 10$

$d_{1}=380 \mathrm{mm}$

Given loading on the cylinder $P=300 \mathrm{kN}$ Helix an gle of the weld form $\theta=20^{\circ}$

(i) Normal stress on the plane at angle $\theta=20^{\circ}$ is

$\sigma=\frac{P \cos ^{2} \theta}{A_{0}}$

$\text { Where } A_{0}=\frac{\pi}{4}\left(d_{0}^{2}-d_{1}^{2}\right)$

$\quad=\frac{\pi}{4}\left(400^{2}-380^{2}\right)$

$=12252.21 \mathrm{mm}^{2}$

$=12.25221 \times 10^{-9} \mathrm{m}^{2}$

$\sigma=\frac{-300 \times 10^{2} \times \cos ^{2} 20}{12.25221 \times 10^{-1}}$

$=-21.6 \mathrm{MPa}$

(ii) Shear stress along an angle of $\theta=20^{\circ}$ is $\tau=\frac{P}{A_{0}} \cos \theta \sin \theta$

$=\frac{-300 \times 10^{-1} \times \cos 20 \times \sin 20}{12.25221 \times 10^{-3}}$

$=-7.86 \mathrm{MPa}$

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2 years ago